Solved Problems In Psychrometry 2zn55
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Introduction The aim of this section of the notes is to allow students to size air conditioning plant such as; cooling coil, heater battery and humidifier. The notes are divided into several sections as follows: PSYCHROMETRY FOR AIR CONDITIONING THE PSYCHROMETRIC CHART EXAMPLES OF PSYCHROMETRIC PROPERTIES AIR CONDITIONING PLANT FOR SUMMER & WINTER BASIC PROCESSES TYPICAL AIR CONDITIONING PROCESSES ANNOTATION AND ROOM RATIO SUMMER AND WINTER CYCLES EXAMPLES The first section deals with Psychrometry for air conditioning and discusses some properties of moist air. A simplified psychrometric chart is shown for familiarisation, and some examples of how to find air properties are provided. A diagram of an air conditioning system is shown in schematic form in the section entitled AIR CONDITIONING PLANT FOR SUMMER & WINTER. Before sizing takes place the student should also understand the processes that take place in air conditioning systems. There are four basic processes for summer and winter air conditioning systems. The following basic processes are explained: 1. Mixing 2. Sensible Cooling and Heating 3. Cooling with Dehumidification 4. Humidification The section on Typical Air Conditioning Processes shows winter and summer schematic diagrams and psychrometric charts. There are some more details that may be useful to the designer of air conditioning systems. Further information is as follows: Annotation, Room ratio When the processes have been superimposed onto a psychrometric chart then calculations may commence.
These are as detailed in the following sections of the notes. Summer and Winter Cycles 1. Summer cycle psychrometrics 2. Summer cycle calculations 3. Winter cycle psychrometrics 4. Winter cycle calculations 5. Duct and Fan gains. The final section is seven examples of plant sizing using psychrometric charts.
Psychrometry for Air Conditioning Psychrometry is the study of air and water vapour mixtures. Air is made up of five main gases i.e. Nitrogen 78.03%, Oxygen 20.99%, Argon 0.94%, Carbon Dioxide 0.03%, and Hydrogen 0.01% by volume. The Ideal Gas Laws are used to determine psychrometric data for air so that the engineer can carry out calculations. To make life easier a chart has been compiled with all the relevant psychrometric data indicated. This is called the Psychrometric Chart. A typical chart is shown below.
Air at any state point can be plotted on the psychrometric chart. The information that can be obtained from a Psychrometric Chart is as follows:
1. 2. 3. 4. 5. 6.
Dry bulb temperature Wet bulb temperature Moisture content Percentage saturation Specific enthalpy Specific volume.
The following is a brief description of each of the properties of air. 1. Dry bulb temperature This is the air temperature measured by a mercury-in-glass thermometer. 2. Wet bulb temperature This is the air temperature measured by a mercury-in-glass thermometer which has the mercury bulb wetted by gauze that is kept moist by a reservoir of water. When exposed to the environment the moisture evaporates from the wetted gauze, which gives a lower reading on the thermometer. This gives an indication of how ‘dry’ or how ‘moist’ the air is, since in ‘dry’ air the water will evaporate quickly from the gauze, which depresses the thermometer reading. 3. Moisture content This is the amount of moisture in air given in kg of moisture per kg of dry air e.g. for room air at 21oC dry bulb and 15oC wet bulb, the moisture content is about 0.008 kg/kg d.a. This is a small mass of moisture ( 0.008 kg = 8 grams) per kg of dry air or 9.5 grams per cubic metre of air. 4. Percentage saturation The Percentage saturation is another indication of the amount of moisture in air. This is the ratio of the moisture content of moist air to the moisture content of saturated air at the same temperature.
When air is saturated it is at 100% saturation and cannot hold any more moisture. 5. Specific enthalpy This is the amount of heat energy (kJ) in air per kg. If heat is added to the air at a heater battery for example, then the amount to be added can be determined from Specific enthalpy change. 6. Specific volume This is the volume of moist air (dry air + water vapour) per unit mass. The units of measurement are m3 per kg. Also specific volume = 1 / density.
The Psychrometric Chart The six properties of air previously discussed can be shown on one chart called a Psychrometric Chart. One of the purposes of the Psychrometric Chart is to size heater batteries, cooling coils and humidifiers. A simplified Psychrometric Chart is shown below.
This chart is only for demonstration purposes. If accurate assessments are to be carried out use a C.I.B.S.E. chart. Using the Psychrometric Chart If any two properties of air are known then the other four can be found from the psychrometric chart.
Examples of Psychrometric Properties EXAMPLE 1 Find the moisture content of air at 25oC dry-bulb temperature and 25oC wetbulb temperature. Referring to the chart below, a vertical line is drawn upwards from 25oC drybulb temperature until it intersects at 25oC wet-bulb temperature. This intersection point happens to be on the 100% saturation line. The intersection point is highlighted and a horizontal line is drawn to the right to find the corresponding moisture content. The moisture content is therefore 0.020 kg/kg dry air.
EXAMPLE 2 Find the specific volume and wet-bulb temperature of air at 20oC dry-bulb temperature and 50% saturation. Referring to the chart below, a vertical line is drawn upwards from 20oC drybulb temperature until it intersects with the 50% saturation curve. The intersection point is sometimes referred to as the state point. The specific volume is found to be 0.84 m3/kg and the wet-bulb temperature is 14oC
EXAMPLE 3 Find the specific volume, percentage saturation and moisture content of air at 15oC dry-bulb temperature and 10oC wet-bulb temperature. Referring to the chart below, a vertical line is drawn upwards from 15oC drybulb temperature until it intersects with the 10oC wet-bulb temperature line. This intersection is the state point. The specific volume is found to be 0.823 m3/kg, the percentage saturation 52% and the moisture content 0.0054 kg/kg d.a.
EXAMPLE 4 Find the specific volume, wet-bulb temperature, moisture content and specific enthalpy of air at 35oC dry-bulb temperature and 30% saturation. Referring to the chart below, a vertical line is drawn upwards from 35oC drybulb temperature until it intersects with the 30% saturation curve. This intersection is the state point.
The specific volume is found to be 0.883 m3/kg, the wet-bulb temperature is 22oC, the moisture content 0.011kg/kg d.a. and the specific enthalpy 65 kJ/kg.
Air Conditioning Plant for Summer and Winter In the summer time when cooling is required by the air conditioning plant it will be necessary to operate the cooling coil, reheater and possibly other plant as well. In winter time the preheater and reheater battery will probably be on to provide warm air to overcome heat losses. Other plant may be switched on as well. These plant items are shown in the diagram below.
Basic Air Conditioning Processes 1. Mixing Where two air streams are mixed the psychrometric process is shown as a straight line between two air conditions on the psychrometric chart, thus points 1 and 2 are ed and the mix point 3 will lie on this line. Two air streams are mixed in air conditioning when fresh air (m1) is brought in from outside and mixed with recirculated air (m2). The resulting air mixture is shown below as (m3). The mixing ratio is fixed by dampers. Sometimes, in more sophisticated plant, modulating dampers are used which are driven by electric motors to control the mixture of air entering the system. The diagrams below show mixing of two air streams.
By the conservation of mass formula: By the conservation of energy formula:
m1 + m2 = m3 m1 h1 + m2 h2 = m3 h3
where: m = mass flow rate of air (kg/s) h = specific enthalpy of air (kJ/kg) found from psychrometric chart.
2. Sensible Cooling and Heating When air is heated or cooled sensibly, that is, when no moisture is added or removed, this process is represented by a horizontal line on a psychrometric chart.
For sensible heating: The amount of heating input to the air approximates to; H1-2 = m x x (t2 - t1) Or more accurately from psychrometric chart: H1-2 = m x (h2 - h1) For sensible cooling: The amount of cooling input to the air approximates to; H2-1 = m x x (t2 - t1) Or more accurately from psychrometric chart: H2-1 = m x (h2 - h1) where: H = Heat or cooling energy (kW) m = mass flow rate of air (kg/s) = Specific heat capacity of air, may be taken as 1.01 kJ/kg degC. t = Dry bulb temperature of air (oC) h = specific enthalpy of air (kJ/kg) found from psychrometric chart.
3. Cooling with Dehumidification The most commonly used method of removing water vapour from air (dehumidification) is to cool the air below its dew point. The dew point of air is when it is fully saturated i.e. at 100% saturation. When air is fully saturated it cannot hold any more moisture in the form of water vapour. If the air is cooled to the dew point air and is still further cooled then moisture will drop out of the air in the form of condensate. This can be shown on a psychrometric chart as air sensibly cooled until it becomes fully saturated (the dew point is reached) and then the air is cooled latently to a lower temperature. This is apparent on the psychrometric chart as a horizontal line for sensible cooling to the 100% saturation curve and then the process follows the 100% saturation curve down to another point at a lower temperature. This lower temperature is sometimes called the Apparatus dew Point (ADP) of the cooling coil. In reality the ADP of the cooling coil is close to the cooling liquid temperature inside the coil. Chilled water or refrigerant may be the cooling liquid. The psychrometric process from state point 1 to 2 to 3 may be shown as a straight line for simplicity as shown above with a yellow line.
The total amount of cooling input to the air approximates to;
H1-3 = m x (h1 - h3) The sensible heat removed is:
H1-2 = m x (h1 - h2) The latent heat removed is:
H2-3 = m x (h2 - h3) where:
H = Cooling energy (kW) m = mass flow rate of air (kg/s) h = specific enthalpy of air (kJ/kg) found from psychrometric chart.
In the absence of a suitable psychrometric chart the following formula may be used; The sensible heat removed is:
H1-2 = m x x (t1 - t2)
The latent heat removed is:
H2-3 = m x hfg x (g2 - g3)
where:
H = Cooling energy (kW) m = mass flow rate of air (kg/s) = Specific heat capacity of air, may be taken as 1.01 kJ/kg
degC. t = Dry bulb temperature of air (oC) hfg = latent heat of evaporisation, may be taken as 2454 kJ/kg @20oC. g = moisture content of air from psychrometric chart (kg/kg dry air)
3.1 Cooling Coil Factor Some of the air going through a cooling coil does not come into with the tubes or fins of the cooling coil and is therefore not cooled to the ADP temperature. A mixing process therefore takes place as two air streams mix downstream of the cooling coil as shown below.
One air stream is cooled down to the ADP and the other air stream by-es the coil surfaces to give an off-coil air temperature (mixed air stream) a little higher than the ADP. This may be looked upon as an inefficiency of the coil and is usually given as the cooling coil factor. The process is shown on the psychrometric chart below.
The factor of a cooling coil may be found from;
Another expression for factor is;
4. Humidification If is it necessary to add some moisture to the supply air then this is best done by injecting steam into the air stream. Humidification can be carried out by spraying a fine mist of water droplets into the air but this is not recommended in rooms occupied by people due to the risk of bacteria carry over. Dry steam may be injected from a steam supply pipe or generated in a local packaged unit as shown in the photograph below. A disadvantage of using an existing steam supply is smells may be carried over into the air.
The steam package unit is situated close to the air duct and is sized to meet the maximum requirements; this is usually in winter in the U.K. A steam pipe ( sometimes hoses are used) es from the packaged unit to the air duct and steam at 100oC is injected into the air stream via. a spargepipe. The un-used steam is drained from the system via a condensate tundish and drain. It is important to layout the steam pipework so that any condensate will drain back to the unit. The psychrometric process is shown below.
See Summer and Winter Cycles section for calculation of amount of moisture added at humidifier.
Typical Air Conditioning Processes The schematic diagram below shows a typical plant system for summer air conditioning.
The psychrometric diagram below shows a typical summer cycle.
The schematic diagram below shows a typical plant system for winter air conditioning.
The psychrometric diagram below shows a typical winter cycle.
Annotation The state points on a psychrometric chart may be given numbers or symbols to identify them. If symbols are used the following system may be adopted:
Room Ratio This is the ratio of sensible to total heat in the room for summer or winter. The total heat gain (summer) or loss (winter) will be determined by adding the Latent and Sensible heat in a room or rooms, i.e. (SUMMER)
Total heat gain
=
Sensible heat gain + Latent heat gain
(WINTER)
Total heat
=
Sensible heat loss + Latent heat gain
The room ratio is used on a psychrometric chart to determine the supply air state point. A room ratio line is superimposed from the protractor on the psychrometric chart onto the main body of the chart by a line ing through the room state point R. An example calculation is as follows:
Sensible heat gain
=
9.0 kW
Latent heat gain
=
2.25 kW
Total heat gain
=
9.0 kW + 2.25 kW
Room ratio
=
Sensible / Total heat
Room ratio
=
9 / 11.25
=
=
11.25 kW.
0.8
The supply air state point must also be somewhere on this room ratio line to meet the room heat gain requirements i.e. the room ratio line always es through points R and S.
Examples of Psychrometric Calculations for Summer and Winter
Example 1. Summer Cycle A room is to be maintained at 22oC dry-bulb temperature, 50% saturation, when the sensible heat gain is 10.8 kW in summer. The latent heat gain is 7.2 kW. Determine the cooling coil and reheater outputs required by using a psychrometric chart if the plant schematic is as shown below. DATA: Outdoor condition is 28oC, 80% saturation. The outdoor air and recirculated air ratio is 20%/80%. The Apparatus Dew Point ADP is 8oC Neglect the cooling coil factor.
Note: The cooling coil output of 38.2 kW is a much higher value than the sensible heat gain of 10.8 kW. It should be ed that the difference is these two values is mostly from the fresh air cooling load. It takes quite a lot of energy in summer to cool fresh air coming into air handling units. This can be minimised by bringing in minimum fresh air but not too little otherwise the building will suffer from lack of oxygen and feel stuffy. Sometimes mistakes are made when sizing cooling apparatus. If a cooling coil or indoor cooling unit is sized on the sensible heat gain only without allowing for fresh air load then it will be grossly undersized. That is why psychrometric charts are required to calculate cooling coil output including fresh air loads.
So, don’t size cooling coil and indoor cooling units on sensible heat gain only if there is fresh air coming into the plant. Size these items of plant using a psychrometric chart.
Example 2. Winter Cycle A room has a 18.0 kW sensible heat loss in winter and a 4.5 kW latent heat gain from the occupants. Determine the supply air temperature and heater battery load using the following information. DATA: Indoor condition: 21oC dry-bulb temperature, 50% saturation. Outdoor condition: -2oC d.b., 80% saturation. The outdoor air and recirculated air ratio is 20%/80%. No preheating or humidification takes place in this simplified example.
Procedure (Winter Cycle) 1. Draw schematic diagram of air-conditioning plant (see above) 2. Plot room condition R on psychrometric chart. 3. Plot outside condition O on psychrometric chart.
4. No Preheater condition P 5. points O and R
6. Find the mix point M by measuring the length of the line O-R and multiply this by the mixing ratio. On a full size CIBSE psychrometric chart this measures 110mm. The ratio of recirculated air is 0. therefore; 110mm x 0.8 = 88mm Measure up the O-R line from point O by 88mm. This determines point M . If there is more recirculated air than outside air at the mix point, then point M will be closer to point R than point O. 7. Find the room ratio. This is the sensible to total heat ratio.
Neglect signs ie. the total heat for the room will be Sensible loss plus Latent gain. Total heat = 18 kW sensible + 4.5 kW latent = 22.5 kW total. Heat ratio = 18 / 22.5 = 0.8 Plot this ratio on the protractor, top segment, on the psychrometric chart and transfer this line onto the chart so that it es through point R.
8. Find the supply air dry bulb temperature by calculation. 9. Plot the supply air condition S on the room ratio line. This is on a horizontal line from point M to the right hand side of the chart, and intersects with the RRL.
The supply air Temperature is found to be 32.5oC.
Supply Air Flow Rate When the sensible heat loss and supply air temperature in winter are known then the mass flow rate of air is calculated from the following formula: Hs = ma x ( ts - tr ) where: Hs = ma =
Sensible heat loss (kW) mass flow rate of air (kg/s)
= degC) tr = o ( C) ts = o ( C)
Specific heat capacity of humid air (approx.1.01 kJ/kg room temperature supply air temperature ..............therefore:
ma = Hs / ( ts - tr ) ma = 18 / 1.01 (32.5 - 21) ma = 18 / 11.615 ma = 1.55 kg/s
Heater Battery Output The heater battery output is as follows: H reheater battery = ma ( hS - hM) where: H reheater battery = Reheater battery output (kW) ma = mass flow rate of air (kg/s) hS = specific enthalpy at condition S (kJ/kg) hM = specific enthalpy at condition M (kJ/kg)
The specific enthalpies at points S and M are shown on the psychrometric chart below.
Hheater battery = ma ( hS - hM) Hheater battery = 1.55 ( 50 - 34) Hheater battery = 24.8 kW Therefore the heater battery load is 24.8 kW.
Example 3. Summer Cycle (Cooling Coil factor) An office is to be maintained at 22oC dry-bulb temperature, 50% saturation in summer. The sensible heat gain is 8.0 kW. The latent heat gain is 2.0 kW. Determine the cooling coil and reheater outputs required by using a psychrometric chart if the plant schematic is as shown below. DATA: Outdoor condition is 28oC, 80% saturation. The outdoor air and recirculated air ratio is; 20% / 80%. The Apparatus Dew Point ADP is 8oC The cooling coil factor is 0.8. (See explanation of factor in page 9)
Procedure (Summer Cycle) 1.
Draw schematic diagram of air-conditioning plant (see above)
2.
Plot room condition R on psychrometric chart.
3.
Plot outside condition O on psychrometric chart.
4.
points O and R.
5. Find the mix point M by measuring the length of the line O-R and multiply this by the mixing ratio. On a full size CIBSE psychrometric chart this measures 85mm. The ratio of recirculated air is 0.8. …….. therefore; 85mm x 0.8 = 68mm Measure down the O-R line from point O by 68mm. This determines point M . 6.
Find the room ratio. This is the sensible to total heat gain ratio. Total heat = 8 kW sensible + 2 kW latent = 10 kW total. Heat ratio = 8 / 10 = 0.8 Plot this ratio on the protractor, bottom segment, on the psychrometric chart and transfer this line onto the chart so that it es through point R. 7.
Plot the Apparatus Dew Point ADP of the cooling coil.
This is on the 100% saturation curve.
8.
The ADP is 80C.
points M and ADP.
9. Find the off-coil condition W by measuring the length of the line MADP and multiply this by the cooling coil factor.. On a full size CIBSE psychrometric chart this measures 75mm. The cooling coil factor is 0.8. ……….. therefore; 75mm x 0.8 = 60mm. Measure down along the line M-ADP by 60mm. This determines point W. 10.
Plot the supply air condition S. The reheater process will be a horizontal line from point W to point S. Point S is on the room ratio line. The supply air temperature is 17oC.
Mass Flow Rate When the supply air temperature has been found from the psychrometric chart then the mass flow rate of air can be calculated from the following formula: ma
=
Hs / ( ( tr - ts ) )
Hs ma
= = =
Sensible heat gain to room (kW) mass flow rate of air (kg/s) Specific heat capacity of humid air (approx.1.01
tr
=
room temperature (oC)
where:
kJ/kg degC)
ts
=
supply air temperature (oC)
The supply air temperature is 17oC. ma
=
8 / ( 1.01 ( 22 - 17 ) )
ma
=
1.584 kg/s
Cooling Coil Output The cooling coil output is as follows: H cooling coil =
ma ( hM - hADP)
where: H cooling coil = Cooling coil output (kW) ma = mass flow rate of air (kg/s) hM = specific enthalpy at condition M (kJ/kg) determined from psychrometric chart. hADP = specific enthalpy at condition ADP (kJ/kg) determined from psychrometric chart
The specific enthalpies at points M and ADP are shown on the psychrometric Chart below. H cooling coil =
1.584 ( 50.5 - 25)
H cooling coil =
40.4 kW
Heater Battery Output The heater battery or reheater output is as follows: H heater battery = where: H heater battery
ma ( hS - hW) =
Heater battery output (kW)
ma = mass flow rate of air (kg/s) hS = specific enthalpy at condition S (kJ/kg) determined from psychrometric chart. hW = specific enthalpy at condition W (kJ/kg) determined from psychrometric chart.
The specific enthalpies at points S and W are shown on the psychrometric Chart below.
H heater battery =
1.584 ( 36.5 - 30.5)
H heater battery =
9.5 kW
Example 4.
Winter Cycle with Humidifier
An conference room is to be maintained at 21oC dry-bulb temperature, 50% saturation in winter. The sensible heat loss for the room is 17.0 kW. The latent heat gain is 40 Watts per person (see Air Conditioning section). Determine the preheater and reheater outputs required and the amount of moisture to be added at the humidifier in litre/hour, by using a psychrometric chart if the plant schematic is as shown below.
DATA: Outdoor condition is -2oC, 80% saturation. The outdoor air and recirculated air ratio is 50%/50%. Maximum occupancy is 250 people. The preheater off coil temperature is 5oC. Supply air quantity is 8 air changes per hour. Room volume is 20 x 12 x 4m high = 960 m3.
Latent Heat Gain
The latent heat gain =
heat gain per person x number of people
The latent heat gain =
40 W/person x 250 =
The latent heat gain =
10 kW
10,000 Watts
Supply Air & Fresh Air Quantities Supply air quantity (m3/h)
=
air change rate x room volume (m3)
Supply air quantity (m3/h)
=
8 x 960(m3)
Supply air quantity (m3/h)
=
7680 (m3/h)
Supply air quantity (m3/s)
=
7680(m3/h) / 3600 = 2.13 (m3/s)
Supply air mass flow rate (kg/s) Volume (m3/kg)
= Supply air quantity (m3/s) / Specific
Supply air mass flow rate (kg/s)
= 2.13 (m3/s) / 0.87 (m3/kg)
Supply air mass flow rate (kg/s)
= 2.45 kg/s
The fresh air flow rate (kg/s)
= 2.45 kg/s x 50% = 1.23 kg/s
Winter Cycle Psychrometrics
1. 2. 3.
4. 5.
6.
Draw schematic diagram of air-conditioning plant (see above). Plot room condition O, M and R on psychrometric chart. Plot the after Preheater condition P. The Preheater process will be a horizontal line from O to P and acts as a frost coil in this case, heating the air to 5oC. points P and R. Find the mix point M by measuring the length of the line P-R and multiply this by the mixing ratio. The line measures 82mm long. 82 x 0.5 = 41 mm Find the room ratio. Plot this ratio on the protractor, so that it es through point R. Total heat is 17 kW sensible + 10 kW latent = 27kW. Ratio is 17/27 = 0.63.
7. Find the supply air dry bulb temperature by calculation. This is found by calculation because we have already calculated the mass flow rate of supply air from information given in the question.
Supply Air Dry Bulb Temperature The temperature of supply air is calculated from the following formula:
Hs where: Hs = ma = =
=
ma x ( ts - tr )
Sensible heat loss from room (kW) mass flow rate of air (kg/s) Specific heat capacity of humid air (approx.1.01 kJ/kg
degC) tr = o ( C) ts =
room temperature supply air temperature (oC)
( ts - tr )
=
Hs /
ma x
( ts - tr )
=
17 / 2.45 x 1.01
( ts - tr )
=
6.87 deg.C
( ts - 21)
=
6.87 deg.C
ts
=
21 + 6.87 deg.C
ts
=
27.87 oC say 28 oC.
8. Plot the supply air condition S on the room ratio line. 9. Plot condition H on the psychrometric chart. This is vertically down from point S, and horizontally across from point M. This is because M-H is the reheater process and thus a horizontal line and H-S is the humidification process and is close to a vertical line if steam is used.
Preheater Battery Output (Or Frost Coil) The preheater battery output is as follows:
H preheater battery = maf ( hP - hO) where: H preheater battery = m af = hP = hO =
Preheater battery output (kW) mass flow rate of fresh air (kg/s) specific enthalpy at condition P (kJ/kg) specific enthalpy at condition O (kJ/kg)
H preheater battery = 1.23 ( 12 - 5.5) H preheater battery = 8.0 kW
Reheater Battery Output The reheater battery output is as follows:
H reheater battery = ma ( hH - hM) where: H reheater battery ma = hH = hM =
= Reheater battery output (kW) mass flow rate of supply air (kg/s) specific enthalpy at condition H (kJ/kg) specific enthalpy at condition M (kJ/kg)
Hreheater battery = 2.45 ( 42 - 27) Hreheater battery = 36.8 kW
Humidifier Output The amount of moisture added to the air may be calculated from the following formula:
m moisture added = ma (msS - msH) where: m moisture added = rate (kg/s)
The amount of moisture or added or steam flow ma m sS
= =
mass flow rate of air (kg/s) moisture content at condition S (kg/kg
m sH
=
moisture content at condition H (kg/kg
d.a.) d.a.)
m moisture added = 2.45 (0.0064 - 0.0054) m moisture added = 2.45 (0.001) m moisture added = 0.00245 kg/s 1 litre of water weights 1 kg, therefore; m moisture added = 0.00245 litre/s
m moisture added = 0.00245 litre/s x 3600 = 8.82 litres/hour
Example 5 Summer Cycle (Air Flows To Be Calculated) A Lecture Theatre measures 15 m x 10 m x 6 m high. It is to be air conditioned in summer so that the room is maintained at 22oC dry-bulb temperature, 50% saturation. Determine the cooling coil and reheater outputs required by using a psychrometric chart if the plant schematic is as shown below. DATA: Outdoor condition is 28oC, 80% saturation. The Apparatus Dew Point ADP is 7.5oC. The latent heat gain is 10.0 kW. The sensible heat gain is 12.0 kW. Maximum occupancy is 200 people. The cooling coil factor is unknown at present and should be calculated. Use CIBSE guide B2 (2001) to determine air flow rates and calculate the mass flow rate of fresh air and supply air to the room. The lecture room may be taken as an Assembly hall. The room is a non-smoking area.
Mass Flow Rates Information from CIBSE Guide B2 (2001) Table 3.3 is as follows (see Ventilation – Ventilation rates section); The recommended outdoor air rate is 8 l/s/person for non-smoking. Information from CIBSE Guide B2 (2001) Table 3.1 is as follows (see Ventilation – Ventilation rates section); Assembly halls and auditorium refers to Table 3.6 (see Ventilation – Ventilation rates section); The recommended total air supply rate is 6 – 10 air changes per hour for high level mechanical strategy. Fresh Air Flow Rate
Fresh air rate = 8 l/s/p x 200 people = 1600 l/s = 1.6 m3/s The specific volume at the outside condition may be determined from a psychrometric chart. It is 0.88 m3/kg. Mass flow rate = Volume flow rate / specific volume Mass flow rate (Fresh Air) = 1.6 / 0.88 = 1.82 kg/s . Supply Air Flow Rate
If the maximum ventilation supply air rate is taken from Table 3.6 to be 10.0 air changes per hour, then the mass flow rate can be calculated. Volume flow rate (m3/h) = Volume of room (m3) x air change rate (ac/h) Volume of room (m3) = 15 x 10 x 6 = 900 m3 Volume flow rate (m3/h) = 900 (m3) x 10 (ac/h) Volume flow rate (m3/h) = 9000 m3/h Volume flow rate (m3/s) = 9000 / 3600 = 2.5 3 m /s. Mass flow rate = Volume flow rate / specific volume The specific volume at the supply condition may be approximated at this stage from a psychrometric chart. It is 0.834 m3/kg. Mass flow rate (Supply Air) = 2.5 / 0.834 = 3.0 kg/s.
Fresh Air And Recirc. Ratio.
The ratio by mass is therefore; Fresh air rate Supply air rate Recirculation air rate
= = =
1.82 kg/s 3.00 kg/s 3.00 - 1.82 = 1.18 kg/s
The ratio of fresh air to total supply air is; 1.82 / 3.00 = 0.6, i.e. 60% fresh air and therefore 40% recirculated air. It is not unusual to have a high percentage of fresh air in a high occupancy room such as a Lecture theatre. The air flows are shown on the schematic diagram below.
Supply Air Temperature by Calculation In this example the supply air temperature will be found by rearranging the following formula:
Hs where: Hs = ma = =
=
ma x ( tr - ts ) Sensible heat gain to room (kW) mass flow rate of air (kg/s) Specific heat capacity of humid air (approx.1.01 kJ/kg
degC) tr = o ( C) ts =
room temperature supply air temperature (oC)
Rearranging the above formula gives: ( tr - ts ) = ( tr - ts ) = ( tr - ts ) = since tr = ts = ts =
Hs / ( ma x ) 12 / ( 3.00 x 1.01 ) 3.96 deg.C 22oC 22 - 3.96 = 18.04 oC 18 oC approx.
The processes can now be plotted on a psychrometric chart as shown below.
1. Points O, M and R can be shown on the chart. 2. Point ADP can be indicated and lines drawn between these points as shown. 3. The room ratio line can be drawn. 4. Point S is then shown on the chart, on the room ratio line at 18oC. 5. A horizontal line is then drawn from point S towards the line O – ADP. 6. Point W can then be found where the horizontal line W - S intersects the line O - ADP. From the psychrometric chart point W is at approximately 9oC dry bulb. The heat ratio is 12 kW sensible / 22 kW total = 0.545.
The specific enthalpies are shown below.
Cooling Coil Output
The cooling coil output is as follows:
H cooling coil
ma (hM - hADP)
=
where: H cooling coil = Cooling coil output (kW) ma = mass flow rate of air (kg/s) hM = specific enthalpy at condition M (kJ/kg) determined from psychrometric chart. hADP = specific enthalpy at condition ADP (kJ/kg) determined from psychrometric chart
The specific enthalpies at points M and ADP are shown on the psychrometric Chart above. H cooling coil =
3.00 ( 64 - 24)
H cooling coil =
120.0 kW
Reheater Battery Output The heater battery or reheater output is as follows:
H heater battery
=
ma ( hS - hW)
where: H heater battery = Heater battery output (kW) ma = mass flow rate of air (kg/s) hS = specific enthalpy at condition S (kJ/kg) determined from psychrometric chart. hW = specific enthalpy at condition W (kJ/kg) determined from psychrometric chart.
The specific enthalpies at points W and S are shown on the psychrometric Chart above. H heater battery = H heater battery =
3.00 ( 36 - 26.5) 28.5 kW
Example 6
Summer Cycle and Winter Cycle
A Concert Hall measures 40 m x 20 m x 8 m high. It is to be air conditioned in summer and winter. Determine the following: o Air flow rates o Supply air temperature by calculation in summer and winter. o Cooling coil factor. o Cooling coil and reheater outputs in summer o Humidifier output in litres/hour. The plant schematic is as shown below.
DATA: Indoor condition all year Outdoor condition summer Outdoor condition winter Maximum occupancy ADP of the cooling coil Fresh air requirement Supply air rate -
22oC dB temperature, 50% saturation. 28oC dB temperature, 80% saturation. -3oC dB temperature, 80% saturation. 1000 people. 8oC 12 l/s/person 8 air changes per hour.
Frost off coil temperature Latent heat gain Sensible heat gain in summer lights, solar & ventilation gains. Sensible heat loss in winter -
7oC. 40 W/person 100 W/person + 20.5 kW fabric, 20 W/m3 air volume (estimated)
Fresh Air Flow Rate
Fresh air rate = 12 l/s/p x 1000 people = 12,000 l/s = 12.0 m3/s The specific volume at the outside condition may be determined from a psychrometric chart. It is approximately 0.88 m3/kg. Mass flow rate = Volume flow rate / specific volume Mass flow rate (Fresh Air) = 12.0 / 0.88 = 13.64 kg/s. Supply Air Flow Rate The ventilation supply air rate is taken from table B2.3 and is given as 8 air changes per hour; the mass flow rate can be calculated. Volume flow rate (m3/h) = Volume of room (m3) x air change rate (ac/h) Volume of room (m3) = 40 x 20 x 8 = 6400 m3 Volume flow rate (m3/h) = 6400 (m3) x 8 (ac/h) Volume flow rate (m3/h) = 51,200 m3/h Volume flow rate (m3/s) = 51,200 / 3600 = 14.22 m3/s. Mass flow rate = Volume flow rate / specific volume The specific volume at the supply condition may be approximated at this stage from a psychrometric chart. It can be taken as 0.834 m3/kg. Mass flow rate (Supply Air) = 14.22 / 0.834 = 17.05 kg/s. Recirculation Air Flow Rate Recirculation air rate Recirculation air rate
= =
Supply air rate - fresh air rate 17.05 - 13.64 = 3.41 kg/s
Fresh Air and Recirc. Ratio. The ratio by mass is therefore; Fresh air rate = 13.64 kg/s Supply air rate = 17.05 kg/s The ratio of fresh air to total supply air is;
13.64 / 17.05 = 0.8, i.e. 80% fresh air and therefore 20% recirculated air. It is not unusual to have a high percentage of fresh air in a high occupancy room such as a Concert Hall. The air flows are shown on the schematic diagram below.
Heat Gains The heat gains are given as: Latent heat gain 40 W/person Sensible heat gain - 100 W/person + 20.5 kW fabric, lights, solar & ventilation gains. Latent gain = 40 W x 1000 people = 40,000 W = 40 kW Sensible gain = 100 W x 1000 people = 100,000 W = 100 kW + 20.5 kW = 120.5 kW. Total heat gain = 120.5 + 40 = 160.5 kW Summer Heat ratio = sensible / total = 120.5 / 160.5 = 0.75 Heat Loss The heat loss in winter is given as 20.0 W/ m3.
Volume of room (m3) = Total heat loss = Total heat (winter) = Winter Heat ratio =
40 x 20 x 8 = 6400 m3 20 x 6400 = 128,000 Watts = 128 kW 128 + 40 (Latent) = 168 kW sensible / total = 128 / 168 = 0.76
Summer Calculations & Process Supply Air Temperature by Calculation In this example the supply air temperature will be found by rearranging the following formula: Hs
=
where: Hs = ma = = kJ/kg degC) tr = ts =
ma x ( tr - ts ) Sensible heat gain to room (kW) mass flow rate of air (kg/s) Specific heat capacity of humid air (approx.1.01 room temperature (oC) supply air temperature (oC)
Rearranging the above formula gives: ( tr - ts ) = Hs / ( ma x ) ( tr - ts ) = 120.5 / ( 17.05 x 1.01 ) ( tr - ts ) = 7.0 deg.C, since tr = 22oC, ts = 22 - 7.0 = 15.0 oC ts = 15 oC The processes can now be plotted on a psychrometric chart as shown below. 1. Points O, M and R can be shown on the chart. 2. Point ADP can be indicated and lines drawn between these points as shown. 3. The room ratio line can be drawn. 4. Point S is then shown on the chart, on the room ratio line at 18oC. 5. A horizontal line is then drawn from point S towards the line O – ADP.
6. Point W can then be found where the horizontal line W - S intersects the line O - ADP.
Cooling Coil Factor On a full size psychrometric chart the length of the line from point M to point ADP is 116mm. The distance from point M to point W is 107mm.
The cooling coil factor is therefore:
The specific enthalpies are shown below.
Cooling Coil Output in Summer The cooling coil output is as follows:
H cooling coil =
ma ( hM - hADP)
where: H cooling coil = Cooling coil output (kW) ma = mass flow rate of air (kg/s) hM = specific enthalpy at condition M (kJ/kg) determined from psychrometric chart. hADP = specific enthalpy at condition ADP (kJ/kg) determined from psychrometric chart
The specific enthalpies at points M and ADP are shown on the psychrometric Chart above. H cooling coil = H cooling coil =
17.05 ( 71 - 25) 784.3 kW
NOTE: The cooling coil output is very high and a lot of energy would be required to provide this amount of cooling. The cooling coil load would probably be spread over several air handling units but the it could be examined with a view to some reduction. The coil output is high because the mass flow rate of supply air is high (17.05 kg/s) and the proportion of fresh air is also high (80%). The mix point M is at approximately 27oC dry-bulb so there is little advantage in recirculation in this instance. It would be advantageous to consider the supply airflow rate to see if a lower rate would be acceptable for this building. If 6 air changes per hour are used as the ventilation rate then this would reduce the mass flow rate of supply air. Also the engineer may consider other methods of air-conditioning a hall with a large volume such as using partial radiant cooling where surfaces are cooled rather than air. Reheater Battery Output in Summer
The heater battery or reheater output is as follows: H heater battery =
ma ( hS - hW)
where: H heater battery = Heater battery output (kW) ma = mass flow rate of air (kg/s) hS = specific enthalpy at condition S (kJ/kg) determined from psychrometric chart. hW = specific enthalpy at condition W (kJ/kg) determined from psychrometric chart.
The specific enthalpies at points W and S are shown on the psychrometric Chart above. H heater battery = 17.05 ( 34 - 28) H heater battery = 102.3 kW
Winter Calculations & Process Winter Cycle Psychrometrics The processes can now be plotted on a psychrometric chart as shown below. 1. Points O, M, P and R can be shown on the chart. 2. points O and P and P and R. 3. Find the mix point M The line measures 81mm long. 81 x 0.8 = 65 mm , the distance from point M to R is 65mm. 4. The room ratio is 0.76 from previous calculation. Draw RRL. 5. Find the supply air dry bulb temperature by calculation.
Supply Air Temperature by Calculation The temperature of supply air is calculated from the following formula: Hs
=
ma x ( ts - tr )
where: Hs = Sensible heat loss from room (kW) ma = mass flow rate of air (kg/s) same as summer rate for constant volume systems. = Specific heat capacity of humid air (approx.1.01 kJ/kg degC) tr = room temperature (oC) ts = supply air temperature(oC)
( ts - tr ) ( ts - tr ) ( ts - tr ) ts ts
= = = = =
Hs / ma x 128 / 17.05 x 1.01 7.43 deg.C 22 + 7.43 deg.C 29.43 oC say 29.5 oC.
6. Plot the condition H at 29.5oC dB. on a horizontal line from M. 7. Plot condition S on a vertical line from H on the RRL. Assume the humidity process is vertical.
Preheater Battery Output (or frost coil) The preheater battery output is as follows: H preheater battery
=
maf ( hP - hO)
where: H preheater battery maf hP (kJ/kg) hO (kJ/kg)
H preheater battery H preheater battery
= = =
Preheater battery output (kW) mass flow rate of fresh air (kg/s) specific enthalpy at condition P
=
specific enthalpy at condition O
= =
13.64 ( 13 - 3) 136.4 kW
Reheater Battery Output in Winter The reheater battery output is as follows: H reheater battery
=
ma ( hH - hM)
where: H reheater battery = ma =
Reheater battery output (kW) mass flow rate of supply air
(kg/s) hH
=
specific enthalpy at condition H
hM
=
specific enthalpy at condition M
(kJ/kg) (kJ/kg)
H reheater battery H reheater battery
= =
17.05 ( 39 - 19) 341 kW
NOTE: The heater battery load may be reduced by using other forms of heating for some of the load, e.g. perimeter convectors or radiators. Humidifier Output in Winter The amount of moisture added to the air may be calculated from the following formula:
m moisture added
=
ma (msS - msH)
where: m moisture added = ma msS msH
= = =
m moisture added m moisture added m moisture added m moisture added m moisture added
The amount of moisture or added or steam flow rate (kg/s) mass flow rate of air (kg/s) moisture content at condition S (kg/kg d.a.) moisture content at condition H (kg/kg d.a.)
= = = = =
17.05 (0.0074 - 0.0035) 17.05 (0.0039) 0.0665 kg/s 0.0665 litre/s 0.0665 litre/s x 3600 = 239.4 litres/hour
Example 7
Summer Cycle with duct and fan gains
A Computer Suite Theatre measures 24 m x 10 m x 3.5 m high. It is to be air conditioned in summer so that the room is maintained at 22oC dry-bulb temperature, 50% saturation. Determine the cooling coil and reheater outputs required. The air conditioning system is shown schematically below. DATA: Outdoor condition is 27oC, 80% saturation. The Apparatus Dew Point ADP is 8oC The internal latent heat gain is 40 W per person, plus additional gain of 5 kW. The internal sensible heat gain is 200 Watts per computer, 100W per person and 15 W/m2 floor area for lights. The solar gain through windows is 6.0 kW Maximum occupancy is 80 people. Number of computers is 80. The cooling coil factor is 0.8. Duct and fans gains are 2oC. Fresh air, recirculated air ratio is 20%/80%.
HEAT GAINS
Sensible (200 x 80) + (100 x 80) + ( 15 x 24 x 10) = 27,600 Watts = 27.6 kW + solar gain 5.0 kW = 32.6 kW Latent (40 x 80) = 3200 Watts = 3.2 kW + other 5kW = 8.2 kW Total heat gain = 32.6 + 8.2 = 40.8 kW Room ratio = 32.6 / 40.8 = 0.8
Mass Flow Rate When the supply air temperature has been found from the psychrometric chart then the mass flow rate of air to offset heat gains can be calculated from the following formula: ma
=
Hs / ( ( tr - ts ) )
where: H s = Sensible heat gain to room (kW) m a = mass flow rate of air (kg/s) C p = Specific heat capacity of humid air (approx.1.01 kJ/kg degC) tr = room temperature (oC) ts = supply air temperature (oC)
The supply air temperature is 14.5oC. ma
=
32.60 / ( 1.01 ( 22 - 14.5 ) )
ma
=
4.3 kg/s
The processes can now be plotted on a psychrometric chart as shown below. From the psychrometric chart point W is at approximately 11oC dry bulb. Point D is 11 oC + 2oC (duct and fan gains given in Data) = 13 oC
Cooling Coil Output The cooling coil output is as follows: H cooling coil =
ma ( hM - hADP)
where: Hcooling coil = Cooling coil output (kW) ma = mass flow rate of air (kg/s) hM = specific enthalpy at condition M (kJ/kg) determined from psychrometric chart. hADP = specific enthalpy at condition ADP (kJ/kg) determined from psychrometric chart
The specific enthalpies at points M and ADP are shown on the psychrometric Chart above. H cooling coil =
4.3 ( 49.5 - 25)
H cooling coil
105.4 kW
=
Reheater Battery Output The heater battery or reheater output is as follows: H heater battery = ma ( hS hD) where: Hheater battery = Heater battery output (kW) ma = mass flow rate of air (kg/s) hS = specific enthalpy at condition S (kJ/kg) determined from psychrometric chart. hD = specific enthalpy at condition D (kJ/kg) determined from psychrometric chart.
The specific enthalpies at points S and D are shown on the psychrometric Chart above. H heater battery =
4.3 ( 33.5 - 32)
H heater battery =
6.5 kW
These are as detailed in the following sections of the notes. Summer and Winter Cycles 1. Summer cycle psychrometrics 2. Summer cycle calculations 3. Winter cycle psychrometrics 4. Winter cycle calculations 5. Duct and Fan gains. The final section is seven examples of plant sizing using psychrometric charts.
Psychrometry for Air Conditioning Psychrometry is the study of air and water vapour mixtures. Air is made up of five main gases i.e. Nitrogen 78.03%, Oxygen 20.99%, Argon 0.94%, Carbon Dioxide 0.03%, and Hydrogen 0.01% by volume. The Ideal Gas Laws are used to determine psychrometric data for air so that the engineer can carry out calculations. To make life easier a chart has been compiled with all the relevant psychrometric data indicated. This is called the Psychrometric Chart. A typical chart is shown below.
Air at any state point can be plotted on the psychrometric chart. The information that can be obtained from a Psychrometric Chart is as follows:
1. 2. 3. 4. 5. 6.
Dry bulb temperature Wet bulb temperature Moisture content Percentage saturation Specific enthalpy Specific volume.
The following is a brief description of each of the properties of air. 1. Dry bulb temperature This is the air temperature measured by a mercury-in-glass thermometer. 2. Wet bulb temperature This is the air temperature measured by a mercury-in-glass thermometer which has the mercury bulb wetted by gauze that is kept moist by a reservoir of water. When exposed to the environment the moisture evaporates from the wetted gauze, which gives a lower reading on the thermometer. This gives an indication of how ‘dry’ or how ‘moist’ the air is, since in ‘dry’ air the water will evaporate quickly from the gauze, which depresses the thermometer reading. 3. Moisture content This is the amount of moisture in air given in kg of moisture per kg of dry air e.g. for room air at 21oC dry bulb and 15oC wet bulb, the moisture content is about 0.008 kg/kg d.a. This is a small mass of moisture ( 0.008 kg = 8 grams) per kg of dry air or 9.5 grams per cubic metre of air. 4. Percentage saturation The Percentage saturation is another indication of the amount of moisture in air. This is the ratio of the moisture content of moist air to the moisture content of saturated air at the same temperature.
When air is saturated it is at 100% saturation and cannot hold any more moisture. 5. Specific enthalpy This is the amount of heat energy (kJ) in air per kg. If heat is added to the air at a heater battery for example, then the amount to be added can be determined from Specific enthalpy change. 6. Specific volume This is the volume of moist air (dry air + water vapour) per unit mass. The units of measurement are m3 per kg. Also specific volume = 1 / density.
The Psychrometric Chart The six properties of air previously discussed can be shown on one chart called a Psychrometric Chart. One of the purposes of the Psychrometric Chart is to size heater batteries, cooling coils and humidifiers. A simplified Psychrometric Chart is shown below.
This chart is only for demonstration purposes. If accurate assessments are to be carried out use a C.I.B.S.E. chart. Using the Psychrometric Chart If any two properties of air are known then the other four can be found from the psychrometric chart.
Examples of Psychrometric Properties EXAMPLE 1 Find the moisture content of air at 25oC dry-bulb temperature and 25oC wetbulb temperature. Referring to the chart below, a vertical line is drawn upwards from 25oC drybulb temperature until it intersects at 25oC wet-bulb temperature. This intersection point happens to be on the 100% saturation line. The intersection point is highlighted and a horizontal line is drawn to the right to find the corresponding moisture content. The moisture content is therefore 0.020 kg/kg dry air.
EXAMPLE 2 Find the specific volume and wet-bulb temperature of air at 20oC dry-bulb temperature and 50% saturation. Referring to the chart below, a vertical line is drawn upwards from 20oC drybulb temperature until it intersects with the 50% saturation curve. The intersection point is sometimes referred to as the state point. The specific volume is found to be 0.84 m3/kg and the wet-bulb temperature is 14oC
EXAMPLE 3 Find the specific volume, percentage saturation and moisture content of air at 15oC dry-bulb temperature and 10oC wet-bulb temperature. Referring to the chart below, a vertical line is drawn upwards from 15oC drybulb temperature until it intersects with the 10oC wet-bulb temperature line. This intersection is the state point. The specific volume is found to be 0.823 m3/kg, the percentage saturation 52% and the moisture content 0.0054 kg/kg d.a.
EXAMPLE 4 Find the specific volume, wet-bulb temperature, moisture content and specific enthalpy of air at 35oC dry-bulb temperature and 30% saturation. Referring to the chart below, a vertical line is drawn upwards from 35oC drybulb temperature until it intersects with the 30% saturation curve. This intersection is the state point.
The specific volume is found to be 0.883 m3/kg, the wet-bulb temperature is 22oC, the moisture content 0.011kg/kg d.a. and the specific enthalpy 65 kJ/kg.
Air Conditioning Plant for Summer and Winter In the summer time when cooling is required by the air conditioning plant it will be necessary to operate the cooling coil, reheater and possibly other plant as well. In winter time the preheater and reheater battery will probably be on to provide warm air to overcome heat losses. Other plant may be switched on as well. These plant items are shown in the diagram below.
Basic Air Conditioning Processes 1. Mixing Where two air streams are mixed the psychrometric process is shown as a straight line between two air conditions on the psychrometric chart, thus points 1 and 2 are ed and the mix point 3 will lie on this line. Two air streams are mixed in air conditioning when fresh air (m1) is brought in from outside and mixed with recirculated air (m2). The resulting air mixture is shown below as (m3). The mixing ratio is fixed by dampers. Sometimes, in more sophisticated plant, modulating dampers are used which are driven by electric motors to control the mixture of air entering the system. The diagrams below show mixing of two air streams.
By the conservation of mass formula: By the conservation of energy formula:
m1 + m2 = m3 m1 h1 + m2 h2 = m3 h3
where: m = mass flow rate of air (kg/s) h = specific enthalpy of air (kJ/kg) found from psychrometric chart.
2. Sensible Cooling and Heating When air is heated or cooled sensibly, that is, when no moisture is added or removed, this process is represented by a horizontal line on a psychrometric chart.
For sensible heating: The amount of heating input to the air approximates to; H1-2 = m x x (t2 - t1) Or more accurately from psychrometric chart: H1-2 = m x (h2 - h1) For sensible cooling: The amount of cooling input to the air approximates to; H2-1 = m x x (t2 - t1) Or more accurately from psychrometric chart: H2-1 = m x (h2 - h1) where: H = Heat or cooling energy (kW) m = mass flow rate of air (kg/s) = Specific heat capacity of air, may be taken as 1.01 kJ/kg degC. t = Dry bulb temperature of air (oC) h = specific enthalpy of air (kJ/kg) found from psychrometric chart.
3. Cooling with Dehumidification The most commonly used method of removing water vapour from air (dehumidification) is to cool the air below its dew point. The dew point of air is when it is fully saturated i.e. at 100% saturation. When air is fully saturated it cannot hold any more moisture in the form of water vapour. If the air is cooled to the dew point air and is still further cooled then moisture will drop out of the air in the form of condensate. This can be shown on a psychrometric chart as air sensibly cooled until it becomes fully saturated (the dew point is reached) and then the air is cooled latently to a lower temperature. This is apparent on the psychrometric chart as a horizontal line for sensible cooling to the 100% saturation curve and then the process follows the 100% saturation curve down to another point at a lower temperature. This lower temperature is sometimes called the Apparatus dew Point (ADP) of the cooling coil. In reality the ADP of the cooling coil is close to the cooling liquid temperature inside the coil. Chilled water or refrigerant may be the cooling liquid. The psychrometric process from state point 1 to 2 to 3 may be shown as a straight line for simplicity as shown above with a yellow line.
The total amount of cooling input to the air approximates to;
H1-3 = m x (h1 - h3) The sensible heat removed is:
H1-2 = m x (h1 - h2) The latent heat removed is:
H2-3 = m x (h2 - h3) where:
H = Cooling energy (kW) m = mass flow rate of air (kg/s) h = specific enthalpy of air (kJ/kg) found from psychrometric chart.
In the absence of a suitable psychrometric chart the following formula may be used; The sensible heat removed is:
H1-2 = m x x (t1 - t2)
The latent heat removed is:
H2-3 = m x hfg x (g2 - g3)
where:
H = Cooling energy (kW) m = mass flow rate of air (kg/s) = Specific heat capacity of air, may be taken as 1.01 kJ/kg
degC. t = Dry bulb temperature of air (oC) hfg = latent heat of evaporisation, may be taken as 2454 kJ/kg @20oC. g = moisture content of air from psychrometric chart (kg/kg dry air)
3.1 Cooling Coil Factor Some of the air going through a cooling coil does not come into with the tubes or fins of the cooling coil and is therefore not cooled to the ADP temperature. A mixing process therefore takes place as two air streams mix downstream of the cooling coil as shown below.
One air stream is cooled down to the ADP and the other air stream by-es the coil surfaces to give an off-coil air temperature (mixed air stream) a little higher than the ADP. This may be looked upon as an inefficiency of the coil and is usually given as the cooling coil factor. The process is shown on the psychrometric chart below.
The factor of a cooling coil may be found from;
Another expression for factor is;
4. Humidification If is it necessary to add some moisture to the supply air then this is best done by injecting steam into the air stream. Humidification can be carried out by spraying a fine mist of water droplets into the air but this is not recommended in rooms occupied by people due to the risk of bacteria carry over. Dry steam may be injected from a steam supply pipe or generated in a local packaged unit as shown in the photograph below. A disadvantage of using an existing steam supply is smells may be carried over into the air.
The steam package unit is situated close to the air duct and is sized to meet the maximum requirements; this is usually in winter in the U.K. A steam pipe ( sometimes hoses are used) es from the packaged unit to the air duct and steam at 100oC is injected into the air stream via. a spargepipe. The un-used steam is drained from the system via a condensate tundish and drain. It is important to layout the steam pipework so that any condensate will drain back to the unit. The psychrometric process is shown below.
See Summer and Winter Cycles section for calculation of amount of moisture added at humidifier.
Typical Air Conditioning Processes The schematic diagram below shows a typical plant system for summer air conditioning.
The psychrometric diagram below shows a typical summer cycle.
The schematic diagram below shows a typical plant system for winter air conditioning.
The psychrometric diagram below shows a typical winter cycle.
Annotation The state points on a psychrometric chart may be given numbers or symbols to identify them. If symbols are used the following system may be adopted:
Room Ratio This is the ratio of sensible to total heat in the room for summer or winter. The total heat gain (summer) or loss (winter) will be determined by adding the Latent and Sensible heat in a room or rooms, i.e. (SUMMER)
Total heat gain
=
Sensible heat gain + Latent heat gain
(WINTER)
Total heat
=
Sensible heat loss + Latent heat gain
The room ratio is used on a psychrometric chart to determine the supply air state point. A room ratio line is superimposed from the protractor on the psychrometric chart onto the main body of the chart by a line ing through the room state point R. An example calculation is as follows:
Sensible heat gain
=
9.0 kW
Latent heat gain
=
2.25 kW
Total heat gain
=
9.0 kW + 2.25 kW
Room ratio
=
Sensible / Total heat
Room ratio
=
9 / 11.25
=
=
11.25 kW.
0.8
The supply air state point must also be somewhere on this room ratio line to meet the room heat gain requirements i.e. the room ratio line always es through points R and S.
Examples of Psychrometric Calculations for Summer and Winter
Example 1. Summer Cycle A room is to be maintained at 22oC dry-bulb temperature, 50% saturation, when the sensible heat gain is 10.8 kW in summer. The latent heat gain is 7.2 kW. Determine the cooling coil and reheater outputs required by using a psychrometric chart if the plant schematic is as shown below. DATA: Outdoor condition is 28oC, 80% saturation. The outdoor air and recirculated air ratio is 20%/80%. The Apparatus Dew Point ADP is 8oC Neglect the cooling coil factor.
Note: The cooling coil output of 38.2 kW is a much higher value than the sensible heat gain of 10.8 kW. It should be ed that the difference is these two values is mostly from the fresh air cooling load. It takes quite a lot of energy in summer to cool fresh air coming into air handling units. This can be minimised by bringing in minimum fresh air but not too little otherwise the building will suffer from lack of oxygen and feel stuffy. Sometimes mistakes are made when sizing cooling apparatus. If a cooling coil or indoor cooling unit is sized on the sensible heat gain only without allowing for fresh air load then it will be grossly undersized. That is why psychrometric charts are required to calculate cooling coil output including fresh air loads.
So, don’t size cooling coil and indoor cooling units on sensible heat gain only if there is fresh air coming into the plant. Size these items of plant using a psychrometric chart.
Example 2. Winter Cycle A room has a 18.0 kW sensible heat loss in winter and a 4.5 kW latent heat gain from the occupants. Determine the supply air temperature and heater battery load using the following information. DATA: Indoor condition: 21oC dry-bulb temperature, 50% saturation. Outdoor condition: -2oC d.b., 80% saturation. The outdoor air and recirculated air ratio is 20%/80%. No preheating or humidification takes place in this simplified example.
Procedure (Winter Cycle) 1. Draw schematic diagram of air-conditioning plant (see above) 2. Plot room condition R on psychrometric chart. 3. Plot outside condition O on psychrometric chart.
4. No Preheater condition P 5. points O and R
6. Find the mix point M by measuring the length of the line O-R and multiply this by the mixing ratio. On a full size CIBSE psychrometric chart this measures 110mm. The ratio of recirculated air is 0. therefore; 110mm x 0.8 = 88mm Measure up the O-R line from point O by 88mm. This determines point M . If there is more recirculated air than outside air at the mix point, then point M will be closer to point R than point O. 7. Find the room ratio. This is the sensible to total heat ratio.
Neglect signs ie. the total heat for the room will be Sensible loss plus Latent gain. Total heat = 18 kW sensible + 4.5 kW latent = 22.5 kW total. Heat ratio = 18 / 22.5 = 0.8 Plot this ratio on the protractor, top segment, on the psychrometric chart and transfer this line onto the chart so that it es through point R.
8. Find the supply air dry bulb temperature by calculation. 9. Plot the supply air condition S on the room ratio line. This is on a horizontal line from point M to the right hand side of the chart, and intersects with the RRL.
The supply air Temperature is found to be 32.5oC.
Supply Air Flow Rate When the sensible heat loss and supply air temperature in winter are known then the mass flow rate of air is calculated from the following formula: Hs = ma x ( ts - tr ) where: Hs = ma =
Sensible heat loss (kW) mass flow rate of air (kg/s)
= degC) tr = o ( C) ts = o ( C)
Specific heat capacity of humid air (approx.1.01 kJ/kg room temperature supply air temperature ..............therefore:
ma = Hs / ( ts - tr ) ma = 18 / 1.01 (32.5 - 21) ma = 18 / 11.615 ma = 1.55 kg/s
Heater Battery Output The heater battery output is as follows: H reheater battery = ma ( hS - hM) where: H reheater battery = Reheater battery output (kW) ma = mass flow rate of air (kg/s) hS = specific enthalpy at condition S (kJ/kg) hM = specific enthalpy at condition M (kJ/kg)
The specific enthalpies at points S and M are shown on the psychrometric chart below.
Hheater battery = ma ( hS - hM) Hheater battery = 1.55 ( 50 - 34) Hheater battery = 24.8 kW Therefore the heater battery load is 24.8 kW.
Example 3. Summer Cycle (Cooling Coil factor) An office is to be maintained at 22oC dry-bulb temperature, 50% saturation in summer. The sensible heat gain is 8.0 kW. The latent heat gain is 2.0 kW. Determine the cooling coil and reheater outputs required by using a psychrometric chart if the plant schematic is as shown below. DATA: Outdoor condition is 28oC, 80% saturation. The outdoor air and recirculated air ratio is; 20% / 80%. The Apparatus Dew Point ADP is 8oC The cooling coil factor is 0.8. (See explanation of factor in page 9)
Procedure (Summer Cycle) 1.
Draw schematic diagram of air-conditioning plant (see above)
2.
Plot room condition R on psychrometric chart.
3.
Plot outside condition O on psychrometric chart.
4.
points O and R.
5. Find the mix point M by measuring the length of the line O-R and multiply this by the mixing ratio. On a full size CIBSE psychrometric chart this measures 85mm. The ratio of recirculated air is 0.8. …….. therefore; 85mm x 0.8 = 68mm Measure down the O-R line from point O by 68mm. This determines point M . 6.
Find the room ratio. This is the sensible to total heat gain ratio. Total heat = 8 kW sensible + 2 kW latent = 10 kW total. Heat ratio = 8 / 10 = 0.8 Plot this ratio on the protractor, bottom segment, on the psychrometric chart and transfer this line onto the chart so that it es through point R. 7.
Plot the Apparatus Dew Point ADP of the cooling coil.
This is on the 100% saturation curve.
8.
The ADP is 80C.
points M and ADP.
9. Find the off-coil condition W by measuring the length of the line MADP and multiply this by the cooling coil factor.. On a full size CIBSE psychrometric chart this measures 75mm. The cooling coil factor is 0.8. ……….. therefore; 75mm x 0.8 = 60mm. Measure down along the line M-ADP by 60mm. This determines point W. 10.
Plot the supply air condition S. The reheater process will be a horizontal line from point W to point S. Point S is on the room ratio line. The supply air temperature is 17oC.
Mass Flow Rate When the supply air temperature has been found from the psychrometric chart then the mass flow rate of air can be calculated from the following formula: ma
=
Hs / ( ( tr - ts ) )
Hs ma
= = =
Sensible heat gain to room (kW) mass flow rate of air (kg/s) Specific heat capacity of humid air (approx.1.01
tr
=
room temperature (oC)
where:
kJ/kg degC)
ts
=
supply air temperature (oC)
The supply air temperature is 17oC. ma
=
8 / ( 1.01 ( 22 - 17 ) )
ma
=
1.584 kg/s
Cooling Coil Output The cooling coil output is as follows: H cooling coil =
ma ( hM - hADP)
where: H cooling coil = Cooling coil output (kW) ma = mass flow rate of air (kg/s) hM = specific enthalpy at condition M (kJ/kg) determined from psychrometric chart. hADP = specific enthalpy at condition ADP (kJ/kg) determined from psychrometric chart
The specific enthalpies at points M and ADP are shown on the psychrometric Chart below. H cooling coil =
1.584 ( 50.5 - 25)
H cooling coil =
40.4 kW
Heater Battery Output The heater battery or reheater output is as follows: H heater battery = where: H heater battery
ma ( hS - hW) =
Heater battery output (kW)
ma = mass flow rate of air (kg/s) hS = specific enthalpy at condition S (kJ/kg) determined from psychrometric chart. hW = specific enthalpy at condition W (kJ/kg) determined from psychrometric chart.
The specific enthalpies at points S and W are shown on the psychrometric Chart below.
H heater battery =
1.584 ( 36.5 - 30.5)
H heater battery =
9.5 kW
Example 4.
Winter Cycle with Humidifier
An conference room is to be maintained at 21oC dry-bulb temperature, 50% saturation in winter. The sensible heat loss for the room is 17.0 kW. The latent heat gain is 40 Watts per person (see Air Conditioning section). Determine the preheater and reheater outputs required and the amount of moisture to be added at the humidifier in litre/hour, by using a psychrometric chart if the plant schematic is as shown below.
DATA: Outdoor condition is -2oC, 80% saturation. The outdoor air and recirculated air ratio is 50%/50%. Maximum occupancy is 250 people. The preheater off coil temperature is 5oC. Supply air quantity is 8 air changes per hour. Room volume is 20 x 12 x 4m high = 960 m3.
Latent Heat Gain
The latent heat gain =
heat gain per person x number of people
The latent heat gain =
40 W/person x 250 =
The latent heat gain =
10 kW
10,000 Watts
Supply Air & Fresh Air Quantities Supply air quantity (m3/h)
=
air change rate x room volume (m3)
Supply air quantity (m3/h)
=
8 x 960(m3)
Supply air quantity (m3/h)
=
7680 (m3/h)
Supply air quantity (m3/s)
=
7680(m3/h) / 3600 = 2.13 (m3/s)
Supply air mass flow rate (kg/s) Volume (m3/kg)
= Supply air quantity (m3/s) / Specific
Supply air mass flow rate (kg/s)
= 2.13 (m3/s) / 0.87 (m3/kg)
Supply air mass flow rate (kg/s)
= 2.45 kg/s
The fresh air flow rate (kg/s)
= 2.45 kg/s x 50% = 1.23 kg/s
Winter Cycle Psychrometrics
1. 2. 3.
4. 5.
6.
Draw schematic diagram of air-conditioning plant (see above). Plot room condition O, M and R on psychrometric chart. Plot the after Preheater condition P. The Preheater process will be a horizontal line from O to P and acts as a frost coil in this case, heating the air to 5oC. points P and R. Find the mix point M by measuring the length of the line P-R and multiply this by the mixing ratio. The line measures 82mm long. 82 x 0.5 = 41 mm Find the room ratio. Plot this ratio on the protractor, so that it es through point R. Total heat is 17 kW sensible + 10 kW latent = 27kW. Ratio is 17/27 = 0.63.
7. Find the supply air dry bulb temperature by calculation. This is found by calculation because we have already calculated the mass flow rate of supply air from information given in the question.
Supply Air Dry Bulb Temperature The temperature of supply air is calculated from the following formula:
Hs where: Hs = ma = =
=
ma x ( ts - tr )
Sensible heat loss from room (kW) mass flow rate of air (kg/s) Specific heat capacity of humid air (approx.1.01 kJ/kg
degC) tr = o ( C) ts =
room temperature supply air temperature (oC)
( ts - tr )
=
Hs /
ma x
( ts - tr )
=
17 / 2.45 x 1.01
( ts - tr )
=
6.87 deg.C
( ts - 21)
=
6.87 deg.C
ts
=
21 + 6.87 deg.C
ts
=
27.87 oC say 28 oC.
8. Plot the supply air condition S on the room ratio line. 9. Plot condition H on the psychrometric chart. This is vertically down from point S, and horizontally across from point M. This is because M-H is the reheater process and thus a horizontal line and H-S is the humidification process and is close to a vertical line if steam is used.
Preheater Battery Output (Or Frost Coil) The preheater battery output is as follows:
H preheater battery = maf ( hP - hO) where: H preheater battery = m af = hP = hO =
Preheater battery output (kW) mass flow rate of fresh air (kg/s) specific enthalpy at condition P (kJ/kg) specific enthalpy at condition O (kJ/kg)
H preheater battery = 1.23 ( 12 - 5.5) H preheater battery = 8.0 kW
Reheater Battery Output The reheater battery output is as follows:
H reheater battery = ma ( hH - hM) where: H reheater battery ma = hH = hM =
= Reheater battery output (kW) mass flow rate of supply air (kg/s) specific enthalpy at condition H (kJ/kg) specific enthalpy at condition M (kJ/kg)
Hreheater battery = 2.45 ( 42 - 27) Hreheater battery = 36.8 kW
Humidifier Output The amount of moisture added to the air may be calculated from the following formula:
m moisture added = ma (msS - msH) where: m moisture added = rate (kg/s)
The amount of moisture or added or steam flow ma m sS
= =
mass flow rate of air (kg/s) moisture content at condition S (kg/kg
m sH
=
moisture content at condition H (kg/kg
d.a.) d.a.)
m moisture added = 2.45 (0.0064 - 0.0054) m moisture added = 2.45 (0.001) m moisture added = 0.00245 kg/s 1 litre of water weights 1 kg, therefore; m moisture added = 0.00245 litre/s
m moisture added = 0.00245 litre/s x 3600 = 8.82 litres/hour
Example 5 Summer Cycle (Air Flows To Be Calculated) A Lecture Theatre measures 15 m x 10 m x 6 m high. It is to be air conditioned in summer so that the room is maintained at 22oC dry-bulb temperature, 50% saturation. Determine the cooling coil and reheater outputs required by using a psychrometric chart if the plant schematic is as shown below. DATA: Outdoor condition is 28oC, 80% saturation. The Apparatus Dew Point ADP is 7.5oC. The latent heat gain is 10.0 kW. The sensible heat gain is 12.0 kW. Maximum occupancy is 200 people. The cooling coil factor is unknown at present and should be calculated. Use CIBSE guide B2 (2001) to determine air flow rates and calculate the mass flow rate of fresh air and supply air to the room. The lecture room may be taken as an Assembly hall. The room is a non-smoking area.
Mass Flow Rates Information from CIBSE Guide B2 (2001) Table 3.3 is as follows (see Ventilation – Ventilation rates section); The recommended outdoor air rate is 8 l/s/person for non-smoking. Information from CIBSE Guide B2 (2001) Table 3.1 is as follows (see Ventilation – Ventilation rates section); Assembly halls and auditorium refers to Table 3.6 (see Ventilation – Ventilation rates section); The recommended total air supply rate is 6 – 10 air changes per hour for high level mechanical strategy. Fresh Air Flow Rate
Fresh air rate = 8 l/s/p x 200 people = 1600 l/s = 1.6 m3/s The specific volume at the outside condition may be determined from a psychrometric chart. It is 0.88 m3/kg. Mass flow rate = Volume flow rate / specific volume Mass flow rate (Fresh Air) = 1.6 / 0.88 = 1.82 kg/s . Supply Air Flow Rate
If the maximum ventilation supply air rate is taken from Table 3.6 to be 10.0 air changes per hour, then the mass flow rate can be calculated. Volume flow rate (m3/h) = Volume of room (m3) x air change rate (ac/h) Volume of room (m3) = 15 x 10 x 6 = 900 m3 Volume flow rate (m3/h) = 900 (m3) x 10 (ac/h) Volume flow rate (m3/h) = 9000 m3/h Volume flow rate (m3/s) = 9000 / 3600 = 2.5 3 m /s. Mass flow rate = Volume flow rate / specific volume The specific volume at the supply condition may be approximated at this stage from a psychrometric chart. It is 0.834 m3/kg. Mass flow rate (Supply Air) = 2.5 / 0.834 = 3.0 kg/s.
Fresh Air And Recirc. Ratio.
The ratio by mass is therefore; Fresh air rate Supply air rate Recirculation air rate
= = =
1.82 kg/s 3.00 kg/s 3.00 - 1.82 = 1.18 kg/s
The ratio of fresh air to total supply air is; 1.82 / 3.00 = 0.6, i.e. 60% fresh air and therefore 40% recirculated air. It is not unusual to have a high percentage of fresh air in a high occupancy room such as a Lecture theatre. The air flows are shown on the schematic diagram below.
Supply Air Temperature by Calculation In this example the supply air temperature will be found by rearranging the following formula:
Hs where: Hs = ma = =
=
ma x ( tr - ts ) Sensible heat gain to room (kW) mass flow rate of air (kg/s) Specific heat capacity of humid air (approx.1.01 kJ/kg
degC) tr = o ( C) ts =
room temperature supply air temperature (oC)
Rearranging the above formula gives: ( tr - ts ) = ( tr - ts ) = ( tr - ts ) = since tr = ts = ts =
Hs / ( ma x ) 12 / ( 3.00 x 1.01 ) 3.96 deg.C 22oC 22 - 3.96 = 18.04 oC 18 oC approx.
The processes can now be plotted on a psychrometric chart as shown below.
1. Points O, M and R can be shown on the chart. 2. Point ADP can be indicated and lines drawn between these points as shown. 3. The room ratio line can be drawn. 4. Point S is then shown on the chart, on the room ratio line at 18oC. 5. A horizontal line is then drawn from point S towards the line O – ADP. 6. Point W can then be found where the horizontal line W - S intersects the line O - ADP. From the psychrometric chart point W is at approximately 9oC dry bulb. The heat ratio is 12 kW sensible / 22 kW total = 0.545.
The specific enthalpies are shown below.
Cooling Coil Output
The cooling coil output is as follows:
H cooling coil
ma (hM - hADP)
=
where: H cooling coil = Cooling coil output (kW) ma = mass flow rate of air (kg/s) hM = specific enthalpy at condition M (kJ/kg) determined from psychrometric chart. hADP = specific enthalpy at condition ADP (kJ/kg) determined from psychrometric chart
The specific enthalpies at points M and ADP are shown on the psychrometric Chart above. H cooling coil =
3.00 ( 64 - 24)
H cooling coil =
120.0 kW
Reheater Battery Output The heater battery or reheater output is as follows:
H heater battery
=
ma ( hS - hW)
where: H heater battery = Heater battery output (kW) ma = mass flow rate of air (kg/s) hS = specific enthalpy at condition S (kJ/kg) determined from psychrometric chart. hW = specific enthalpy at condition W (kJ/kg) determined from psychrometric chart.
The specific enthalpies at points W and S are shown on the psychrometric Chart above. H heater battery = H heater battery =
3.00 ( 36 - 26.5) 28.5 kW
Example 6
Summer Cycle and Winter Cycle
A Concert Hall measures 40 m x 20 m x 8 m high. It is to be air conditioned in summer and winter. Determine the following: o Air flow rates o Supply air temperature by calculation in summer and winter. o Cooling coil factor. o Cooling coil and reheater outputs in summer o Humidifier output in litres/hour. The plant schematic is as shown below.
DATA: Indoor condition all year Outdoor condition summer Outdoor condition winter Maximum occupancy ADP of the cooling coil Fresh air requirement Supply air rate -
22oC dB temperature, 50% saturation. 28oC dB temperature, 80% saturation. -3oC dB temperature, 80% saturation. 1000 people. 8oC 12 l/s/person 8 air changes per hour.
Frost off coil temperature Latent heat gain Sensible heat gain in summer lights, solar & ventilation gains. Sensible heat loss in winter -
7oC. 40 W/person 100 W/person + 20.5 kW fabric, 20 W/m3 air volume (estimated)
Fresh Air Flow Rate
Fresh air rate = 12 l/s/p x 1000 people = 12,000 l/s = 12.0 m3/s The specific volume at the outside condition may be determined from a psychrometric chart. It is approximately 0.88 m3/kg. Mass flow rate = Volume flow rate / specific volume Mass flow rate (Fresh Air) = 12.0 / 0.88 = 13.64 kg/s. Supply Air Flow Rate The ventilation supply air rate is taken from table B2.3 and is given as 8 air changes per hour; the mass flow rate can be calculated. Volume flow rate (m3/h) = Volume of room (m3) x air change rate (ac/h) Volume of room (m3) = 40 x 20 x 8 = 6400 m3 Volume flow rate (m3/h) = 6400 (m3) x 8 (ac/h) Volume flow rate (m3/h) = 51,200 m3/h Volume flow rate (m3/s) = 51,200 / 3600 = 14.22 m3/s. Mass flow rate = Volume flow rate / specific volume The specific volume at the supply condition may be approximated at this stage from a psychrometric chart. It can be taken as 0.834 m3/kg. Mass flow rate (Supply Air) = 14.22 / 0.834 = 17.05 kg/s. Recirculation Air Flow Rate Recirculation air rate Recirculation air rate
= =
Supply air rate - fresh air rate 17.05 - 13.64 = 3.41 kg/s
Fresh Air and Recirc. Ratio. The ratio by mass is therefore; Fresh air rate = 13.64 kg/s Supply air rate = 17.05 kg/s The ratio of fresh air to total supply air is;
13.64 / 17.05 = 0.8, i.e. 80% fresh air and therefore 20% recirculated air. It is not unusual to have a high percentage of fresh air in a high occupancy room such as a Concert Hall. The air flows are shown on the schematic diagram below.
Heat Gains The heat gains are given as: Latent heat gain 40 W/person Sensible heat gain - 100 W/person + 20.5 kW fabric, lights, solar & ventilation gains. Latent gain = 40 W x 1000 people = 40,000 W = 40 kW Sensible gain = 100 W x 1000 people = 100,000 W = 100 kW + 20.5 kW = 120.5 kW. Total heat gain = 120.5 + 40 = 160.5 kW Summer Heat ratio = sensible / total = 120.5 / 160.5 = 0.75 Heat Loss The heat loss in winter is given as 20.0 W/ m3.
Volume of room (m3) = Total heat loss = Total heat (winter) = Winter Heat ratio =
40 x 20 x 8 = 6400 m3 20 x 6400 = 128,000 Watts = 128 kW 128 + 40 (Latent) = 168 kW sensible / total = 128 / 168 = 0.76
Summer Calculations & Process Supply Air Temperature by Calculation In this example the supply air temperature will be found by rearranging the following formula: Hs
=
where: Hs = ma = = kJ/kg degC) tr = ts =
ma x ( tr - ts ) Sensible heat gain to room (kW) mass flow rate of air (kg/s) Specific heat capacity of humid air (approx.1.01 room temperature (oC) supply air temperature (oC)
Rearranging the above formula gives: ( tr - ts ) = Hs / ( ma x ) ( tr - ts ) = 120.5 / ( 17.05 x 1.01 ) ( tr - ts ) = 7.0 deg.C, since tr = 22oC, ts = 22 - 7.0 = 15.0 oC ts = 15 oC The processes can now be plotted on a psychrometric chart as shown below. 1. Points O, M and R can be shown on the chart. 2. Point ADP can be indicated and lines drawn between these points as shown. 3. The room ratio line can be drawn. 4. Point S is then shown on the chart, on the room ratio line at 18oC. 5. A horizontal line is then drawn from point S towards the line O – ADP.
6. Point W can then be found where the horizontal line W - S intersects the line O - ADP.
Cooling Coil Factor On a full size psychrometric chart the length of the line from point M to point ADP is 116mm. The distance from point M to point W is 107mm.
The cooling coil factor is therefore:
The specific enthalpies are shown below.
Cooling Coil Output in Summer The cooling coil output is as follows:
H cooling coil =
ma ( hM - hADP)
where: H cooling coil = Cooling coil output (kW) ma = mass flow rate of air (kg/s) hM = specific enthalpy at condition M (kJ/kg) determined from psychrometric chart. hADP = specific enthalpy at condition ADP (kJ/kg) determined from psychrometric chart
The specific enthalpies at points M and ADP are shown on the psychrometric Chart above. H cooling coil = H cooling coil =
17.05 ( 71 - 25) 784.3 kW
NOTE: The cooling coil output is very high and a lot of energy would be required to provide this amount of cooling. The cooling coil load would probably be spread over several air handling units but the it could be examined with a view to some reduction. The coil output is high because the mass flow rate of supply air is high (17.05 kg/s) and the proportion of fresh air is also high (80%). The mix point M is at approximately 27oC dry-bulb so there is little advantage in recirculation in this instance. It would be advantageous to consider the supply airflow rate to see if a lower rate would be acceptable for this building. If 6 air changes per hour are used as the ventilation rate then this would reduce the mass flow rate of supply air. Also the engineer may consider other methods of air-conditioning a hall with a large volume such as using partial radiant cooling where surfaces are cooled rather than air. Reheater Battery Output in Summer
The heater battery or reheater output is as follows: H heater battery =
ma ( hS - hW)
where: H heater battery = Heater battery output (kW) ma = mass flow rate of air (kg/s) hS = specific enthalpy at condition S (kJ/kg) determined from psychrometric chart. hW = specific enthalpy at condition W (kJ/kg) determined from psychrometric chart.
The specific enthalpies at points W and S are shown on the psychrometric Chart above. H heater battery = 17.05 ( 34 - 28) H heater battery = 102.3 kW
Winter Calculations & Process Winter Cycle Psychrometrics The processes can now be plotted on a psychrometric chart as shown below. 1. Points O, M, P and R can be shown on the chart. 2. points O and P and P and R. 3. Find the mix point M The line measures 81mm long. 81 x 0.8 = 65 mm , the distance from point M to R is 65mm. 4. The room ratio is 0.76 from previous calculation. Draw RRL. 5. Find the supply air dry bulb temperature by calculation.
Supply Air Temperature by Calculation The temperature of supply air is calculated from the following formula: Hs
=
ma x ( ts - tr )
where: Hs = Sensible heat loss from room (kW) ma = mass flow rate of air (kg/s) same as summer rate for constant volume systems. = Specific heat capacity of humid air (approx.1.01 kJ/kg degC) tr = room temperature (oC) ts = supply air temperature(oC)
( ts - tr ) ( ts - tr ) ( ts - tr ) ts ts
= = = = =
Hs / ma x 128 / 17.05 x 1.01 7.43 deg.C 22 + 7.43 deg.C 29.43 oC say 29.5 oC.
6. Plot the condition H at 29.5oC dB. on a horizontal line from M. 7. Plot condition S on a vertical line from H on the RRL. Assume the humidity process is vertical.
Preheater Battery Output (or frost coil) The preheater battery output is as follows: H preheater battery
=
maf ( hP - hO)
where: H preheater battery maf hP (kJ/kg) hO (kJ/kg)
H preheater battery H preheater battery
= = =
Preheater battery output (kW) mass flow rate of fresh air (kg/s) specific enthalpy at condition P
=
specific enthalpy at condition O
= =
13.64 ( 13 - 3) 136.4 kW
Reheater Battery Output in Winter The reheater battery output is as follows: H reheater battery
=
ma ( hH - hM)
where: H reheater battery = ma =
Reheater battery output (kW) mass flow rate of supply air
(kg/s) hH
=
specific enthalpy at condition H
hM
=
specific enthalpy at condition M
(kJ/kg) (kJ/kg)
H reheater battery H reheater battery
= =
17.05 ( 39 - 19) 341 kW
NOTE: The heater battery load may be reduced by using other forms of heating for some of the load, e.g. perimeter convectors or radiators. Humidifier Output in Winter The amount of moisture added to the air may be calculated from the following formula:
m moisture added
=
ma (msS - msH)
where: m moisture added = ma msS msH
= = =
m moisture added m moisture added m moisture added m moisture added m moisture added
The amount of moisture or added or steam flow rate (kg/s) mass flow rate of air (kg/s) moisture content at condition S (kg/kg d.a.) moisture content at condition H (kg/kg d.a.)
= = = = =
17.05 (0.0074 - 0.0035) 17.05 (0.0039) 0.0665 kg/s 0.0665 litre/s 0.0665 litre/s x 3600 = 239.4 litres/hour
Example 7
Summer Cycle with duct and fan gains
A Computer Suite Theatre measures 24 m x 10 m x 3.5 m high. It is to be air conditioned in summer so that the room is maintained at 22oC dry-bulb temperature, 50% saturation. Determine the cooling coil and reheater outputs required. The air conditioning system is shown schematically below. DATA: Outdoor condition is 27oC, 80% saturation. The Apparatus Dew Point ADP is 8oC The internal latent heat gain is 40 W per person, plus additional gain of 5 kW. The internal sensible heat gain is 200 Watts per computer, 100W per person and 15 W/m2 floor area for lights. The solar gain through windows is 6.0 kW Maximum occupancy is 80 people. Number of computers is 80. The cooling coil factor is 0.8. Duct and fans gains are 2oC. Fresh air, recirculated air ratio is 20%/80%.
HEAT GAINS
Sensible (200 x 80) + (100 x 80) + ( 15 x 24 x 10) = 27,600 Watts = 27.6 kW + solar gain 5.0 kW = 32.6 kW Latent (40 x 80) = 3200 Watts = 3.2 kW + other 5kW = 8.2 kW Total heat gain = 32.6 + 8.2 = 40.8 kW Room ratio = 32.6 / 40.8 = 0.8
Mass Flow Rate When the supply air temperature has been found from the psychrometric chart then the mass flow rate of air to offset heat gains can be calculated from the following formula: ma
=
Hs / ( ( tr - ts ) )
where: H s = Sensible heat gain to room (kW) m a = mass flow rate of air (kg/s) C p = Specific heat capacity of humid air (approx.1.01 kJ/kg degC) tr = room temperature (oC) ts = supply air temperature (oC)
The supply air temperature is 14.5oC. ma
=
32.60 / ( 1.01 ( 22 - 14.5 ) )
ma
=
4.3 kg/s
The processes can now be plotted on a psychrometric chart as shown below. From the psychrometric chart point W is at approximately 11oC dry bulb. Point D is 11 oC + 2oC (duct and fan gains given in Data) = 13 oC
Cooling Coil Output The cooling coil output is as follows: H cooling coil =
ma ( hM - hADP)
where: Hcooling coil = Cooling coil output (kW) ma = mass flow rate of air (kg/s) hM = specific enthalpy at condition M (kJ/kg) determined from psychrometric chart. hADP = specific enthalpy at condition ADP (kJ/kg) determined from psychrometric chart
The specific enthalpies at points M and ADP are shown on the psychrometric Chart above. H cooling coil =
4.3 ( 49.5 - 25)
H cooling coil
105.4 kW
=
Reheater Battery Output The heater battery or reheater output is as follows: H heater battery = ma ( hS hD) where: Hheater battery = Heater battery output (kW) ma = mass flow rate of air (kg/s) hS = specific enthalpy at condition S (kJ/kg) determined from psychrometric chart. hD = specific enthalpy at condition D (kJ/kg) determined from psychrometric chart.
The specific enthalpies at points S and D are shown on the psychrometric Chart above. H heater battery =
4.3 ( 33.5 - 32)
H heater battery =
6.5 kW
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