Heat Release Rate 646vd

HEAT RELEASE RATE CALCULATIONS

COMBUSTION THERMODYNAMICS (Assignment)

Harish.C Gopinath.M

2013602004 2013602030

Need for “Heat release analysis” • It is very easy to obtain pressure data of an engine with good accuracy and resolution using piezoelectric sensors • By doing heat release analysis, the pressure data can be used directly related to quantify the heat released by combustion • It can be used to study about cycle to cycle comparison • It can be used to evaluate the performance of engine for various biodiesel blends

Assumptions • All the processes are Quasi Static, i.e. the properties will be same throughout the control volume at any given instant • Suction starts at TDC and ends when piston is at BDC • Compression begins when piston is at BDC and ends at TDC • Mass of fuel injected is not considered • Expansion starts at TDC and ends at BDC

• Suction and Exhaust are assumed to follow Pvg = constant • Working medium is an ideal gas

Heat release analysis • To start with, First aw of thermodynamics is applied for the closed period (from start of compression to end of expansion stroke) dQch = dUs + dQht + dW + Shidmi • • • • •

dQchdUs dQht dW Shidmi -

Chemical energy released by combustion Change in sensible energy of the charge Wall heat transfer Piston work Mass flux term ( Fuel injection and crevice flows)

• This is the equation for Gross heat release rate • Gross heat release is the total heat energy released by the combustion process

Net heat release rate • Net heat release rate = Gross heat release rate – ( Heat transferred to walls + Crevice effects + Fuel Vaporisation & heat up ) • Crevice effects, Fuel vaporisation and heat up are neglected which makes Net heat release rate = Sensible internal energy change of charge + Piston work ie

dQ net = dUs + dW

(1)

dUs =

m Cv dt

(2)

dW =

P dV

(3)

PV =

mRT

(4)

P dV + V dP = mR dT

(5)

Differentiating (4) gives

Cont …. • Substituting (2), (3) & (4) in (1) and simplifying, we get dQ net = (γ/γ-1)P(dV/dt)+(1/γ-1)V(dP/dt)

(5)

Calculations Given data Kirloskar diesel engine 80% 4 Stroke, single cylinder Bore X Stroke = 80mm X 110mm Compression Ratio = 16:1

Connecting rod length = 130mm Speed = 1500 rpm

Calculations (cont …) • From the given engine geometry, instantaneous volume for cylinder is found out using formula • Rate of change of volume with respect to crank angle is also computed • From the given Pressure data, change of pressure wrt crank angle is computed • Initially when the piston is at TDC, temperature is assumed to be 400K • Using ideal gas equation, mass of residual gas is found out • Instantenious mass is found out by applying mass flow rate formula up to crank angle 194 degrees ( end of suction) • For the exhaust stroke also, mass flow rate formula is used to find mass inside the cylinder

Cont ….. • Temperature for the whole cycle is found out by using ideal gas equation PV = mRT • Instantaneous area of cylinder is calculated from volume data and the engine geometry details

• Instantaneous pressure change was calculated using • Instantaneous cylinder volume was calculated using

Cont ….. • Gamma is a function of temperature, It is found out by

Cont…. Mass flow rate calculations

Assumed Valve diameter = 20mm Maximum valve lift = 8mm

Heat transfer rate calculations Net heat release rate The datas are substituted in following equation and Qnet is found out dQ net = (γ/γ-1)PdV+(1/γ-1)VdP Wall heat transfer dQw = h Awall (Tw – T) (60/1500 X 2Pi) Correlation used here is Eichelbergs correlation h = 0.00767 Sp0.333 (PT)0.5 Sp= Specific speed (2 (stroke) N/60)(m/s) Gross heat release rate dQGross = dQ net + dQw

RESULTS This graph shows dp/dt

This graph shows instantaneous volume

This graph shows dv/dt

This graph shows instantaneous γ

Curtain area calculated

This graph shows instantaneous mass

Mass flow rate

This graph shows instantaneous temperature

This graph shows the heat release comparison (Qnet,Qloss & Qgross)

Inference from graph •

•

dp/dt

Max Volume = 6×10^-4 m^3

dv/dt

Min Volume = 0.4×10^-4 m^3

•

Net heat release (Qnet) Max value = 1450 (J/rads)

Max value = 1.41

Min value = -400 (J/rads) •

Gross heat release

Instantaneous Mass

Max value = 1450 (J/rads)

Max = 6.8×10^-4 Kg

Min value = -400 (J/rads)

Instantaneous Pressure Max = 6×10^6 N/m^2

•

•

Instantaneous γ

Min value = 1.31 •

Instantaneous Volume

Max Pressure=1.55×10^7 (N/m^2)

Max Volume=300 (cm^3/rad)

•

•

•

Heat loss Max value = 140 (J/rads)

Instantaneous Temperature initial temp = 400 K (assumed)

Max temp = 1650 K Exhaust temp = 650 K

All the values are obtained from graph for Kriloskar 80% data

Reference • “ Fundamentals of Internal combustion engines” by J.B.Haywood • “engine heat release via spread sheet” by C. E. Goering • “Simulation of SI engines” By V Ganesan

Thank you