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COMBUSTION THERMODYNAMICS (Assignment)
HEAT RELEASE RATE
CALCULATIONS
Harish.C 20
Gopinath.M 20
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Need for Heat release analysis
It is very easy to obtain pressure data of an engine with good accuracy and repiezoelectric sensors
By doing heat release analysis, the pressure data can be used directly related to qureleased 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
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Assumptions
All the processes are Quasi Static, i.e. the properties will be same throughout the conany 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
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Heat release analysis
To start with, First aw of thermodynamics is applied for the closed period (frcompression to end of expansion stroke)
dQch
= dUs+ dQ
ht+ dW + Sh
idm
i
dQch
- Chemical energy released by combustion
dUs
- Change in sensible energy of the charge
dQht
- Wall heat transfer
dW - Piston work
Shidm
i- Mass flux term ( Fuel injection and crevice
This is the equation for Gross heat release rate
Gross heat release is the total heat energy released by the combustion process
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Net heat release rate
Net heat release rate = Gross heat release rate( Heat transferred to walls + Crevice 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 dQnet
= dUs+ dW (1)
dUs= m Cv dt (2)
dW = P dV (3)
PV = mRT (4)Differentiating (4) gives
P dV + V dP = mR dT (5)
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Cont .
Substituting (2), (3) & (4) in (1) and simplifying, we get
dQnet
= (/-1)P(dV/dt)+(1/-1)V(dP/dt) (5)
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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
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Calculations (cont )
From the given engine geometry, instantaneous volume for cylinder is found out usin
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 ang
( end of suction) For the exhaust stroke also, mass flow rate formula is used to find mass inside the cy
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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 geomet
Instantaneous pressure change was calculated using
Instantaneous cylinder volume was calculated using
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Cont ..
Gamma is a function of temperature, It is found out by
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Cont.
Assumed Valve diameter = 20mm
Maximum valve lift = 8mm
Mass flow rate calculations
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Heat transfer rate calculations
Net heat release rate
The datas are substituted in following equation and Qnet is found out
dQnet
= (/-1)PdV+(1/-1)VdP
Wall heat transfer
dQw =h Awall(TwT) (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
= dQnet
+ dQw
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RESULTS
This graph shows dp/dt
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This graph shows instantaneous volume
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This graph shows dv/dt
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This graph shows instantaneous
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Curtain area calculated
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This graph shows instantaneous mass
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Mass flow rate
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This graph shows instantaneous temperature
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This graph shows the heat release comparison (Qnet,Qloss & Qgross)
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Inference from graph
dp/dt
Max Pressure=1.5510^7 (N/m^2)
dv/dt
Max Volume=300 (cm^3/rad)
Instantaneous
Max value = 1.41
Min value = 1.31
Instantaneous Mass
Max = 6.810^-4 Kg
Instantaneous Pressure
Max = 610^6 N/m^2
Instantaneous Temperature
initial temp = 400 K (assumed)
Max temp = 1650 K
Exhaust temp = 650 K
Instantaneous Volume
Max Volume = 610^-4 m^3
Min Volume = 0.410^-4 m^3
Net heat release (Qnet)
Max value = 1450 (J/rads)
Min value = -400 (J/rads)
Gross heat release
Max value = 1450 (J/rads)
Min value = -400 (J/rads)
Heat loss
Max value = 140 (J/rads)
All the values are obtained from graph for Krilosk
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Reference
Fundamentals of Internal combustion enginesby J.B.Haywood
engineheat release via spread sheetby C. E. Goering
Simulationof SI enginesBy V Ganesan
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Thank