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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