CONNECT FOR THERMODYNAMICS: AN ENGINEERI
CONNECT FOR THERMODYNAMICS: AN ENGINEERI
9th Edition
ISBN: 9781260048636
Author: CENGEL
Publisher: MCG
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Chapter 9.12, Problem 153P

(a)

To determine

The maximum temperature in the cycle.

(a)

Expert Solution
Check Mark

Answer to Problem 153P

The maximum temperature in the cycle is 2308K_.

Explanation of Solution

Draw the Pv diagram of four stroke as in Figure (I).

CONNECT FOR THERMODYNAMICS: AN ENGINEERI, Chapter 9.12, Problem 153P

Refer to Table A-2b, obtain the properties of air at 1000 K.

constantpressurespecificheat,cp=1.142kJ/kgKconstantvolumespecificheat,cv=0.855kJ/kgKGasconstant,R=0.287kJ/kgKspecific heatratio,k=1.336

Express the compression ratio.

r=vc+vdvc (I)

Here, clearance volume is vc and displacement volume is vd.

Express the total volume of the engine at the beginning of compression process (state 1).

v1=vc+vd (II)

Process 1-2: Isentropic compression

Calculate the temperature at state 2.

T2=T1(v1v2)k1=T1(r)k1 (III)

Here, temperature at state 1 is T1, volume at state 1 and 2 is v1andv2 respectively.

Calculate the pressure at state 2.

P2=P1(v1v2)k=P1(r)k (IV)

Here, pressure at state 1 is P1.

Process 2-x and x-3: Constant-volume and constant pressure heat addition processes

Calculate the temperature at x state.

Tx=T2PxP2 (V)

Here, pressure at x state is Px.

Calculate the heat addition to the process 2-x.

q2x=cv(TxT2) (VI)

Here, constant volume specific heat is cv.

Calculate the heat addition to the process x-3.

q2x=qx3q2x=cp(T3Tx) (VII)

Here, constant pressure specific heat is cp.

Conclusion:

Substitute 16 for r and 1.8 L for vd in Equation (I).

16=vc+1.8Lvc16=vc+1.8L×0.001m3Lvcvc=0.00012m3

The value of vc=v2=vx.

Substitute 0.00012m3 for vc and 1.8 L for vd in Equation (II).

v1=0.00012m3+1.8L=0.00012m3+1.8L×0.001m31L=0.00192m3=v4

Substitute 343 K for T1, 16 for r, and 1.336 for k in Equation (III).

T2=(343K)(16)1.3361=870.7K

Substitute 95 kPa for P1, 16 for r, and 1.336 for k in Equation (IV).

P2=(95kPa)(16)1.336=3859kPa

Substitute 3859 kPa for P2, 7500 kPa for Px, and 870.7 K for T2 in Equation (V).

Tx=(870.7K)7500kPa3859kPa=1692K

Substitute 0.855kJ/kgK for cv, 1692 K for Tx, and 870.7 K for T2 in Equation (VI).

q2x=0.855kJ/kgK(1692K870.7K)=702.6kJ/kg

Substitute 1.142kJ/kgK for cp, 1692 K for Tx, and 702.6 kJ/kg for q2x in Equation (VII).

702.6kJ/kg=1.142kJ/kgK(T31692K)T3=2308K

Thus, the maximum temperature in the cycle is 2308K_.

(b)

To determine

The net work output.

The thermal efficiency.

(b)

Expert Solution
Check Mark

Answer to Problem 153P

The net work output is 835.8kJ/kg_.

The thermal efficiency is 59.5%_.

Explanation of Solution

Express the total heat addition to the process.

qin=q2x+qx3 (VIII)

Calculate the volume at state 3.

v3=vxT3Tx (IX)

Here, volume at state x is vx, temperature at state 3 and x are T3andTx respectively.

Process 3-4: Isentropic expansion

Calculate the temperature at state 4.

T4=T3(v3v4)k1 (X)

Here, volume at state 3 and 4 are v3 and v4 respectively.

Calculate the pressure at state 4.

P4=P3(v3v4)k (XI)

Here, pressure at state 3 and 4 are P3 and P4 respectively.

Process 4-1: Constant volume heat rejection

Calculate the heat rejection.

qout=cv(T4T1) (XII)

Calculate the net work output.

wnet,out=qinqout (XIII)

Calculate the thermal efficiency.

ηth=wnet,outqin (XIV)

Conclusion:

Substitute 702.6 kJ/kg for q2x and 702.6 kJ/kg for qx3 in Equation (VIII).

qin=702.6kJ/kg+702.6kJ/kg=1405kJ/kg

Substitute 0.00012m3 for vx, 2308 K for T3, and 1692 K for Tx in Equation (IX).

v3=(0.00012m3)2308K1692K=0.0001636m3

Substitute 2308 K for T3, 0.0001636m3 for v3, 1.336 for k, and 0.00192m3 for v4 in Equation (X).

T4=2308K(0.0001636m30.00192m3)1.3361=1.009K

Substitute 7500 kPa for P3, 0.0001636m3 for v3, 1.336 for k, and 0.00192m3 for v4 in Equation (XI).

P4=7500kPa(0.0001636m30.00192m3)1.336=279.4kPa

Substitute 0.855kJ/kgK for cv, 1009 K for T4, and 343 K for T1 in Equation (XII).

qout=0.855kJ/kgK(1009K343K)=569.3kJ/kg

Substitute 569.3 kJ/kg for qout and 1405 kJ/kg for qin in Equation (XIII).

wnet,out=1405kJ/kg569.3kJ/kg=835.8kJ/kg

Thus, the net work output is 835.8kJ/kg_.

Substitute 835.8 kJ/kg for wnet,out and 1405 kJ/kg for qin in Equation (XIV).

ηth=835.8kJ/kg1405kJ/kg=0.5948=59.5%

Thus, the thermal efficiency is 59.5%_.

(c)

To determine

The mean effective pressure.

(c)

Expert Solution
Check Mark

Answer to Problem 153P

The mean effective pressure is 860.4kPa_.

Explanation of Solution

Calculate the mass.

m=P1v1RT1 (XV)

Calculate the mean effective pressure.

MEP=mwnet,outv1v2 (XVI)

Conclusion:

Substitute 95 kPa for P1, 0.00192m3 for v1, 0.287kPam3kgK for R, and 343 K for T1 in Equation (XV).

m=95kPa(0.00192m3)(0.287kPam3kgK)343K=0.001853kg

Substitute 0.001853 kg for m, 835.8kJ/kg for wnet,out, 0.00192m3 for v1, and 0.00012m3 for v2 in Equation (XVI).

MEP=0.001853kg(835.8kJ/kg)0.00192m30.00012m3=0.001853kg(835.8kJ/kg)0.00192m30.00012m3(kPam3kJ)=860.4kPa

Thus, the mean effective pressure is 860.4kPa_.

(d)

To determine

The power for engine speed of 3500 rpm.

(d)

Expert Solution
Check Mark

Answer to Problem 153P

The power for engine speed of 3500 rpm is 28.39kW_

Explanation of Solution

Calculate the power for engine speed of 3500 rpm.

W˙net=mwnetn˙2 (XVII)

Here, engine speed is n˙.

Conclusion:

Substitute 2200 rev/min for n˙, 0.001853 kg for m, and 835.8kJ/kg for wnet,out in Equation (XVII).

W˙net=0.001853kg(835.8kJ/kg)2200rev/min2=0.001853kg(835.8kJ/kg)2200rev/min2(1min60s)=28.39kW

Thus, the power for engine speed of 3500 rpm is 28.39kW_.

(e)

To determine

The second law efficiency of the cycle.

The rate of exergy output with the exhaust gases.

(e)

Expert Solution
Check Mark

Answer to Problem 153P

The second law efficiency of the cycle is 68.3%_.

The rate of exergy output with the exhaust gases is 10.30kW_.

Explanation of Solution

Express the maximum thermal efficiency of the cycle.

ηmax=1T0T3 (XVIII)

Here, dead state temperature is T0.

Express the second law efficiency of the cycle.

ηII=ηthηmax (XIX)

Calculate the rate of exergy of the exhaust gases.

X˙4=mx4n˙2=m[(u4u0+P0(v4v0)T0(s4s0))]n˙2=m[cv(T4T0)+P0(v4v0)T0[cplnT4T0RlnP4P0]]n˙2=m[cv(T4T0)+P0(V4mRT0P0)T0[cplnT4T0RlnP4P0]]n˙2 (XX)

Here, specific internal energy at state 4, dead state is u4,u0, specific entropy at state 4 and dead state is s4,s0, specific volume at dead state is v0, and pressure at dead state is P0.

Conclusion:

Substitute 25°C for T0 and 2308 K for T3 in Equation (XVIII).

ηmax=125°C2308K=1(25+273)K2308K=0.8709

Substitute 0.8709 for ηmax and 0.5948 for ηth in Equation (XIX).

ηII=0.59480.8709=0.683=68.3%

Thus, the second law efficiency of the cycle is 68.3%_.

Substitute 0.855kJ/kgK for cv, 2200 rev/min for n˙, 1009 K for T4, 298 K for T0, 100 kPa for P0, 0.00192m3 for V4, 0.001853 kg for m, 0.287kJ/kgK for R, 279.4kPa for P4, and 1.142kJ/kgK for cp in Equation (XX).

X˙4=0.001853kg{0.855kJ/kgK(1009K298K)+100kPa(0.00192m30.001853kg(0.287kJ/kgK)298K100kPa)298K[1.142kJ/kgKln1009K298K0.287kJ/kgKln279.4kPa100kPa]}2200rev/min2=0.001853kg(303.3kJ/kg)2200rev/min2(1min60s)=10.30kW

Thus, the rate of exergy output with the exhaust gases is 10.30kW_.

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Chapter 9 Solutions

CONNECT FOR THERMODYNAMICS: AN ENGINEERI

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