Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
9th Edition
ISBN: 9781259822674
Author: Yunus A. Cengel Dr., Michael A. Boles
Publisher: McGraw-Hill Education
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Chapter 9.12, Problem 164RP

a)

To determine

The net power output of the cycle.

a)

Expert Solution
Check Mark

Answer to Problem 164RP

The net power output of the cycle is 6488kW.

Explanation of Solution

Draw the Pν for an ideal Otto cycle as shown in Figure (1).

Thermodynamics: An Engineering Approach, Chapter 9.12, Problem 164RP

Write the expression for compression ratio to calculate the clearance volume or one cylinder.

r=νc+νd/4νc (I)

Here, clearance volume is νc, displacement volume for four cylinders νd.

Write the expression to calculate the volume at state 1.

ν1=νc+νd4 (II)

Write the expression to calculate the mass of the air.

P1ν1=mRT1m=P1ν1RT1 (III)

Write temperature and specific volume relation for the isentropic compression process 1-2.

T2=T1(v1v2)k1 (IV)

Write the pressure, temperature, and specific volume relation for isentropic compression process 1-2.

P2v2T2=P1v1T1P2=(v1v2)T2T1×P1 (V)

Here, the temperature at state 1 is T1, the temperature at state 2 is T2, specific volume at state 1 is v1, specific volume at state 2 is v2, and the specific heat ratio is k.

Write the expression for heat addition process 2-3(Q23,in).

Q23,in=mcv(T3T2) (VI)

Here, temperature at state 3 is T3 and specific heat at constant volume is cv.

Write the temperature, pressure, and specific volume relation for the constant volume heat addition process 2-3.

P3v3T3=P2v2T2

For process 2-3, v2 and v3 are equal.

P3=T3T2P2 (VII)

Conclusion:

From Table A-1, “Ideal-gas specific heats of various common gases”, obtain the value of gas constant (R) of air as 0.287kPam3/kgK.

Substitute 11 for r and 1.8L for νd in Equation (I).

11=νc+1.8L4νc11=νc+1.8L(1m31000L)4νc11=νc+0.0018m34νcνc=ν2=0.000045m3

Substitute 0.000045m3 for νc and 1.8L for νd in Equation (II).

ν1=0.000045m3+1.8L4=0.000045m3+1.8L4(1m31000L)=0.000495m3

Substitute 90kPa for P1 , 0.000495m3 for ν1 , 0.287kPam3/kgK for R and 50°C for T1 in Equation (III).

m=90kPa×0.000495m30.287kPam3/kgK×50°C=90kPa×0.000495m30.287kPam3/kgK×(50+273)K=0.0004805kg

Substitute 50°C for, 11 for (v1v2), and 0.35 for k in Equation (IV).

T2=50°C(11)1.351=(50+273)K(11)1.351=747.6K

Substitute 11 for (v1v2), 90kPa for P1, 747.6K for T2, and 50°C for T1 to find P2 in Equation (V).

P2=(11)(747.6K50°C)(90kPa)=(11)(747.6K(50+273)K)(90kPa)=2292kPa

Substitute 0.5kJ for Q23,in, 0.0004805kg for m , 0.821kJ/kgK for cv, 747.6K for T2 in Equation(VI).

0.5kJ=(0.0004805kg)(0.821kJ/kgK)(T3747.6K)T3=2015K

Substitute 2015K for T3, 747.6K for T2, and 2292kPa for P2 in Equation (VIII).

P3=2015K747.6K(2292kPa)=6176kPa

b)

To determine

The net work per cycle per cylinder and the thermal efficiency of the cycle.

b)

Expert Solution
Check Mark

Answer to Problem 164RP

The net work per cycle per cylinder is 0.284kJ.

The thermal efficiency of the cycle is 56.8%.

Explanation of Solution

Write the temperature and specific volume relation for isentropic expansion process 3-4

T4=T3(v3v4)k1 (VIII)

Here, temperature at state 4 is T4, and specific volume at state 4 is v4.

Write the expression for heat rejection process 4-1, (qout).

Q41,out=mcv(T4T1) (IX)

Write the expression to calculate the net power output Wnet.

Wnet=Q23,inQ41,out (X)

Write the expression to calculate the thermal efficiency of the cycle (ηth).

ηth=WnetQin (XI)

Conclusion:

Substitute 1734K for T3 , 111 for (v3v4), and 1.35 for k in Equation (VIII).

T4=(1734K)(111)1.351=1734K×(111)0.35=870.4K

Substitute 0.0004805kg for m , 0.821kJ/kgK for cv, 870.5K for T4, 50°C for T1 in Equation(IX).

Q41,out=0.0004805kg×0.821kJ/kgK(870.5K50°C)=0.0004805kg×0.821kJ/kgK(870.5K(50+273)K)=0.216kJ

Substitute 0.5kJ for Q23,in and 0.216kJ for Q41,out in Equation (X).

Wnet=0.5kJ0.216kJ=0.284kJ

Thus, the net work per cycle per cylinder is 0.284kJ.

Substitute 0.284kJ for Wnet, and 0.5kJ for Qin in Equation(XI).

ηth=0.284kJ0.5kJ=0.568×100%=56.8%

Thus, the thermal efficiency of the cycle is 56.8%.

c)

To determine

The mean effective pressure of the cycle.

c)

Expert Solution
Check Mark

Answer to Problem 164RP

The mean effective pressure of the cycle is 631.1kPa.

Explanation of Solution

Write the expression to calculate the mean effective pressure for an ideal otto cycle (MEP).

MEP=Wnetv1v2 (XII)

Here, the compression ratio is r.

Conclusion:

Substitute, 0.284kJ for Wnet, 0.000045m3 for ν2, and 0.000495m3 for v1 in Equation (XII).

MEP=0.284kJ0.000495m30.000045m3=0.284kJ(1kPam3kJ)0.000495m30.000045m3=631.1kPa

Thus, the mean effective pressure of the cycle is 631.1kPa.

d)

To determine

The power output for an engine speed of 3000 rpm.

d)

Expert Solution
Check Mark

Answer to Problem 164RP

The power output for an engine speed of 3000 rpm is 28.4W.

Explanation of Solution

Write the expression to calculate the power produced by the engine (W˙net).

W˙net=(ncyl×Wnet)n˙nrev (XIII)

Here, speed of the engine is n˙, and number of revolutions per cycle in a four-stroke engine is nrev.

Here, the compression ratio is r.

Conclusion:

In one cycle there are two revolutions in four stroke engines.

Substitute 4cylinder for ncyl , 0.284kJ/cylinder-cycle for Wnet , 3000rev/min for n˙ and 2rev/cycle in Equation (XIII).

W˙net=(4cylinder×0.284kJ/cylinder-cycle)3000rev/min2rev/cycle=(4cylinder×0.284kJ/cylinder-cycle)3000rev/min(1min60s)2rev/cycle=28.4W

Thus, the power output for an engine speed of 3000 rpm is 28.4W.

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Chapter 9 Solutions

Thermodynamics: An Engineering Approach

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