CONNECT FOR THERMODYNAMICS: AN ENGINEERI
CONNECT FOR THERMODYNAMICS: AN ENGINEERI
9th Edition
ISBN: 9781260048636
Author: CENGEL
Publisher: MCG
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Question
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Chapter 9.12, Problem 167RP

(a)

To determine

The temperature at the end of expansion process.

(a)

Expert Solution
Check Mark

Answer to Problem 167RP

The temperature at the end of expansion process is 1098K_.

Explanation of Solution

Determine the state 2 temperature in the polytropic compression process 1-2.

T2=T1(v1v2)n1 (I)

Here, the state 1 temperature is T1, the specific volume at state 1 is v1, the specific volume at state 2 is v2, and the polytropic constant is n.

Determine the state 2 pressure in the polytropic compression process 1-2.

P2=P1(v1v2)n (II)

Here, the state 1 pressure is P1, the specific volume at state 1 is v1, and the specific volume at state 2 is v2.

Determine the work per unit mass in the polytropic compression process 1-2.

w12=R(T2T1)1n (III)

Here, the universal gas constant is R.

Determine the state 3 temperature in the constant volume heat addition process 2-3.

T3=T2(P3P2) (IV)

Here, the state 2 temperature is T2 and the state 3 pressure is P3.

Determine the heat transfer per unit mass in the constant volume heat addition process 2-3.

qin=u3u2=cv(T3T2) (V)

Here, the specific heat of constant volume is cv.

Determine the state 4 temperature in the polytropic expansion process 3-4.

T4=T3(v3v4)n1 (VI)

Here, the specific volume at state 3 is v3 and the specific volume at state 4 is v4.

Determine the state 4 pressure in the polytropic expansion process 3-4.

P4=P3(v3v4)n (VII)

Here, the state 3 pressure is P3.

Determine the work per unit mass in the polytropic compression process 3-4.

w34=R(T4T3)1n (VIII)

Here, the universal gas constant is R.

Conclusion:

From the Table A-2 (a), “Ideal-gas specific heats of various common gases”, obtain the value of universal gas constant of air is 0.287kJ/kgK.

Refer to Table A-2 (b), “Ideal-gas specific heats of various common gases”, obtain the below properties at the average temperature of 850 K using interpolation method of two variables.

Write the formula of interpolation method of two variables.

y2=(x2x1)(y3y1)(x3x1)+y1 (I)

Here, the variables denote by x and y are temperature and specific heat of constant pressure.

Show the temperature at 800 K and 900 K as in Table (1).

S. No

Temperature, K

(x)

specific heat of constant pressure, kJ/kgK

(y)

1800 K1.099
2850 Ky2=?
3900 K1.121

Calculate specific heat of constant pressure at an average temperature of 850 K for liquid phase using interpolation method.

Substitute 800 K for x1, 850 K for x2, 900 K for x3, 1.099kJ/kgK for y1, and 1.121kJ/kgK for y3 in Equation (I).

y2=(850K800K)(1.121kJ/kgK1.099kJ/kgK)(900K800K)+1.099kJ/kgK=(50K)(0.022kJ/kgK)(100K)+1.099kJ/kgK=1.110kJ/kgK

From above calculation the specific heat of constant pressure is 1.110kJ/kgK at a temperature of 850 K.

Similarly repeat the interpolation method for specific heat of constant volume and ratio of specific heat as:

cv=0.823kJ/kgK.

k=1.349.

Substitute 37°C for T1, 11 for v1/v2, and 1.3 for n in Equation (I).

T2=(37°C)×(11)1.31=(37°C+273)×(11)0.3=(310K)×(2.053136)=636.47K

    636.5K

Substitute 100kPa for P1, 11 for v1/v2, and 1.3 for n in Equation (II).

P2=(100kPa)×(11)1.3=(100kPa)×(22.5845)=2258K

Substitute 0.287kJ/kgK for R, 636.5K for T2, 37°C for T1, and 1.3 for n in Equation (III).

w12=(0.287kJ/kgK)(636.5K37°C)11.3=(0.287kJ/kgK)(636.5K(37°C+273))11.3=(0.287kJ/kgK)(636.5K310K)11.3=(0.287kJ/kgK)(326.5K)0.3

      =312.3kJ/kg

Substitute 636.5K for T2, 8 MPa for P3, and 2258kPa for P2 in Equation (IV).

T3=(636.5K)×(8MPa2258kPa)=(636.5K)×(8MPa×(1000kPa1MPa)2258kPa)=(636.5K)×(8000kPa2258kPa)=2255K

Substitute 0.823kJ/kgK for cv, 2255 K for T3, and 636.5 K for T2 in Equation (V).

qin=(0.823kJ/kgK)×(2255636.5)K=(0.823kJ/kgK)×(1618.5K)=1332kJ/kg

Substitute 2255K for T3, 1/11 for v3/v4, and 1.3 for n in Equation (VI).

T4=(2255K)×(111)1.31=(2255K)×(111)0.3=(2255K)×(0.48706)=1098K

Thus, the temperature at the end of expansion process is 1098K_.

Substitute 8MPa for P3, 1/11 for v2/v1, and 1.3 for n in Equation (VII).

P4=(8MPa)×(111)1.3=(8MPa×(1000kPa1MPa))×(0.044278)=354.2kPa

Substitute 0.287kJ/kgK for R, 1098K for T4, 2255K for T3, and 1.3 for n in Equation (VIII).

w34=(0.287kJ/kgK)(1098K2255K)11.3=(0.287kJ/kgK)(1157K)0.3=1106kJ/kg

(b)

To determine

The net-work output at the constant volume heat rejection.

The thermal efficiency at the constant volume heat rejection.

(b)

Expert Solution
Check Mark

Answer to Problem 167RP

The net-work output at the constant volume heat rejection is 794kJ/kg_.

The thermal efficiency at the constant volume heat rejection is 59.6%_.

Explanation of Solution

Determine the net-work output at the constant volume heat rejection.

wnet,out=w34w12 (IX)

Determine the thermal efficiency at the constant volume heat rejection.

ηth=wnet,outqin (X)

Conclusion:

Substitute 1106kJ/kg for w34 and 312.3kJ/kg for w12 in Equation (IX).

wnet,out=(1106kJ/kg)(312.3kJ/kg)=793.7kJ/kg794kJ/kg

Thus, the net-work output at the constant volume heat rejection is 794kJ/kg_.

Substitute 794kJ/kg for wnet,out and 1332kJ/kg for qin in Equation (X).

ηth=794kJ/kg1332kJ/kg=0.596=0.596×10059.6%

Thus, the thermal efficiency at the constant volume heat rejection is 59.6%_.

(c)

To determine

The mean effective pressure at the constant volume heat rejection.

(c)

Expert Solution
Check Mark

Answer to Problem 167RP

The mean effective pressure at the constant volume heat rejection is 982kPa_.

Explanation of Solution

Determine the initial volume at the constant volume heat rejection.

ν1=RT1P1 (XI)

Determine the mean effective pressure at the constant volume heat rejection.

MEP=wnet,outν1ν2=wnet,outν1(11/r) (XII)

Here, the compression ratio is r.

Note: νmin=ν2=νmaxr.

Conclusion:

Substitute 0.287kJ/kgK for R, 100kPa for P1, 37°C for T1 in Equation (XI).

ν1=(0.287kJ/kgK)×(37°C)(100kPa)=(0.287kJ/kgK)×(1kPam31kJ)×(37°C+273)(100kPa)=(0.287kPam3/kgK)×(310K)(100kPa)=0.8897m3/kg

Substitute 794kJ/kg for wnet,out, 11 for r, and 0.8897m3/kg for v1 in Equation (XII).

MEP=794kJ/kg(0.8897m3/kg)(11/11)=794kJ/kg(0.8897m3/kg)(0.909091)=794kJ/kg(0.808818m3/kg)×(1kPam31kJ)=981.6kPa

        982kPa

Thus, the mean effective pressure at the constant volume heat rejection is 982kPa_.

(d)

To determine

The engine speed for a given net power.

(d)

Expert Solution
Check Mark

Answer to Problem 167RP

The engine speed for a given net power is 3820rev/min_.

Explanation of Solution

Determine the clearance volume at the beginning of compression process.

r=νc+vdνc (XIII)

Here, the volume of the gasoline engine is νd.

Determine the initial volume.

ν1=νc+νd (XIV)

Determine the total mass contained in the cylinder.

mt=P1ν1RT1 (XV)

Determine the engine speed for a net power output of 50 kW.

n˙=2×W˙netmtwnet (XVI)

Note: the two revolutions in one cycle in four-stroke engines.

Conclusion:

Substitute 1.6L for νd and 11 for r in Equation (XIII).

11=νc+1.6Lνc11=νc+1.6L×(1m31000L)νc11=νc+0.0016m3νcνc=0.00016m3

Substitute 0.00016m3 for νc and 1.6L for νd in Equation (XIV).

ν1=(0.00016m3)×(1.6L)=(0.00016m3)×(1.6L×(1m31000L))=0.00176m3

Substitute 100kPa for P1, 0.00176m3 for ν1, 0.287kJ/kgK for R, 37°C for T1 in Equation (XV).

mt=(100kPa)(0.00176m3)(0.287kJ/kgK)(37°C)=(100kPa)(0.00176m3)(0.287kJ/kgK)×(1kPam31kJ)(37°C+273)=(100kPa)(0.00176m3)(0.287kPam3/kgK)(310K)=0.001978kg

Substitute 50kW for W˙net, 0.001978kg for mt, 749kJ/kg in Equation (XVI).

n˙=2(rev/cycle)×(50kW)(0.001978kg)(749kJ/kgcycle)=2(rev/cycle)×(50kW×(1kJ/s1kW))(0.001978kg)(749kJ/kgcycle)=2(rev/cycle)×50kJ/s(0.001978kg)(749kJ/kgcycle)×(60s1min)=3820rev/min

Thus, the engine speed for a given net power is 3820rev/min_.

(e)

To determine

The specific fuel consumption.

(e)

Expert Solution
Check Mark

Answer to Problem 167RP

The specific fuel consumption is 267g/kWh_.

Explanation of Solution

Determine the mass of fuel burned during one cycle.

AF=mamf=mtmfmf (XVII)

Here, the air-fuel ratio is AF.

Determine the specific fuel consumption.

sfc=mfmtwnet (XVIII)

Conclusion:

Substitute 16 for AF and 0.001978kg for mt in Equation (XVII).

16=(0.001978kg)mfmfmf=0.0001164kg

Substitute 0.0001164kg for mf, 0.001978kg for mt, and 794kJ/kg for wnet in Equation (XVIII).

sfc=0.0001164kg(0.001978kg)(794kJ/kg)=0.0001164kg(1.570532kJ)×(1000g1kg)(3600kJ1kWh)=2668g/kWh267g/kWh

Thus, the specific fuel consumption is 267g/kWh_.

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Chapter 9 Solutions

CONNECT FOR THERMODYNAMICS: AN ENGINEERI

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