Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
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Chapter 22, Problem 24PQ

(a)

To determine

The efficiency of a diesel engine ediesel=11γ(rγRγr1R1)

(a)

Expert Solution
Check Mark

Answer to Problem 24PQ

The efficiency of a diesel engine is given by ediesel=11γ(rγRγr1R1) is shown.

Explanation of Solution

Write the expression to calculate efficiency of the diesel engine.

    ediesel=1|Qc|Qh                                                                                              (I)

Here, ediesel is the efficiency of the diesel engine, Qc is the heat energy corresponding to cold reservoir, and Qh is the heat energy corresponding to hot reservoir.

Write the expression for the energy when heat enters the engine from the hot reservoir through isobaric process BC .

    Qh=nCP(TCTB)                                                                                      (II)

Here, n is the number of moles, Cp is the specific heat isobaric process, TC is the temperature at C, and TB is the temperature at B.

Write the expression for the energy when heat leaves the engine to the cold reservoir through isochoric process DA.

    |Qc|=nCv(TDTA)                                                                                     (III)

Here, Cv is the specific heat at isochoric process, TD is the temperature at D, and TA is the temperature at A.

For an adiabatic process at AB ,

    TATB=(VmaxVmin)(1γ)=R1γTA=TBR1γ                                                                                    (IV)

Here, Vmax is the maximum volume, Vmin is the minimum volume, and γ is the specific heat ratio.

For an adiabatic process at CD ,

    TDTC=(VmaxVint)(1γ)=r1γTD=TCr1γ                                                                                    (V)

Here, Vint is the intermediate volume.

Write the expression for specific heat capacity.

    γ=CpCv                                                                                              (VI)

Substitute (II), (III) in (I).

    ediesel=1nCv(TDTA)nCP(TCTB)                                                                           (VII)

Substitute (IV), (V), (VI) in (VII).

    ediesel=11(TCr1γTBR1γ)γ(TCTB)                                                                   (VIII)

At B and C pressure is same PC=PB, Hence

    TCVint=TBVminTC=VintVminTB                                                                                       (IX)

Conclusion:

Substitute (IX) in (VIII).

    ediesel=11γ[(VintVminTBr1γTBR1γ)(VintVminTBTB)]=11γ[(VintVminrrγRRγ)(VintVmin1)]=11γ[(VintVmaxVintrγVminVmaxVminRγ)(VintVmin)]=11γ(rγRγVintVmaxVminVmax)=11γ(rγRγr1R1)                                         (X)

Hence, the efficiency of a diesel engine is given by ediesel=11γ(rγRγr1R1) is shown.

(b)

To determine

The efficiency of the diesel engine represented by Figure P22.24.

(b)

Expert Solution
Check Mark

Answer to Problem 24PQ

The efficiency of the diesel engine is 56%

Explanation of Solution

Rewrite (X).

    ediesel=11γ(rγRγr1R1)                                                                     (X)

Use VmaxVint for r and VmaxVmin for R in (X).

    ediesel=11γ((VmaxVint)γ(VmaxVmin)γ(VmaxVint)1(VmaxVmin)1)                                                    (XI)

Conclusion:

Substitute 1.4 for γ, 1050cm3 for Vmax , 50cm3 for Vmin , and 230cm3 for Vint in (XI) to calculate ediesel.

    ediesel=111.4((1050cm3230cm3)1.4(1050cm350cm3)1.4(1050cm3230cm3)1(1050cm350cm3)1)=111.4((4.57)1.4(21.0)1.4(4.57)1(21.0)1)=0.56=56%

Therefore, the efficiency of the diesel engine is 56%

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Chapter 22 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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