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Unformatted text preview: be used to find QH , the magnitude of the input
heat. The magnitude QC of the exhaust heat can then be found from Equation 15.12,
QH = W + QC . 800 THERMODYNAMICS SOLUTION
a. The maximum efficiency is e = 1− TC
TH = 1− 323 K
= 0.360
505 K b. Since the power output of the power plant is P = 84 000 kW, the required heat input
QH for a 24 hour period is
QH = W = e Pt (8.4 ×107 J/s)(24 h) 3600 s 13
= = 2.02 ×10 J
e
0.360 1h Therefore, solving Equation 15.12 for QC , we have
QC = QH − W = 2.02 × 1013 J − 7.3 × 1012 J = 1.3 × 1013 J 60. REASONING AND SOLUTION The temperature of the gasoline engine input is
T1 = 904 K, the exhaust temperature is T2 = 412 K, and the air temperature is T3 = 300 K.
The efficiency of the engine/exhaust is
e1 = 1 − (T2/T1) = 0.544 The efficiency of the second engine is
e2 = 1 − (T3/T2) = 0.272 The magnitude of the work done by each segment is W1 = e1 QH1 and W2 = e2 QH2 = e2 QC1 since QH2 = QC1 ( ) Now examine W1 + W2 / W1 to find the ratio of the total work produced by both engines
to that produced by the first engine alone.
W1 + W2
W1 ( ) = e1 QH1 + e2 QC1
e1 QH1 e Q
= 1 + 2 C1 e1 QH1 But, e1 = 1 − QC1 / QH1 , so that QC1 / QH1 = 1 − e1 . Therefore, Chapter 15 Problems W1 =1+ e2
e
(1 − e1 ) = 1 + e2 − e2 = 1 + 0.500 − 0.272 = 1.23
e1
1 61. SSM REASONING The expansion from point
a to point b and the compression from point c to
point d occur isothermally, and we will apply the
first law of thermodynamics to these parts of the
cycle in order to obtain expressions for the input
and rejected heats, magnitudes QH and QC ,
respectively. In order to simplify the resulting
expression for QC / QH , we will then use the a
Pressure → W1 + W2 801 b TC TH d
c fact that the expansion from point b to point c and
the compression from point d to point a are
adiabatic. Volume → SOLUTION According to the first law of thermodynamics, the change in internal energy
∆U is given by ∆U = Q − W (Equation 15.1), where Q is the heat and W is the work. Since
the internal energy of an ideal gas is proportional to the Kelvin temperature and the
temperature is constant for an isothermal process, it follows that ∆U = 0 J for such a case.
The work of isothermal expansion or compression for an ideal gas is W = nRT ln (Vf / Vi )
(Equation 15.3), where n is the number of moles, R is the universal gas constant, T is the
Kelvin temperature, Vf is the final volume of the gas, and Vi is the initial volume. We have,
then, that
V V ∆U = Q − W or 0 = Q − nRT ln f or Q = nRT ln f V i Vi Applying this result for Q to the isothermal expansion (temperature = TH) from point a to
point b and the isothermal compression (temperature = TC) from point c to point d, we have
V
QH = nRTH ln b
V
a and V
QC = nRTC ln d
V
c where Vf = Vb and Vi = Va for the isotherm at TH and Vf = Vd and Vi = Vc for the isotherm
at TC. In this problem, we are interested in the magnitude of the heats. For QH, this poses no problem, since Vb > Va , ln (Vb / Va ) is positive, and we have 802 THERMODYNAMICS V (1)
QH = nRTH ln b V a
However, for QC, we need to be careful, because Vc > Vd and ln (Vd / Vc ) is negative. Thus, we write for the magnitude of QC that
V
QC = − nRTC ln d
V
c Vc = nRTC ln V d (2) According to Equations (1) and (2), the ratio of the magnitudes of the rejected and input
heats is
V V nRTC ln c TC ln c V V QC d= d
(3)
= Vb Vb QH
nRTH ln TH ln V V a a
We now consider the adiabatic parts of the Carnot cycle. For the adiabatic expansion or
compression of an ideal gas the initial pressure and volume (Pi and Vi) are related to the
final pressure and volume (Pf and Vf) according to
Pi Viγ = Pf Vfγ (15.5) where γ is the ratio of the specific heats at constant pressure and constant volume. It is also
true that P = nRT / V (Equation 14.1), according to the ideal gas law. Substituting this
expression for the pressure into Equation 15.5 gives nRTi γ nRTf
V = V i V i
f γ Vf TViγ −1 = Tf Vfγ −1
i or Applying this result to the adiabatic expansion from point b to point c and to the adiabatic
compression from point d to point a, we obtain
γ
THVb −1 = TCVcγ −1 and TCVdγ −1 = THVaγ −1 Dividing the first of these equations by the second shows that
THVbγ −1
γ −1 THVa = TCVcγ −1
γ −1 TCVd or With this result, Equation (3) becomes Vbγ −1
γ −1 Va = Vcγ −1
γ −1 Vd or Vb
Va = Vc
Vd Chapter 15 Problems QC
QH = V
TC ln c
V
d V
TH ln b
V
a = 803 TC
TH 62. REASONING AND SOLUTION The efficiency e of the power plant is threefourths its
Carnot efficiency so, according to Equation 15.15,
T
e = 0.75 1 − C
T H 40 K + 273 K = 0.75 1 − = 0.33 285 K + 273 K The power output of the plant is 1.2 ×109 watts. According to Equation 15.11,
e = W / QH = ( Power ⋅ t ) / QH . Therefore, at 33% efficiency, the magnitude of the heat
input per unit time is QH
t Power 1.2 ×109 W
=
=
= 3.6 ×109 J/s
e
0.33 From the principle of conservation of energy, the heat output per unit time must be
QC...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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