chapter8 b
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7/30/2019 Chapter8 B
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Chapter 8 - Section B - Non-Numerical Solutions
8.12 (a) Because Eq. (8.7) for the efficiency ηDiesel includes the expansion ratio, r e ≡ V B / V A, we relate
this quantity to the compression ratio, r ≡ V C / V D, and the Diesel cutoff ratio, r c ≡ V A/ V D. Since
V C = V B , r e = V C / V A. Whence,
r
r e=
V C / V D
V C / V A=
V A
V D= r c or
1
r e=
r c
r
Equation (8.7) can therefore be written:
ηDiesel = 1 −1
γ
(r c/r )γ
− (1/r )γ
r c/r − 1/r
= 1 −
1
γ
(1/r )γ
1/r
r
γ c − 1
r c − 1
or ηDiesel = 1 − 1
r
γ −1r
γ c − 1
γ (r c − 1)
(b) We wish to show that:
r γ c − 1
γ (r c − 1)> 1 or more simply
x a− 1
a( x − 1)> 1
Taylor’s theorem with remainder, taken to the 1st derivative, is written:
g = g(1) + g(1) · ( x − 1) +R
where, R ≡g[1 + θ ( x − 1)]
2!
· ( x − 1)2 (0 < θ < 1)
Then, x a= 1 + a · ( x − 1) +
12
a · (a − 1) · [1 + θ ( x − 1)]a−2· ( x − 1)2
Note that the final term is R. For a > 1 and x > 1, R > 0. Therefore:
x a > 1 + a · ( x − 1) x a− 1 > a · ( x − 1)
andr
γ c − 1
γ (r c − 1)> 1
(c) If γ = 1.4 and r = 8, then by Eq. (8.6):
ηOtto = 1 −
1
8
0.4
and ηOtto = 0.5647
• r c = 2 ηDesiel = 1 −
1
8
0.421.4
− 1
1.4(2 − 1)and ηDiesel = 0.4904
• r c = 3 ηDesiel = 1 −
1
8
0.431.4
− 1
1.4(3 − 1)and ηDiesel = 0.4317
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8.15 See the figure below. In the regenerative heat exchanger, the air temperature is raised in step B → B∗,
while the air temperature decreases in step D → D∗. Heat addition (replacing combustion) is in step
B∗→ C .
By definition, η ≡−W AB − W C D
Q B∗C
where, W AB = ( H B − H A) = C P (T B − T A)
W C D = ( H D − H C ) = C P (T D − T C )
Q B∗C = C P (T C − T B∗ ) = C P (T C − T D)
Whence, η =T A − T B + T C − T D
T C − T D= 1 −
T B − T A
T C − T D
By Eq. (3.30b),
T B = T A P B
P A
(γ −1)/γ
and T D = T C P D
PC
(γ −1)/γ
= T C P A
P B
(γ −1)/γ
Then, η = 1 −
T A
P B
P A
(γ −1)/γ
− 1
T C
1 −
P A
P B
(γ −1)/γ
Multiplication of numerator and denominator by (P B / P A)(γ −1)/γ gives:
η = 1 −
T A
T C
P B
P A
(γ −1)/γ
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8.21 We give first a general treatment of paths on a P T diagram for an ideal gas with constant heat capacities
undergoing reversible polytropic processes. Equation (3.35c), p. 78, may be rewritten as
P = K T δ/(δ−1) ln P = ln K +δ
δ − 1ln T
d P
P=
δ
δ − 1
d T
T
d P
d T =
δ
δ − 1
P
T ( A) Sign of d P/d T is that of δ − 1, i.e., +
Special cases
δ = 0 −→ d P/d T = 0 Constant P
δ = 1 −→ d P/d T = ∞ Constant T
By Eq. ( A),d 2 P
d T 2=
δ
δ − 1
1
T
d P
d T −
P
T 2
=
δ
δ − 1
1
T
δ
δ − 1
P
T −
P
T
d 2 P
d T 2=
δ
(δ − 1)2
P
T 2( B) Sign of d 2 P/d T 2 is that of δ, i.e., +
For a constant-V process, P varies with T in accord with the ideal-gas law: P = RT / V or P = K T
With respect to the initial equation, P = K T δ/(δ−1), this requires δ = ∞. Moreover, d P/d T = K
and d 2 P/d T 2 = 0. Thus a constant-V process is represented on a P T diagram as part of a straight
line passing through the origin. The slope K is determined by the initial P T coordinates.
For a reversible adiabatic process (an isentropic process), δ = γ . In this case Eqs. ( A) and ( B) become:
d P
d T =
γ
γ − 1
P
T
d 2 P
d T 2=
γ
(γ − 1)2
P
T 2
We note here that γ/(γ − 1) and γ/(γ − 1)2 are both > 1. Thus in relation to a constant-V process
the isentropic process is represented by a line of greater slope and greater curvature for the same T and P . Lines characteristic of the various processes are shown on the following diagram.
T
P
δ = 1
δ = 0
δ = γ
δ = ∞
00
The required sketches appear on the following page. (Courtesy of Prof. Mark T. Swihart, State Uni-
versity of New York at Buffalo.)
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T
P
00
Figure 1: The Carnot cycle
T
P
00
Figure 2: The Otto cycle
T
P
00
Figure 3: The Diesel cycle
T
P
00
Figure 4: The Brayton cycle
8.23 This is a challenging and open-ended problem for which we offer no solution. Problem 8.21 may offer
some insight.
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