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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

670



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

671



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.)

672



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.

673

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