chapter7 b

7/30/2019 Chapter7 B

1/8

Chapter 7 - Section B - Non-Numerical Solutions

7.2 (a) Apply the general equation given in the footnote on page 266 to the particular derivative of interest

here: T P

S

=

T S

P

S P

T

The two partial derivatives on the right are found from Eqs. (6.17) and (6.16); thus,

T

P

S

= TCP

V

T

P

For gases, this derivative is positive. It applies to reversible adiabatic expansions and compressions

in turbines and compressors.

(b) Application of the same general relation (page 266) yields:

T

V

U

=

T

U

V

U

V

T

The two partial derivatives on the right are found from Eqs. (2.16) and (6.31); thus,

T

V

U

= 1CV

P T

P

T

V

For gases, this may be positive or negative, depending on conditions. Note that it is zero for an

ideal gas. It applies directly to the Joule expansion, an adiabatic expansion of gas confined in a

portion of a container to fill the entire container.

7.3 The equation giving the thermodynamic sound speed appears in the middle of page 257. As written,

it implicitly requires that V represent specific volume. This is easily confirmed by a dimensional

analysis. IfV is to be molar volume, then the right side must be divided by molar mass:

c2 = V2

M

P

V

S

(A)

Applying the equation given in the footnote on page 266 to the derivative yields: P

V

S

=

P

S

V

S

V

P

This can also be written: P

V

S

=

P

T

V

T

S

V

S

T

P

T

V

P

=

T

S

V

S

T

P

P

T

V

T

V

P

Division of Eq. (6.17) by Eq. (6.30) shows that the first product in square brackets on the far right is

the ratio CP /CV. Reference again to the equation of the footnote on page 266 shows that the second

product in square brackets on the far right is ( P/ V)T, which is given by Eq. (3.3).

Therefore,

P

V

S

= CPCV

P

V

T

= CPCV

1 V

662


2/8

Substitute into Eq. (A): c2 = V CPMCV

or c =

V CP

MCV

(a) For an ideal gas, V = RT/P and = 1/P . Therefore, cig = RTM CPCV(b) For an incompressible liquid, V is constant, and = 0, leading to the result: c = . This of

course leads to the conclusion that the sound speed in liquids is much greater than in gases.

7.6

As P2 decreases from an initial value of P2 = P1, both u2 and m steadily increase until the critical-pressure ratio is reached. At this value of P2, u2 equals the speed of sound in the gas, and further

reduction in P2 does not affect u2 or m.

7.7 The mass-flow rate m is of course constant throughout the nozzle from entrance to exit.

The velocity u rises monotonically from nozzle entrance (P/P1 = 1) to nozzle exit as P and P/P1decrease.

The area ratio decreases from A/A1 = 1 at the nozzle entrance to a minimum value at the throat andthereafter increases to the nozzle exit.

7.8 Substitution of Eq. (7.12) into (7.11), with u1 = 0 gives:

u2throat =2P1V1

1

1 2

+ 1

= P1V1

2

+ 1

where V1 is specific volume in m3kg1 and P1 is in Pa. The units ofu2throat are then:

Pa m3 kg1 = Nm2

m3 kg1 = N m kg1 = kg m s2 m kg1 = m2 s2

With respect to the final term in the preceding equation, note that P1V1 has the units of energy per unit

mass. Because 1 N m = 1 J, equivalent units are Jkg1. Moreover, P1V1 = RT1/M; whence

u2throat =RT1

M

2

+ 1

With R in units of J(kg mol)1K1, RT1/M has units of Jkg1 or m2s2.

663


3/8

7.16 It is shown at the end of Ex. 7.5 that the Joule/Thomson inversion curve is the locus of states for which

(Z/ T)P = 0. We apply the following general equation of differential calculus:x

y

z

=

x

y

w

+

x

w

y

w

y

z

Z

T

P

=

Z

T

+

Z

T

T

P

Whence,

Z

T

=

Z

T

P

Z

T

T

P

Because P = Z RT, = PZ RT

and

T

P

= PR

1(Z T)2

Z + T

Z

T

P

Setting (Z/ T)P = 0 in each of the two preceding equations reduces them to:

Z T

=

Z

T

T

P

and T

P =

P

Z RT2 =

T

Combining these two equations yields:

T

Z

T

=

Z

T

(a) Equation (3.42) with van der Waals parameters becomes:

P = RTV b

a

V2

Multiply through by V/RT, substitute Z = P V/RT, V = 1/, and rearrange:Z = 1

1 b a

RT

In accord with Eq. (3.51), define q a/b RT. In addition, define b. Then,

Z = 11 q (A)

Differentiate:

Z

T

=

Z

T

= dqd T

By Eq. (3.54) with (Tr) = 1 for the van der Waals equation, q = / Tr. Whence,dq

d T=

1T2r

d Tr

d T=

1

T2r Tc=

1

T Tr= q

T

Then,

Z

T

= ( ) q

T

= q

T

In addition,

Z

T

= b

Z

T

= b(1 )2 qb

664


4/8

Substitute for the two partial derivatives in the boxed equation:

Tq

T= b

(1 )2 qb or q =

(1 )2 q

Whence, =

1

1

2q(B)

By Eq. (3.46), Pc = RTc/b. Moreover, P = ZRT. Division of the second equation by thefirst gives Pr = ZbT/ Tc. Whence

Pr =ZTr

(C)

These equations allow construction of a Tr vs. Pr inversion curve as in Fig. 7.2. For a given value

of Tr, calculate q. Equation (B) then gives , Eq. (A) gives Z, and Eq. (C) gives Pr.

(b) Proceed exactly as in Part (a), with exactly the same definitions. This leads to a new Eq. (A):

Z = 11

q1

+

(A)

By Eq. (3.54) with (Tr) = T0.5r for the Redlich/Kwong equation, q = / T1.5r . This leads to:dq

d T= 1.5 q

Tand

Z

T

= 1.5 qT(1 + )

Moreover,

Z

T

= b(1 )2

bq

(1 + )2

Substitution of the two derivatives into the boxed equation leads to a new Eq. (B):

q

= 1 + 1

2

12.5 + 1.5 (B)

As in Part (a), for a given Tr, calculate q, and solve Eq. (B) for , by trial or a by a computer

routine. As before, Eq. (A) then gives Z, and Eq. (C) of Part (a) gives Pr.

7.17 (a) Equal to. (b) Less than. (c) Less than. (d) Equal to. (e) Equal to.

7.28 When a saturated liquid is expanded in a turbine some of the liquid vaporizes. A turbine properly

designed for expansion of liquids cannot handle the much larger volumes resulting from the formation

of vapor. For example, if saturated liquid at 5 bar expands isentropically to 1 bar, the fraction of the

original liquid that vaporizes is found as follows:

S2 = Sl2 + x v2 (Sv2 Sl2) = S1

or x v2 =S1 Sl2Sv2 Sl2

= 1.8604 1.30277.3598 1.3027 = 0.0921

Were the expansion irreversible, the fraction of liquid vaporized would be even greater.

7.33 Apply Eq. (2.29) to this non-steady-state process, with n replacing m, with the tank as control volume,

and with a single inlet stream. Since the process is adiabatic and the only work is shaft work, this

equation may be multiplied by dt to give:

d(nU)tank H dn = d Ws

665


5/8

Because the inlet stream has constant properties, integration from beginning to end of the process

yields:

Ws = n2U2 n1U1 n Hwhere the subscripted quantities refer to the contents of the tank and n and H refer to the inlet stream.

Substitute n

=n2

n1 and H

=U

+P V

=U

+RT:

Ws = n2U2 n1U1 (n2 n1)(U + RT) = n2(U2 U RT) n1(U1 U RT)With U = CVT for an ideal gas with constant heat capacities, this becomes:

Ws = n2[CV(T2 T) RT] n1[CV(T1 T) RT]However, T = T1, and therefore:

Ws = n2[CV(T2 T1) RT1] + n1RT1

By Eq. (3.30b), T2 =

P2

P1

(1)/)

Moreover, n1 = P1VtankRT1

and n2 = P2VtankRT2

With = 1.4, T2 = 573.47 K. Then, with R = 8.314 m3 kPa kmol1 K1,

n1 =(101.33)(20)

(8.314)(298.15)= 0.8176 kmol and n2 =

(1000)(20)

(8.314)(573.47)= 4.1948 kmol

Substitution of numerical values into the boxed equation, with R = 8.314 kJ kmol1 K1, gives:

Ws = 15, 633 kJ

7.40 Combine Eqs. (7.13) and (7.17):

Ws = n H = n(H)S

By Eq. (6.8), (H)S =

V d P = VP

Assume now that P is small enough that V, an average value, can be approximated by V1 =RT1/P1. Then

(H)S =RT1

P1P and Ws = n

RT1

P1P

Equation (7.22) is the usual equation for isentropic compression of an ideal gas with constant heat

capacities. For irreversible compression it can be rewritten:

Ws =nCP T1

P2

P1

R/CP 1

For P sufficiently small, the quantity in square brackets becomes:P2

P1

R/CP 1 =

1 + P

P1

R/CP 1

1 + R

CP

P

P1

1

The boxed equation is immediately recovered from this result.

666


6/8

7.41 The equation immediately preceding Eq. (7.22) page 276 gives T2 = T1 . With this substitution,Eq. (7.23) becomes:

T2 = T1 +T1 T1

= T1

1 + 1

The entropy generation SG is simply S for the compression process, for which Eq. (5.14) may be

rewritten:S

R= CP

Rln

T2

T1 ln P2

P1= CP

Rln

T2

T1 CP

Rln

P2

P1

R/CPCombine the two preceding equations:

S

R= CP

R

ln

1 + 1

ln

= CP

Rln

1 + 1

Whence,SG

R= CP

Rln

+ 1

7.43 The relevant fact here is that CP increases with increasing molecular complexity. Isentropic compres-

sion work on a mole basis is given by Eq. (7.22), which can be written:

Ws = CP T1( 1) where

P2

P1

R/CP

This equation is a proper basis, because compressor efficiency and flowrate n are fixed. With all

other variables constant, differentiation yields:

d Ws

dCP =T1 (

1)

+CP

d

dCP

From the definition of ,

ln = RCP

lnP2

P1whence

dln

dCP= 1

d

dCP= R

CP2

lnP2

P1

Then,d

dCP= R

CP2

lnP2

P1

andd Ws

dCP= T1

1 R

CPln

P2

P1

= T1( 1 ln )

When = 1, the derivative is zero; for > 1, the derivative is negative (try some values). Thus, thework of compression decreases as CP increases and as the molecular complexity of the gas increases.

7.45 The appropriate energy balance can be written: W = H Q. Since Q is negative (heat transfer isout of the system), the work of non-adiabatic compression is greater than for adiabatic compression.

Note that in order to have the same change in state of the air, i.e., the same H, the irreversibilities of

operation would have to be quite different for the two cases.

7.46 There is in fact no cause for concern, as adiabatic compression sends the steam further into the super-

heat region.

667


7/8

7.49 (a) This result follows immediately from the last equation on page 267 of the text.

(b) This result follows immediately from the middle equation on page 267 of the text.

(c) This result follows immediately from Eq. (6.19) on page 267 of the text.

(d) Z V

P =

Z

T

P T

V

P

but by (a), this is zero.

(e) Rearrange the given equation:V

T= ( P/ T)V

( P/ V)T=

V

P

T

P

T

V

=

V

T

P

For the final equality see footnote on p. 266. This result is the equation of (c).

7.50 From the result of Pb. 7.3: c =

V

M CP

CV 1

where = 1

V

V

P

T

With V

=

RT

P +B then

V

P

T =

RT

P

2Also, let

=

CP

CV

Then c = P V

MRT= (RT + B P)

MRT=

1 + B PRT

RT

M

c =

RT

M+ B

RT

RT

M P

A value for B at temperature T may be extracted from a linear fit ofc vs. P .

7.51 (a) On the basis of Eq. (6.8), write:

H

ig

S = Vig d P = RTP d P (const S)HS =

V d P =

Z RT

Pd P (const S)

HS

Hig

S

=

Z RT

Pd P (const S)

RT

Pd P (const S)

Z

By extension, and with equal turbine efficiencies,H

Hig =

...W...

Wig = Z

7.52 By Eq. (7.16), H = (H)S For CP = constant, T2 T1 = [(T2)S T1]

For an ideal gas with constant CP , (T2)S is related to T1 by (see p. 77): (T2)S = T1

P2

P1

R/CP

Combine the last two equations, and solve for T2: T2 = T1

1 +

P2

P1

R/CP 1

668


8/8

From which =T2

T1 1

P2

P1

R/CP 1

Note that < 1

Results: For T2 = 318 K, = 1.123; For T2 = 348 K, = 1.004; For T2 = 398 K, = 0.805.Only T2 = 398 K is possible.

7.55 The proposal of Pb. 7.53, i.e., pumping of liquid followed by vaporization. The reason is that pumping

a liquid is much less expensive than vapor compression.

7.56 What is required here is the lowest saturated steam temperature that satisfies the T constraint. Data

from Tables F.2 and B.2 lead to the following:

Benzene/4.5 bar; n-Decane/17 bar; Ethylene glycol/33 bar; o-Xylene/9 bar

669

chapter7 b

Documents