This

**preview**shows pages 1-3. to view the full**12 pages of the document.**Solutions, Chapter 10

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10.1* Q=A⋅L

t=10 in2⋅5in⋅1

s=50 in3

s⋅gal

231 in3⋅60 s

min =13.0 gpm = 8.2 ⋅10-4 m

s

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10.2 Po =QΔP=500 gal

min

⎛

⎝

⎜

⎞

⎠

⎟ 25 lbf

in2

⎛

⎝

⎜

⎞

⎠

⎟ ⋅ 231 in3

gal ⋅ft

12 in ⋅hp s

550 ft lbf ⋅min

60 s

=

7.29 hp = 5.44 kW

This answer is independent of fluid properties for a constant-density fluid.

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10.3 (a) The amount of energy converted to friction heating is

F=2.8 hp ⋅1−0.73

(

)

Ý

m =Δu

ΔT=Δu

CV

=F

CV

=

1−0.73

()280 hp

(

)

416.5 lbm

min

⎛

⎝

⎜ ⎞

⎠

⎟

1Btu

lbm oF

⋅2545 Btu

hp hr ⋅hr

60 min =0.077oF = 0.042°C

(b)

ρ

ρ

0

=1−

α

ΔT+

β

ΔP=1−0.11

⋅

10

−

3

°F⋅0.077°F+0.3

⋅

10-5

psi 70 psi

=1−8.5 ⋅10−6+2.1⋅10

−

4

=

1.0002

In this case the friction heating is very small, so its effect is less than that of the increase

in pressure. The two effects act in opposite directions. The resulting change in density is

certainly negligible.

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10.4 See Ex. 10.2. At room temperature the vapor pressure of mercury is ≈ 10-6 atm ≈

0, so that

Δz=ΔP

ρ

g=

14.7 lbf

in2

13.6 ⋅62.3 lbm

f

t

3⋅32.2 ft

s2

⋅32.2 lbm ft

lbf s2⋅144 in2

ft2=2.50 ft = 0.76 m

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10.5 One regularly sees Fig 10.6 or its equivalent, but rarely sees the same figure for

an incompressible fluid, because it is so simple. It is shown below. Asking students to

sketch it helps them understand Fig. 10.6.

Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 10, page 1

Only pages 1-3 are available for preview. Some parts have been intentionally blurred.

P

V

in

out

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10.6 Differential thermal expansion between the various parts of the compressor make it

impractical. The cylinder walls and head are cooled, the piston is generally not. Some

clearance room must be allowed for this.

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10.7* (a)Po =Ý

n RT ln P2

P

1

=20 lbmol

hr ⋅1.987 BT

U

lbmol oR⋅528 oRln10 hp h

r

2545 BTU =18.98 hp =14.16 kW

(b)

=20()1.987()528()

1.667

0.667

⎛

⎝

⎜

⎞

⎠

⎟ 10

()

0.667

1.667 −1

⎡

⎣

⎢

⎤

⎦

⎥ ⋅1

2545 =31.16 hp =23.2 kW

(c) Po =above⋅

10

()

k−1

k

⎛

⎝

⎜

⎜

⎞

⎠

⎟

⎟ −1+10

()

k−1

k−1

⎡

⎣

⎢

⎢

⎤

⎦

⎥

⎥

10

()

k−1

k−1

⎡

⎣

⎢ ⎤

⎦

⎥

=24.11 hp = 18.0 kW

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10.8 For an adiabatic compressor

ΔW

n=RT

1k

k−1

P

2

P

1

⎛

⎝

⎜

⎜

⎞

⎠

⎟

⎟

(k−1)/k

−1

⎡

⎣

⎢

⎢

⎤

⎦

⎥

⎥ =1.4

0.4 ⋅1.987 Btu

lbmol °R⋅528°RP

2

P

1

⎛

⎝

⎜

⎜

⎞

⎠

⎟

⎟

0.4 /1.4

−1

⎡

⎣

⎢

⎢

⎤

⎦

⎥

⎥

While for an isothermal compressor

ΔW

n=RT ln P

2

P

1

=1.987 Btu

lbmol °R⋅528°RlnP2

P

1

=1043 Btu

lbmol ln P

2

P

1

For

P

2/

P

1=20

Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 10, page 2

Only pages 1-3 are available for preview. Some parts have been intentionally blurred.

ΔW

n

⎛

⎝

⎜

⎞

⎠

⎟

adiabatic

=1.4

0.4 ⋅1.987 Btu

lbmol °R⋅528°R20()

0.4/1.4 −1

[

]=4968 Btu

lbmol

Δ

W

n

⎛

⎝

⎜

⎞

⎠

⎟

isothermal

=RT ln P

2

P

1

=1043 Btu

lbmol ln20 =3141 Btu

lbmol

Using a spreadsheet we compute the corresponding values for intermediate pressures and

plot them as shown below. As expected, the isothermal always has a lower power

requirement than the adiabatic.

0

1000

2000

3000

4000

5000

0 5 10 15 20 25

P / P

21

W/n, Btu

/lb

mo

l

Adiabatic

Isothermal

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10.9 ΔW

η

RT

1

k

k

−1

=P

i

P

1

⎛

⎝

⎜

⎜

⎞

⎠

⎟

⎟

k−1

k−1+P

2

P

i

⎛

⎝

⎜

⎜

⎞

⎠

⎟

⎟

k

−

1

k−1

⎡

⎣

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

detc.()

dP

i

=k+1

k

P

i

P

1

⎛

⎝

⎜

⎜

⎞

⎠

⎟

⎟

−1

k1

P

1

⎛

⎝

⎜

⎜

⎞

⎠

⎟

⎟ +P

2

Pi

⎛

⎝

⎜

⎜

⎞

⎠

⎟

⎟

−1

k−P

2

P

i

2

⎛

⎝

⎜

⎜

⎞

⎠

⎟

⎟

⎡

⎣

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

We set this equal to zero, and

Pi

−1

k

P

1

k−1

k

=P

2

k−1

k

P

i

2k−1

k

; P

i

2k−2

k=P

1P

2

()

k

−

1

k; P

i

=

P

1P

2

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Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 10, page 3

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