This

**preview**shows pages 1-3. to view the full**52 pages of the document.**Solutions Chapter 7

In the problems in Chapter 6 it says to assume pipes are Sch. 40. In most problems in

this chapter pipe inside dimensions are given, normally in integral inches. The solutions

here are based on those integral inch diameters, rather than Sch. 40 diameters. I hope

this causes no confusion.

In all solutions, except as noted, it is assumed that the flow is in the positive x-direction,

and the direction subscripts are dropped unless it would cause confusion. Where a sketch

is made, the positive x-direction is from left to right.

______________________________________________________________________

7.1* (a) System: earth, open system dmV()=−Vout dmout ,ΔV=−Vout

dmout

mearth

ΔV=−20 f

t

s⋅1 lbm

1025 lbm =−2⋅10−24 f

t

s

(b) System: earth-ball, closed system

dmV()=0; Vf=V

i

It is instructive to plot the earth's velocity and trajectory (perpendicular to the main

direction in its orbit), as shown below

time

di

stance perpen

di

cu

l

ar

to orbital path

ball thrown

top of flight

ball hits earth

time

perpendicular

displacement from

orbital path

______________________________________________________________________

7.2 (a) dm ; mVV()

sys =−Vout dmout f

−

mVi

(

)

=

−

Vout dmout

Vf=−Vout

dmout

mgun

=− 1500 ft

s

⎛

⎝

⎜

⎞

⎠

⎟ 0.05 lbm

5 lbm

⎛

⎝

⎜

⎜

⎞

⎠

⎟

⎟ =−15 ft

s=−4.57 m

s

(b) ; ΔmV()=0=mV()

gun +mV()

bullet V

gun =−V

bullet

mbullet

mgun

⎛

⎝

⎜

⎜

⎞

⎠

⎟

⎟

which is the same as in part (a).

______________________________________________________________________

Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 7, page 1

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

7.3 System: duck-bullet, closed system ΔΣmu+V2

2

⎛

⎝

⎜

⎜

⎞

⎠

⎟

⎟

⎡

⎣

⎢

⎢

⎤

⎦

⎥

⎥ =0; Δmu =−ΔmV2

2

mV2

2

∑

⎛

⎝

⎜

⎜ ⎞

⎠

⎟

⎟

i

=

3 lbm ⋅15 ft

s

⎛

⎝

⎜

⎞

⎠

⎟

2

2+

0.05 lbm 1000 ft

s

⎛

⎝

⎜

⎞

⎠

⎟

2

2=337.5 +25000 =25337.5 ft2

s2

mV2

2

⎛

⎝

⎜

⎜ ⎞

⎠

⎟

⎟

f

=

3.05lbm 1.6 ft

s

⎛

⎝

⎜

⎞

⎠

⎟

2

2=3.90 ft2

s2

99,985% of the original KE is converted to internal energy. If we had chosen the initial

masses and velocities so that the final velocity of the duck-bullet system were zero, then

all of the initial KE would have been converted to internal energy.

______________________________________________________________________

7.4 If the bullet stays in the shootee, then the momentum transfer to the shootee is the

same as that to the shooter. We all know that guns "kick", meaning that they transfer as

much momentum to the shooter as they do to the bullet. For artillery pieces there are

large and complicated shock absorbers to accept this force without either damaging the

gun or its mount, or moving it from its location, thus requiring re-sighting.

The following eyewitness account describes the physical response of Mata Hari, whom

the French executed as a spy, (which she denied), to having 11 rifle bullets hit her chest

simultaneously, .(Rifle bullets almost certainly pass through a human body, and exit with

much of their momentum; so the total momentum transferred to her body was certainly

less than the total momentum in the entering bullets.)

".. Mata Hari fell. She did not die as actors and moving-picture stars would have us

believe that people die when they are shot. She did not throw up her hands nor did she

plunge straight forward or straight back.

Instead she seemed to collapse. Slowly, inertly, she settled to her knees, her head

up always, and without the slightest change of expression on her face. For a fraction of a

second it seemed she tottered there, on her knees, gazing directly as those who had taken

her life. Then she fell backward, bending at the waist, with her legs doubled up beneath

her. ...." Wales, Henry G. International News Service, 19 October, 1917.

Movies and TV are a poor place to learn one's physics.

______________________________________________________________________

7.5* System: as sketched at the right, open

0=

Ý

m

V

in −0−

F

wall : F

wall

=

−

Ffluid

=

−

Ý

m

V

in

F

fluid =+Ý

m V

in =50 kg

s⋅80 m

s⋅Ns

2

kg m =4000 N =899.2 lbf

Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 7, page 2

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

Ask the students how one would do this by Bernoulli's equation? Answer; determine the

force by an integral of PdA, determine P by Bernoulli's equation. That can be done by

the methods shown in Chap. 10. It is much more difficult than the method shown here.

______________________________________________________________________

7.6 F

fluid =+Ý

m V

in =200 kg

s⋅400 m

s⋅Ns

2

kg m =80 kN =1.8⋅104lbf

______________________________________________________________________

7.7 (a) By material balance,

Ý

m

3

=

Ý

m

1

−

Ý

m

2 Substituting this in Eq. 7.BK and

simplifying, we find

Ý

m

2

Ý

m

1

=1+cos

θ

2

(b) By momentum balance in the r direction

−

Fwall, r direction

=

Ý

m

1

V

r,1 −

Ý

m

2

V

r,2 −

Ý

m

3

V

r,3

The two rightmost velocities are zero, and

V

r,1

=

−

V

1sin

θ

, minus because it is flow in the

minus r direction, so that

F

wall, r direction

=

V

1sin

θ

One may make plausibility checks by setting

θ

= 0 and 90°, finding that the answers are

what we expect.

______________________________________________________________________

7.8 We make an x-directed momentum balance for steady flow

0= Ý

m V

in −Vout

()+F=

ρ

in AV

in Vin

−

Vout

(

)

+

AP

in

−

P

out

(

); Vout =Vin

ρ

in

ρ

out

⎛

⎝

⎜

⎜

⎞

⎠

⎟

⎟

P

in −P

out =

ρ

inVin

2

ρ

in

ρ

out

−1

⎛

⎝

⎜

⎜

⎞

⎠

⎟

⎟ ;

ρ

in

ρ

out

≅3660oR

528oR=6.93

P

in −P

out =0.075⋅28

29

lbm

ft3

⎛

⎝

⎜

⎞

⎠

⎟ 2ft

s

⎛

⎝

⎜

⎞

⎠

⎟

2

6.93 −1()⋅lbf s2

32.2 lbm ft ⋅ft2

144in2

=3.7⋅10−4psi =2.56 Pa

=

0.019 torr

=

0.01 in H2O

Contrary to our intuition, in an unconfined steady-state flame the pressure decreases as

the gas flows through the flame. The magnitude of this pressure change is quite small,

but it plays a role in shaping the flame.

______________________________________________________________________

7.9 We make a z-directed momentum balance for a steady flow system consisting of the

elevator and jet inside turning vane

−

Felevator acting on jet

=

Ý

m

in (

V

in

−

V

out )

By BE,

V

in =

V

out and Vin =V0−2gz so that −F

elevator acting on jet =Ý

m

in ⋅2V0

2−2gz

For ground level, z = 0, we have

−F

elevator acting on jet =500 lbm

s⋅2 200 ft

s

⎛

⎝

⎜ ⎞

⎠

⎟

2

⋅lbf s2

32.2 lbm ft =6,211 lbf = 27.6 kN

For zero weight of the elevator and its cargo, we have V

0

2−2gz =0

Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 7, page 3

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