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Final

# Ch 07 09Exam

Department
Chemical Engineering
Course Code
CHE 3101
Professor
All
Study Guide
Final

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
s1 lbm
1025 lbm =−21024 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=
m
V
in 0
F
wall : F
wall
=
Ffluid
=
Ý
m
V
in
F
fluid =+Ý
m V
in =50 kg
s80 m
sNs
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
s400 m
sNs
2
kg m =80 kN =1.8104lbf
______________________________________________________________________
7.7 (a) By material balance,
m
3
=
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 =0.07528
29
lbm
ft3
2ft
s
2
6.93 1()lbf s2
32.2 lbm ft ft2
144in2
=3.7104psi =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 =V02gz so that F
elevator acting on jet =Ý
m
in 2V0
22gz
For ground level, z = 0, we have
F
elevator acting on jet =500 lbm
s2 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
22gz =0
Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 7, page 3