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4-12.
Turpentine density = 55 lbm/ft3
a. Compute the total volume spilled by the change in liquid level:
2
t
3
4
= 759 ft 5,680 gallons
D
b. Determine the maximum spill rate from Equation 4-12:
cg
mo L
== 2 +
gP
QuAAC gh
mmm
= 0.00471 lb / sec 2.82 lb /min = 169 l
b
/hr
c. Use Equation 4-18 to determine the time required for the level to drop to 13 ft above
the bottom.
L
15 ft
13 ft
17.9 ft
7 ft
Leak, 0.1 in diam.
15
LL o L
tt
2
AA
Substituting:
52
2
5.45 10 ft
Solve for t with units of seconds. There are two roots:
4-13. The best approach is to use the steam tables for an isenthalpic flash.
Solve for the fraction flashed by mass:
At the initial conditions.
33
mm
The quantity flashed is then:
16
If we use Equation 4-87 instead,
Po b
CT T
We can also use Equation 4-91,
vPobv
From the steam tables
v
oo
ob
1.01
2
405 F, 212 F
C
TT
Substituting into Equation 4-91:
This is close to the other values.
4-14. a. The pipe is threaded into the tank. If the pipe breaks off, it usually breaks off just
b. Use Equation 4-12:
gP
mm
Substituting into Equation 4-12:
322
m
22
m
mm
0.698 lb / ft 2 13.51 63.0 ft / s
8.64 lb / s
Q
c. Need to determine the total volume of the vessel.
2
3
28 in
3.14 18 in
12 in/ft 6.41 ft
D
mm
Then the total draining time, assuming a constant draining rate, is:
347 lb 40.2 s
d. For a pool on the floor of 1-cm depth,
The area of the pool is then,
e. Use Equation 3-12,
18
1/3 1/3
2
o
18
M
f
2
mm
4.32 10 lb / s
Q
f. Use Equation 3-9 or Equation 3-14.
QRT
v74,300 ft 6 / hr 1 hr/3600 s 124 ft / sQ
Substituting into Equation 3-9,
23oo
4-15.
1
4.1 m
2.1 m
19
Then it is seen that:
The mechanical energy balance then reduces to:
cc
2
gg m
The various terms are:
s
2
22 2
3
2.1 4.1 6.2 m
1000 J/s
kW
3.14 0.05 m
1600 kg 3.14 kg/m
m4
z
muA u u
The friction term must consider entrance, pipe friction and exit effects. It is given by,
2
ENT PIPE EXIT
c
2
u
From Equation 4-39,
160 0.5
K
PIPE
0.05 m
d
Substituting,
c
Re 2
g
Returning to the ME balance above,
22
22
160 1000
uu
20
PROCEDURE:
22
4-20. This is a complicated problem involving a mechanical energy balance. For part a the
velocity can be determined and the solution is direct. However, part b is trial and error
since the velocity is not known.
The geometry for this problem is shown below:
For the crude oil:
4
v
The final height of oil in the destination vessel, if all oil transferred, is:
12
22
12
12
22
1
21
2
44
30 m
9 m 5.06 m
40 m
VV
DD
hh
D
hh
D
a. The required pipe velocity to move all the oil in 1 hour is:
33
p
23
We need to size the pump for the worst case pumping situation. In this case, the
b. A 100 HP pump is 74.57 kW. The spreadsheet solution is shown below. All the
c. The conclusion is that this is not a viable method for emergency transfer of liquid
26
4-17. a. From a mechanical energy balance for incompressible flow:
c2
c
gg m
The maximum velocity will occur if: a) the pressure drop across the pump is small, b)
no change in height, and c) no friction. Then the ME balance reduces to:
s
c
2
gm
But
2222
21
muA
uuuu
Then,
2c
guA
Solving for u,
2gW
A
b. 50 mm diameter = 0.05 m
2232
3.14 0.05 m 1.96 10 m
d
Substituting into the equation:
2
The mass flow is:
32
1000 kg 10.1 m/s 1.96 10 m 19.8 kg/s
