QUESTION 3 A vertical cylindrical tank (diameter 1 m) is being filled with cold water (4 °C) at a rate of 5 kg/s. Water is also escaping from the tank via a 5 cm diameter tube located at the bottom; the velocity (m/s) of water in the tube is given by √2gh where h is the height of water above the tube. a. Using the cylindrical tank as the control volume, write the general mass balance equation for the water in the tank. b. Determine the height of water in the tank (H*) at which the system is at steady-state and thus there is no change with time. Inlet flow c. At some time (tor), the inlet flow of water is stopped but the water continues to escape out the bottom. Derive the appropriate equation for the rate at which the water level in the tank is falling (dh/dt). State the appropriate initial condition that is necessary to solve this equation. NOTE: You do NOT need to solve the differential equation in part c)

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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QUESTION 3
A vertical cylindrical tank (diameter = 1 m) is being filled with cold water (4 °C) at a rate of 5 kg/s.
Water is also escaping from the tank via a 5 cm diameter tube located at the bottom; the velocity (m/s) of
water in the tube is given by √√2gh where h is the height of water above the tube.
a. Using the cylindrical tank as the control
volume, write the general mass balance
equation for the water in the tank.
b. Determine the height of water in the tank (H*)
at which the system is at steady-state and thus
there is no change with time.
c. At some time (tom), the inlet flow of water is
stopped but the water continues to escape out
the bottom. Derive the appropriate equation for
the rate at which the water level in the tank is
falling (dh/dt). State the appropriate initial
condition that is necessary to solve this
equation.
NOTE: You do NOT need to solve the
differential equation in part c)
5cm
I
2
A = 2 (50) ²
v= √2gh
Q (volumetric flowrate) = 1.96*10³ Nagh
3
m (mass flow rate) = P (density) Q
1000
= 1.96 √2gh
a.
*= volume
=
dt
act
I D² h = x 1²h
4
4
*PRIJ
water in tank
h
(7000
(1000 - Ih) =
4
= 1-96 x 10-3
250 X
=
250x
250
Inlet
flow
4
rate of accumulation of water = ± (P.*)| p= 1000
dt
3
kg/m ³
dh
dt
z dh
dt
m²
dh
dt
1 m
1-96*103 √zgn
IN
minlet
5-1.96 √2gh
OUT
+1.96 №2gh
moutlet
5 = 100
kg/s
b.
Im
di
T = 4°C
minket = 5 kg/s
dt
OUT
d₂ = 5cm = 0.05m
2
v=12gh
A = x.
Ø
Ø
250x5
2504
IN
5
=
5 =
S
1.96
1/12/201
2²
-1.96√2gh
250 x
volume of
a cylinder
=
ड
- 1.96√2h
250x
250 x
1.96 12gh.280*
250x
1.96 Nigh
1zgh
2
) ²
0.33m=h
¥=x√²h
= *.1. 1². n
и
4
Nzg(1.96)
૧૮
h
C.
dh
√zgh
t
0
dt
LOO
h
t
- (2√5) - £0
(স)
=
100
√2g
№2g (√n-√no)
= t
h(t) =
=
LOO
+ √2gho
100
2gh = (1 + √2gh) ²
2/3 (10/10 + √2gh) ²
29
idk مرد ((
Transcribed Image Text:QUESTION 3 A vertical cylindrical tank (diameter = 1 m) is being filled with cold water (4 °C) at a rate of 5 kg/s. Water is also escaping from the tank via a 5 cm diameter tube located at the bottom; the velocity (m/s) of water in the tube is given by √√2gh where h is the height of water above the tube. a. Using the cylindrical tank as the control volume, write the general mass balance equation for the water in the tank. b. Determine the height of water in the tank (H*) at which the system is at steady-state and thus there is no change with time. c. At some time (tom), the inlet flow of water is stopped but the water continues to escape out the bottom. Derive the appropriate equation for the rate at which the water level in the tank is falling (dh/dt). State the appropriate initial condition that is necessary to solve this equation. NOTE: You do NOT need to solve the differential equation in part c) 5cm I 2 A = 2 (50) ² v= √2gh Q (volumetric flowrate) = 1.96*10³ Nagh 3 m (mass flow rate) = P (density) Q 1000 = 1.96 √2gh a. *= volume = dt act I D² h = x 1²h 4 4 *PRIJ water in tank h (7000 (1000 - Ih) = 4 = 1-96 x 10-3 250 X = 250x 250 Inlet flow 4 rate of accumulation of water = ± (P.*)| p= 1000 dt 3 kg/m ³ dh dt z dh dt m² dh dt 1 m 1-96*103 √zgn IN minlet 5-1.96 √2gh OUT +1.96 №2gh moutlet 5 = 100 kg/s b. Im di T = 4°C minket = 5 kg/s dt OUT d₂ = 5cm = 0.05m 2 v=12gh A = x. Ø Ø 250x5 2504 IN 5 = 5 = S 1.96 1/12/201 2² -1.96√2gh 250 x volume of a cylinder = ड - 1.96√2h 250x 250 x 1.96 12gh.280* 250x 1.96 Nigh 1zgh 2 ) ² 0.33m=h ¥=x√²h = *.1. 1². n и 4 Nzg(1.96) ૧૮ h C. dh √zgh t 0 dt LOO h t - (2√5) - £0 (স) = 100 √2g №2g (√n-√no) = t h(t) = = LOO + √2gho 100 2gh = (1 + √2gh) ² 2/3 (10/10 + √2gh) ² 29 idk مرد ((
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