2. A cylindrical tank 5 ft high has a constant diameter of 4 ft is full of water. The tank has a hole in its bottom that measures 0.1 ft². All losses are insignificant. How long will it take for the tank to empty? Note that you can NOT assume a static head and constant flow rate because the water head in the tank decreases as the water flows out of the tank.
2. A cylindrical tank 5 ft high has a constant diameter of 4 ft is full of water. The tank has a hole in its bottom that measures 0.1 ft². All losses are insignificant. How long will it take for the tank to empty? Note that you can NOT assume a static head and constant flow rate because the water head in the tank decreases as the water flows out of the tank.
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Transcribed Image Text:2. A cylindrical tank 5 ft high has a constant diameter of 4 ft is full of water. The tank has a hole
in its bottom that measures 0.1 ft². All losses are insignificant. How long will it take for the tank
to empty? Note that you can NOT assume a static head and constant flow rate because the
water head in the tank decreases as the water flows out of the tank.
(A) 35 sec
(B) 70 sec
(C) 105 sec
(D) 140 sec
![From the conservation of mass.
min-mout=mcv
=mcv
0-m
mout
out
= -mcv
pxaxV=-px
Where, V = √ (2gH)
d
ax√(2gH):
dt 4
axV (2g) XH1/
ax√(2g) xdt=
After integrating,
ax√ (29) × /dt = - xD²×₁
11
x
==
d (7 ×D² xH)
==
-7
t = 70.0296 s
t≈70 s
t=70 s
π
XD² xH)
품
0.1 x√ (2x32.2) xt=
XD² x
XD² x.
(2g) xt=
XD²
Putting the known values
dH
H1/2
π
dH
dt
ex√ (29) Xt=-=xD³² x [01/2-1/2]
--
H1/2
1/2
میں
O
dH
H H1/2
51/2
·x4² x
x
1/2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5504642f-0145-4427-b5f2-66a9310e79bc%2F966d7b62-aa72-4e7a-b609-8a5374897ea2%2Fxxsa6jn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:From the conservation of mass.
min-mout=mcv
=mcv
0-m
mout
out
= -mcv
pxaxV=-px
Where, V = √ (2gH)
d
ax√(2gH):
dt 4
axV (2g) XH1/
ax√(2g) xdt=
After integrating,
ax√ (29) × /dt = - xD²×₁
11
x
==
d (7 ×D² xH)
==
-7
t = 70.0296 s
t≈70 s
t=70 s
π
XD² xH)
품
0.1 x√ (2x32.2) xt=
XD² x
XD² x.
(2g) xt=
XD²
Putting the known values
dH
H1/2
π
dH
dt
ex√ (29) Xt=-=xD³² x [01/2-1/2]
--
H1/2
1/2
میں
O
dH
H H1/2
51/2
·x4² x
x
1/2
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