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
icon
Related questions
Topic Video
Question
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
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
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
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps with 1 images

Blurred answer
Knowledge Booster
Fluid Statics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Elements Of Electromagnetics
Elements Of Electromagnetics
Mechanical Engineering
ISBN:
9780190698614
Author:
Sadiku, Matthew N. O.
Publisher:
Oxford University Press
Mechanics of Materials (10th Edition)
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:
9780134319650
Author:
Russell C. Hibbeler
Publisher:
PEARSON
Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:
9781259822674
Author:
Yunus A. Cengel Dr., Michael A. Boles
Publisher:
McGraw-Hill Education
Control Systems Engineering
Control Systems Engineering
Mechanical Engineering
ISBN:
9781118170519
Author:
Norman S. Nise
Publisher:
WILEY
Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:
9781337093347
Author:
Barry J. Goodno, James M. Gere
Publisher:
Cengage Learning
Engineering Mechanics: Statics
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:
9781118807330
Author:
James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:
WILEY