Tank & Pipes Question

@maria-flor-8cAkif • Oct 23, 2024

Oct 23, 2024

2.8K

A tank is supplied by two pipes A and B and emptied by a third pipe C. If the tank is empty and all pipes are opened, the tank can be filled in 20 hours. If the tank is full and A and C are opened, the tank can be emptied in 4 hours. If the tank is full and B and C are opened, the tank can be emptied in 2 hours. Pipe A supplies 50 liters per minute more than B.
*Find the rate of pipe A in liters per minute.
*Find the rate of pipe C in liters per minute.
*find the capacity of the tank in liters.

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ISHAN TOPRE

@ishan-nohePN • Mar 23, 2011

Re: ?????

The volume of tank=[C-(a+b)]*20*60=[(b+50+b)-c]*20*60=[2b-c+50]20*60---------1

If A and C are opened volume of tank= (c-a)*4*60--------2
If B and C are opened volume of tank=(c-b)*2*60------------3

Equating 2 and 3 with 1 we get two equations as

21b-11c+500=0 and 11b-6c+300=0
a=110lit/min, b=60 lit/min c=100 lit/min
vol of tank= 1200liters.

Am I right?
0
maria flor

@maria-flor-8cAkif • Mar 23, 2011

Re: ?????

I think your correct....but something wrong on your answer in pipe B.
Find the rate of pipe A in liters per minute.
a. 120 b. 130 c.110 d. 140
Find the rate of pipe C in liters per minute.
a. 170 b. 160 c. 150 d. 140
find the capacity of the tank in liters.
a. 1200 b. 12500 c. 11500 d. 13000
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ISHAN TOPRE

@ishan-nohePN • Mar 23, 2011

Yup got it Maria. 😀
a=110lit/min, c=160lit/min volume of tank= 1200 liters. It was just a calculation mistake. 😁
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maria flor

@maria-flor-8cAkif • Mar 23, 2011

Find the rate of pipe A in liters per minute. Answer: 110
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maria flor

@maria-flor-8cAkif • Mar 23, 2011

Here's the next question:
A cylindrical tank of radius 1.5m. and height of 5m. is filled with water to depth of of 2m.
*If one spherical ball having a diameter of 2m. is dropped into tank slowly, by how much is the level of water raised?
*If two spherical balls having diameter of 2m. each is dropped into the tank slowly, by how much is the level of the water raised.
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ISHAN TOPRE

@ishan-nohePN • Mar 23, 2011

Yup Maria.
Volume of water initially in tank= Pi*1.5*1.5*2=14.137m3
Volume of ball=4*Pi*8/3=33.510m3
When the ball is dropped in tank Vol=vol of tank initially+vol of ball.

Dividing this total volume by Pi*1.5*1.5 we will get the vertical height.
Vertical height-2 will give total rise in water level.
The rise in tank level will be 4.7407m and 9.481m respectively.

Am I right?
0
maria flor

@maria-flor-8cAkif • Mar 23, 2011

ishutopre
Yup Maria.
Volume of water initially in tank= Pi*1.5*1.5*2=14.137m3
Volume of ball=4*Pi*8/3=33.510m3
When the ball is dropped in tank Vol=vol of tank initially+vol of ball.

Dividing this total volume by Pi*1.5*1.5 we will get the vertical height.
Vertical height-2 will give total rise in water level.
The rise in tank level will be 4.7407m and 9.481m respectively.

Am I right?
It's wrong Ishu...
0
maria flor

@maria-flor-8cAkif • Mar 25, 2011

Here's another question but not related to pipe or pipe. This is just easy.
Two circles are tangent to a third circle internally and are tangent to each other externally. The distance between their centers are 10 inches, 7 inches and 5 inches. Find their radii.
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ISHAN TOPRE

@ishan-nohePN • Mar 25, 2011

Oh sorry I calculated the answers yesterday but forgot to post them. 😔

If you will draw the diagram you will get the three equations.
x+y=10
y+z=5
x+z=7

Solving them we get the radii of those circle as x=6, y=4, z=1.
Am I right?
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maria flor

@maria-flor-8cAkif • Mar 26, 2011

Here's next:
A frog is trapped in a 24m well. Every minute it jumps 3m and slips back 2m. After how many jumps will it reach the top of the well????
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ISHAN TOPRE

@ishan-nohePN • Mar 26, 2011

The frog jumps 1m on each jump. So he will jump 22 times. Because after the 21st jump, the frog will get out on 22nd jump.
How's that? 😀
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maria flor

@maria-flor-8cAkif • Mar 26, 2011

OK. Pls. show your solution
Next:
The perimeter of a triangle with integral sides is 24 units. If it is a right triangle, what must be its are? Show your solution.😀
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ISHAN TOPRE

@ishan-nohePN • Mar 26, 2011

As you say, if the triangle is an isosceles right angled triangle.
Now hypotenuse=(underroot2)*length of side.

hence let length of side be x.

x+x+(root 2)*x=24
x=7.0294
Now area of triangle=0.5*base*height.
Area=0.5*7.0294*7.0294
=24.706 square unit
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maria flor

@maria-flor-8cAkif • Mar 26, 2011

Next question:
A mixture compound from equal parts of two liquids, water and oil was placed in a hemispherical bowl. The total depth of the two liquids is 6cm. After standing for a short time in the mixture separated. Thethickness of the segment of oil is 2cm.Find the base radius of the segment containing water.
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maria flor

@maria-flor-8cAkif • Mar 26, 2011

ishutopre
Yup Maria.
Volume of water initially in tank= Pi*1.5*1.5*2=14.137m3
Volume of ball=4*Pi*8/3=33.510m3
When the ball is dropped in tank Vol=vol of tank initially+vol of ball.

Dividing this total volume by Pi*1.5*1.5 we will get the vertical height.
Vertical height-2 will give total rise in water level.
The rise in tank level will be 4.7407m and 9.481m respectively.

Am I right?
Just to remind you, your answer is wrong.
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CIVILPRINCESS

@civilprincess-308hDv • Mar 27, 2011

OK Maria. Assuming the hemispherical container is filled full. The original depth is full. So the radius of hemispherical container is 8 cm. Agreed?

Now the upper 2 cm is the oil layer. So if you draw the diagram and chop off above 2 cm from hemisphere, then you get the radius of immediate base radius of segment.as 7.746 cm. 😀

Also the diameter is 15.492 cm. 😀

Am I right?
0
maria flor

@maria-flor-8cAkif • Mar 27, 2011

ishutopre
OK Maria. Assuming the hemispherical container is filled full. The original depth is full. So the radius of hemispherical container is 8 cm. Agreed?

Now the upper 2 cm is the oil layer. So if you draw the diagram and chop off above 2 cm from hemisphere, then you get the radius of immediate base radius of segment.as 7.746 cm. 😀

Also the diameter is 15.492 cm. 😀

Am I right?
your answer is wrong..
0