How Do You Calculate Mass Flow Rate in a Pressurized Tank?

[Soaring Crane]

A pressurized cylindrical tank, 5.0 m in diameter, contains water which emerges from the pipe at point C, with velocity 29 m/s. Point A is 10 m above point B and point C is 3 m above point B. The area of the pipe at point B is 0.08 m^2 and the pipe narrows to an area of 0.04 m^2 at point C. Assume water is an ideal fluid in laminar fluid. The density of water is 1000 kg/m^3. The mass flow rate in the pipe is closest to:
a. 1000 kg/s
b. 810 kg/s
c. 1200 kg/s
d. 700 kg/s
e. 930 kg/s

I know that the mass flow rate = p*A*v, p = density and v = velocity.

p*A*v = constant

However, the density would be the same since the fluid is water, so wouldn't it be the flow rate that is constant?

A*v = constant

Exactly how do I apply the flow rate equation to this problem?


Thanks.

 

[OlderDan]

There does seem to be a lot of unnecessary information in the problem. If the area and velocity are given at point C, then your mass flow rate equation should be fine unless the velocity that is given is the velocity at one point, perhaps in the middle of the pipe, and you are supposed to be assuming a velocity profile. Take a look at this and see if you think this applies to your problem.

http://hyperphysics.phy-astr.gsu.edu/hbase/pfric.html#veff

 

[Soaring Crane]

Well, the velocity 29 m/s is specifically at point C, where the water flows out of the pipe. Point C is at the end of the pipe. Point B is inside the pipe; it is near the pipe's middle, but it is not quite the middle. Point B is closer to where the cylinder's opening breaches into the pipe.

The order is:
------------pipe
Cylinder ==============
-----------B-------------C

Point A is at the cylinder's top. Point C is higher than B. The pipe is not horizontally straight as depicted above.

Upon reading the website, it mentions that the velocity increases in the pipe's middle if it is a laminar fluid?

At point B, would I have to find the velocity there?

A_b*v_b = A_c*v*c

v_b = (29 m/s)*(.04 m^2)/[.08 m^2] = 14.5 m/s

Or do I just use the info for Point C:

mass flow rate = 1000 kg/m^3*(29 m/s)*(.04 m^2) = 1160 kg/s?

 

[Soaring Crane]

I don't have to apply Bernoulli's equation, do I?

 

[OlderDan]

I don't have to apply Bernoulli's equation, do I?

The mass flow rate is the thing that will be constant throughout the pipe. That is why sections of different diameters have different velocities. Bernoulli's equation would be used to find the pressure at different points, and the velocity (squared) is a contributing factor.

I suspect the laminar flow is the key to the problem. According to the link I posted, the average velocity of the fluid is only half the velocity at the center, so the mass flow is only half what you calculated. I can't be sure. You need to put it in the context of the material you are studying.

 

[Cyrus]

You have all the necessary information,

mdot=rho*a*v=1000*29*0.04=1160

ans. c)

Whats so hard about this?

 

[OlderDan]

You have all the necessary information,

mdot=rho*a*v=1000*29*0.04=1160

ans. c)

Whats so hard about this?

There is nothing hard about it if you assume the velocity is constant across the diameter of the pipe. Without knowing the context of the problem, can we be sure that is a valid assumption?