6 replies to this topic

[vhpk]

Hi everyone,
I've got a question needed your help.
A 22% sodium chloride solution is to be pumped up from a feed tank into a header tank at the top of a building. If the feed tank is 40 m lower than the header and the pipe is 1.5 cm in diameter, find
a/ e velocity head of the solution flowing in the pipe
b/ the power required to pump the solution at a rate of 8.1 cubic metres per hour. Assume that the solution is at 10°C, pipeline losses can be ignored, the pump is 68% efficient, and that the density of the sodium chloride solution is 1160 kg m^-3
For part a, is that true to say that all the PE will be converted to KE so we can deduce the velocity ?
For part b, I take the energy per unit of mass multiply with the mass flow rate and 1/density to get the final answer.
However, I need your help with part a since it's related to part b. Thank you a lot .

[vhpk]

Can anyone help me with this

[latexman]

vhpk,

I doubt you will get much help. It appears you want someone to do your entire homework problem. That is pitiful for many reasons, but the most important one is, it will detract from your education. Suck it up and give it a try, or get out of Engineering.

[vhpk]

No, it's not my homework, it's a problem I'm trying to answer. As I said above, I got my own way to deal with it. However, I don't get the answer in the solution, so what I want is people take a look at my problem and confirm whether my manner is true or false. If true, I can believe that the answer is wrong and unless, plz point at my fault. I'm old enough and self respected enough not to ask anyone to solve the homework for me. Sorry but if you only look at anyone ask for their problem they meet in their questions and you think they want the others to solve the problem for them, you should re-view your view point.

Edited by vhpk, 10 March 2010 - 10:19 AM.

[latexman]

It's in the Student Forum and it reads like a textbook problem to me, therefore I classified it as "homework".

"I can believe that the answer is wrong" What answer is given? That information would have helped us.

If you ignore the flow rate in b/ there is not enough information. If the flow rate is ~ 0 (no kinetic energy) and there is ~ 0 friction, then the discharge pressure of the pump (potential energy) will be converted to the higher elevation (potential energy) of the fluid. If the flow rate is significant and there is friction, then the discharge pressure of the pump (potential energy) will be converted to the higher elevation (potential energy) of the fluid and overcome the friction and the kinetic energy of the fluid in the pipe will be lost as it goes into the tank.

[kkala]

A 22% sodium chloride solution is to be pumped up from a feed tank into a header tank at the top of a building. If the feed tank is 40 m lower than the header and the pipe is 1.5 cm in diameter, find
a/ the velocity head of the solution flowing in the pipe
b/ the power required to pump the solution at a rate of 8.1 cubic metres per hour. Assume that the solution is at 10°C, pipeline losses can be ignored, the pump is 68% efficient, and that the density of the sodium chloride solution is 1160 kg m^-3
For part a, is that true to say that all the PE will be converted to KE so we can deduce the velocity ?
For part b, I take the energy per unit of mass multiply with the mass flow rate and 1/density to get the final answer.

Suggestion, assuming that "header" is the upper tank:
a/ Q=8.1 m3/h = 0.00225 m3/s. Pipe section s=π*0.015^2/4=0.000176714 m2. v=Q/s=12.73 m/s. Velocity head= v2/2g=12.73^2/(2*9.81)=8.26 m of liquid.
Velocity is unrealistically high for the case.
b/ Weight rate of liquid transferred W=Qrg=25.604 Nt/s. Theoretical power Wh=25.604*40 Ntm/s=1024 W (pipeline frictional losses ignored). Pump power 1024/0.68=1506 W.
Usually frictional losses are not neglected in such cases. Even though "velocity head" is well defined in Chemical Engineering hydraulics, can it have here another meaning?
According to above thinking, reply to the questions are negative, noting the following in addition.
a. Velocity of liquid is zero at both tanks. The liquid receives potential energy (PE) from the pump to be lifted. It acquires temporal kinetic energy of 1/2*2.61kg/s*12.73^2(m/s)^2=211.5 W, transferred into potential energy as approaches upper tank (of course this under the unrealistic assumption of no energy loss due to friction).
b. energy/mass * mass rate = power (without involvement of 1/density). Indeed mass rate = 2.61 kg/s, work/mass=9.81m/s2*40m=392.4 (m/s)^2, product=2.61kg/s*392.4m2/s2=1024 (kgm/s2)(m/s)=1024 W.

Edited by kkala, 12 March 2010 - 10:14 AM.

[vhpk]

Thank you both of you, latexman and kkala for the useful direction you gave me.
For the question a/, firstly I thought the data 8.1m3 was only for question b so I didn't use it, therefor I didn't get the correct answer. Thank you for the confirmation