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Fluid Flow Rate
#1
Posted 09 March 2010 - 10:10 AM
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 .
#2
Posted 10 March 2010 - 07:25 AM
#3
Posted 10 March 2010 - 09:13 AM
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.
#4
Posted 10 March 2010 - 10:07 AM
Edited by vhpk, 10 March 2010 - 10:19 AM.
#5
Posted 10 March 2010 - 11:17 AM
"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.
#6
Posted 11 March 2010 - 04:00 PM
Suggestion, assuming that "header" is the upper tank: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.
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.
#7
Posted 19 March 2010 - 04:08 AM
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
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