Thermodynamics problem -- Water pumped into an enclosed tank....

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Homework Help Overview

The discussion revolves around a thermodynamics problem involving the work done by a pump to move water from an open well to a closed tank under pressure. The original poster presents a scenario with specific flow rates and pressures, seeking to compute the work required for this process.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the equation for work in relation to pressure differences and volume changes. There is confusion regarding the terms used in the equations, particularly the definitions of 'v' and its relevance to the problem. Some participants attempt to clarify the relationship between flow rate, volume, and time.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and questioning the validity of the equations presented. Some guidance has been offered regarding the definitions of variables, but no consensus has been reached on the correct approach to the problem.

Contextual Notes

There is a noted confusion regarding the terminology used in the problem statement, particularly the use of 'pa' instead of 'MPa', and the implications this has for the calculations. Additionally, the participants are navigating the constraints of the problem as presented, including the flow rate and the need to compute work over a specified time period.

danny12345
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Homework Statement


Q3.1 (a) A pump forces 1m^3 /min of water horizontally from an open well to a closed tank where the pressure is 0.9 Mpa.compute the work the pump must do upon the water in an hour just to force the water into the tank against the pressure.sketch the system upon which the work iss done before and after the process.

Homework Equations

The Attempt at a Solution


here flow rate is 1m^3/min
p1=0.101325 pa
p2=0.9 pa
we don't know v.
but equation formed will be as:
w=dpdv
now I'm where i can get v.
 
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This is not a thermodynamics question. Titling it so may have deterred some potential responders.
Where you wrote pa in your attempted solution, I guess you meant MPa.
You don't need to know the velocity.
I don't understand your w=dpdv equation. Please explain the terms.
 
it's w=(p2-p1)(v2-v1)
 
it's w=(p2-p1)(v2-v1)
 
danny12345 said:
it's w=(p2-p1)(v2-v1)
I asked you to explain the terms. What are the v's? If they're velocities, and the p's are pressures, and w is work, then the equation is dimensionally wrong. If the v's are volumes then the dimensions work, but I don't see the relevance to problem. [And anyway, I think you would mean W= d(PV)=PdV+VdP.]
 
v=volume
 
danny12345 said:
v=volume
Ok.
You know the flow rate, and you know the time, so...
 
so what then?
 
danny12345 said:
so what then?
So you can find the volume.
 
  • #10
how?
 
  • #11
danny12345 said:
how?
danny12345 said:
A pump forces 1m^3 /min of water ... the work the pump must do upon the water in an hour...
A cubic metre of water per minute, for one hour, gives...?
 

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