Chestermiller said:Have you taken thermo yet? If so, do you remember the form of the first law for a flow system, involving shaft work and change in enthalpy of the flow stream? How much of a temperature rise do you expect through the pump? (I expect negligible).
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Yes, this is exactly the equation I was referring to, and yes, ΔU = 0. Also, Q is equal to zero (if you assume that the pump is adiabatic). So, what do you get for the shaft work?Undergrad1147 said:I'm currently taking thermo. I think our terminology may be different to yours, or else I'm not following. A flow system, as in a control volume at steady state? One form of the first law which comes to mind that may be applicable to the pump is is: $$Q - W + \Sigma m_{in}h_{in} - \Sigma m_{out}h_{out} = \Delta U$$
The pump would be steady flow so ##\Delta U## would be 0, yes? Is this the equation that you're referring to or have I misunderstood you?
It's just one stream so ##m_{in}## will equal ##m_{out}##, yes? If so, then: $$W = m(h_{in} - h_{out})$$Chestermiller said:Yes, this is exactly the equation I was referring to, and yes, ΔU = 0. Also, Q is equal to zero (if you assume that the pump is adiabatic). So, what do you get for the shaft work?
OK so far.Undergrad1147 said:It's just one stream so ##m_{in}## will equal ##m_{out}##, yes? If so, then: $$W = m(h_{in} - h_{out})$$
We're given the mass flow rate so we'll end up with the equation: $$\dot{W} = \dot{m} (h_{in} - h_{out})$$ Is this correct?
The water is initially at 20 C and atmospheric presure, thus it's enthalpy is 83.95 kJ/kg and after leaving the pump it's at 20 C and 1.69 MPa with an enthalpy of 85.5 kJ/kg. Thus the Shaft Work Rate is equal to: $$\frac{36(1000) (83.95 - 85.5)}{3600} = -15.5 J/s = -15.5 W$$
Does this sound correct?
I was thinking it was a little low alright. Why isn't it ##C_v## as opposed to ##C_p## in the above equation though? h = u + Pv, so ##\Delta h = \Delta u + \Delta Pv## and ##\Delta u = C_v \Delta T## no?Chestermiller said:OK so far.
No. The change in specific enthalpy for an incompressible fluid is Δh=CpΔT+vΔP, where v is the specific volume = 1/ρ, where ρ is the density. Try again. (15.5 Watts is a little low for a pump that is pumping all that water to a high pressure).
For an incompressible liquid, Cp and Cv are the same.Undergrad1147 said:I was thinking it was a little low alright. Why isn't it ##C_v## as opposed to ##C_p## in the above equation though? h = u + Pv, so ##\Delta h = \Delta u + \Delta Pv## and ##\Delta u = C_v \Delta T## no?
Chestermiller said:For an incompressible liquid, Cp and Cv are the same.
Incidentally, that result you got probably was correct, except off by a factor of 1000. Those specific enthalpies were in kJ, not J. So the work ought to be 15.5 kW, not 15.5 W. Try doing it the way I suggested also to see that the result comes out the same.
The 15.5 kW is the mechanical work that must be delivered by the pump. But the pump will not be 100% efficient in converting electrical energy from the motor to mechanical energy. Pick what you consider to be a reasonable efficiency, and divide the 15.5 kW by that to get the actual pump requirement.
Very nice. I would round that result off to 20 kW.Undergrad1147 said:Yeah. Using your way you get -15.9 kW. So, they're fairly close. If you were to say the pump is about 80% efficient then, 19.875 kJ/s needs to be supplied to the pump for the water to be pumped from ##P_{atm}## to 1.69 MPa.
Chestermiller said:Very nice. I would round that result off to 20 kW.
I'm hoping you might consider continuing with the heat exchanger design. My suggestion: assume 100 tubes of 4 cm diameter, and use the equations in Chapter 14 of BSL to determine the heat transfer coefficients on the water side and on the steam side of the tubes, and then to determine the required length L. It's pretty much cook book. At the beginning of the chapter, they tell you to use the "film temperature" to determine the required physical properties of water. The film temperature is arithmetic average of the inlet and the outlet water temperatures, then averaged with the steam temperature. You need the water viscosity at this temperature, the heat capacity, the density, and the thermal conductivity. You need to determine the Reynolds number in each tube and the Prantdl number for water at the film temperature. There is then a forced convection graph from which you can determine the heat transfer coefficient on the water side. There is also an equation later in the chapter for getting the heat transfer coefficient for condensing vapor on a horizontal tube. From this, you can get a much much more accurate estimate of the overall heat transfer coefficient than the 680 value which you used earlier, and that it was too crude an approximation.
My pleasure. It was very nice working with you. You showed great determination and tenaciousness.Undergrad1147 said:Thanks for the help. I'll have to hold off on the heat exchanger for now until I free up some time. I'll definitely get back to it at some point and I'll post about it. Thanks for the help with the problem in general.