Using Steam Tables to Solve Part (b) - Find Initial and Final Temps

[theBEAST]

Homework Statement

The Attempt at a Solution

I thought you could use steam tables to find the initial and final temperatures in order to solve part (b). The solution assumes rho = 1000kg/m^3 (or in other words all liquid). So in the steam tables:

This tells me that at 100kPa the temperature is around 100C and at 300kPa the temperature is 134C. Thus the difference is 34C and not 0.028C (which is the answer). Why does this method not work?

 

[SteamKing]

You are using what looks like saturated liquid tables. The temperatures you see in the table represent the point at which the vapor pressure of water is equal to the ambient pressure, or, the point at which boiling can occur.

For liquid water at the inlet and outlet pressures of the pump, temperature can be a variety of values. The liquid water is not necessarily boiling.

The temperature rise in the liquid between the inlet and outlet of the pump can be analyzed by using the first law of thermodynamics with the control volume drawn around the pump.

 

[Chestermiller]

In a steady flow system like this, what is the rate at which work is being done to push 18 l/s from the inlet pipe at 100 kPa into the outlet pipe at 300 kPa? How does this compare with the work being done by the electric motor in running the pump? In a steady flow system like this, how is the change in enthalpy per unit mass of fluid flowing through the system related to the rate of shaft work, assuming adiabatic operation of the pump?

 

[theBEAST]

Should I be using the steady state steady flow equation from the first law: [Qcv_dot + m1_dot * (h1+v1^2/2+gz1)=m2_dot * (h2+v2^2/2+gz2)+Wcv_dot]? but in this case we are missing enthalpy.

So essentially the efficiency would be the change in enthalpy over the work done by the electric motor. But once again we don't have enthalpy.

 

[Astronuc]

Note that the water goes in as a liquid, probably cold at 100 kPa and exits at a higher pressure, so it is a compressed liquid. One would not use saturated steam tables.

The work from the pump goes into raising the pressure and some slight increase in water temperature, so one is looking for the change(s) in the liquid.

 

[Chestermiller]

Should I be using the steady state steady flow equation from the first law: [Qcv_dot + m1_dot * (h1+v1^2/2+gz1)=m2_dot * (h2+v2^2/2+gz2)+Wcv_dot]? but in this case we are missing enthalpy.

So essentially the efficiency would be the change in enthalpy over the work done by the electric motor. But once again we don't have enthalpy.

You are aware that you can calculate the change in enthalpy without using enthalpy tables, correct? In terms of internal energy, pressure, and volume, what is the definition of enthalpy. Can you calculate the change in internal energy of a parcel of liquid?

You also need to go back to the derivation of the above equation in your textbook, and get a better fundimental understanding of where the various terms come from, and how this expression of the first law for a steady flow system is related to the expression of the first law for a closed system.

chet

 

[theBEAST]

Thanks for the help everyone!

I was able to find the change in temperature using the first law but I am not sure how to find the efficiency.

Is it just the work done by the fluid over the work in from the electric motor? Since we are looking at mechanical efficiency...? So (m_dot*vΔP)/W_dot?

 

[Chestermiller]

Thanks for the help everyone!

I was able to find the change in temperature using the first law but I am not sure how to find the efficiency.

Is it just the work done by the fluid over the work in from the electric motor? Since we are looking at mechanical efficiency...? So (m_dot*vΔP)/W_dot?

You have the right idea. You are looking for the work done on the fluid by the pump. Your equation is correct, provided v is the specific volume. Also note that m_dot*v = Q, where Q is the volumetric throughput rate (18 l/s).

Good progress!