Energy Analysis of Control Volumes at Steady State

[blackkeys]

Homework Statement

Steam enters a well insulated nozzle at...

Pressure1 =300lbf/in^2
Temp1 =600 degrees F
Velocity1 =100 ft/s

The steam exits the nozzle at...

Pressure2 =40lbf/in^2
Temp2 =?
Velocity2 =1800ft/s

For steady-state operation, and neglecting potential energy effects, determine the exit temperature in F

Homework Equations

Energy rate balance
Mass rate

The Attempt at a Solution

I began by stating that heat transfer is zero since the nozzle is well insulated. Then I stated that work is also zero since the steam is traveling through a nozzle and not some kind of paddle wheel or device as such. Next I stated that PE was zero since this was a given.

Then, I set up my conservation of energy for a control volume equation.

0 = (mass rate)[(h1-h2)+(1/2)(vel(1)^2-vel(2)^2)]

The next thing I did was to consult my tables. Due to having two quantities I could use at the entrance of the nozzle I found that h1 was 1314.5 Btu/lb as was stated in the properties of superheated water vapor table.

So now I have,

0 = (mass rate)[(1314.5-h2)+(1/2)((100)^2-(1800)^2)]

My two unknowns are the mass rate and h2. Now if I could somehow solve for the mass rate I could determine h2 and then use my tables and linear interpolation to solve for the temp at exit, but I can't figure out how to solve for the mass rate without having an area of the nozzle or knowing the density of steam in these given conditions. This leads me to think that I may be I'm using the wrong equation and don't need to find the mass rate after all?

So my question is, how would I solve for mass rate or am I using the right equation?

Thanks.

 

[KataKoniK]

You are on the right track with using the conservation of energy equation, but there are a few things to consider. First, you are correct in assuming that work is zero since it is a nozzle and there are no moving parts. However, potential energy should also be taken into account, as it affects the total energy of the system. In this case, the potential energy at both the entrance and exit of the nozzle can be assumed to be zero, but it should still be included in the equation.

Next, you are correct in using the specific enthalpy values from the superheated water vapor table. However, the specific enthalpy at the exit of the nozzle (h2) cannot be determined from the given information. This is because the velocity at the exit (1800 ft/s) is supersonic, meaning the steam expands and undergoes a phase change from superheated vapor to a mixture of vapor and liquid. This change in phase also affects the specific enthalpy, so it cannot be determined solely from the given information.

To solve for the exit temperature, you will need to use the isentropic expansion process for steam. This process takes into account the change in phase and the decrease in pressure, and can be found in steam tables. You will need to use the given pressure at the exit (40 lbf/in^2) and the specific enthalpy at the entrance (1314.5 Btu/lb) to find the specific enthalpy at the exit. From there, you can use the specific enthalpy and temperature at the exit to find the exit temperature.

In summary, the correct equation to use would be:

0 = (mass rate)[(h1+hPE1)-(h2+hPE2)+(1/2)(vel(1)^2-vel(2)^2)]

Where hPE1 and hPE2 are the specific enthalpies due to potential energy at the entrance and exit, respectively. You will need to use the isentropic expansion process for steam to find h2.