Problem solving state change exercise

[M Wierckx]

Hi everybody,

I'm having a problem with the following question:

A piston-cylinder device originally contains 5 kg of saturated-liquid water at 100 kPa. Determine the heat addition required to bring the fluid to a saturated-vapor state.

Using the first law.. i was thinking the following:

1Q2 = 1W2 + (U2 - U1)

U2 and U1 could be found in the table of liquid water with a pressure of 100 kPa, with U2 being Saturated vapor (Ug) and U1 saturated liquid (Uf).

The W (work) part I was thinking about the P x (V2-V1) again using the table, V2 being sat. vapor (vg) and V1 being sat. liquid (vf)


Doing this I'm not finding the right answer.. can anybody help me solving this and telling me which step I'm taking wrong?

 

[RTW69]

If the piston isn't rising there isn't any work being done so w=0 and q= ΔU. See if that works. I would pick Internal Energy values off of a Saturated Water: Pressure Table.

 

[M Wierckx]

I think the piston can be moved, or else they would have made it a rigid tank.. I tried leaving the W out and making Q equal to U2-U1, but this is not correct.. The answer has to be eleven dot something (11,...kJ/kg).. but thanks for the reply!

 

[RTW69]

For a constant batch system Q=ΔU+W where W= P(V_g-V_f)

 

[asteriatic]

I understand that problem solving can be a challenging and complex process. In this case, it seems that you are on the right track by using the first law of thermodynamics and the properties of water at different states. However, there are a few things that could be causing your answer to be incorrect.

First, make sure that you are using the correct units for all the values in your calculations. The properties of water in the table may be listed in different units than what is needed for the equation. Double check to ensure consistency in units.

Secondly, be sure to consider the energy balance in the system. In this case, the heat added to the system is equal to the change in internal energy plus the work done on the system. It is important to consider both the internal energy and work terms in your equation.

Lastly, it is always helpful to double check your calculations and make sure that you are using the correct values from the table. It is possible that a small error in inputting the values could result in a significant difference in the final answer.

I hope this helps guide you in the right direction and leads to a successful solution to the problem. Don't be discouraged, problem solving is a skill that takes practice and patience. Keep at it and you will find the correct answer.