Saturated water tables:use temp or pressure

[kdorsel]

Homework Statement

In my class we use two Saturated Water (Liquid-Vapor) tables. One is tabulated based on temperature and the other based on pressure.


For example if a state is defined by p=10 kPa and T=40 C and I need to find h_f I can get either 167.57 or 191.82 depending on which table I use.

I am unsure which table to use.

The Attempt at a Solution

Looking in terms of the temperature table at 40 C the given pressure is 7.384 kPa which is less than the state's pressure.

Looking in terms of the pressure table at 10 kPa the given temperature is 45.81 C which is more than the state's temperature.

I have looked at a few problems and whenever this happens one table will yield results were the two properties will be less then the two properties from the other table. In these questions the values needed are taken from the table where the two property values are smaller.

Thus h_f=167.57.

Is my logic correct in using the table were the properties have the lowest values. Could I run into a state where one value is higher and the other one is lower ? Then what ?

 

[rude man]

As long as you're in the mixed-phase region (saturated water to saturated vapor, or 0% < quality < 100%) there is only one possible temperature for a given pressure, or only one possible pressure for a given temperature. The combination you cite (10 kPa and 40C) is not in that region, so it is not saturated water.

 

[kdorsel]

It seems that if you look at the saturated water tables @T you get P_sat. If P_sat < P than the substance is actually sub cooled.

Since the internal energy is mostly a function of temperature using the values found using the table @T gives you the same answers (with the table precision).

You can find the actual values using the table @ T using:

h_subcooled(T) = h_sat + v_sat(P_subcooled-P_sat)

Where h_sat, v_sat and P_sat are found from the table @T_subcooled

Hopefully that makes sense.