2D4 :  Determine Properties Using Thermodynamic Tables  8 pts 

Complete the following table by determining the values of all the blank entries. The system contains only H_{2}O.  


Read :  The key to this problem is to recognize that all of the variables in the table are state variables (except maybe x), or properties, and that they are all intensive properties. It is also important to assume that either one or two phases exist. The triple point of water does not appear in this table. Also, since we have no data availabe about solid water, we can assume that we have either a subcooled liquid, a superheated vapor or an equilibrium mixture of saturated vapor and saturated liquid in the system. Gibbs Phase Rule tells us that for a pure substance in a single phase there are 2 degrees of freedom. If two phases are present, then there is just 1 degree of freedom. In either case, the two values of intensive properties given in each part of this problem will be sufficient to completely determine the values of all of the other intensive properties of the system. So, we are in good shape to move forward on solving this problem.  
Given:  T (^{o}C) 
P (kPa) 
V (m^{3}/kg) 
U (kJ/kg) 
H (kJ/kg) 
x (kg vap/kg tot) 

a.)  50  12.35  2125  
b.)  273  187  
Find:  a.)  V  ???  m^{3}/kg  b.)  V  ???  m^{3}/kg  
U  ???  kJ/kg  U  ???  kJ/kg  
H  ???  kJ/kg  H  ???  kJ/kg  
x  ???  kg vap/kg  
Assumptions:  1 No ice exists in the system in either part of the problem.  
Equations / Data / Solve:  
Part a.)  Given :  T  50  ^{o}C  
P  12.35  kPa  
The first step in solving each part of this problem is to determine the state of the system. Is it subcooled liquid, superheated vapor or a twophase VLE mixture.  
We could do this by determining the boiling point or saturation temperature at the system pressure. But, since 12.35 kPa does not appear in the Saturation Pressure Table for water, this would require an interpolation. It is easier to determine the saturation pressure or vapor pressure based on the system temperature because 50^{o}C does appear in the Saturation Temperature Table and therefore does not require an interpolation.  
P*(kPa)  T_{sat} (^{o}C)  V_{sat liq} (m^{3}/kg)  V_{sat vap} (m^{3}/kg)  U_{sat liq} (kJ/kg)  U_{sat vap} (kJ/kg)  H_{sat liq} (kJ/kg)  H_{sat vap} (kJ/kg)  
#VALUE!  50  #VALUE!  #VALUE!  #VALUE!  #VALUE!  #VALUE!  #VALUE!  
P*(50^{o}C)  #VALUE!  kPa  
Since the actual system pressure is EQUAL TO the vapor pressure, the system is at saturation.  
In order to determine the values of the other properties of the system using the following equation, we will need to know the quality, x.  

Eqn 1  
We can determine x from the saturation data and the known value of u for the system using :  

Eqn 2  
x  #VALUE!  kg vap/kg  
Now, we can plug x back into Eqn 1 and apply it to the unknown properties, V and U.  
V  #VALUE!  m^{3}/kg  U  #VALUE!  kJ/kg  
Part b.)  Given :  T  273  ^{o}C  P  187  kPa  
We again begin by determining the state of the system. Unfortunately the system temperature is not listed in the Saturation Temperature Table and the system pressure is is not listed in the Saturation Pressure Table. Either way we go, interpolation is required.  
P*(kPa)  T_{sat} (^{o}C)  V_{sat liq} (m^{3}/kg)  V_{sat vap} (m^{3}/kg)  U_{sat liq} (kJ/kg)  U_{sat vap} (kJ/kg)  H_{sat liq} (kJ/kg)  H_{sat vap} (kJ/kg)  
#VALUE!  270  #VALUE!  #VALUE!  #VALUE!  #VALUE!  #VALUE!  #VALUE!  
#VALUE!  275  #VALUE!  #VALUE!  #VALUE!  #VALUE!  #VALUE!  #VALUE!  
175  #VALUE!  #VALUE!  #VALUE!  #VALUE!  #VALUE!  #VALUE!  #VALUE!  
200  #VALUE!  #VALUE!  #VALUE!  #VALUE!  #VALUE!  #VALUE!  #VALUE!  
We could interpolate to determine the saturation properties at 187 kPa, but there isn't much point! Since the system temperature is higher than the saturation temperature at EITHER 175 kPa or 200 kPa, the system temperature must also be higher than the interpolated value of T_{sat}(187 kPa).  
Since the system temperature is greater than the saturation temperature at the system pressure, the system contains a superheated vapor. Therefore, we must use data from the Superheated Vapor Table to determine the unknown properties of the system.  
x  N/A  Superheated  
The Superheated Vapor Table includes tables for pressure of 100 and 200 kPa, but not 187 kPa. These two tables include rows for 250^{o}C and 300^{o}C, but not for 273^{o}C. Consequently a double interpolation is required for each unknown system propert, V, U and H.  
The double interpolation can be done with the aid of tables like the ones developed in Lesson 2C on page 18.  
The data required for the double interpolation tables are :  
P*(kPa)  T (^{o}C)  V (m^{3}/kg)  U (kJ/kg)  H (kJ/kg)  
100  250  #VALUE!  #VALUE!  #VALUE!  
100  300  #VALUE!  #VALUE!  #VALUE!  
200  250  #VALUE!  #VALUE!  #VALUE!  
200  300  #VALUE!  #VALUE!  #VALUE!  
Here is the double interpolation table for V :  
Pressure (kPa)  
T( ^{o}C )  100  187  200  
250  #VALUE!  #VALUE!  #VALUE!  
273  #VALUE!  #VALUE!  #VALUE!  
300  #VALUE!  #VALUE!  #VALUE!  
I chose to interpolate on pressure first and then to interpolate on temperature. If you do the interpolations in the opposite order, you will get a slightly different answer. Either method is satisfactory.  
V  #VALUE!  m^{3}/kg  
Here is the double interpolation table for U :  
Pressure (kPa)  
T( ^{o}C )  100  187  200  
250  #VALUE!  #VALUE!  #VALUE!  
273  #VALUE!  #VALUE!  #VALUE!  
300  #VALUE!  #VALUE!  #VALUE!  
I chose to interpolate on pressure first and then to interpolate on temperature. If you do the interpolations in the opposite order, you will get a slightly different answer. Either method is satisfactory.  
U  #VALUE!  kJ/kg  
Here is the double interpolation table for H :  
Pressure (kPa)  
T( ^{o}C )  100  187  200  
250  #VALUE!  #VALUE!  #VALUE!  
273  #VALUE!  #VALUE!  #VALUE!  
300  #VALUE!  #VALUE!  #VALUE!  
I chose to interpolate on pressure first and then to interpolate on temperature. If you do the interpolations in the opposite order, you will get a slightly different answer. Either method is satisfactory.  
H  #VALUE!  kJ/kg  
Verify:  The assumption that there is no ice in the is confirmed by the fact that the the states in part (a) and part (b) are both located in the Steam Tables. The Steam Tables only consider states in which no ice can exist at equilibrium.  
Answers :  T (^{o}C) 
P (kPa) 
V (m^{3}/kg) 
U (kJ/kg) 
H (kJ/kg) 
x (kg vap/kg tot) 

a.)  50  12.35  #VALUE!  #VALUE!  2125  #VALUE!  
b.)  273  187  #VALUE!  #VALUE!  #VALUE!  N/A  Superheated 