Understanding how to use steam tables

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The discussion focuses on using steam tables to find missing values for energy and steam quality at a given temperature of 75°C and pressure of 500 KPa. The user initially misinterprets the state of water, identifying it as a compressed liquid due to the pressure being below the saturated pressure for that temperature. They struggle to find relevant data in standard steam tables, which typically start at higher pressures. A suggestion is made to consult the NIST website for more comprehensive tables that include lower pressures. The conversation concludes with clarifications on energy values and the minor effect of pressure on the internal energy of compressed liquids.
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Homework Statement


I have a table for 5 sets of data. Each set includes data from one ore more of the following properties - temperature, pressure, energy, and steam quality. I have to look at steam tables for water to find the missing values. For the data set in question I'm given T = 75 C, and P = 500 KPa. I need to find the energy and steam quality, and from steam quality I get number of phases present.

Homework Equations


for finding x or energy if calculation is required - u = x*(u liquid) + ((1-x)*(u gas))
just rearrange the above to find x if needed

The Attempt at a Solution


One of the data sets gives me a temperature and pressure of 75 C and 500 KPa, nothing else. I looked at water's saturated steam tables for this temperature and pressure. I found that P>Saturated Pressure and T<Saturated Temperature (at these given values), so I'm assuming that I'm actually looking at a compressed liquid. I then looked at the steam tables for compressed liquid, but I can't find a data set for 75 C and 500 KPa. The temperature range includes 75 C, but all the pressures are in the multiple MPa's. Can someone help me clarify how to properly approach these tables? I'm a bit confused here. Any and all help is appreciated, thank you.
 
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reed2100 said:

Homework Statement


I have a table for 5 sets of data. Each set includes data from one ore more of the following properties - temperature, pressure, energy, and steam quality. I have to look at steam tables for water to find the missing values. For the data set in question I'm given T = 75 C, and P = 500 KPa. I need to find the energy and steam quality, and from steam quality I get number of phases present.

Homework Equations


for finding x or energy if calculation is required - u = x*(u liquid) + ((1-x)*(u gas))
just rearrange the above to find x if needed

The Attempt at a Solution


One of the data sets gives me a temperature and pressure of 75 C and 500 KPa, nothing else. I looked at water's saturated steam tables for this temperature and pressure. I found that P>Saturated Pressure and T<Saturated Temperature (at these given values), so I'm assuming that I'm actually looking at a compressed liquid. I then looked at the steam tables for compressed liquid, but I can't find a data set for 75 C and 500 KPa. The temperature range includes 75 C, but all the pressures are in the multiple MPa's. Can someone help me clarify how to properly approach these tables? I'm a bit confused here. Any and all help is appreciated, thank you.
What is 500 KPa expressed in units of MPa? This should be a simple conversion. What do the prefixes K and M stand for in the SI system?
 
SteamKing said:
What is 500 KPa expressed in units of MPa? This should be a simple conversion. What do the prefixes K and M stand for in the SI system?
Yes, 500 KPa is .5 MPa unless I'm mistaken. I kept the conversions in mind. On the compressed liquid table the lowest pressure available is 5 MPa.
 
reed2100 said:
Yes, 500 KPa is .5 MPa unless I'm mistaken. I kept the conversions in mind. On the compressed liquid table the lowest pressure available is 5 MPa.
Then get another table.

You can get tables from the NIST website which cover a greater range of pressures for compressed liquid.

http://www.nist.gov/srd/upload/NISTIR5078-Tab3.pdf
 
The internal energy can't be varying much with pressure for a compressed liquid because water is nearly incompressible. So just use the saturated value.
 
SteamKing's link seems to exactly answer the question, i see an entry for 0.5mPa and 75C
but the number is strange to me because I'm used to ASME 1968 tables in BTU's and Fahrenheit... and that table doesn't say...
314.4 ? I suppose it's an SI value for something ? Looks about right to be kJ per kg , around 135 BTU/lb ,
per http://www.endmemo.com/sconvert/kj_kgbtu_lb.php
 
jim hardy said:
SteamKing's link seems to exactly answer the question, i see an entry for 0.5mPa and 75C
but the number is strange to me because I'm used to ASME 1968 tables in BTU's and Fahrenheit... and that table doesn't say...
314.4 ? I suppose it's an SI value for something ? Looks about right to be kJ per kg , around 135 BTU/lb ,
per http://www.endmemo.com/sconvert/kj_kgbtu_lb.php
Yes. My steam tables give 313.9 kj per kg at saturation pressure
 
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Chestermiller said:
Yes. My steam tables give 313.9 kj per kg at saturation pressure
Thanks Chestermiller ! Makes sense now.

With Imperial steam tables enthalpy of compressed water is almost intuitive, ~a BTU/lb per degF above 32.
75C is on Fahrenheit scale 135 degrees above freezing, so around 135BTU/lb . And as you said pressure has a very minor effect.

old jim
 
Last edited:
Thank you everyone for your help! I appreciate it.
 

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