Where can I find the specific volume of steam at 179.91 deg Celsius?

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SUMMARY

The specific volume of steam at 179.91 degrees Celsius can be calculated using the formula for total volume in a closed tank. Given the specific volumes of water and steam, v_f = 0.001108 m³/kg and v_g = 0.2729 m³/kg, the total volume is determined to be V_total = 0.1949 m³. This volume remains constant after the heat addition process, leading to a specific volume of steam of approximately 0.1949 m³/kg at the specified temperature.

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Please see attachment below. Thanks
 

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Hi,

Here's a quick fix:

At 7b,
v_f = 0.001108 m^3/kg
v_g = 0.2729 m^3/kg

Since the masses of steam and water are given, we can find the total volume of the closed tank to be

V_total = (0.287 * 0.001108) + (0.713 * 0.2729) m^3

= 0.1949 m^3

We know that the volume remains the same after the heat addition process. And things get simplified, when all that's left is steam. You'll just have to look into the pressure states where steam has the specific volume of

v_g = V_total / total mass = 0.1949 / (0.287 + 0.713) = 0.1949 m^3 / kg

which is roughly 10 b, at 179.91 deg Celsius.


Hope this helps. ;)
 
Last edited:

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