How to calculate the power consumption during steam compression

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To calculate the power consumption during steam compression, the gross enthalpy of steam at both the initial and final states must be determined and subtracted to find the enthalpy change. For compressing saturated steam from 50°C to 100°C at 1 bar, the ideal case shows a power requirement of 84.9 J/g, which increases to about 121 J/g with a 70% efficient compressor. However, the discussion raises doubts about applying gas laws to steam below its critical temperature, as steam behaves differently than gas under compression. It is clarified that during adiabatic and reversible compression, the steam becomes superheated, and the final temperature can be calculated using both enthalpy and ideal gas methods, yielding similar results. Ultimately, both methods validate the use of gas laws for vapor under certain conditions, despite the differences between gas and vapor.
  • #31
pranj5 said:
As per wikipedia (https://en.wikipedia.org/wiki/Water_(data_page)#Water.2Fsteam_equilibrium_properties), density of steam at 20C saturated level is 0.01728 kg/m3 and at 100C it's 0.5974 kg/m3. Density in inversely proportional to volume and kindly count the compression ratio by volume.
The final temperature is not 100 C, so you can't use the data at 100 C for anything. The final temperature is 450 C, and that results in a much lower volume compression ratio. (For some reason, you seem obsessed with 100 C).
 
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  • #32
I haven't said anything about temperature, but rather pressure. The exhaust steam will be at 1 bar pressure. And, as far as I know, the higher the temperature, lesser will be density i.e. bigger will be the volume.
 
  • #33
pranj5 said:
I haven't said anything about temperature, but rather pressure. The exhaust steam will be at 1 bar pressure. And, as far as I know, the higher the temperature, lesser will be density i.e. bigger will be the volume.
The higher the final temperature, the lower the volume compression ratio.

Are you saying that you think your answer is correct and mine is incorrect? I solved this problem by three different methods and got the same answer.
 
Last edited:
  • #34
"The higher the final temperature, the lower the volume compression ratio."
How?
 
  • #35
pranj5 said:
"The higher the final temperature, the lower the volume compression ratio."
How?
From the ideal gas law,
$$volume\ compression\ ratio\ =\frac{V_1}{V_2}=\frac{T_1}{T_2}\frac{P_2}{P_1}$$
For fixed values of T1, P1, V1 and P2, the higher the value of T2, the higher the value of V2 and the lower the value of the volume compression ratio.
 

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