Archived Mixing steam at different pressures

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The discussion focuses on calculating the mass of throttling steam needed to achieve a final mixture of dry saturated steam at 800 kPa. Given a steam flow of 3.5 kg/s at 650 kPa and 90% dryness, the enthalpy rates and specific enthalpies for different pressures are utilized. The heat balance equation is established to relate the enthalpy of the incoming steam and the throttled steam to the desired output. The solution reveals that approximately 48.7 kg/s of throttling steam is required for the mixture. This calculation is essential for understanding steam mixing processes in thermodynamic systems.
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Homework Statement


steam flow through a line is 3.5 kg/s at 650 kps and .9 dry. It enters a mixing line where it exits at 800kpa, dry and saturated. The mixing steam is throttle from 2000kps to 800kpa. What is the mass of throttling steam added to have a final mixture of 800 kpa dry steam.


Homework Equations


steam tables
q=mcΔT


The Attempt at a Solution



@650: hf=684 hfg=2076 hg=2552
@800: hf=721 hfg=2048 hg=2769
@2000:hf=908 hfg=1890 hg=2798
 
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To do this, we need to assume that the throttled steam was saturated. Otherwise the problem can't be solved.

Enthalpy rate of entering 650 kpa stream = 3.5((0.1)(684)+(0.9)(2552))=8278.2 kJ/s
Specific enthalpy of 2000 kPa steam = 2798 kJ/kg
Specific enthalpy of 800 kPa product steam =2769 kJ/kg

From heat balance on system, 8278.2 + 2798 w = (3.5+w) 2769

w = 48.7 kg/s
 

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