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Typical Steam Jet Refrigeration System: Surface Condenser

This document summarizes a typical steam jet refrigeration system used in a pulp and paper plant application. The system uses 100 psig saturated steam from a co-generation unit to power a steam ejector that creates a vacuum. This pulls air and water through a surface condenser which cools it to 72F. The cooled fluid is then fed into a flash drum where 0.9% flashes to vapor at 0.178 psia and 10,898 lb/hr condenses. The condensate is returned to the co-gen unit or used as make-up water, while the vapor is used for refrigeration in the plant. Calculations show the flash percentage is determined through a mass and energy balance in the flash

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Hsein Wang
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141 views1 page

Typical Steam Jet Refrigeration System: Surface Condenser

This document summarizes a typical steam jet refrigeration system used in a pulp and paper plant application. The system uses 100 psig saturated steam from a co-generation unit to power a steam ejector that creates a vacuum. This pulls air and water through a surface condenser which cools it to 72F. The cooled fluid is then fed into a flash drum where 0.9% flashes to vapor at 0.178 psia and 10,898 lb/hr condenses. The condensate is returned to the co-gen unit or used as make-up water, while the vapor is used for refrigeration in the plant. Calculations show the flash percentage is determined through a mass and energy balance in the flash

Uploaded by

Hsein Wang
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as XLS, PDF, TXT or read online on Scribd
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Typical Steam Jet Refrigeration System

Quantity of steam from ejector vendor


100 psig sat'd steam from co-gen unit
92F air / H2O to atm
Steam Ejector

10,898 lb/hr make-up water 72F vacuum pump


Surface
Condenser
0.178 psia
V =10,898 lb/hr cond 1.6927 psia
User isbl 120F
60F Feed
Flash Drum Condensate Drum

Pulp & Paper Plant Application


Cond return to co-gen
50F 120F
Condensate Pump
1,000 tons L =1,200,000 lb/hr
refrigeration
V / Feed = 10898 / (1.2E+06 + 10898) = 0.009 see calc'ns below

Calculations Ref Perry, CEHB, p- 12-37

Assume 100 lbs water at 60F to flash drum

Mass balance: 100 = V + L

V = 100 - L substitute into heat balance

Heat Balance: (100)(HL 60F) = (V)(HV 50F ) - (L)(HL 50F)

(100)(HL 60F) = (100 - L)(HV 50F ) - (L)(HL 50F)

enthalpy values from steam tables

(100)(28.06) = (100 - L)(1083.4) + (L)(18.05)

solve for L L= 99.1


V= 0.9 % flash at 0.178 psia
100

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