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Condensate Load Calculation

I just read some pages on Handbook of Chemical Engineering Calculations 2nd edition, written by Nicholas P. Chopey and got interested in calculation of condensate load in order to determine the capacity of steam trap. I will take one example, where the steam indirectly heats a liquid through metallic surface, as in heat exchanger and kettles.

Case I

For example, we have 1892.5 L of water heated in 30 min from 22.2 to 100 oC with 50-psig (344.7-kPa) steam in the jacket of kettle. How much condensate formed?

For normal warm up load, the quantity of condensate can be calculated by the following equation:

http://www.velansteamtraps.com

 

 

 

 

 

 

Where

Q = quantity of condensate (kg/hr)

W = total weight of pipes (kg)

T = saturation steam trap temperature (oC)

t = initial temperature of pipe (oC) usually surrounding air temperature

m = minutes of heating up system

L = latent heat of steam (Btu/hr)

For the case above:

W = 500 L x 1 (m3 / 1000 L) x 1000 (kg/m3) = 500 kg

T = 100 oC

t = 22.2 oC

m = 30 min

Spec heat = 2.44 kJ/kg oC

At 50-psig (64.69 psia), latent heat, L = 901 Btu/lb = 2095.73 kJ/kg

So, we calculate by using the above equation :

Q = (500 kg) x (100-22) oC x (2.44 kJ/kg oC) x 60 / (2095.73 kJ/kg) x 30 min = 90.58 kg/hr

So, the condensate formed at condition above is 90.58 kg/hr.

Case II

Data:

Ambient temperature =  70 oF

Working temperature = 366 oF (150 psig)

Warming up time = 720 menit

1000 feet of 10 inch schedule 40 pipe weighs = 40483 lbs

Latent heat = 857 Btu/lb

Q = 472 lbs/hr

References:

  1. Steam characteristics (0-30 bar). http://www.thermexcel.com/english/tables/vap_eau.htm

 

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