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 ^{o}C 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:

Where

Q = quantity of *condensate* (kg/hr)

W = total weight of pipes (kg)

T = saturation steam trap temperature (^{o}C)

t = initial temperature of pipe (^{o}C) 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/m^{3}) = 500 kg

T = 100 ^{o}C

t = 22.2 ^{o}C

m = 30 min

Spec heat = 2.44 kJ/kg ^{o}C

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 ^{o}C) 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 ^{o}F

Working temperature = 366 ^{o}F (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:**

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