Example #1: Calculate a preliminary estimate of the heat exchanger area needed to cool 55,000 lb/hr of a light oil (specific heat = 0.74 Btu/lb-oF) from 190oF to 140oF using cooling water that is available at 50oF. The cooling water can be allowed to heat to 90oF. An initial estimate of the overall heat transfer coefficient is 120 Btu/hr-ft2-oF. Also estimate the required mass flow rate of cooling water.
Solution: First calculate the required heat transfer rate based on the required light oil cooling:
Q = (55,000 lb/hr)(0.74 Btu/lb-oF)(190 – 140)oF = 2,035,000 Btu/hr.
Next calculate the log mean temperature difference:
ΔTlm = [(190 – 90) – (140 – 50)]/ln[(190 – 90)/(140 – 50)] = 94.9oF
The preliminary area estimate can now be calculated as:
A = Q/(U ΔTlm) = 2,035,000/(120)(79.58) = 178.7 ft2 = A
The required mass flow rate of water can be calculated from Q = m Cp ΔT.
Rearranging: m = Q/Cp ΔT = (2,035,000 Btu/hr)/(1 Btu/lb-oF)(40oF) = 50,875 lb/hr
Example #2: A shell and tube heat exchanger is to be used for the light oil cooling described in Example #1. How many tubes of 3 inch diameter and 10 ft length should be used?
Solution: The surface area per tube will be πDL = π(3/12)(10) ft2 = 7.854 ft2. The number of tubes required would thus be:
n = 178.7 ft2/7.854 ft2/tube = 22.7 tubes (round up to 23 tubes).
The next step would be to check on the pressure drop for this tube configuration and the specified flow. If the pressure drop is acceptable, then the overall heat transfer coefficient could be re-estimated for this heat exchanger configuration. These topics will be dealt with in a future article.