Heat Exchanger Sizing
Finds the heat-transfer area a heat exchanger needs for a given duty, from the inlet and outlet temperatures of both streams and the overall heat transfer coefficient. It uses the log mean temperature difference (LMTD) method for a counter-flow or parallel-flow exchanger. Process and HVAC engineers use it for a first estimate of exchanger size before asking a vendor for a rating.
Equations
A = q / (U·ΔTlm)
ΔTlm = (ΔT1 − ΔT2) / ln(ΔT1/ΔT2)
A is the heat-transfer area, q the heat duty, U the overall heat transfer coefficient and ΔTlm the LMTD [1]. ΔT1 and ΔT2 are the temperature differences between the streams at the two ends of the exchanger. In counter flow, ΔT1 = Th,in − Tc,out and ΔT2 = Th,out − Tc,in. In parallel flow, both streams enter at the same end, so ΔT1 = Th,in − Tc,in and ΔT2 = Th,out − Tc,out. When the two differences are equal, the LMTD is simply that difference.
U combines the film coefficients on each side, the wall and any fouling. Pick a Service to fill in the midpoint of the typical range for it [2], or enter a U from your own calculation.
Limits and common mistakes
- The page doesn’t check the energy balance. Make sure the duty matches both streams: q = m·cp·ΔT for the hot side and for the cold side.
- Shell-and-tube units with more than one tube pass, and cross-flow units, need the LMTD multiplied by a correction factor F, which is less than 1 [1]. The page doesn’t apply it, so it gives too small an area for them.
- The typical U ranges are wide. Water to oil runs from 110 to 350 W/(m²·K), so the area can be off by a factor of two. Add a margin, or calculate U.
- In parallel flow the cold outlet can’t reach the hot outlet. Counter flow allows a temperature cross and needs less area for the same duty.
Worked example
A hydraulic oil cooler must remove 60 kW, cooling the oil from 80 °C to 50 °C. Cooling water enters at 25 °C and leaves at 35 °C. Estimate the area of a counter-flow unit.
- Choose Flow Arrangement Counter-Flow.
- Enter Hot Fluid Inlet 80 °C, Hot Fluid Outlet 50 °C, Cold Fluid Inlet 25 °C and Cold Fluid Outlet 35 °C.
- Enter Heat Duty 60 kW.
- Choose Service Water to oil. U fills in as 230 W/(m²·K). Press Solve. The Solution shows:
LMTD: 34.026 °C
Required Area (A): 7.66678 m²
Check: the end differences are 45 and 25 °C, so the LMTD is 20 / ln(1.8) = 34.0 °C, a little below their average of 35. 60,000 / (230 × 34.0) = 7.67 m². The same job in parallel flow needs 8.47 m².
