OpenSpecEngineering

Finds the log mean temperature difference (LMTD) between a hot and a cold stream from their four end temperatures, and the heat transfer area an exchanger needs to carry a given duty. Process and HVAC engineers use it for a first sizing of an exchanger, or to check whether an existing one has enough surface.

Equations

LMTD = (ΔT₁ − ΔT₂) / ln(ΔT₁/ΔT₂)

A = Q / (U·F·LMTD)

ΔT₁ and ΔT₂ are the temperature differences between the streams at the two ends of the exchanger [1], [2]. In countercurrent flow they are T_h,in − T_c,out and T_h,out − T_c,in. In co-current flow both streams enter at the same end, so they are T_h,in − T_c,in and T_h,out − T_c,out. Q is the heat duty in W, U the overall heat transfer coefficient in W/(m²·K) and A the area in m². F is the correction factor for multipass and crossflow exchangers [1], [3]. It is 1 for pure countercurrent or co-current flow and less than 1 otherwise.

Inputs

• Choosing a Service fills U with the midpoint of the typical range for that duty [4]. The ranges are wide, so replace it with a vendor or calculated value when you have one.
• Read F from the TEMA or textbook charts for your shell and pass layout [1], [3]. Below about 0.75, another shell in series is usually better.
• Enter a heat duty of 0 to get the LMTD alone.

Limits and common mistakes

• The method assumes constant U and constant specific heats along the exchanger. For a condensing stream with subcooling, or a large change of U, split the exchanger into zones.
• Pairing the wrong temperatures is the usual mistake. If a stream’s outlet is on the wrong side of the other’s inlet, the page reports a crossed approach.
• The four temperatures and Q must agree with the heat balance of both streams (Q = m·c_p·ΔT). The page does not check this.
• The area is the clean area. Add a fouling margin per your design practice.

Worked example

A lube oil cooler must remove 250 kW, cooling the oil from 120 to 70 °C with cooling water that warms from 25 to 45 °C. Roughly how much surface does a countercurrent exchanger need?

1. Choose Flow Arrangement Countercurrent.
2. Enter Hot Fluid Inlet 120 °C, Hot Fluid Outlet 70 °C, Cold Fluid Inlet 25 °C and Cold Fluid Outlet 45 °C.
3. Enter Heat Duty 250 kW.
4. Choose Service Water to oil. U fills in as 230 W/(m²·K).
5. Leave LMTD Correction Factor at 1 and press Solve. The Solution shows: Terminal ΔT (end 1): 75 °C
   Terminal ΔT (end 2): 45 °C
   LMTD: 58.7285 °C
   Corrected MTD = F·LMTD: 58.7285 °C
   Required surface area A: 18.5082 m²

Check by hand: (75 − 45)/ln(75/45) = 30/0.5108 = 58.73 °C, a little below the arithmetic mean of 60 °C, as it always is. Then A = 250,000/(230 × 58.73) = 18.5 m².