OpenSpecEngineering

Rates an exchanger whose size is known: from its conductance UA, the two stream capacity rates and the two inlet temperatures, it finds the effectiveness, the heat duty and both outlet temperatures. Use it to check how an exchanger performs, or responds to a change in flow, without trial-and-error on the LMTD.

Equations

C_r = C_min/C_max     NTU = UA/C_min

q = ε·C_min·(T_h,in − T_c,in)

Each stream’s capacity rate C is its mass flow times its specific heat, ṁ·c_p, in W/K. C_min is the smaller of the two and C_max the larger. The product C_min·(T_h,in − T_c,in) is q_max, the most heat any exchanger could pass between the two inlets. The effectiveness ε is the fraction of q_max actually transferred. It depends on NTU, C_r and the flow arrangement [1], [2]. For countercurrent flow:

ε = (1 − e^{−NTU(1−C_r)}) / (1 − C_r·e^{−NTU(1−C_r)})

The page uses the matching relations for co-current flow, for one shell pass with an even number of tube passes, and for crossflow with both fluids unmixed [2]. The outlets then follow from the heat balance: T_h,out = T_h,in − q/C_h and T_c,out = T_c,in + q/C_c.

Inputs

• Enter UA directly, or as U times the area. Choosing a Service fills U with the midpoint of its typical range [3].
• Which Stream Is Cmin? only decides which outlet gets which temperature change.

Limits and common mistakes

• U and the specific heats are taken as constant. The method doesn’t apply to a stream that boils or condenses over part of the exchanger.
• The crossflow relation for both fluids unmixed is an approximate fit [2].
• Entering the capacity rates in kW/K while UA is in W/K throws NTU off by a factor of 1000.

Worked example

A 4 m² shell-and-tube exchanger heats 2.5 kg/s of water from 15 °C with 1.5 kg/s of hot water at 90 °C. With c_p = 4180 J/(kg·K), the hot stream’s C is 6270 W/K and the cold stream’s is 10,450 W/K. What are the outlets?

1. Choose Flow Arrangement 1 shell pass, 2+ tube passes.
2. Choose Conductance UA U × area, then Service Water to water. U fills in as 1275 W/(m²·K).
3. Enter Heat Transfer Area 4 m².
4. Enter Minimum Capacity Rate 6270 W/K and Maximum Capacity Rate 10450 W/K.
5. Enter Hot Inlet Temperature 90 °C and Cold Inlet Temperature 15 °C.
6. Choose Which Stream Is Cmin? Hot stream and press Solve. The Solution shows: UA = U·A: 5,100 W/K
   Capacity ratio Cr = Cmin/Cmax: 0.6
   NTU = UA/Cmin: 0.813397
   Effectiveness ε: 0.471644
   Maximum possible duty qmax: 470.25 kW
   Heat duty q: 221.791 kW
   Hot outlet temperature: 54.6267 °C
   Cold outlet temperature: 36.224 °C

Check the balance: the hot side drops 221.791/6.27 = 35.37 °C and the cold side rises 221.791/10.45 = 21.22 °C. The countercurrent formula above gives ε = 0.490 for the same NTU and C_r, so the shell-and-tube unit comes out a little lower, as it should.