Heat exchanger effectiveness–NTU calculator

Introduction

The effectiveness–NTU method predicts how much heat an existing exchanger transfers when you know only the inlet temperatures. The effectiveness ε is the actual duty divided by the largest duty the two streams could exchange, C_min (T_h,in − T_c,in). It depends only on the number of transfer units, NTU = UA / C_min, the capacity ratio C_r and the flow arrangement. An oil cooler with UA = 197 W/K, oil at 100 °C and cooling water at 30 °C transfers 8.52 kW in counterflow, an effectiveness of 57 %.

This calculator gives C_min, C_r, NTU, the effectiveness, the duty and both outlet temperatures for counterflow, parallel flow, shell-and-tube, three cross-flow arrangements and condensing or boiling streams. It is for engineers rating an exchanger at new flow rates or temperatures, and for students working the relations of Incropera Table 11.3.

Formulas

Heat capacity rates, NTU and the duty:

C_h = ṁ_h c_p,h,   C_c = ṁ_c c_p,c,   C_r = C_min / C_max
NTU = U A / C_min
q = ε C_min (T_h,in − T_c,in)
T_h,out = T_h,in − q / C_h,   T_c,out = T_c,in + q / C_c

Effectiveness (Incropera, Table 11.3):

Parallel:      ε = [1 − exp(−NTU (1 + C_r))] / (1 + C_r)
Counterflow:   ε = [1 − exp(−NTU (1 − C_r))] / [1 − C_r exp(−NTU (1 − C_r))],   ε = NTU / (1 + NTU) at C_r = 1
One shell:     ε₁ = 2 {1 + C_r + √(1 + C_r²) [1 + exp(−NTU₁ √(1 + C_r²))] / [1 − exp(−NTU₁ √(1 + C_r²))]}⁻¹
n shells:      ε = [((1 − ε₁ C_r)/(1 − ε₁))ⁿ − 1] / [((1 − ε₁ C_r)/(1 − ε₁))ⁿ − C_r],   NTU₁ = NTU / n
Cross, both unmixed:   ε ≈ 1 − exp{(1/C_r) NTU^0.22 [exp(−C_r NTU^0.78) − 1]}
Cross, C_max mixed:    ε = (1/C_r) {1 − exp[−C_r (1 − exp(−NTU))]}
Cross, C_min mixed:    ε = 1 − exp{−(1/C_r) [1 − exp(−C_r NTU)]}
Any, C_r = 0:          ε = 1 − exp(−NTU)

• ṁ — mass flow; c_p — specific heat; U — overall heat transfer coefficient; A — area
• C_min, C_max — the smaller and larger of C_h and C_c
• n — number of shell passes, each with 2, 4 or more tube passes

Assumptions and limitations

• The exchanger runs steadily, loses no heat to the surroundings, and has constant specific heats and a uniform U. Large property changes call for splitting it into zones.
• The cross-flow relation with both fluids unmixed is an approximation. It is within 4 % of the exact series from NTU 0.25 to 5 and is exact as C_r approaches zero.
• A condensing or boiling stream stays at its inlet temperature, so C_r = 0. Superheat and subcooling zones are not included.
• U must already include fouling, wall resistance and fin efficiency. Enter a service value rather than a clean one when sizing for the long term.
• For the mixed cross-flow cases, choose which fluid is mixed. The calculator then picks the C_max mixed or C_min mixed relation from the flow rates.

Worked example

Incropera Example 11.1 is a counterflow oil cooler. Oil flows at 0.1 kg/s with c_p = 2 131 J/(kg·K) and enters at 100 °C. Water flows at 0.2 kg/s with c_p = 4 178 J/(kg·K) and enters at 30 °C. U = 38.1 W/(m²·K) on an area of 5.176 m². What are the duty and the outlet temperatures?

1. Capacity rates: C_h = 0.1 × 2 131 = 213.1 W/K and C_c = 0.2 × 4 178 = 835.6 W/K, so the oil is C_min.
2. C_r = 213.1 / 835.6 = 0.255.
3. UA = 38.1 × 5.176 = 197.2 W/K, so NTU = 197.2 / 213.1 = 0.925.
4. Counterflow: ε = (1 − e^−0.689) / (1 − 0.255 e^−0.689) = 0.571.
5. q = 0.571 × 213.1 × (100 − 30) = 8.52 kW.
6. Oil out: 100 − 8 521 / 213.1 = 60.0 °C. Water out: 30 + 8 521 / 835.6 = 40.2 °C.

These are the temperatures the example started from, so the LMTD method gives the same exchanger. With the same UA, one shell pass would transfer 8.34 kW and parallel flow 8.17 kW.

FAQ

Related calculators

• Heat exchanger LMTD — the sizing method from four temperatures.
• Heat transfer calculator — conduction, convection and U-values.
• Specific heat calculator — the heat to warm a fluid.
• Pipe insulation — heat loss from the pipes in between.