Training Heat Exchangers — ε-NTU Method ε-NTU for Shell-and-Tube and Cross-Flow
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ε-NTU for Shell-and-Tube and Cross-Flow

24 min Heat Exchangers — ε-NTU Method

Shell-and-tube and cross-flow heat exchangers have more complex ε-NTU relations that account for multiple tube passes and mixed or unmixed flow conditions. While the formulas are longer, the approach is identical: given NTU and Cr, look up or compute ε, then find the actual heat transfer. Familiarity with these relations lets you handle the full range of industrial exchanger types using the ε-NTU framework.

ε-NTU for Shell-and-Tube and Cross-Flow

Beyond simple double-pipe exchangers, the ε-NTU method also has relations for shell-and-tube and cross-flow configurations, which are the most common in industrial practice.

1-Shell, 2-Tube-Pass (Even Number of Passes)

$$\varepsilon = 2\left[1 + C_r + \sqrt{1+C_r^2}\,\frac{1+e^{-NTU\sqrt{1+C_r^2}}}{1-e^{-NTU\sqrt{1+C_r^2}}}\right]^{-1}$$

Cross-Flow (Both Unmixed)

$$\varepsilon = 1 - \exp\left[\frac{NTU^{0.22}}{C_r}\left(e^{-C_r\,NTU^{0.78}} - 1\right)\right]$$

This is an approximation valid for most engineering purposes.

Cross-Flow (One Mixed, One Unmixed)

$C_{\max}$ mixed: $\varepsilon = \frac{1}{C_r}\left(1 - e^{-C_r(1-e^{-NTU})}\right)$

$C_{\min}$ mixed: $\varepsilon = 1 - e^{-(1/C_r)(1-e^{-C_r\,NTU})}$

Example 1 — 1-Shell 2-Tube-Pass

$NTU = 2.0$, $C_r = 0.5$, 1-shell 2-tube-pass. Find $\varepsilon$.

  1. $\sqrt{1 + C_r^2} = \sqrt{1.25} = 1.118$
  2. $NTU \cdot 1.118 = 2.236$
  3. $e^{-2.236} = 0.1065$
  4. $\frac{1 + 0.1065}{1 - 0.1065} = \frac{1.1065}{0.8935} = 1.238$
  5. $\varepsilon = 2/(1 + 0.5 + 1.118 \times 1.238)^{-1}$
  6. Wait — let’s apply carefully:
  7. $\varepsilon = 2[1 + 0.5 + 1.118 \times 1.238]^{-1} = 2[1.5 + 1.384]^{-1} = 2/2.884 = 0.693$
  8. Compare: counter-flow at same NTU and $C_r$ gave $\varepsilon = 0.775$.
  9. The multi-pass penalty is clear.
Example 2 — Cross-Flow (Both Unmixed)

A car radiator (cross-flow, both unmixed): $NTU = 1.5$, $C_r = 0.6$. Estimate $\varepsilon$.

  1. $NTU^{0.22} = 1.5^{0.22} = 1.091$
  2. $NTU^{0.78} = 1.5^{0.78} = 1.375$
  3. $C_r \cdot NTU^{0.78} = 0.6 \times 1.375 = 0.825$
  4. $e^{-0.825} = 0.438$
  5. $\frac{NTU^{0.22}}{C_r}(e^{-0.825} - 1) = \frac{1.091}{0.6}(0.438 - 1) = 1.818 \times (-0.562) = -1.022$
  6. $\varepsilon = 1 - e^{-1.022} = 1 - 0.360 = 0.640$
Example 3 — Rating a Real HX

A 1-shell, 2-tube-pass exchanger has $U = 400$ W/(m²·K), $A = 20$ m². Hot oil ($C = 3000$ W/K, $T_{h,\text{in}} = 130$°C). Cold water ($C = 6000$ W/K, $T_{c,\text{in}} = 25$°C). Find outlet temperatures.

  1. $C_{\min} = 3000$, $C_r = 3000/6000 = 0.5$
  2. $NTU = 400 \times 20/3000 = 2.667$
  3. Using the 1-shell formula: $\sqrt{1+0.25} = 1.118$, $NTU \times 1.118 = 2.981$
  4. $e^{-2.981} = 0.0507$
  5. $(1+0.0507)/(1-0.0507) = 1.107$
  6. $\varepsilon = 2/(1 + 0.5 + 1.118 \times 1.107) = 2/(1.5 + 1.238) = 2/2.738 = 0.731$
  7. $\dot{Q} = 0.731 \times 3000 \times (130-25) = 0.731 \times 315{,}000 = 230{,}300$ W
  8. $T_{h,\text{out}} = 130 - 230{,}300/3000 = 53.2$°C
  9. $T_{c,\text{out}} = 25 + 230{,}300/6000 = 63.4$°C

Practice Problems

1. 1-shell 2-tube: $NTU = 1.0$, $C_r = 0.8$. Find $\varepsilon$.
2. Cross-flow (both unmixed): $NTU = 2.0$, $C_r = 0.4$. Find $\varepsilon$.
3. Cross-flow ($C_{\max}$ mixed): $NTU = 1.5$, $C_r = 0.7$. Find $\varepsilon$.
4. Which arrangement gives highest $\varepsilon$ for the same NTU and $C_r$?
5. A 1-shell HX has $\varepsilon = 0.65$. What NTU is needed for $C_r = 0.5$?
6. For all arrangements with $C_r = 0$, what is $\varepsilon$?
Show Answer Key

1. $\sqrt{1+0.64} = 1.281$. $e^{-1.281} = 0.278$. $(1.278)/(0.722) = 1.770$. $\varepsilon = 2/(1+0.8+1.281 \times 1.770) = 2/(1.8+2.267) = 2/4.067 = 0.492$

2. $NTU^{0.22} = 1.17$, $NTU^{0.78} = 1.74$, $C_r \cdot 1.74 = 0.696$, $e^{-0.696} = 0.499$. Exponent $= (1.17/0.4)(0.499-1) = 2.925 \times (-0.501) = -1.466$. $\varepsilon = 1 - e^{-1.466} = 1 - 0.231 = 0.769$

3. $\varepsilon = (1/0.7)(1 - e^{-0.7(1-e^{-1.5})})$. $e^{-1.5} = 0.223$, $1-0.223 = 0.777$. $0.7 \times 0.777 = 0.544$. $e^{-0.544} = 0.580$. $\varepsilon = (1/0.7)(1-0.580) = 1.429 \times 0.420 = 0.600$

4. Counter-flow

5. NTU $\approx 1.6$ (requires iterative or graphical solution of the 1-shell formula).

6. $\varepsilon = 1 - e^{-NTU}$ for all arrangements.

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