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

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 $\varepsilon$.

$\sqrt{1 + C_r^2} = \sqrt{1.25} = 1.118$

$NTU \cdot 1.118 = 2.236$

$e^{-2.236} = 0.1065$

$\frac{1 + 0.1065}{1 - 0.1065} = \frac{1.1065}{0.8935} = 1.238$

$\varepsilon = 2/(1 + 0.5 + 1.118 \times 1.238)^{-1}$

Wait — let’s apply carefully:

$\varepsilon = 2[1 + 0.5 + 1.118 \times 1.238]^{-1} = 2[1.5 + 1.384]^{-1} = 2/2.884 = 0.693$

Compare: counter-flow at same NTU and $C_r$ gave $\varepsilon = 0.775$. 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 $\varepsilon$.

$NTU^{0.22} = 1.5^{0.22} = 1.091$

$NTU^{0.78} = 1.5^{0.78} = 1.375$

$C_r \cdot NTU^{0.78} = 0.6 \times 1.375 = 0.825$

$e^{-0.825} = 0.438$

$\frac{NTU^{0.22}}{C_r}(e^{-0.825} - 1) = \frac{1.091}{0.6}(0.438 - 1) = 1.818 \times (-0.562) = -1.022$

$\varepsilon = 1 - e^{-1.022} = 1 - 0.360 = 0.640$

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.

$C_{\min} = 3000$, $C_r = 3000/6000 = 0.5$

$NTU = 400 \times 20/3000 = 2.667$

Using the 1-shell formula: $\sqrt{1+0.25} = 1.118$, $NTU \times 1.118 = 2.981$

$e^{-2.981} = 0.0507$

$(1+0.0507)/(1-0.0507) = 1.107$

$\varepsilon = 2/(1 + 0.5 + 1.118 \times 1.107) = 2/(1.5 + 1.238) = 2/2.738 = 0.731$

$\dot{Q} = 0.731 \times 3000 \times (130-25) = 0.731 \times 315{,}000 = 230{,}300$ W

$T_{h,\text{out}} = 130 - 230{,}300/3000 = 53.2$°C

$T_{c,\text{out}} = 25 + 230{,}300/6000 = 63.4$°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.

ε-NTU Relations: Parallel and Counter-Flow LMTD vs. ε-NTU — When to Use Which