Training Heat Exchangers — ε-NTU Method ε-NTU Relations: Parallel and Counter-Flow
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ε-NTU Relations: Parallel and Counter-Flow

24 min Heat Exchangers — ε-NTU Method

For parallel-flow and counter-flow arrangements, closed-form ε-NTU relations express effectiveness as a function of NTU and the capacity ratio. Counter-flow always achieves higher effectiveness than parallel-flow for the same NTU, and in the special case where Cr = 0 (one fluid undergoes phase change), both configurations give ε = 1 − e−NTU. These formulas are the workhorses of the ε-NTU method and appear in every heat-exchanger textbook.

ε-NTU Relations: Parallel and Counter-Flow

Each flow arrangement has its own ε-NTU formula. These let you find effectiveness from NTU (rating problem) or NTU from effectiveness (sizing problem) without any iteration.

Parallel-Flow

$$\varepsilon = \frac{1 - e^{-NTU(1+C_r)}}{1 + C_r}$$

As $NTU \to \infty$: $\varepsilon_{\max} = \frac{1}{1+C_r}$ (always $< 1$ unless $C_r = 0$).

Counter-Flow

$$\varepsilon = \frac{1 - e^{-NTU(1-C_r)}}{1 - C_r\,e^{-NTU(1-C_r)}} \qquad (C_r < 1)$$

$$\varepsilon = \frac{NTU}{1 + NTU} \qquad (C_r = 1)$$

As $NTU \to \infty$: $\varepsilon \to 1$ (unlike parallel-flow!). This is why counter-flow can theoretically achieve 100% effectiveness.

Condenser or Evaporator ($C_r = 0$)

$$\varepsilon = 1 - e^{-NTU}$$

Same formula regardless of flow arrangement (phase change dominates).

Example 1 — Counter-Flow Rating

Counter-flow HX: $NTU = 2.0$, $C_r = 0.5$. Find $\varepsilon$.

  1. $\varepsilon = \frac{1 - e^{-2.0(1-0.5)}}{1 - 0.5\,e^{-2.0(1-0.5)}} = \frac{1 - e^{-1}}{1 - 0.5e^{-1}} = \frac{1 - 0.3679}{1 - 0.1839} = \frac{0.6321}{0.8161} = 0.775$
  2. 77.5% effectiveness.
Example 2 — Parallel-Flow Comparison

Same $NTU = 2.0$, $C_r = 0.5$, but parallel-flow. Find $\varepsilon$.

  1. $\varepsilon = \frac{1 - e^{-2.0(1+0.5)}}{1 + 0.5} = \frac{1 - e^{-3.0}}{1.5} = \frac{1 - 0.0498}{1.5} = \frac{0.9502}{1.5} = 0.633$
  2. 63.3% — significantly lower than counter-flow (77.5%) for the same NTU and $C_r$.
Example 3 — Condenser

A condenser ($C_r = 0$) has $NTU = 3.0$. Find $\varepsilon$ and the outlet water temperature if $T_{\text{sat}} = 100$°C, $T_{c,\text{in}} = 20$°C.

  1. $\varepsilon = 1 - e^{-3} = 1 - 0.0498 = 0.950$
  2. $\dot{Q} = \varepsilon \cdot C_{\min}(T_{h,\text{in}} - T_{c,\text{in}})$
  3. $T_{c,\text{out}} = T_{c,\text{in}} + \varepsilon(T_{\text{sat}} - T_{c,\text{in}}) = 20 + 0.95 \times 80 = 96$°C
Example 4 — Sizing (Finding NTU)

Counter-flow, $C_r = 0.6$, required $\varepsilon = 0.80$. Find NTU, then area if $U = 600$ and $C_{\min} = 4000$.

  1. Invert the counter-flow formula:
  2. $$NTU = \frac{1}{C_r - 1}\ln\frac{\varepsilon - 1}{\varepsilon C_r - 1} = \frac{1}{0.6-1}\ln\frac{0.80-1}{0.80 \times 0.6 - 1}$$
  3. $= \frac{1}{-0.4}\ln\frac{-0.20}{-0.52} = -2.5 \times \ln(0.3846) = -2.5 \times (-0.9555) = 2.39$
  4. $A = NTU \cdot C_{\min}/U = 2.39 \times 4000/600 = 15.9$ m²

Practice Problems

1. Counter-flow: $NTU = 1.5$, $C_r = 0.8$. Find $\varepsilon$.
2. Parallel-flow: $NTU = 1.5$, $C_r = 0.8$. Find $\varepsilon$.
3. Balanced counter-flow ($C_r = 1$): $NTU = 3$. Find $\varepsilon$.
4. Evaporator ($C_r = 0$): $\varepsilon = 0.90$. Find NTU.
5. Counter-flow with $NTU \to \infty$, $C_r = 0.5$. $\varepsilon \to$ ?
6. Parallel-flow with $NTU \to \infty$, $C_r = 0.5$. $\varepsilon_{\max}$?
Show Answer Key

1. $\varepsilon = (1 - e^{-1.5 \times 0.2})/(1 - 0.8 e^{-0.3}) = (1-0.7408)/(1 - 0.8 \times 0.7408) = 0.2592/0.4074 = 0.636$

2. $\varepsilon = (1 - e^{-1.5 \times 1.8})/1.8 = (1 - e^{-2.7})/1.8 = (1-0.0672)/1.8 = 0.518$

3. $\varepsilon = 3/(1+3) = 0.75$

4. $0.9 = 1 - e^{-NTU}$, $e^{-NTU} = 0.1$, $NTU = \ln 10 = 2.303$

5. $\varepsilon \to 1.0$ (counter-flow can achieve 100%)

6. $\varepsilon_{\max} = 1/(1+0.5) = 0.667$ (parallel-flow is limited)

📈 ε-NTU: Parallel vs Counter-Flow
ε counter-flow
ε parallel-flow
Advantage of counter-flow
Effectiveness, NTU, and Capacity Ratio ε-NTU for Shell-and-Tube and Cross-Flow