Training Heat Exchangers — ε-NTU Method Effectiveness, NTU, and Capacity Ratio
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Effectiveness, NTU, and Capacity Ratio

24 min Heat Exchangers — ε-NTU Method

Effectiveness, NTU, and Capacity Ratio

The ε-NTU method is the preferred approach when outlet temperatures are unknown. Instead of iterating on LMTD, we compute effectiveness directly from the heat exchanger’s size and flow parameters.

Heat Exchanger Effectiveness

$$\varepsilon = \frac{\dot{Q}}{\dot{Q}_{\max}} = \frac{\text{actual heat transfer}}{\text{maximum possible heat transfer}}$$

$$\dot{Q}_{\max} = C_{\min}(T_{h,\text{in}} - T_{c,\text{in}})$$

$C_{\min} = \min(C_h, C_c)$ where $C = \dot{m}c_p$. The fluid with $C_{\min}$ undergoes the maximum temperature change.

$$\varepsilon = \frac{C_h(T_{h,\text{in}} - T_{h,\text{out}})}{C_{\min}(T_{h,\text{in}} - T_{c,\text{in}})} = \frac{C_c(T_{c,\text{out}} - T_{c,\text{in}})}{C_{\min}(T_{h,\text{in}} - T_{c,\text{in}})}$$

Number of Transfer Units (NTU)

$$NTU = \frac{UA}{C_{\min}}$$

Dimensionless measure of heat exchanger size. Higher NTU = more area (relative to flow rate) = higher effectiveness.

Capacity Ratio

$$C_r = \frac{C_{\min}}{C_{\max}}$$

$0 \le C_r \le 1$. $C_r = 0$ for phase change (condenser/evaporator). $C_r = 1$ when both streams have equal capacity rates ("balanced" HX).

Example 1 — Computing Effectiveness

Hot oil ($C_h = 2000$ W/K) cools from 120°C to 80°C. Cold water ($C_c = 5000$ W/K) enters at 20°C. Find $\varepsilon$ and $T_{c,\text{out}}$.

$\dot{Q} = C_h(120 - 80) = 2000 \times 40 = 80{,}000$ W

$C_{\min} = 2000$ W/K (oil)

$\dot{Q}_{\max} = 2000 \times (120 - 20) = 200{,}000$ W

$\varepsilon = 80{,}000/200{,}000 = 0.40$

$T_{c,\text{out}} = 20 + 80{,}000/5000 = 36$°C

Example 2 — NTU Calculation

$U = 500$ W/(m²·K), $A = 12$ m², $C_{\min} = 3000$ W/K. Find NTU.

$$NTU = \frac{UA}{C_{\min}} = \frac{500 \times 12}{3000} = 2.0$$

Example 3 — Maximum Possible Heat Transfer

Steam condenses at 100°C ($C_h \to \infty$). Cold water: $\dot{m} = 2$ kg/s, $c_p = 4180$, enters at 15°C. Find $\dot{Q}_{\max}$ and $C_r$.

$C_c = 2 \times 4180 = 8360$ W/K

$C_h \to \infty$ (condensing), so $C_{\min} = C_c = 8360$ W/K

$\dot{Q}_{\max} = 8360 \times (100 - 15) = 710{,}600$ W = 710.6 kW

$C_r = C_{\min}/C_{\max} = 8360/\infty = 0$

Practice Problems

1. $C_h = 4000$, $C_c = 6000$ W/K. Which is $C_{\min}$? What is $C_r$?
2. $T_h$: 150→100°C, $T_c$: 30→?°C. $C_h = 3000$, $C_c = 5000$. Find $\varepsilon$, $T_{c,\text{out}}$.
3. $U = 800$, $A = 6$ m², $C_{\min} = 4000$. NTU?
4. $\varepsilon = 0.75$, $C_{\min} = 2000$, $T_h = 90$°C, $T_c = 10$°C. Find $\dot{Q}$.
5. Can $\varepsilon > 1$? Why or why not?
6. A balanced HX has $C_r =$ ?
Show Answer Key

1. $C_{\min} = 4000$ (hot); $C_r = 4000/6000 = 0.667$

2. $\dot{Q} = 3000 \times 50 = 150{,}000$ W. $\varepsilon = 150{,}000/(3000 \times 120) = 0.417$. $T_{c,o} = 30 + 150{,}000/5000 = 60$°C

3. $NTU = 800 \times 6/4000 = 1.2$

4. $\dot{Q} = 0.75 \times 2000 \times 80 = 120{,}000$ W = 120 kW

5. No — $\varepsilon = \dot{Q}/\dot{Q}_{\max}$, and actual heat transfer cannot exceed the thermodynamic maximum.

6. $C_r = 1$

ε-NTU Relations: Parallel and Counter-Flow