--- title: "Effectiveness-NTU analysis of a heat exchanger" subject: "Heat Transfer" chapter: 6 key: "semester_5/heat_transfer:effectiveness-ntu-counterflow-oil-water" source: https://www.bctengineer.com/q/semester_5/heat_transfer/effectiveness-ntu-counterflow-oil-water site: BCT Engineering --- # Effectiveness-NTU analysis of a heat exchanger Heat Transfer · Chapter 6: Applications of Heat Transfer - Asked in: year not given, 8 marks - Video: [Heat Transfer L33 p1 - Effectiveness-NTU Method](https://www.youtube.com/watch?v=Rqkgb_qiz0c) — Ron Hugo - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/heat_transfer/effectiveness-ntu-counterflow-oil-water - Whole chapter: https://www.bctengineer.com/md/semester_5/heat_transfer/chapter-6 ## Question Explain the effectiveness-NTU method and state when it is preferred to the LMTD method. A counterflow heat exchanger has a surface area of 3 m^2 and U = 350 W/m^2 K. Hot oil (cp = 2100 J/kg K) at 0.8 kg/s enters at 140 °C and cold water (cp = 4180 J/kg K) at 0.6 kg/s enters at 25 °C. Find the effectiveness, the heat transfer rate and the outlet temperatures of both fluids. The counterflow effectiveness is e = [1 - exp(-NTU(1 - Cr))] / [1 - Cr exp(-NTU(1 - Cr))]. ## Short answer Effectiveness ε = Q/Qmax; Qmax = Cmin(Th,in−Tc,in). NTU = UA/Cmin, Cr = Cmin/Cmax. The ε-NTU method is preferred when outlet temperatures are unknown. Cmin = 1680 W/K, Cr = 0.67, NTU = 0.625. ε = [1−exp(−0.625(0.33))]/[1−0.67exp(−0.625(0.33))] = 0.41. Q = 0.41×1680×115 = 79.2 kW. Th,out = 140−79160/1680 = 92.9°C, Tc,out = 25+79160/2508 = 56.6°C. ## Answer #### The effectiveness-NTU method **Effectiveness** $\varepsilon$ is the ratio of the actual heat transfer to the maximum possible: $$ \varepsilon=\frac{Q}{Q_{max}},\qquad Q_{max}=C_{min}\,(T_{h,in}-T_{c,in}) $$ where $C = \dot m c_p$ is the heat capacity rate. **Number of transfer units** $NTU = UA/C_{min}$ and capacity ratio $C_r = C_{min}/C_{max}$. For each flow arrangement, $\varepsilon = f(NTU, C_r)$. It is preferred when the **outlet temperatures are unknown** (rating or performance problems): the LMTD method would need trial and error, while the NTU method is direct. LMTD is easier for sizing when all four temperatures are known. #### Solution Heat capacity rates: $$ C_h = 0.8\times2100 = 1680\ \text{W/K},\qquad C_c = 0.6\times4180 = 2508\ \text{W/K} $$ So $C_{min} = C_h = 1680$ W/K (oil), $C_{max} = 2508$ W/K. $$ C_r = \frac{1680}{2508} = 0.670,\qquad NTU = \frac{UA}{C_{min}} = \frac{350\times3}{1680} = 0.625 $$ **Effectiveness (counterflow):** $$ \varepsilon = \frac{1-e^{-0.625(1-0.670)}}{1-0.670\,e^{-0.625(1-0.670)}} = \frac{1-0.8134}{1-0.5449} = \frac{0.1866}{0.4551}= 0.410 $$ **Heat transfer rate:** $$ Q = \varepsilon C_{min}(T_{h,in}-T_{c,in}) = 0.4097\times1680\times(140-25) = 79\,160\ \text{W}\approx79.2\ \text{kW} $$ **Outlet temperatures:** $$ T_{h,out} = 140 - \frac{79\,160}{1680} = 92.9\ ^\circ\text{C},\qquad T_{c,out} = 25 + \frac{79\,160}{2508} = 56.6\ ^\circ\text{C} $$ (For comparison, a parallel-flow exchanger of the same size would have $\varepsilon = \frac{1-e^{-NTU(1+C_r)}}{1+C_r} = 0.388$.) **Answer:** $\varepsilon \approx 0.41$; $Q \approx 79.2$ kW; $T_{h,out} \approx 92.9$ °C; $T_{c,out} \approx 56.6$ °C. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Effectiveness of a heat exchanger is defined as A. Q divided by U A B. Q divided by Q_max, where Q_max = C_min (Th,in - Tc,in) C. Q divided by C_max (Th,in - Tc,in) D. NTU divided by Cr Answer: B. The maximum possible heat transfer is limited by the fluid with the smaller heat capacity rate. (Key point: Effectiveness definition) 2. The effectiveness-NTU method is preferred when A. all four terminal temperatures are known B. outlet temperatures are unknown, as in rating problems C. only sizing is needed D. there is no heat exchange Answer: B. LMTD would need trial and error when outlets are unknown, but NTU gives them directly. (Key point: When NTU method is preferred) 3. Here C_h = 1680 W/K and C_c = 2508 W/K. Which values are right? A. Cmin = 2508, Cr = 1.49, NTU = 0.42 B. Cmin = 1680, Cr = 0.67, NTU = 0.625 C. Cmin = 1680, Cr = 1.49, NTU = 0.625 D. Cmin = 2508, Cr = 0.67, NTU = 0.42 Answer: B. Oil has the smaller capacity rate, so Cr = 1680/2508 = 0.67 and NTU = 350 x 3 / 1680 = 0.625. (Key point: Cmin, Cr and NTU) 4. With Q about 79.2 kW, the outlet temperatures are found from A. T_h,out = 140 - Q/C_h = 92.9 C and T_c,out = 25 + Q/C_c = 56.6 C B. T_h,out = 140 - Q/C_c and T_c,out = 25 + Q/C_h C. T_h,out = 25 and T_c,out = 140 D. T_h,out = 140 x (1 - epsilon) Answer: A. Each fluid's temperature change equals Q divided by its own heat capacity rate. 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