Heat Exchangers

Abstract

Heat exchangers have been classified in various ways. Double pipe and shell-and-tube heat exchangers are most widely used. This chapter deals with heat exchanger fundamentals and design of heat exchangers. Heat transfer equation for double pipe heat exchanger has been presented and log mean temperature difference (LMTD) has been defined in Part 1 of the chapter followed by derivations of equations of LMTD for parallel flow and counterflow exchangers. In Sect. 14.4, graphs of correction factor for calculation of LMTD for other flow arrangements have been presented. Basic treatment of effectiveness-NTU method for counterflow and parallel flow heat exchangers is given in Sect. 14.5. Graphs for determination of effectiveness for other flow arrangements have been given as a function of NTU in Sect. 14.6. In the second part of the chapter, basic considerations for heat exchanger design have been presented followed by defining of fouling factor, and clean and design overall heat transfer coefficients. An illustrative example of double pipe heat exchanger design has been given.

Appendices

Review Questions

• 14.1 Classify heat exchangers and draw diagrams to show the temperature distributions along their length or at the outlet.

• 14.2 Describe with the help of figures the double pipe, and shell-and-tube heat exchangers . Discuss the effect of baffles on the flow pattern in the shell-and-tube heat exchanger .

• 14.3 Derive the equation of overall heat transfer coefficient for tube flow. State the condition when you can neglect the pipe wall resistance to heat transfer.

• 14.4 Establish the expressions of LMTD for parallel flow and counterflow heat exchangers. State clearly the simplifying assumptions made in the derivation of the LMTD equation. What is the effect of these assumptions on the estimate of LMTD?

• 14.5 Define the correction factor for the LMTD and compare the performance of the counterflow, parallel flow and cross-flow arrangements.

• 14.6 Compare arithmetic mean and log mean temperature differences .

• 14.7 What is fouling factor? Explain with proper comments.

• 14.8 Define clean and design overall heat transfer coefficients .

• 14.9 What do you mean by controlling film coefficient? Explain giving a suitable example.

• 14.10 Define clearly the effectiveness of heat exchangers and develop an expression for effectiveness in terms of NTU and heat capacity ratio for a counterflow heat exchanger.

• 14.11 What advantage does the effectiveness-NTU method has over the LMTD method?

• 14.12 Using the effectiveness—NTU charts, compare the performance of parallel and counterflow heat exchangers.

• 14.13 Describe with suitable sketch the constructional features of a double pipe heat exchanger. What are its merits and demerits?

• 14.14 For a boiler or condenser, show that

  $$\varepsilon = 1 - \exp ( - NTU)$$

• 14.15 For a gas turbine heat exchanger (C_min/C_max ≈ 1), show that

  • (i) for the parallel flow arrangement,

    $$\varepsilon = (1/2)[1 - \exp ( - 2NTU)]$$

  • (ii) for the counterflow arrangement,

    $$\varepsilon = NTU/(1 + NTU).$$

    Which arrangement will you prefer?

Problems

• 14.1 Water [c_p = 4.1868 kJ/(kg K)] flowing at the rate of 2.0 kg/min is heated from 15°C to 45°C in a concentric, double pipe, parallel flow heat exchanger . The hot fluid is oil [c_p = 1.9 kJ/(kg K)], flowing at 2.5 kg/min, which enters the exchanger at 200°C. Determine the heat exchanger surface area required for an overall heat transfer coefficient U = 250 W/(m² K). Also calculate the effectiveness and NTU of the exchanger.

  [Ans. (T_o)_oil = 147.11°C from heat balance; LMTD= 139.47°C; A = 0.12 m²; ε = 0.286; NTU= 0.38.]

• 14.2 A counterflow double pipe heat exchanger is to be used to cool hot oil [c_p = 2.0 kJ/(kg K)] from 200°C to 65°C. The cold stream oil [c_p = 1.8 kJ/(kg K)] at 50°C enters at a flow rate of 1.0 kg/s. If the flow rate of the hot oil is 0.7 kg/s, determine the required heat exchanger surface area [U = 300 W/(m² K)].

  [Ans. t_o = 155°C from heat balance equation; LMTD = 27.31°C; A = q/(LMTD × U) = 23.07 m².]

• 14.3 Determine the surface area required in a cross-flow heat exchanger with both streams unmixed to cool 300 m³ of air per min from 50°C to 30°C. Water is available at 15°C for cooling of the air. The flow rate of water is 300 kg/min and the overall heat transfer coefficient is 150 W/(m² K). Given: ρ_air = 1.1 kg/m³, c_p of air = 1.006 kJ/(kg K), c of water = 4.1868 kJ/(kg K).

  [Ans. q = 110.66 kW; (t_o)_w = 20.3°C; LMTD = 21.52°C; P = 0.1514; R = 3.77; F_T ≈ 0.97; A = q/(UF_T ∆t_m) = 35.34 m².]

• 14.4 Air at 100°C [c_p = 1.006 kJ/(kg K)] flows into a cross-flow heat exchanger (both fluids unmixed) at a flow rate of 10 kg/s. Water [c_p = 4.1868 kJ/(kg K)] enters the exchanger at 15°C and at a flow rate of 5 kg/s. If the heat exchanger surface area is 100 m² and the overall heat transfer coefficient is 200 W/(m² K), determine the exit temperature of the air.

  [Ans. NTU = UA/C_min = 1.99; C* = C_min/C_max = 0.4806; effectiveness ≈ 0.74 (Fig. 14.20a); (T_o)_air = 37.1°C; (T_o)_w = 45.2°C from heat balance; Check: (LMTD)_counter = 36.0°C; P = 0.355; R = 2.08; F_T = 0.88; q = UAF_T(LMTD)_counter = 633600 W ≈ mc_p∆t.]

• 14.5 Hot water [c_p = 4.1868 kJ/(kg K)] at 90°C is used to heat liquid ammonia [c_p = 4.8 kJ/(kg K)] from 40°C in a 1-shell pass and 2-tube passes (water in the tube). The water outlet temperature is 60°C and flow rate is 12 kg/min. If U = 100 W/(m² K) and A = 9.8 m², determine the outlet temperature of the ammonia.

  [Ans. Assuming C_h = C_min, NTU= UA/C_min = 1.17; ε = (90 – 60)/(90–40) = 0.6; from Fig. 14.20c, C* = C_min/C_max ≈ 0.25; m_Ammonia = C_max/c_Ammonia = 41.87 kg/min; (t_o)_Ammonia = 47.5°C from heat balance.]

• 14.6 Check the result of Example 14.19 using the LMTD approach.

  [Ans. P = 0.887; R = 0.2437; F_t = 0.6; LMTD = 64.07°C; q = 53819 W ≈ mc_p∆t = C_max ∆t = 4.1 × 325 × (55 −15).]

• 14.7 Check the result of Example 14.6 using ε-NTU approach.

  [Ans. C_h = 14250, C_c = 10080, C*= C_min/C_max = 0.7074, ε = (724.1–300)/(800–300) = 0.8482, using relevant equation from Table 14.2 gives NTU = 3.311 and A = (NTU × C_min)/U = 111.25 m².]

• 14.8 Saturated steam at 100°C is condensing in a shell-and-tube heat exchanger with a UA value of 3650 W/K. Cooling water enters the tubes at 30°C. Determine cooling water flow rate required to maintain a heat rate of 200 kW.

  [Ans. For the condensing fluid, C_max = ∞. Hence, C^* = C_min/C_max = 0 and effectiveness relation is

  $\varepsilon = 1 - \exp \left( { - NTU} \right)$, or $\frac{q}{{q_{{max} } }} = 1 - \exp \left( { - \frac{UA}{{C_{{min} } }}} \right)$, or $\frac{200 \times 1000}{{m_{c} c_{c} (100 - 30)}} = 1 - \exp \left( { - \frac{3650}{{m_{c} c_{c} }}} \right)$; assuming mean temperature of water as 40°C, c_c = 4179 J/(kg K). Substitution gives $\frac{200 \times 1000}{{m_{c} \times 4179 \times (100 - 30)}} = 1 - \exp - \left( {\frac{3650}{{m_{c} \times 4179}}} \right)$. Solution by trial and error gives m_c = 1.7 kg/s. From I law equation, $q = m_{c} \times c_{c} \times (t_{\text{co}} - t_{\text{ci}} )$ or $200 \times 1000 = 1.7 \times 4179 \times (t_{\text{co}} - 30)$, which gives t_co =58.15°C. So t_cm = (30 + 58.15)/2 = 44.07°C. Retrial is not required as c_c will change only marginally.]

• 14.9 Steam is condensed in the shell of a shell-and-tube (one shell, two tube passes) heat exchanger, which consists of 160 tubes of 25 mm diameter. Steam pressure is 1 atm and heat transfer coefficient on the condensation side is 8000 W/(m² K). If water flow rate is 12 kg/s and it is heated from 20°C to 60°C, determine the tube length.

  [Ans. Thermophysical properties of water at mean temperature [= (t_ci + t_co)/2] = 40°C are (Table A4) c = 4179 J/(kg K), k = 0.631 W/(m K), μ = 651 × 10⁻⁶ N s/m² and Pr = 4.3; water flow rate per tube is (flow rate/no. of tubes in single pass) = 12/80 = 0.15 kg/s (refer to Fig. 14.2); Reynolds number of the flow in the tube, $\text{Re} = \frac{{md_{i} }}{{(\pi /4)d_{i}^{2} \mu }} = \frac{4m}{{\pi d_{i} \mu }} = \frac{4 \times 0.15}{{\pi \times 0.025 \times 651 \times 10^{ - 6} }} = 11735$; flow is turbulent. Dittus–Boelter equation gives $h_{i} = \frac{k}{{d_{i} }} \times 0.024\text{Re}^{0.8} \Pr^{0.4}$$= \frac{0.631}{0.025} \times 0.024 \times 11735^{0.8} \times 4.3^{0.4}$$= 1955.6$ W/(m² K); $U = \frac{{h_{i} h_{o} }}{{h_{i} + h_{o} }} = \frac{1955.6 \times 8000}{1955.6 + 8000} =$$1571.5$ W/(m² K);$\varepsilon = \frac{q}{{q_{{max} } }} = \frac{{m_{c} c_{c} (t_{\text{co}} - t_{\text{ci}} )}}{{m_{c} c_{c} (t_{\text{hi}} - t_{\text{ci}} )}} =$$\frac{{t_{\text{co}} - t_{\text{ci}} }}{{t_{\text{hi}} - t_{\text{ci}} }} = \frac{60 - 20}{100 - 20} = 0.5;$ for the condensing fluid,$C_{{max} } = \infty$. Hence, $C^{*} = \frac{{C_{{min} } }}{{C_{{max} } }} = 0$ and the applicable NTU-ε relation gives $NTU = - \ln (1 - \varepsilon ) = - \ln (1 - 0.5) = 0.693$; $C_{{min} } = m_{c} c_{c} =$$12 \times 4179 = 50148$ W/K; hence, $NTU = \frac{UA}{{C_{{min} } }}$ gives $A = \frac{{NTU \times C_{{min} } }}{U} = \frac{0.693 \times 50148}{1571.5} = 22.11$ m²; from $A = N\pi DL,$ $L = \frac{A}{N\pi D} = \frac{22.11}{160 \times \pi \times 0.025} = 1.76$ m.]