Special Topic 5 of the Registered Chemical Engineer Professional Examination: Heat Transfer
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This post was last edited by zhanghp30 on 2019-9-29 at 14:22. 1. Professional knowledge: 1) Basic concepts: heat flux density, heat flux, Fourier’s law, temperature gradient. Special attention should be paid to the factors that affect the thermal conductivity of different materials; multiple-choice questions on this topic appear almost every year. For more details, refer to the textbooks used at Tianjin University; there is no need to memorize anything—just draw it out clearly! 2) Formulas for heat conduction in flat walls and cylinders, as well as the concept of insulation layers. The relationship between temperature difference and thermal resistance in steady-state heat conduction processes. 3) Formula for forced turbulence in circular straight pipes: lalufa*d/lanbuda=0.023*Re^0.8*Pr^n. It is important to know how to use this formula for qualitative analysis! 4). Concepts and graphs of film boiling and nucleate boiling: Refer to the first volume of the textbook on Chemical Engineering Principles published by Tianjin University; multiple-choice questions on this topic are asked almost every year! 5). Heat transfer balance equation, calculation of the overall heat transfer coefficient, formula for calculating the heat exchange area, concept of the number of heat transfer units. See the textbook for details. 1). Which of the following statements about the thermal conductivity coefficient is correct? (ABC) A. Generally, metals have higher thermal conductivity than non-metals. B. The thermal conductivity of pure copper is dozens of times higher than that of stainless steel. C. The thermal conductivity of metals decreases as temperature increases. D. The thermal conductivity of gases is independent of temperature.2). For steady-state heat conduction in multi-layer flat walls, which of the following statements is correct? (BCD) A. The greater the temperature difference, the higher the thermal conductivity. B. The layer with the largest temperature difference has the greatest thermal resistance. C. The heat transfer per unit area is the same in all layers. D. The amount of heat transferred in each layer is the same.
3). Regarding the thermal conductivity coefficient for forced convection inside pipes, which of the following statements is incorrect? (A) A. Straight pipes have higher thermal conductivity than curved pipes. B. Turbulent flow has higher thermal conductivity than laminar flow. C. Short pipes have higher thermal conductivity than long pipes. D. Special-shaped pipes have higher thermal conductivity than circular pipes.
4). To calculate the thermal conductivity coefficient for forced convection inside pipes, which value should be used as the characteristic temperature? (C) A. The arithmetic average of the fluid inlet temperature and the wall temperature. B. The wall temperature. C. The arithmetic average of the fluid inlet and outlet temperatures. D. The geometric mean of the fluid inlet and outlet temperatures.
5). To improve the heat transfer coefficient during steam condensation, which of the following measures is correct? (ABC) A. Changing the orientation from vertical to horizontal. B. Reducing the wettability of water on the condensation surface. C. Increasing the steam flow rate. D. Increasing the roughness of the condensation surface.
6). In a conventional shell-and-tube heat exchanger, saturated steam is used to heat light oil. If it is desired to increase the overall heat transfer coefficient K, which measure should be taken? (B) A. Changing from co-current to counter-current flow. B. Increasing the flow rate of the light oil inside the tubes. C. Increasing the pressure of the saturated steam. D. Increasing the flow rate of the saturated steam.
7). When using circulating water to cool oil at 200°C, which type of heat exchanger is suitable? (AC) A. Floating-head type. B. Fixed-tube-sheet type. C. U-tube type. D. Air cooler.
8). When using a shell-and-tube heat exchanger to heat a high-viscosity fluid with medium-pressure steam, which arrangement is appropriate? (A) A. Steam flows through the tubes, while the fluid to be heated flows through the shell. B. Steam flows through the shell, while the fluid to be heated flows through the tubes. C. There is no significant difference between the two arrangements. D. A shell-and-tube heat exchanger cannot be used.
9). In an evaporator, medium-pressure steam is used to evaporate an inorganic salt solution. Under ideal conditions, what is the average temperature difference? (D) A. The temperature difference between the steam inlet and outlet. B. The temperature difference between the solution inlet and outlet. C. The logarithmic mean temperature difference between the steam and the solution at the inlet and outlet. D. The temperature difference between the steam and the solution.
10)* What is the correct order of the convective heat transfer coefficients? (B) (1) Alfa value when steam condenses in droplets; (2) Alfa value when air flows at 15 m/s; (3) Alfa value when water flows at 1.5 m/s. A. (1) > (2) > (3) B. (1) > (3) > (2) C. (3) > (1) > (2) D. (3) > (2) > (1)
11)* Benzene flows turbulently in a circular straight pipe with an inner diameter of 20 mm, and the convective heat transfer coefficient is 1270 W/m²·K. If the flow rate and properties remain unchanged, but the inner diameter is changed to 30 mm, what will be the new convective heat transfer coefficient? (C) A. It remains unchanged. B. 2540 W/m²·K. C. 612 W/m²·K. D. It cannot be determined.
12). A furnace is constructed using 0.3 m thick refractory bricks, 0.15 m thick insulating bricks, and 0.25 m thick ordinary bricks. The inner surface temperature is 1530 K, and the outer surface temperature is 325 K. The heat loss per unit area is (A) W: A 937.2 B 1028 C 789 D 1560. 13) In steam condensation heat transfer, the effect of non-condensable gases on the condensation heat transfer coefficient is (A): A Decrease B Increase C No effect D Indeterminable. 14) For boiling in large containers, it is always operated under (B) condition in industry: A Natural convection B Nucleate boiling C Film boiling D Unstable film boiling. 15) Generally speaking, for the same fluid, the correct statement regarding the convective heat transfer coefficient is (C): A It is smaller in forced convection than in natural convection; it is also smaller when there is phase change compared to when there is no phase change. B It is larger in forced convection than in natural convection; it is smaller when there is phase change compared to when there is no phase change. C It is larger in forced convection than in natural convection; it is larger when there is phase change compared to when there is no phase change. D It is smaller in forced convection than in natural convection; it is larger when there is phase change compared to when there is no phase change. 16) When saturated water vapor condenses outside a circular tube with an outer diameter of 20 mm and a length of 2 m, and all other conditions remain the same but the tube is changed from a vertical position to a horizontal one, the change in the external condensation heat transfer coefficient is (A): A Doubles B Reduces by half C Slightly decreases D Remains unchanged. 17) To reduce fouling thermal resistance, one should (D): A Reduce the flow velocity B Use tubes with high roughness C Use tube walls with high thermal conductivity D Clean them regularly. 18) In shell-and-tube heat exchangers, if the thermal resistance due to the tube wall and fouling can be ignored, and the heat transfer coefficient of the fluid outside the tube is much higher than that inside the tube, then to increase the overall heat transfer coefficient, the key is (B): A Increase the flow velocity of the fluid outside the tube B Increase the flow velocity of the fluid inside the tube C Reduce fouling D Reduce the wall thickness. 19) In industry, finned heating pipes are used instead of steel pipes for the purpose of (C): A Increasing thermal resistance B Saving steel and improving aesthetics C Increasing the heat transfer area and enhancing heat transfer efficiency D Reducing heat loss. 20) Spiral plate heat exchangers are suitable for (D) applications. A High pressure B High temperature C Regular maintenance D Low pressure, with the two fluids flowing in strict counterflow
This post was last edited by zhanghp30 on 2016-3-31 at 22:06. 2. Case knowledge: 1) In a double-pass shell-and-tube heat exchanger, saturated steam is introduced on the shell side to heat the air inside the tubes. The saturated steam at 110°C condenses into water at the same temperature, thereby heating the air from 20°C to 80°C. What is the outlet temperature of the first pass? For a double-pass system, Q = Wc*Cp*(t2-t1) = KS*deltaT; thus, deltaT = 54.6°C. Also, WC*CP*(80-20) = KS*54.6 ......(1) For the first pass: WCCP*(t1-20) = KS/2*(t1-20)/Ln((11-20)/(110-t1)).......(2) t1 = 58°C. Pay attention to the problem-solving techniques! 2). In a co-current heat exchanger, oil is cooled with water. The inlet and outlet temperatures of water are 25°C and 40°C respectively, while those of oil are 150°C and 100°C. Due to production requirements, the outlet temperature of oil needs to be reduced to 80°C. Assuming that the flow rates of oil and water, as well as their inlet and outlet temperatures and physical properties remain unchanged, if the original length of the heat exchanger tubes is 1.2 meters, determine what length the tubes need to have in order to meet this requirement. Heat loss is ignored. The original average temperature difference was detat1 = 88.6°C. When T2’ = 80°C, the outlet temperature of the cold fluid is given by whcph/wccpc = (t2 – t1) / (T1 – T2) = (t2’ – t1) / (T1 – T2’); since t2’ = 46°C, deta t2’ = 69.9°C. Therefore, WhCPh(150–100) = K * (N * PAI * d * L1) * 88.6……(1), and WhCph(150–80) = K * (N * PAI * d * L2) * 69.9……(2). By dividing equation (2) by equation (1), we obtain L2 = 2.13 m. 3) In a single-pass shell-and-tube heat exchanger, a certain solution flows turbulently inside the tubes, with a flow rate of 10,800 kg/h and an average specific heat capacity of 4.18 kJ/kg·°C. The solution is heated from 15°C to 100°C. The convective heat transfer coefficient inside the tubes is 600 W/m²·°C, while at 110°C, saturated water vapor condenses in the shell side to form water at the same temperature, with a convective heat transfer coefficient of 12,000 W/m². The tube specifications are Q25*2.5 mm, with 160 tubes in total. The thermal conductivity of the tubes is 17 W/m²·°C. Ignoring fouling and heat losses, determine the length of the tubes for a 4-shell-side configuration. 4-pass, ai=4^0.8ai=1819 W/m2·C, K0’=1156 W/m2·C, ΔTm=37.8°C. Q=WcCp(t2-t1)=10800/3600*4.18*(100-15)=1065.9 KW. S0’=Q/K0’, ΔTm=1065.9*1000/1156/37.8=24.4 m2. L’=S0’/πn/d0=24.4/160/3.14/0.025=1.94 m. 4) There is a 15 m2 single-pass shell-and-tube heat exchanger, in which saturated steam flows in the shell side to heat the air inside the tubes. 160C water vapor condenses to water at the same temperature. The air flow rate is 2.8 kg/s, the inlet temperature is 30°C, the specific heat capacity is 1 KJ/KG·°C, and the convective heat transfer coefficient is 87 W/M2·°C. Determine the outlet temperature of the air. a steam >> a air; thus K~~a air = 87. Q = WcCp(t2 – t1) = KS deta tm. Therefore, 2.8*1000*(t2 – 30) = 87*15*(t2 – 30)/Ln((160 – 30)/(160 – t2)), which gives t2 = 78.4°C. 5) In a heat exchanger with a heat transfer surface area of 50 m2, cooling water at 20°C and a flow rate of 33,000 kg/h is used to cool acetic acid with an inlet temperature of 110°C. The two fluids flow counterflowwise to each other. Upon initial operation after the heat exchanger has been cleaned, the outlet temperatures of the cooling water and acetic acid are 45°C and 40°C respectively. After operating for some time, while the flow rates and inlet temperatures of the hot and cold fluids remain unchanged, the outlet temperature of the cooling water drops to 38°C. What is the fouling thermal resistance at this point? The specific heat of water is 4200 J/KG/K; changes in physical properties are ignored. W1CP1/W2CP2 = t2 – t1 / (T1 – T2) = 45 – 20 / (110 – 40) = 0.357.
ΔTm = (T1 – T2) – (T2 – T1) / ln(T1 – T2) / (T2 – T1) = 38.2°C.
K = W2CP2 * (t2 – t1) / A.
ΔTm = 33000 / 3600 * 4200 * (45 – 20) / 50 / 38.2 = 504 W/M²·K.
After operation, T2’ = T1 – W2CP2 * (t2 – t1) / W1CP1 = 110 – (38 – 20) / 0.357 = 59.6°C.
ΔTm’ = 54.2°C; K’ = W2CP2 * (t2 – t1) / A. ΔTm’ = 255.7 W/M²·K.
1/504 = 1/a1 + 1/a2; 1/255.7 = 1/a1 + 1/a2 + R. Subtracting these equations gives R = 1.93*10^-3 M²·K/W.
6) A tubular heat exchanger consists of tubes with dimensions Q57MM*3.5MM and Q89MM*4.5MM. A solution flows inside these tubes at a flow rate of 5 tons per hour, and its temperature is reduced from 65°C to 35°C. The convective heat transfer coefficient for this solution is a2 = 1510 W/M²·°C. The cooling water flows counterflow between the tubes, with an inlet temperature of 25°C and an outlet temperature of 40°C. The heat transfer coefficient for the cooling water is a1 = 3270 W/M²·°C. Heat losses due to the tube walls and fouling are ignored. Given that the average heat capacity of the solution is 2.6 KJ/kg·°C, what is the theoretical length of the tubes in this heat exchanger? Q = WhCph(T1 – T2) = 5000/3600 * 2.6 * 1000 * (65 – 35) = 1.083 * 10^5 W. Since T2 > T1, the flow is countercurrent; therefore, deta tm = 16.37°C. K = 942.71 W/m²·°C. L = Q/K * deta tm = PAI * d = 39.19 m. When a fluid flows through a circular tube with an inner diameter of 50 mm, the convective heat transfer coefficient is 100 W/m²·°C, and the Reynolds number inside the tube is 10^5, indicating turbulent heating conditions. If a rectangular flat tube with a perimeter equal to that of the circular tube and a height-to-width ratio of 1:3 is used, with the fluid velocity remaining unchanged, what will be the new convective heat transfer coefficient? Let the height of the rectangle be a and its width be b, with a/b = 1/3; 2(a+b) = πd. Thus, a = πd/8 and b = 3πd/8. de = 2ab/(a+b) = 0.0294 m. Re’ = de·pu/u = 0.0294·pu/u, and Re = 0.05·pu/u. a’/a = 0.023·lanbuda/(de^(0.8)·Pr^(0.4)) = 0.023·lanbuda/(0.0294^(0.8)·Pr^(0.4)) = 0.05/0.0294·(0.0294/0.05)^(0.8) = 1.112. Therefore, a’ = 1.112·100 = 111.2 W/m². The walls of a vertical fireplace are constructed with an inner layer of some fire-resistant material 120 mm thick and an outer layer of some building material 230 mm thick. The thermal conductivity of both materials is unknown. The temperature of the inner wall of the furnace was measured at 800°C, while that of the outer wall was 113°C. Later, asbestos with a thickness of 50 mm and a thermal conductivity of 0.2 W/M·K was wrapped around the building materials to reduce heat loss. After wrapping, the temperature on the inner wall of the furnace is 800°C, at the interface between the refractory material and the building materials it is 686°C, at the interface between the building materials and asbestos it is 405°C, and on the outside of the asbestos it is 77°C. What is the percentage reduction in heat loss after wrapping the asbestos? Before wrapping: Q/A = (t1 – t3) / (s1/lanbuda1 + s2/lanbuda2); Q’/A = (t1 – t3’) / (s1/lanbuda1 + s2/lanbuda2). Percentage reduction: (Q/A – Q’/A) / Q/A = (t3’ – t3) / (t1 – t3) = 405 – 113 / (800 – 113) = 42.5%. 9) A steam pipe with an outer diameter of 100 mm is covered with a 50 mm thick insulating material A, whose thermal conductivity is LANBUDA A = 0.05 W/m²·K. Outside this layer, there is another 25 mm thick insulating material B, with a thermal conductivity of LANBUDA B = 0.075 W/m²·K. The temperatures on the inner surface of material A and the outer surface of material B are 170°C and 38°C respectively. Determine the temperature at the interface between A and B. The heat loss per meter of pipe length is given by Q/L = 2PAI((t1 – t3) / (1/lanbuda) + A·Ln(r2/r1) + (1/lanbuda)·B·Ln(r3/r2)) = 49.3 W/m. Substituting the values: 2PAI((170 – t2) / 0.05·Ln(100/50)) = 49.3; therefore, t2 = 61.3°C. 10) For a row of tube condensers with specifications Q25*2.5 MM and an effective length of 3.0 m, the coolant flows through the tubes at a velocity of 0.7 m/s, with its temperature rising from 20°C to 50°C. The flow rate is 5000 kg/h, and the temperature is 75°C. Saturated steam condenses into water at the same temperature; the latent heat is 310 kJ/kg. The heat transfer coefficient for steam condensation has been measured to be 800 W/m2·K, while the convective heat transfer coefficient of the coolant is 2500 W/m2·K. The fouling thermal resistance on the coolant side is 0.00055 m2·K/W; the fouling thermal resistance and wall thermal resistance on the steam side are negligible. Calculate the heat transfer area of this heat exchanger, and determine the total number of heat transfer tubes as well as the number of tube banks in it. The specific heat of the coolant is 2.5 KJ/KG·°C, and its density is 860 kg/m3. The heat released by the steam is given by Q = qm·r = 5000/3600·310·1000 = 4.31·10^5 W. The temperature difference is ΔTm = 38°C. The value of K is calculated as 1/K = 1/a1 + 1/a2(d1/d2) + Rs²·d1/d2; thus, K = 410 W/m²·°C. The area required for heat exchange is A = Q/K·ΔTm = 27.2 m². The amount of coolant required per second is qm2 = Q/Cp·(t2 – t1) = 4.31·10^5 / 2.5·10^3·(50 – 20) = 5.75 kg/s. The number of heat exchange tubes in each pass is determined by dividing the total amount of coolant by the amount of coolant flowing through each tube: ni = qm2/0.785·d2. Substituting the values, we get up2 = 5.75/0.785/0.02·0.02/0.7/860 = 30. The heat exchange area for each pass is Ai = ni·PAi. With L = 30·3.14·0.025·3 = 7.07 m², the number of passes required is N = A/Ai = 27.2/7.07 = 3.9. Rounding up, N = 4. The total number of tubes is therefore 30·4 = 120 tubes.
ΔTm = (T1 – T2) – (T2 – T1) / ln(T1 – T2) / (T2 – T1) = 38.2°C.
K = W2CP2 * (t2 – t1) / A.
ΔTm = 33000 / 3600 * 4200 * (45 – 20) / 50 / 38.2 = 504 W/M²·K.
After operation, T2’ = T1 – W2CP2 * (t2 – t1) / W1CP1 = 110 – (38 – 20) / 0.357 = 59.6°C.
ΔTm’ = 54.2°C; K’ = W2CP2 * (t2 – t1) / A. ΔTm’ = 255.7 W/M²·K.
1/504 = 1/a1 + 1/a2; 1/255.7 = 1/a1 + 1/a2 + R. Subtracting these equations gives R = 1.93*10^-3 M²·K/W.
6) A tubular heat exchanger consists of tubes with dimensions Q57MM*3.5MM and Q89MM*4.5MM. A solution flows inside these tubes at a flow rate of 5 tons per hour, and its temperature is reduced from 65°C to 35°C. The convective heat transfer coefficient for this solution is a2 = 1510 W/M²·°C. The cooling water flows counterflow between the tubes, with an inlet temperature of 25°C and an outlet temperature of 40°C. The heat transfer coefficient for the cooling water is a1 = 3270 W/M²·°C. Heat losses due to the tube walls and fouling are ignored. Given that the average heat capacity of the solution is 2.6 KJ/kg·°C, what is the theoretical length of the tubes in this heat exchanger? Q = WhCph(T1 – T2) = 5000/3600 * 2.6 * 1000 * (65 – 35) = 1.083 * 10^5 W. Since T2 > T1, the flow is countercurrent; therefore, deta tm = 16.37°C. K = 942.71 W/m²·°C. L = Q/K * deta tm = PAI * d = 39.19 m. When a fluid flows through a circular tube with an inner diameter of 50 mm, the convective heat transfer coefficient is 100 W/m²·°C, and the Reynolds number inside the tube is 10^5, indicating turbulent heating conditions. If a rectangular flat tube with a perimeter equal to that of the circular tube and a height-to-width ratio of 1:3 is used, with the fluid velocity remaining unchanged, what will be the new convective heat transfer coefficient? Let the height of the rectangle be a and its width be b, with a/b = 1/3; 2(a+b) = πd. Thus, a = πd/8 and b = 3πd/8. de = 2ab/(a+b) = 0.0294 m. Re’ = de·pu/u = 0.0294·pu/u, and Re = 0.05·pu/u. a’/a = 0.023·lanbuda/(de^(0.8)·Pr^(0.4)) = 0.023·lanbuda/(0.0294^(0.8)·Pr^(0.4)) = 0.05/0.0294·(0.0294/0.05)^(0.8) = 1.112. Therefore, a’ = 1.112·100 = 111.2 W/m². The walls of a vertical fireplace are constructed with an inner layer of some fire-resistant material 120 mm thick and an outer layer of some building material 230 mm thick. The thermal conductivity of both materials is unknown. The temperature of the inner wall of the furnace was measured at 800°C, while that of the outer wall was 113°C. Later, asbestos with a thickness of 50 mm and a thermal conductivity of 0.2 W/M·K was wrapped around the building materials to reduce heat loss. After wrapping, the temperature on the inner wall of the furnace is 800°C, at the interface between the refractory material and the building materials it is 686°C, at the interface between the building materials and asbestos it is 405°C, and on the outside of the asbestos it is 77°C. What is the percentage reduction in heat loss after wrapping the asbestos? Before wrapping: Q/A = (t1 – t3) / (s1/lanbuda1 + s2/lanbuda2); Q’/A = (t1 – t3’) / (s1/lanbuda1 + s2/lanbuda2). Percentage reduction: (Q/A – Q’/A) / Q/A = (t3’ – t3) / (t1 – t3) = 405 – 113 / (800 – 113) = 42.5%. 9) A steam pipe with an outer diameter of 100 mm is covered with a 50 mm thick insulating material A, whose thermal conductivity is LANBUDA A = 0.05 W/m²·K. Outside this layer, there is another 25 mm thick insulating material B, with a thermal conductivity of LANBUDA B = 0.075 W/m²·K. The temperatures on the inner surface of material A and the outer surface of material B are 170°C and 38°C respectively. Determine the temperature at the interface between A and B. The heat loss per meter of pipe length is given by Q/L = 2PAI((t1 – t3) / (1/lanbuda) + A·Ln(r2/r1) + (1/lanbuda)·B·Ln(r3/r2)) = 49.3 W/m. Substituting the values: 2PAI((170 – t2) / 0.05·Ln(100/50)) = 49.3; therefore, t2 = 61.3°C. 10) For a row of tube condensers with specifications Q25*2.5 MM and an effective length of 3.0 m, the coolant flows through the tubes at a velocity of 0.7 m/s, with its temperature rising from 20°C to 50°C. The flow rate is 5000 kg/h, and the temperature is 75°C. Saturated steam condenses into water at the same temperature; the latent heat is 310 kJ/kg. The heat transfer coefficient for steam condensation has been measured to be 800 W/m2·K, while the convective heat transfer coefficient of the coolant is 2500 W/m2·K. The fouling thermal resistance on the coolant side is 0.00055 m2·K/W; the fouling thermal resistance and wall thermal resistance on the steam side are negligible. Calculate the heat transfer area of this heat exchanger, and determine the total number of heat transfer tubes as well as the number of tube banks in it. The specific heat of the coolant is 2.5 KJ/KG·°C, and its density is 860 kg/m3. The heat released by the steam is given by Q = qm·r = 5000/3600·310·1000 = 4.31·10^5 W. The temperature difference is ΔTm = 38°C. The value of K is calculated as 1/K = 1/a1 + 1/a2(d1/d2) + Rs²·d1/d2; thus, K = 410 W/m²·°C. The area required for heat exchange is A = Q/K·ΔTm = 27.2 m². The amount of coolant required per second is qm2 = Q/Cp·(t2 – t1) = 4.31·10^5 / 2.5·10^3·(50 – 20) = 5.75 kg/s. The number of heat exchange tubes in each pass is determined by dividing the total amount of coolant by the amount of coolant flowing through each tube: ni = qm2/0.785·d2. Substituting the values, we get up2 = 5.75/0.785/0.02·0.02/0.7/860 = 30. The heat exchange area for each pass is Ai = ni·PAi. With L = 30·3.14·0.025·3 = 7.07 m², the number of passes required is N = A/Ai = 27.2/7.07 = 3.9. Rounding up, N = 4. The total number of tubes is therefore 30·4 = 120 tubes.
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