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Unformatted text preview: Name: 1 ME 315: Heat and Mass Transfer Spring 2008 EXAM 3 Thursday, 17 April 2008 7:00 to 8:00 PM Instructions: This is an openbook exam . You may refer to your course textbook, your class notes and your graded homework assignments. Show all work clearly for maximum credit. CIRCLE ALL ANSWERS . Problem Score 1 / 20 2 / 22 3 / 18 Total / 60 Name: 2 1. (20 points) A concentric tube heat exchanger is used to cool water at a mass flow rate of 1 kg/s from 80 C to 40 C. The coolant enters at 0 C and exits at 50 C. The specific heat for water and coolant are c p,water = 4200 J/(kg . K) and c p,cool = 1500 J/(kg . K), respectively. The surface area of the thinwalled tube separating the water and coolant is 50 m 2 . Water Coolant h o h o h i = 150 W/(m 2. K) h i = 150 W/(m 2. K) Wall resistance, R wall = 0.002 (m 2. K)/W Area of wall, A wall = 50 m 2 a. (2 points) In order to achieve the desired inlet and outlet temperatures, which concentric tube heat exchanger configuration must be used? Circle the correct configuration below. Parallel Flow Counter Flow b. (4 points) Find the rate of heat transferred from the water to the coolant. c. (4 points) What must the mass flow rate of the coolant be in order to remove the desired heat from the water? d. (4 points) What is the effectiveness of the heat exchanger? e. (6 points) If the heat transfer coefficient on the coolant side is h i = 150 W/(m 2. K) and the thin wall between the water and coolant has a resistance of " wall R = 0.002 (m 2. K)/W, then what is the heat transfer coefficient on the water side (h o )? c ) C c q T co T ci = C c 3.36 10 3 W K = m c.dot C c c pc = m c.dot 2.24 kg s = d ) Therefore, C min C c = q max C min T hi T ci ( 29 = q q max = 0.625 = Note that epsilon could also be found using the epsilon  NTU method if they solve for U to get NTU and then use equation 11.29b or Figure 11.11 but this is much more involved since they know q and q_max....
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 Fall '08
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