Explanation Given Information: Mass flow of water m c = 2.5kg/s Inlet temperature of water T c , i = 15°C. Outlet temperature of water T c , o = 85°C. Inlet temperature of engine oil T h , j = 160°C. Outlet temperature of engine oil T h , o = 100°C. Average convection coefficient of the oil , h o = 400W/m 2 .K Diameter of the tube D = 25mm Number of shells n=2 Number of tubes in each shell N 1 =8 Number of tube passes in each shell P 1 = 8 Available floor space length l= 8m Calculating the mean temperature of water as m h C p h ( T h , j − T h , o ) = m c C p c ( T c , o − T c , j ) m h × 2.350 × 10 3 ( 160 − 100 ) = 2.5 × 4.181 × 10 3 ( 85 − 15 ) m h = 5.18 k g / s T m = T c , j + T c , o 2 = 15 + 85 2 = 50 ∘ C =323 K The properties of water at 323K are as follows ρ c = 987.9 k g / m 3 C p , c = 4.181 × 10 3 J / k g . K μ c = 547.6 × 10 − 6 N . s / m 2 K c = 643 × 10 − 3 W / m . K Pr = 3.56 Now, calculating the temperature of engine oil is Since, C c < C h C min = C c C max = C h Heat capacity ratio is C r = C min C max = 10454.5 12173 = 0.858 Properties of engine oil at 403K are as follows C p , h = 2.350 × 10 3 J / k g . K Based on the energy balance, Heat assumed by the hot fluid = Heat occupied by the cold fluid m h C p h ( T h , j − T h , o ) = m c C p c ( T c , o − T c , j ) m h × 2.350 × 10 3 ( 160 − 100 ) = 2.5 × 4.181 × 10 3 ( 85 − 15 ) m h = 5.18 k g / s Heat capacity rate of water is C c = m c C p c = 2.5 × 4.181 × 10 3 = 10454.5 W / K Heat capacity rate of oil is C h = m h C p h = 5.18 × 2.350 × 10 3 = 12173 W / K Since, C c < C h C min = C c C max = C h Heat capacity ratio is C r = C min C max = 10454.5 12173 = 0.858 Effectiveness of the heat exchanger is d = L s h e l l × 2.5 = 2.07 × 2.5 = 5.17 m For n shell passes, F = ( ε C r − 1 ε − 1 ) 1 / n = [ ( 0.48 × 0.858 ) − 1 0.48 − 1 ] 1 / 2 = 1

7th Edition

ISBN: 9780470501979

Author: Frank P. Incropera, David P. DeWitt, Theodore L. Bergman, Adrienne S. Lavine

Publisher: Wiley, John & Sons, Incorporated

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Chapter 11, Problem 11.80P

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The tube length for all 10 passes in both shells.

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Chapter 11, Problem 11.79P

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Question 2 En. Karim recently install a hot water system operates by a solar energy. The system consists of double pipe counter flow heat exchanger. Cold waters enters a tube at 22°C at a rate of 0.1 kg/s, while hot air enters the heat exchanger at 90°C at a rate of 0.3 kg/s. The specific heat for both cold water and hot air is cp = 4180 J/kg.K and c, = 1010 J/kg.K , respectively. The overall heat transfer coefficient based on the inner side of the tube is 80 W/m².K. The length of the tube is 12 m and the internal diameter of the tube is 1.2 cm. En. Karim assigned you do the complete analysis on this hot water system including to calculate the effectiveness of the heat exchanger. As an engineer, you have to determine: i. the heat capacity rates of both fluids, ii. the maximum rate of heat transfer (kW), iii. the effectiveness of the heat exchanger (NTU method), iv. the actual rate of heat transfer (kW), and v. the outlet temperatures of both cold water and hot air.

Chapter 11 Solutions

Fundamentals of Heat and Mass Transfer

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The tube length for all 10 passes in both shells.

Solution Summary: The author calculates the total tube length for all 10 passes in both shells, which is 33.18 mtrs.

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The tube length for all 10 passes in both shells.

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Chapter 11 Solutions