FTFS Chap14 P090 - Chapter 14 Flow in Pipes Review Problems...

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Chapter 14 Flow in Pipes Review Problems 14-90 A compressor takes in air at a specified rate at the outdoor conditions. The useful power used by the compressor to overcome the frictional losses in the duct is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The entrance effects are negligible, and thus the flow is fully developed. 3 Air is an ideal gas. 4 The duct involves no components such as bends, valves, and connectors, and thus minor losses are negligible. 5 The flow section involves no work devices such as fans or turbines. Properties The properties of air at 1 atm = 101.3 kPa and 15 ° C are ρ 0 = 1.225 kg/m 3 and μ = 1.802 × 10 -5 kg/m s (Table A-22). The roughness of galvanized iron surfaces is ε = 0.00015 m (Fig. A-32). The dynamic viscosity is independent of pressure, but density of an ideal gas is proportional to pressure. The density of air at 95 kPa is 3 3 0 0 kg/m 149 . 1 ) kg/m 225 . 1 )( 3 . 101 / 95 ( ) / ( = = = ρ P P . Analysis The mean velocity and the Reynolds number are m/s 594 . 8 4 / m) (0.20 /s m 0.27 4 / 2 3 2 = = = = π D V A V c m V 5 5 3 10 096 . 1 s kg/m 10 802 . 1 m) m/s)(0.20 )(8.594 kg/m (1.149 Re × = × = = - μ h m D V which is greater than 4000. Therefore, the flow is turbulent. The relative roughness of the pipe is 10 5 . 7 m 20 . 0 m 10 5 . 1 / 4 4 - - × = × = D ε The friction factor can be determined from the Moody chart, but to avoid the reading error, we determine it from the Colebrook equation using an equation solver (or an iterative scheme), × + × - = + - = - f f f D f h 5 4 10 096 . 1 51 . 2 7 . 3 10 5 . 7 log 0 . 2 1 Re 51 . 2 7 . 3 / log 0 . 2 1 It gives f = 0.02109. Then the pressure drop in the duct and the required pumping power become Pa 8 . 35 N/m 1 Pa 1 m/s kg 1 N 1 2 m/s) 594 . 8 )( kg/m 149 . 1 ( m 0.20 m 8 02109 . 0 2 2 2 2 3 2 = = = = m L D L f P P V W 9.66 = = = /s m Pa 1 W 1 ) Pa 0 . 8 . 35 )( /s m 27 . 0 ( 3 3 u pump, P V W Discussion Note hat the pressure drop in the duct and the power needed to overcome it is very small (relative to 150 hp), and can be disregarded. The friction factor could also be determined easily from the explicit Haaland relation. It would give f = 0.02086, which is very close to the Colebrook value. Also, the power input determined is the mechanical power that needs to be imparted to the fluid. The shaft power will be more than this due to fan inefficiency; the electrical power input will be even more due to motor inefficiency (but probably no more than 20 W). 14-64 8 m 20 cm Air compressor 150 hp 0.27 m 3 /s 95 kPa
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Chapter 14 Flow in Pipes 14-91 Air enters the underwater section of a circular duct. The fan power needed to overcome the flow resistance in this section of the duct is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The entrance effects are negligible, and thus the flow is fully developed. 3 Air is an ideal gas. 4 The duct involves no components such as bends, valves, and connectors. 5 The flow section involves no work devices such as fans or turbines. 6 The pressure of air is 1 atm.
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This homework help was uploaded on 04/18/2008 for the course EML 3007 taught by Professor Chung during the Spring '08 term at University of Florida.

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FTFS Chap14 P090 - Chapter 14 Flow in Pipes Review Problems...

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