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Unformatted text preview: Chapter 8 Internal Forced Convection Review Problems 861 Geothermal water is supplied to a city through stainless steel pipes at a specified rate. The electric power consumption and its daily cost are to be determined, and it is to be assessed if the frictional heating during flow can make up for the temperature drop caused by heat loss. Assumptions 1 The flow is steady and incompressible. 2 The entrance effects are negligible, and thus the flow is fully developed. 3 The minor losses are negligible because of the large lengthtodiameter ratio and the relatively small number of components that cause minor losses. 4 The geothermal well and the city are at about the same elevation. 5 The properties of geothermal water are the same as fresh water. 6 The fluid pressures at the wellhead and the arrival point in the city are the same. Properties The properties of water at 110 ° C are ρ = 950.6 kg/m 3 , μ = 0.255 × 103 kg/m ⋅ s, and C p = 4.229 kJ/kg ⋅° C (Table A9). The roughness of stainless steel pipes is 2 × 106 m (Table 83). Analysis ( a ) We take point 1 at the wellhead of geothermal resource and point 2 at the final point of delivery at the city, and the entire piping system as the control volume. Both points are at the same elevation ( z 2 = z 2 ) and the same velocity ( V 1 = V 2 ) since the pipe diameter is constant, and the same pressure ( P 1 = P 2 ). Then the energy equation for this control volume simplifies to 2 2 u pump, turbine 2 2 2 2 u pump, 1 2 1 1 L L h h h h z g g P h z g g P = → + + + + = + + + V V ρ ρ That is, the pumping power is to be used to overcome the head losses due to friction in flow. The mean velocity and the Reynolds number are 7 3 3 2 3 2 10 186 . 1 s kg/m 10 255 . m) m/s)(0.60 305 . 5 )( kg/m 6 . 950 ( V Re m/s 305 . 5 4 / m) (0.60 /s m 1.5 4 / V × = ⋅ × = μ ρ = = π = π = = D D V A V m c m which is greater than 10,000. Therefore, the flow is turbulent. The relative roughness of the pipe is 10 33 . 3 m 60 . m 10 2 / 6 6 × = × = 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 7 6 10 187 . 1 51 . 2 7 . 3 10 33 . 3 log . 2 1 Re 51 . 2 7 . 3 / log . 2 1 ε It gives f = 0.00829. Then the pressure drop, the head loss, and the required power input become kPa 2218 kN/m 1 kPa 1 m/s kg 1000 kN 1 2 m/s) 305 . 5 )( kg/m 6 . 950 ( m 0.60 m 000 , 12 00829 . 2 2 2 3 2 V = ⋅ = = ∆ m D L f P ρ kW 5118 = ⋅ = ∆ = = /s m kPa 1 kW 1 0.65 ) kPa 2218 )( /s m (1.5 3 3 motor pump motor pump u pump, elect η η P V W W Therefore, the pumps will consume 5118 kW of electric power to overcome friction and maintain flow....
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This homework help was uploaded on 03/19/2008 for the course ME 410 taught by Professor Benard during the Spring '08 term at Michigan State University.
 Spring '08
 BENARD
 Heat Transfer

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