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Heat Chap08-061

# Heat Chap08-061 - Chapter 8 Internal Forced Convection...

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Chapter 8 Internal Forced Convection Review Problems 8-61 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 length-to-diameter 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 = 950.6 kg/m 3 , = 0.255 10 -3 kg/m s, and C p = 4.229 kJ/kg  C (Table A-9). The roughness of stainless steel pipes is 2 10 -6 m (Table 8-3). Analysis ( a ) We take point 1 at the well-head 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 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 . 0 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 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 . 0 m 10 2 / 6 6 D 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 0 . 2 1 Re 51 . 2 7 . 3 / log 0 . 2 1 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 . 0 2 2 2 3 2 V m D L f P 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 P V W W Therefore, the pumps will consume 5118 kW of electric power to overcome friction and maintain flow.

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