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FTFS Chap12 P075

# FTFS Chap12 P075 - Chapter 12 Bernoulli and Energy...

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Chapter 12 Bernoulli and Energy Equations Review Problems 12-75 Water discharges from the orifice at the bottom of a pressurized tank. The time it will take for half of the water in the tank to be discharged and the water level after 10 s are to be determined. Assumptions 1 The flow is uniform and incompressible, and the frictional effects are negligible. 2 The tank air pressure above the water level is maintained constant. Properties We take the density of water to be 1000 kg/m 3 . Analysis We take point 1 at the free surface of the tank, and point 2 at the exit of orifice. We take the positive direction of z to be upwards with reference level at the orifice ( z 2 = 0). Fluid at point 2 is open to the atmosphere (and thus P 2 = P atm ) and the velocity at the free surface is very low ( V 1 0). Then, / 2 2 2 2 2 gage , 1 1 2 2 2 1 1 2 2 2 2 1 2 1 1 P gz g g P z g P z g g P z g g P atm V V V V or, / 2 2 gage , 1 2 P gz V where z is the water height in the tank at any time t . Water surface moves down as the tank drains, and the value of z changes from H initially to 0 when the tank is emptied completely. We denote the diameter of the orifice by D , and the diameter of the tank by D o . The flow rate of water from the tank is obtained by multiplying the discharge velocity by the orifice cross-sectional area, / 2 2 4 gage , 1 2 2 orifice P gz D A V V Then the amount of water that flows through the orifice during a differential time interval dt is dt P gz D dt V dV / 2 2 4 gage , 1 2 (1) which, from conservation of mass, must be equal to the decrease in the volume of water in the tank, dz D dz A dV 4 ) ( 2 0 tank (2) where dz is the change in the water level in the tank during dt . (Note that dz is a negative quantity since the positive direction of z is upwards. Therefore, we used –dz to get a positive quantity for the amount of water discharged). Setting Eqs. (1) and (2) equal to each other and rearranging, / 2 2 1 4 / 2 2 4 gage , 1 2 2 0 2 0 gage , 1 2 dz P gz D D dt dz D dt P gz D The last relation can be integrated since the variables are separated. Letting t f be the discharge time and integrating it from t = 0 when z = z 0 to t = t when z = z gives t D D g P g z g P g z gage gage 2 2 0 2 , 1 2 , 1 0 2 2 2 2 where 2 2 2 2 3 2 2 gage , 1 s 274 . 7 kN 1 m/s kg 1000 ) m/s 81 . 9 )( kg/m 1000 ( kN/m ) 100 450 ( 2 2 g P The time for half of the water in the tank to be discharged ( z = z 0 /2) is t 2 2 2 2 2 2 m) 2 ( m) 1 . 0 ( s 274 . 7 m/s 81 . 9 m) 5 . 1 ( 2 s 274 . 7 m/s 81 . 9 m) 3 ( 2 t = 22.0 s ( b ) Water level after 10s is s) 10 ( m) 2 ( m) 1 . 0 ( s 274 . 7 m/s 81 . 9 2 s 274 . 7 m/s 81 . 9 m) 3 ( 2 2 2 2 2 2 2 z z = 2.31 m Discussion Note that the discharging time is inversely proportional to the square of the orifice diameter.

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