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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 2245 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 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 2 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 to t = t when z = z gives t D D g P g z g P g z gage gage 2 2 2 , 1 2 , 1 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 . ( 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 . ( 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....
View Full Document

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.

Page1 / 11

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online