Question

10 peles A 3-m-diameter tank is initially filled with water 2 m above the center of a sharp-edged 10 cm diameter orifice. The tank water surface is open to the other, and the orice drains to the atmosphere through a 100-m-long pipe. The initial discharge velocity is 4 m/s. The friction coefficient of the can be when to be 0.015 and the effect of the energy correction the factor can be neglected. Determine (a) the initial velocity from the tank and to the time required to empty the tank. -996kg/m 30°C and ay - 1 Pipe Inter Reentrant: K = 0.80 (to Dand / -0.10 Sharp edged K = 0.50 0 Well rounded HD > 0.2 K = 0.03 Slightly rounded VID = 0.1% - 0.12 f 10 cm 100 4 ms Pump V V2 P1 + 01 29 P2 pg + 2y + hpump, u - Woumpu - VAP + 22 + Harbie e + ha – (5 + 2x) + 02 29 pg ("Note the first blank is value and the second blank is unit use this sign for raise to the power a) Find the mass flow rate: "A" for the unit) ('Note, the first blank is value and the second blank is unit) b) Find the useful pumping head:

Answer 1

Answer:

We will take the point 1 at the free surface of the tank and point 2 at the exit of the pipe. after this centerline is taken at the centerline of the orific where Z2=0. Positive direction of z is taken upwards

Fluid velocity at free surface is very low and almost

0

Also the fluid at both the points is open to atmosphere so in that case

P1 P2 Pa atm

$\frac{P_{1}}{\rho g}+\alpha _{1}\frac{V_{1}^{2}}{2g}+z_{1}+h_{pump,u}=\frac{P_{2}}{\rho g}+\alpha _{2}\frac{V_{2}^{2}}{2g}+z_{2}+h_{turbine}+h_{L}$

After putting the above mentioned conditions

$z_{1}+h_{pump,u}=\alpha _{2}\frac{V_{2}^{2}}{2g}+h_{L}..................1$

where Given

10

So from here the total head can be defined as

$h_{L}=h_{L,total}=h_{L,major}+h_{L,minor}=\left ( f\frac{L}{D}+\sum K_{L} \right )\frac{V^{2}}{2g}$

$h_{L}=\left ( f\frac{L}{D}+ K_{L} \right )\frac{V^{2}}{2g}..................2$

Since the dia of the piping system is constant and the initial velocity is given as 4 m/s so from here required mass flow rate and head can be calculated as

$\dot{m}=\rho A_{c}V_{2}$

$\dot{m}=\rho (\frac{\Pi D^{2}}{4})V_{2}$

$\dot{m}=996 kg/m^{3} (\frac{\Pi .1m^{2}}{4})4m/s$

$\dot{m}=31.3 Kg/m^{3}$ Answer

Now from equation 1 and 2

$h_{pump,u}=\left ( 1+f\frac{L}{D}+ K_{L} \right )\frac{V^{2}}{2g}-z_{1}$

$h_{pump,u}=\left (1+ (.015)\frac{100m}{.1m}+ .5 \right )\frac{4m/s^{2}}{2\times 9.81m/s^{2}}-2m$

Answer