3. (8 pts) Water (?=1000 kg/m2, ?=0.001 kg/ms) is discharged to atmosphere as in

Question

3. (8 pts) Water (?=1000 kg/m2, ?=0.001 kg/ms) is discharged to atmosphere as in the figure below. Determine H, i.e. the level of water in the tank, required to maintain a flow rate of 0.600 m/min. Along with the major frictional losses, consider also all minor/fitting losses. Pipe is smooth, globe valve is fully open, 90° elbows are standard elbows, use the velocity in smaller diameter pipe for hu in the case of a fitting between sections with different diameters, g-9.81 m/s) 3 m vaive D 5 cm loo m

Explanation / Answer

flowrate = 0.6 m3/min = .6/3600 = .00016 m3/s

Cross sectional area of pipe = 3.14/4*(Pipe diameter)^2

For 8 cm diameter pipe, Cross sectional area = 3.14/4*(8*10^-2)^2=.005 m2

Velocity in pipe = Flowrate/Cross sectional area

=> Velocity in 8 cm pipe = .00016/.005=.032 m/s

Reynold number, Re = Density*velocity*Diameter/viscosity

For Water,

Density = 1000 kg/m3

viscosity = 8.9*10^-4 Pa s

=> for 8 cm diameter pipe, Re = 1000*.032*8*10^-2/(8.9*10^-4)=2876

For smooth pipe, roughness = 0.12 mm

Relative roughness,RR = roughness/Pipe diameter

for 8 cm pipe, RR=.12*10^-3/(8*10^-2)=.0015

Using Moody's chart,

for 8 cm pipe, Darcy Friction factor, FF = 0.058

Pressure drop in pipe, Pd = FF*(.5*density*velocity^2*length of pipe)/pipe diameter

Length of 8 cm diamter pipe = 10 m

=> For 8 cm pipe, Pd = .058*(.5*1000*.032^2*10)/(8*10^-2)=3.7 Pa

Head loss, HL = Pressure drop/density/g

g=10 m2/s

=>  For 8 cm pipe, HL=3.7/1000/10=3.7*10^-4 m

For 5 cm diameter pipe, Cross sectional area = 3.14/4*(5*10^-2)^2=.0019 m2

Velocity in pipe = Flowrate/Cross sectional area

=> Velocity in 5 cm pipe = .00016/.0019=.084 m/s

Reynold number, Re = Density*velocity*Diameter/viscosity

=> for 5 cm diameter pipe, Re = 1000*.084*5*10^-2/(8.9*10^-4)=4719

For smooth pipe, roughness = 0.12 mm

Relative roughness,RR = roughness/Pipe diameter

for 5 cm pipe, RR=.12*10^-3/(5*10^-2)=.0024

Using Moody's chart,

for 5 cm pipe, Darcy Friction factor, FF = 0.06

Pressure drop in pipe, Pd = FF*(.5*density*velocity^2*length of pipe)/pipe diameter

for 90 degree elbow, Equivalent length = 30*Pipe diameter

In our case, two 90 degree elbows

Diameter = 5 cm = .05 m

=> equivalent length in elbows=2*30*.05=3 m

for Globe valve Equivalent length = 340*Pipe diameter

Diameter = 5 cm = .05 m

=> equivalent length for valve=340*.05=17 m

Length of 5 cm diamter pipe = 3+100+elbow equivalent length+Globe valve equivalent length=103+3+17 = 123 m

=> For 5 cm pipe, Pd = .06*(.5*1000*.084^2*123)/(5*10^-2)=520 Pa

Head loss, HL = Pressure drop/density/g

g=10 m2/s

=>  For 5 cm pipe, HL=520/1000/10=.052 m

Total friction loss, FL=Head loss in 5 cm diameter pipe+head loss in 8 cm diameter pipe

=>Total friction loss, FL= .052+3.7*10^-4=.05237 m

Effective Potential head for system = 10+H-4=6+H m

Kinetic head of system = .5*velocity^2/g=.5*.082^2/10=.00033 m

By Bernouli Theorem,

Potential head =kinetic head +friction loss

=>6+H=.05237+.00033

=> H is negative.

Hence for the given data, already available Potential head without considering H m is sufficient to overcome friction head loss.

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