EXERCISE 4 A closed conduit model of 14 scale is constructed

EXERCISE 4: A closed conduit model of 1/4 scale is constructed in the laboratory. The prototype has a diameter, D, of 100 mm and a length, L, of 100 m.

Determine the discharge of the model, for a flow of 10 lt×s^-1 in the prototype.

Determine the velocity of the flow in the model.

Determine the regimes of the flow in the prototype and model, respectively (u=2.5×10^-4 m²×s^-1).

Solution

discharge of a closed conduit = area * velocity

dia of the model = 100/4 = 25mm i.e 0.025m

length of the model= 100/4 = 25m

area of model = (2*3.14*0.025*25)/2 = 1.9625m²

volume of the model = (3.14*0.025²*25)/4 = 0.0122m³

discharge of the model = 10/4 = 2.5lt per sec

velocity of the flow = discharge / area

i.e (2.5*1000)( for m³)/ 1.96 = 1275m/sec

regime of the flow describes weather the flow is turbulant of laminar

this can be known by knowing the reynolds no.

re. no.of model= (density*velocity*Length)/Viscosity

i.e (1*1275*25)/ 2.5*10^-4

= 12750 that is > 4000

the flow is turbulant.

the prototype shall also be having turbulant flow