Question

Water flow steadily from a large closed tank. The deflection in mercury manometer is 1 inches or ( 0.0254-m) and viscous effects are negligible.

1). If the tank of water is maintained at 8.50 c(celsius) and the ambient air temperature is 34 c (celsius), determine the volume flow rate in (m^3/s) leaving the 3-inch or (0.0762-m) diameter outlet from the tank, the pressure differential (delta P) between the pitot tube entrance (across from the discharge outlet) & the fluid pressure in the 1-ft or (0.3048-m) diameter discharge chamber in (kpa), and the gauge pressure of the air in (kpag) in the space above the water in the tank

2). A nozzle with a throat diameter of 3.850-mm is placed on the tank discharge to slow down the outlet flow velocity to 11 nanoliters/minute (nL/min) with a nozzle exit pressure of 3.4 kpa, determine the new pressure above the tank water in (kpa) and the new mass flow rate in (kg/s) of the water.

3). For part 2, estimate the new mercury deflection for the pressure differential (delta P) between the pitot tube entrance (across from the discharge outlet) & the fluid pressure in the 1-ft or (0.3048-m) diameter discharge chamber in (kpa)

Answer 1

1. In the question it is given that,

Diameter of the outlet, D_o= 0.0762 m

Therefore, crosssectional area of the outlet will be,

=>

Now, Velocity at the outlet, ...............................(1)

Where = Head difference of manometer=

=manometer deflection= 0.0254 m

= density of Hg= 13600 kg/m³

= density of water= 1000 kg/m³

also, g is taken as 9.8 m/s²

therefore, (1)=>

Volumetric flow rate will be,

=>

Also for the manometer we know that

=>

=>

=>

where, Pressure difference between the pitot tube entrance and fluid pressure in discharge chamber.

Now let the gauge pressure of air above the water level is 1 atm=101.325 kPa