) Water flows steadily up the vertical-1-in-diameter pipe and out the nozzle, which is 0.5 in. in diameter, discharging to atmospheric pressure. The stream velocity at the nozzle exit must be 30 f t/s. Calculate the minimum gage pressure required at the section (1). If the device were inverted, what would be the required minimum pressure at section (1) to maintain the nozzle exit velocity at 30 f t/s.

a) 9.995 psi

b) 1.3725 psi

Explanation:

Given:-

- The diameter of the pipe at inlet, d1 = 1 in

- The diameter of the pipe at exit, de = 0.5 in

- The exit velocity, Ve = 30 ft/s

- The exit discharge pressure Pe = 0 ( gauge )

- The density of water ρ = 1.940 slugs/ft3

Find:-

Calculate the minimum gage pressure required at the section (1)

Solution:-

- The mass flow rate m ( flow ) for the fluid remains constant via the continuity equation applies for all steady state fluid conditions.

                 m  ( flow ) = ρ*An*Vn = constant

Where,

           An: the area of nth section

           Vn: the velocity at nth section

- Consider the point ( 1 ) and exit point. Determine the velocity at point ( 1 ) via continuity equation.

- The cross sectional area of the pipe at nth point is given by:

                 An = π*dn^2 / 4

- The continuity equation becomes:

                 ρ*A1*V1 = ρ*Ae*Ve

Note: Water is assumed as incompressible fluid; hence, density remains constant.

                V1 = ( Ae / A1 ) * Ve

                V1 = ( (π*de^2 / 4 ) / (π*d1^2 / 4) ) *Ve

                V1 = ( de / d1 ) ^2 * Ve

                V1 = ( 0.5 / 1 )^2 * 30

                V1 =7.5 ft/s

- The required velocity at section ( 1 ) is V1 = 7.5 ft/s.

- Apply the bernoulli's principle for the point ( 1 ) and exit. Assuming the frictional losses are minimal.

        P1 + 0.5*ρ*V1^2 + ρ*g*h1 = Pe + 0.5*ρ*Ve^2 + ρ*g*he

- We will set " h1 " as datum; hence, h1 = 0. The elevation of exit nozzle from point (1) is at he = 10 ft.

- The bernoulli's equation expressed above is in " gauge pressure ". So the gauge pressure of exit Pe = 0 ( Patm ).

Therefore the simplified equation becomes:

        P1 + 0.5*ρ*V1^2  =  0.5*ρ*Ve^2 + ρ*g*he

        P1 = 0.5*ρ* ( Ve^2 - V1^2 ) + ρ*g*he

        P1  = 0.5*1.940*( 30^2 - 7.5^2 ) + 1.940*32*10

        P1 = 818.4375 + 620.8

        P1 = 1439.2375 lbf / ft^2 = 9.995 psi

- If the device is inverted then the velocity at the inlet " V1 " would remain same as there is no change in continuity equation - ( Diameters at each section remains same ).

- The only thing that changes is the application of bernoulli's equation as follows:

         P1 + 0.5*ρ*V1^2 + ρ*g*h1 = Pe + 0.5*ρ*Ve^2 + ρ*g*he

- We will set " he " as datum; hence, he = 0. The elevation of point ( 1 ) from exit nozzle is at h1 = 10 ft.

- The bernoulli's equation expressed above is in " gauge pressure ". So the gauge pressure of exit Pe = 0 ( Patm ).

Therefore the simplified equation becomes:

        P1 + 0.5*ρ*V1^2 + ρ*g*h1 =  0.5*ρ*Ve^2

        P1 = 0.5*ρ* ( Ve^2 - V1^2 ) - ρ*g*h1

        P1  = 0.5*1.940*( 30^2 - 7.5^2 ) - 1.940*32*10

        P1 = 818.4375 - 620.8

        P1 = 197.6375 lbf / ft^2 = 1.3725 psi

answer:

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