Answer:
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
