Question

The fresh water intake in a reservoir has a cross-sectional area of 0.800 m² and receives water flowing at a speed of 0.500 m/s. At the outlet, a vertical distance of 200 m below the intake, the cross-sectional area is 0.020 m². Calculate the pressure difference between the inlet and outlet.

Answer 1

Area of cross section at inlet = A_i=0.8 m²

velocity of flow at inlet=V_i=0.5 m/s

Flow rate =Q= 0.8*0.5=0.4 m³/s

cross sectional area at outlet =A_o= 0.02 m²

velocity of flow at outlet = V_o=0.4/0.02=20 m/s

Let elevation of inlet=h_i

Elevation of outlet =h_i-200

Let pressure at inlet be P_i and pressure at outlet be P_o

Let us apply Bernoulli's equation to determine the pressure difference between inlet and outlet

P_i+gh_i+0.5v_i² = P_o+gh_o+0.5v_o²

Or, P_i+1000*9.8*h_i+0.5*1000*0.5² = P_o+1000*9.8*(h_i-200)+0.5*1000*20²

P_i-P_o =1000*9.8*(h_i-200)+0.5*1000*20² -  (1000*9.8*h_i+0.5*1000*0.5² )

= -1760125 N/m²

Pressure at inlet is lower than pressure at outlet and the pressure difference is 1760125 N/m²