Answer:
1.58 MPa
163 m
Explanation:
Given:
- Diameter @ inlet of nozzle D_i = 0.09 m
- Diameter @ exit of nozzle D_f = 0.03 m
- Flow rate of water through nozzle Q = 40 L /s
- Density of water p = 997 kg/m^3
Find:
a) Pressure drop in the nozzle (P_i - P_f)
- Compute Velocities at the two points V_i and V_f:
Q = A*V
40 / 1000 = pi*D_i^2 / 4 * V_i
V_i = (4*40) / (1000*pi*(0.09)^2)
V_i = 6.2876 m /s
V_f = (4*40) / (1000*pi*(0.03)^2)
V_i = 56.588 m /s
- Apply the Bernoulli's Equation:
(P_i - P_f) = 0.5*p* (V_f^2 - V_i^2)
dP = 0.5*997*(56.588^2 - 6.2876^2)
dP = 1.58 MPa
b) The maximum height the water can rise:
Assuming the water exits the nozzle as a free stream dP = 0
- Apply the Bernoulli's Equation:
p*g*h = 0.5*p* (V_f^2)
h = V^2_f / 2*g
h = (56.588)^2 / 2*9.81
h = 163 m
