Question

Water with a density of 1000 kg/m³ comes out of an industrial condenser through a horizontal pipe of 15 cm radius at the flow rate of 4.5 m³/min. The outlet of the pipe is connected to a coaxial diffuser of 0.5 m length using a flange to raise the pressure of water to atmospheric condition without any backflow. The inner radius (\(r\), in m) of the diffuser cross-section is expressed as
\[ r = 0.15 + 0.4x^2 \]
where, \(x\) represents the axial distance of the diffuser in m, from its inlet. Considering frictionless flow, the magnitude of the force exerted by the diffuser on the flange is ________ N (rounded off to 2 decimal places).

Hint:

Apply Bernoulli's equation to find the inlet pressure, then use the momentum equation on the diffuser's fluid control volume.

Answer 1

Correct Answer: 16.3

Step 1: Find the diffuser's inlet and outlet areas.
The inlet radius matches the pipe, \(r_1 = 0.15\) m, so \(A_1 = \pi r_1^2 = \pi(0.15)^2 = 0.07069\) m\(^2\). At the outlet, \(x = 0.5\) m, so \(r_2 = 0.15 + 0.4(0.5)^2 = 0.15+0.1 = 0.25\) m, giving \(A_2 = \pi(0.25)^2 = 0.19635\) m\(^2\).

Step 2: Get the velocities from the flow rate.
\(Q = 4.5\) m\(^3\)/min \(= 0.075\) m\(^3\)/s. Then \(V_1 = Q/A_1 = 0.075/0.07069 = 1.061\) m/s and \(V_2 = Q/A_2 = 0.075/0.19635 = 0.382\) m/s.

Step 3: Use Bernoulli's equation to get the inlet gauge pressure.
Since the diffuser is frictionless and horizontal and the outlet reaches atmospheric pressure (\(p_2 = 0\) gauge), \(p_1 + \tfrac12\rho V_1^2 = p_2 + \tfrac12\rho V_2^2\), so \(p_1 = \tfrac12\rho(V_2^2-V_1^2) = 500(0.382^2-1.061^2) = 500(0.1459-1.1258) = -489.9\) Pa. The negative sign shows the inlet pressure sits below atmospheric, as expected since the diffuser is raising the pressure up to atmospheric.

Step 4: Apply the linear momentum equation on the water inside the diffuser.
Taking the flow direction as positive, the net force \(F_R\) that the flange applies on the fluid, plus the inlet pressure force, minus the outlet pressure force, equals the momentum change: \(p_1 A_1 - p_2 A_2 + F_R = \dot{m}(V_2-V_1)\), where \(\dot m = \rho Q = 1000(0.075) = 75\) kg/s.
\(\dot m (V_2-V_1) = 75(0.382-1.061) = -50.93\) N.
\(F_R = \dot m(V_2-V_1) - p_1A_1 + p_2A_2 = -50.93 - (-489.9)(0.07069) + 0 = -50.93+34.63 = -16.30\) N.

Step 5: Apply Newton's third law.
\(F_R\) is the force the flange applies on the fluid to hold the diffuser in place. By Newton's third law, the diffuser pushes back on the flange with the same magnitude in the opposite sense.

Final Answer:
The magnitude of the force the diffuser exerts on the flange comes straight from the momentum balance across the diverging section. \[ \boxed{F = 16.30 \ \text{N}} \]