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: Get the two cross-sectional areas.
At the diffuser inlet, $r_1 = 0.15$ m (same as the pipe), so $A_1 = \pi(0.15)^2 = 0.0707$ m$^2$. At the outlet, $x=0.5$ m gives $r_2 = 0.15+0.4(0.25) = 0.25$ m, so $A_2 = \pi(0.25)^2 = 0.1963$ m$^2$.

Step 2: Convert the flow rate and find both velocities.
$Q = 4.5$ m$^3$/min $= 0.075$ m$^3$/s. By continuity, $V_1 = Q/A_1 = 1.061$ m/s and $V_2 = Q/A_2 = 0.382$ m/s; the flow slows down as the area grows, which is exactly what a diffuser does.

Step 3: Get the inlet pressure from energy conservation.
With no losses and the outlet fixed at atmospheric gauge pressure, Bernoulli gives $p_1 = \tfrac12\rho(V_2^2-V_1^2)$. Numerically, $p_1 = 500(0.146-1.126) = -490$ Pa, below atmospheric, since pressure recovers as the flow decelerates through the diffuser.

Step 4: Draw a free body of the water and solve for the support force.
Forces on the water are the inlet pressure push $p_1A_1$, the outlet back push $-p_2A_2$ (zero here), and the force $F_R$ the mounting flange applies on the water through the diffuser walls. Newton's second law for steady flow gives $p_1A_1 + F_R = \dot m(V_2-V_1)$ with $\dot m = \rho Q = 75$ kg/s.
$\dot m(V_2-V_1) = 75(0.382-1.061) = -50.93$ N, and $p_1 A_1 = (-490)(0.0707) = -34.64$ N.
$F_R = -50.93 - (-34.64) = -16.29$ N.

Step 5: Flip the sign for the reaction on the flange.
The water pushes on the flange with the equal and opposite force, so the flange feels about $16.29$ N pushing outward along the flow direction.

Final Answer:
The diffuser passes this net force on to its mounting flange as it slows and pressurizes the water stream. \[ \boxed{F = 16.29 \ \text{N}} \]