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+0.4x²

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. (Round off to two decimal places)

The correct answer is 16.29.

1. Parameters and Identification of Flow States
We are given the following properties of the system:

• Density of water, $\rho =1000{\text{kg/m}}^3$
• Inlet pipe radius, $r_1=15\text{cm}=0.15\text{m}$
• Volumetric flow rate, $Q=4.5{\text{m}}^3\text{/min}=\frac{4.5}{60}{\text{m}}^3\text{/s}=0.075{\text{m}}^3\text{/s}$
• Length of the coaxial diffuser, $L=0.5\text{m}$
• The inner radius of the diffuser as a function of axial distance $x$ from the inlet is given by:
  $r\left(x\right)=0.15+0.4x^2$

2. Computing Cross-sectional Areas and Velocities
At the inlet of the diffuser ($x=0$):
$r_1=0.15\text{m}$
$A_1=\pi r_1^2=\pi {\left(0.15\right)}^2=0.0225\pi {\text{m}}^2$
The inlet velocity is:
$V_1=\frac{Q}{A_1}=\frac{0.075}{0.0225\pi }=\frac{10}{3\pi }\text{m/s}\approx 1.061\text{m/s}$

At the outlet of the diffuser ($x=L=0.5\text{m}$):
$r_2=0.15+0.4{\left(0.5\right)}^2=0.15+0.10=0.25\text{m}$
$A_2=\pi r_2^2=\pi {\left(0.25\right)}^2=0.0625\pi {\text{m}}^2$
The outlet velocity is:
$V_2=\frac{Q}{A_2}=\frac{0.075}{0.0625\pi }=\frac{1.2}{\pi }\text{m/s}\approx 0.382\text{m/s}$

3. Determining Pressures via Bernoulli's Equation
Since the flow is horizontal and frictionless, we can apply Bernoulli's equation between the inlet (1) and outlet (2) sections:
$P_1+\frac{1}{2}\rho V_1^2=P_2+\frac{1}{2}\rho V_2^2$
Given that the outlet discharges to atmospheric condition without backflow, we take the gauge pressure at the outlet to be zero ($P_2=0$). This gives the gauge pressure at the inlet as:
$P_1=\frac{1}{2}\rho (V_2^2-V_1^2)$

4. Calculating the Force Balance
We apply the linear momentum equation in the horizontal x-direction to a control volume containing the water within the diffuser:
$\sum F_x=\dot{m}(V_2-V_1)$
Substituting the forces acting on the water control volume:
$P_1{A}_1-P_2{A}_2+R_x=\rho Q(V_2-V_1)$
where $R_x$ is the force exerted by the diffuser walls on the fluid. Using gauge pressures:
$R_x=\rho Q(V_2-V_1)-P_1{A}_1$
Substituting $P_1$ into this expression:
$R_x=\rho Q(V_2-V_1)-\frac{1}{2}\rho (V_2^2-V_1^2)A_1$

Now, substitute the exact values:
$\rho Q(V_2-V_1)=1000·0.075·(\frac{1.2}{\pi }-\frac{10}{3\pi })=75·\left(\frac{3.6-10}{3\pi }\right)=-\frac{160}{\pi }\text{N}$
$P_1{A}_1=\frac{1}{2}\left(1000\right)[{\left(\frac{1.2}{\pi }\right)}^2-{\left(\frac{10}{3\pi }\right)}^2]\left(0.0225\pi \right)$
$P_1{A}_1=500·(\frac{1.44}{{\pi }^2}-\frac{100}{9{\pi }^2})·0.0225\pi =11.25·\left(\frac{1.44-11.111}{\pi }\right)=-\frac{108.8}{\pi }\text{N}$
Thus, the force of the wall on the fluid is:
$R_x=-\frac{160}{\pi }-(-\frac{108.8}{\pi })=-\frac{51.2}{\pi }\text{N}\approx -16.297\text{N}$

5. Force on the Flange
By Newton's third law, the force exerted by the fluid on the diffuser is equal and opposite to $R_x$, which is transmitted to the flange. Therefore, the magnitude of the force exerted by the diffuser on the flange is:
$\left|F_{\text{flange}}\right|=16.297\text{N}$

Rounding to two decimal places, the magnitude is 16.29 N (or 16.30 N).