Question

V1. A converging elbow (see to the right) turns water through an angle of 135 ∘

in a vertical plane. The flow cross sectional diameter is 400 mm at the elbow inlet, section (1) and 200 mm at the elbow outlet, section (2). The elbow flow passage volume is 0.2 m 3
between sections (1) and (2). The water volume flowrate is 0.4 m 3
/s and the elbow inlet and outlet pressures are 150kPa and 90kPa, respectively. The elbow mass is 12 kg. Calculate the horizontal and vertical anchoring forces required to hold the elbow in place.

Answer 1

To calculate the horizontal and vertical anchoring forces required to hold the elbow in place, we can use the principle of conservation of momentum. The horizontal force is zero since the elbow is at rest, but the vertical anchoring force can be calculated using the density of water and the volume flow rate.

Step-by-step explanation:

To calculate the horizontal and vertical anchoring forces required to hold the elbow in place, we can use the principle of conservation of momentum.

First, let's calculate the change in momentum of the water as it flows through the elbow. The initial momentum can be calculated as:

P1 = m1 * v1

Where m1 is the mass of the water at the inlet and v1 is its velocity. Similarly, the final momentum can be calculated as:

P2 = m2 * v2

Where m2 is the mass of the water at the outlet and v2 is its velocity. Since the elbow is at rest, the change in momentum is equal to zero:

P2 - P1 = 0

Substituting the values, we can find the velocities:

(m2 * v2) - (m1 * v1) = 0

We can re-arrange the equation to solve for v2:

v2 = (m1 * v1) / m2

Using the given values, we find that the velocity at the outlet is v2 = 2 * v1.

Now, let's calculate the horizontal and vertical anchoring forces. The horizontal force can be calculated using the equation:

Fhorizontal = m * ahorizontal

Where m is the mass of the water and ahorizontal is the acceleration in the horizontal direction. Since the elbow is at rest, the horizontal force is equal to zero:

Fhorizontal = 0

The vertical force can be calculated using the equation:

Fvertical = m * avertical

Where avertical is the acceleration in the vertical direction. Substituting the values, we find that:

Fvertical = m * g

Where g is the acceleration due to gravity. Using the given mass of the elbow, we can calculate the vertical anchoring force. Since we know the density of water, we can calculate its mass using the volume flow rate:

m = ρ * V = ρ * A * v

Substituting the values, we find that:

Fvertical = (ρ * A * v) * g

Using the given values, we find that the vertical anchoring force is Fvertical = 117.6 N.

Answer 2

The Horizontal anchoring force is
`(Fx): $15974.35 \mathrm{~N}$` and Vertical anchoring force is
`$(\mathrm{Fz}):-5600.56 \mathrm{~N}$`

Let's calculate the horizontal and vertical anchoring forces required to hold the elbow in place, step by step.

Step 1: Calculate the Cross-Sectional Areas at Inlet and Outlet

• The cross-sectional area A of a circle is calculated using the formula
  `$A=(\pi D^2)/(4)$`.
• 
  `$$\begin{aligned}& A_1=\frac{\pi *(0.4 \mathrm{~m})^2}{4} \\& A_2=\frac{\pi *(0.2 \mathrm{~m})^2}{4}\end{aligned}$$`
• 
  `$A_1=0.1257 \mathrm{~m}^2$` (Area at the elbow inlet)
• 
  `$A_2=0.0314 \mathrm{~m}^2$` (Area at the elbow outlet)

Step 2: Calculate the Velocities at Inlet and Outlet

• The velocity V is found using the volume flow rate Q and the cross-sectional area A, where
  `$V=(Q)/(A)$`.
• 
  `$$\begin{aligned}& V_1=\frac{0.4 \mathrm{~m}^3 / \mathrm{s}}{A_1} \\& V_2=\frac{0.4 \mathrm{~m}^3 / \mathrm{s}}{A_2}\end{aligned}$$`

• 
  `$V_1=3.183 \mathrm{~m} / \mathrm{s}$` (Velocity at the elbow inlet)
• 
  `$V_2=12.732 \mathrm{~m} / \mathrm{s}$` (Velocity at the elbow outlet)

Step 3: Calculate the Mass Flow Rate

• The mass flow rate
  `$\dot{m}$` is calculated using the density
  `$\rho$` of water and the volume flow rate Q, where
  `$\dot{m}=\rho Q$`.
• 
  `$$\dot{m}=1000 \mathrm{~kg} / \mathrm{m}^3 * 0.4 \mathrm{~m}^3 / \mathrm{s}$$`

• 
  `$\dot{m}=400 \mathrm{~kg} / \mathrm{s}$` (Mass flow rate of the water)

Step 4: Apply the Conservation of Momentum Equation

• The momentum change in the horizontal (x-axis) and vertical (z-axis) directions due to the flow is given by:
• 
  `$$\begin{aligned}& \Delta \text { Momentum }_x=\dot{m} *\left(V_(2 x)-V_(1 x)\right) \\& \Delta \text { Momentum }_z=\dot{m} *\left(V_(2 z)-V_(1 z)\right)\end{aligned}$$`
• where
  `$V_(2 x)$` and
  `$V_(2 z)$` are the horizontal and vertical components of the outlet velocity, respectively, and
  `$V_(1 x)$` is the horizontal inlet velocity (since the inlet is horizontal,
  `$V_(1 z)=$`
  `0`.

• Momentum change in the x-direction:
  `$-4874.50 \mathrm{~N}$`

• Momentum change in the z-direction:
  `$-3601.27 \mathrm{~N}$`

Step 5: Calculate the Pressure Forces

• The pressure force
  `$F_{\text {pressure }}$` is the product of pressure P and cross-sectional area A.
• 
  `$$\begin{aligned}& F_(P 1)=P_1 * A_1 \\& F_(P 2 x)=P_2 * A_2 * \cos (\theta) \\& F_(P 2 z)=P_2 * A_2 * \sin (\theta)\end{aligned}$$`

• Pressure force at the inlet:
  `$F_(P 1)=18849.56 \mathrm{~N}$`
• Pressure force in the x-direction at the outlet:
  `$F_(P 2 x)=-1999.30 \mathrm{~N}$` (Negative due to the direction)
• Pressure force in the z-direction at the outlet:
  `$F_(P 2 z)=1999.30 \mathrm{~N}$`

Step 6: Solve for the Anchoring Forces

• The horizontal anchoring force
  `$F_x$` and the vertical anchoring force
  `$F_z$` are the sums of the pressure forces and the momentum changes in their respective directions:
• 
  `$$\begin{aligned}& F_x=\dot{m} *\left(V_(2 x)-V_1\right)+\left(F_(P 1)-F_(P 2 x)\right) \\& F_z=\dot{m} *\left(-V_(2 z)\right)-F_(P 2 z)\end{aligned}$$`

• Horizontal anchoring force (Fx):
  `$15974.35 \mathrm{~N}$`
• Vertical anchoring force (Fz):
  `$-5600.56 \mathrm{~N}$` (The negative sign indicates that the force is directed upwards)

These anchoring forces are required to hold the elbow in place against the momentum changes and pressure forces due to the flowing water.

The complete question is here:

A converging elbow shown in FIGURE Q3 turns water through an angle of a vertical plane. The flow cross section diameter is
`$400 \mathrm{~mm}$` at the elbow inlet, section (1), and
`$200 \mathrm{~mm}$` at the elbow outlet, section (2). The elbow flow passage volume is
`$0.2 \mathrm{~m}^3$` between sections (1) and (2). The water volume flowrate is
`$0.4 \mathrm{~m}^3 / \mathrm{s}$` and the elbow inlet and outlet pressures are
`$150 \mathrm{kPa}$` and
`$90 \mathrm{kPa}$`, respectively. The elbow mass is
`$12 \mathrm{~kg}$`.