Figure 8.9 shows an industrial storage tank from which a horizontal 100 m long pipe carries water at 25°C to a process where a bulk food product is being prepared. The pipe is a DN 50 Schedule 40 steel pipe and the design delivery rate to the process is 520 L/min. Determine the amount of pressure drop that occurs in the pipe from the storage tank to the processing system.

First, lay out a plan for completing this problem.

Question

Figure 8.9 shows an industrial storage tank from which a horizontal 100 m long pipe carries water at 25°C to a process where a bulk food product is being prepared. The pipe is a DN 50 Schedule 40 steel pipe and the design delivery rate to the process is 520 L/min. Determine the amount of pressure drop that occurs in the pipe from the storage tank to the processing system.

First, lay out a plan for completing this problem.

Accepted Answer

Here is one approach:

1. Define point A in the pipe where it exits the storage tank and point B where the tank delivers the water to the processing system. The objective of the problem is to calculate [latex] p_A - p_B [/latex], the pressure drop between points A and B.

2. Use the energy equation to determine the pressure drop, considering the energy loss due to friction in the pipe.

3. Compute the energy loss using Darcy’s equation.

Write the energy equation between points A and B now, and solve algebraically for the pressure drop.

The energy equation is:

[latex] p_A/\gamma +Z_A +v^2_A/2g -h_L = p_B/\gamma +Z_B +v^2_B/2g[/latex]

Notice that [latex] Z_A = Z_B [/latex] and [latex] v_A = v_B [/latex] and therefore those terms can be cancelled from the equation. Now we can solve for the pressure drop.

[latex] p_A - p_B = \gamma h_L[/latex]

How can you calculate the energy loss due to friction in the pipe?

Darcy’s equation can be used:

[latex] h_L = f imes \frac{L}{D} imes \frac{v^2}{2ga} [/latex]

Determine the data needed to complete the calculation.

From the given data, we can show that L = 100 m, Q = 520 L/min, the pipe is a DN 50 Schedule 40 steel pipe, [latex]g = 9.81 m/s^2 [/latex], and the fluid is water at 25°C for which Appendix A gives

[latex]\gamma = 9.78 kN/m^3[/latex] and the kinematic viscosity is, [latex] u = 8.94 imes 10^{-7} m^2/s[/latex]

For a DN 50 steel pipe, from Appendix F, D = 52.5 mm = 0.0525 m, and [latex] A = 2.168 imes 10^{-3} m^2 [/latex] .

Now the velocity of flow can be computed.

[latex] v = \frac{Q}{A} = \frac{520 L/min}{2.168 imes 10^{-3} m^2} imes \frac{1.0m^3/s}{60000L/min} =4.00m/s [/latex]

To determine the friction factor, f, we need to compute the Reynolds number.

[latex] N_R = \frac{vD}{ u} = \frac{\left(4.00m/s ight)\left(0.0525m ight) }{8.94 imes 10^{-7}m^2/s} = 2.349 imes 10^5[/latex]

Then the flow is turbulent and we use the Moody diagram to find f. From Table 8.2 we find the roughness for a clean steel pipe to be [latex] \epsilon = 4.6 imes 10^{-5} m[/latex]

TABLE 8.2 Pipe roughness—design values

Material	Roughness [latex]\epsilon[/latex] (m)	Roughness [latex]\epsilon[/latex] (ft)
Glass	Smooth	Smooth
Plastic	[latex]3.0 imes 10^{-7}[/latex]	[latex]1.0 imes 10^{-6}[/latex]
Drawn tubing; copper, brass, steel	[latex]1.5 imes 10^{-6}[/latex]	[latex]5.0 imes 10^{-6}[/latex]
Steel, commercial or welded	[latex]4.6 imes 10^{-5}[/latex]	[latex]1.5 imes 10^{-4}[/latex]
Galvanized iron	[latex]1.5 imes 10^{-4}[/latex]	[latex]5.0 imes 10^{-4}[/latex]
Ductile iron—coated	[latex]1.2 imes 10^{-4}[/latex]	[latex]4.0 imes 10^{-4}[/latex]
Ductile iron—uncoated	[latex]2.4 imes 10^{-4}[/latex]	[latex]8.0 imes 10^{-4}[/latex]
Concrete, well made	[latex]1.2 imes 10^{-4}[/latex]	[latex]4.0 imes 10^{-4}[/latex]
Riveted steel	[latex]1.8 imes 10^{-3}[/latex]	[latex]6.0 imes 10^{-3}[/latex]

From Moody’s diagram, we can read f = 0.0203. Complete the calculation for [latex] h_L [/latex] now.

Here is the result:

[latex] h_L = f imes \frac{L}{D} imes \frac{v^2}{2g} = 0.0203 imes \frac{100}{0.0525} imes \frac{\left(4.00 ight)^2 }{2\left(9.81 ight) } =31.53 m [/latex]

Now complete the calculation for the pressure drop in the pipe.

You should get [latex] p_A - p_B = 308.4kPa .[/latex] Here are the details:

[latex] p_A - p_B = \gamma h_L = \left(9.78 kN/m^3 ight) \left(31.53 ight) = 308.4 kN/m^2 = 308.4kPa [/latex]

This example problem is completed.

Question 8.8: Figure 8.9 shows an industrial storage tank from which a hor...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions