# 65 Worked Examples: Internal Flows

These worked examples have been fielded as homework problems or exam questions.

Worked Example #1

Oil with a density kg m and kinematic viscosity  m s, flows at 0.2 ms  through a 500 m length of 300 mm diameter cast iron pipe. The average roughness of the pipe’s surface is = 0.26 mm. Calculate (a) The average flow velocity in the pipe. (b) The Reynolds number of the flow and explain if the flow is laminar or turbulent. (c) The pressure drop and head loss along the pipe. (d) The minimum power of the pump that is needed to move the oil.

(a) The average flow velocity is calculated from the volume flow rate, i.e.,

so the average flow velocity is given by

(b) The Reynolds number of the flow in the pipe is

so the pipe flow will be turbulent because the Reynolds number is greater than 2,000 and we will need to use the Moody chart to find the friction factor .

(c) We now need the pressure drop and head loss along the pipe. To use the Moody chart, we also need the relative surface roughness, which is

From the Moody chart for a Reynolds number of 84,900 and a relative roughness of 0.00087 (using interpolation), we have . Therefore, the pressure loss over the length of the pipe is

and inserting the values gives

The corresponding head loss over this pipe is

(d) The pumping power required will be

Worked Example #2

You are tasked with the design of the fuel delivery system. The system requires a flow through a smooth pipe of 200 m in length and 15 mm in diameter. The required fuel flow rate is 125 kg hr. The fluid properties of the fuel are given as: 800 kg m and 0.00164 kg m s. All entrance effects can be disregarded. (a) What is the pressure drop along the length of the pipe? (b) What pressure capability (in terms of head) is required of the pump? (c) What are the pumping power requirements?

(a) The cross-sectional area of the pipe is

The mass flow rate is given as 125 kg hr = 0.0347~kg s, i.e.,

So, the average flow velocity in the pipe is

The Reynolds number based on pipe diameter is

Notice that this Reynolds number is the laminar regime, so the friction factor is given by

The pressure drop is given by

and inserting the values gives

(b) The equivalent head loss will be

(c) The pumping power can be determined from

Worked Example #3

Oil at 20 C, with density = 888.0 kg m and viscosity = 0.800 kg m s, respectively, is flowing steadily through a 6 cm diameter pipe that is 40 m long. The pressure at the inlet and outlet of the pipe is 745 kPa and 97 kPa, respectively. Determine the flow rate of oil through the pipe.

This pressure drop is a direct consequence of the action of viscous effects. The fluid is oil (dense and thick), so we expect the flow to be at a very low Reynolds number (i.e., laminar). One of the first goals in pipe flow problems is to find the Reynolds number to verify the laminar assumption or to show it is turbulent ( > 2,000). Based on the information provided in this problem, the only unknown is . From laminar pipe flow theory, we have that

Notice that for a given length of pipe, the pressure drop is proportional to the fluid’s viscosity and flow speed (so the faster we try to move a given fluid, the larger the pressure drop will be) but inversely proportional to the square of the pipe diameter. The average flow velocity is

We can now check to see that the flow is laminar by calculating the pipe Reynolds number, i.e.,

so it is very low, much less than 2,000, and the expectation of laminar flow is verified, confirming the correct equation (Eq. ??) for the pressure drop. The volume flow rate through the pipe will be

Substituting in the known values gives

The corresponding mass flow rate of oil will be

Worked Example #4

A coolant type of fluid used in the air-conditioning system of an aircraft is flowing through a smooth 0.12-inch diameter 30 ft long horizontal pipe steadily at an average flow speed of 3 ft s. The fluid has a temperature of 40F, = 1.93 slug ft, and = 3.32610 slug fts. Determine for this pipe flow: (a) The Reynolds number of the flow based on the pipe diameter and whether the flow is laminar or turbulent; (b) The pressure drop along the length of the pipe; (c) The pumping power required to overcome this pressure drop; (d) Because the pump manufacturer provides pumps measured in pressure units of “head of inches of water,” what is the minimum head needed?

The pressure losses associated with the pipe flow are determined using the equation

where is the Darcy-Welsbach friction factor. The friction factor depends on the flow Reynolds number and the roughness of the pipe material. Hence, the pressure loss in a pipe is also a function of the Reynolds number and the roughness of the pipe. Remember that the frictional effects of viscosity cause this pressure loss, which is irrecoverable and subsequently appears as heat.

(a) The Reynolds number of the flow can be calculated using the equation

(b) The pressure loss can be determined using the equation

where the friction factor for the laminar flow in a smooth pipe is

Therefore

and so

(c) The pumping power required to overcome this pressure drop is

where the volume flow rate is

Therefore

(d) The head loss can be determined using

So, we would likely need to specify a pump with a head capability of at least 185 .

Worked Example #5

Write MATLAB code to calculate the friction factor from the Colebrook-White equation.

title(‘Friction Factor vs Reynolds Number’);
function f = frictionFactor(Re, D, epsilon)
% Re: Reynolds number (can be an array)
% D: Pipe diameter
% epsilon: Pipe roughness
% Initial guess for friction factor
f0 = 0.01;
% Iterative solution using the Colebrook-White equation
error = 1e-6;
maxIterations = 100;
f = zeros(size(Re));
for i = 1:length(Re)
f_current = f0;
for iter = 1:maxIterations
leftSide = 1 / sqrt(f_current);
rightSide = -2 * log10((epsilon / D) / 3.7 + 2.51 / (Re(i) * sqrt(f_current)));
f_new = 1 / (leftSide + rightSide^2);
if abs(f_new – f_current) < error
f(i) = f_new;
break;
end
f_current = f_new;
end
end
end

Worked Example #6

In 1865, Van Syckle revolutionized petroleum oil delivery by using an iron oil pipeline. His company transported petroleum oil over a five-statute mile-long pipe with a 4-inch diameter. If the company delivered 200 barrels per hour, what is the minimum power (in horsepower) required to pump the oil along the full length of the pipe? Assume = 1.6046 slug ft and = slug fts. The roughness = 0.09 mm for the pipe. Assume 1 barrel = 5.615 ft.

We must find the pressure loss to calculate the power. We need to know whether the flow is turbulent or laminar, and so we need to know the Reynolds number, i.e.,

For the above equation, the average flow velocity can be calculated based on the flow rate of the oil. The flow rate of the oil is

Therefore,

Now the Reynolds number can be calculated, i.e.,

This value of the Reynolds number is greater than 2,000. Hence, the flow is turbulent. For a turbulent flow, the pressure loss can be determined using the equation

Here, the friction factor is a function of relative roughness and Reynolds number. For the wrought iron pipe, the relative roughness

Consulting the Moody chart for and , we can find the friction factor, which is about . The pressure drop in the pipe can now be calculated, i.e.,

and inserting the values give

The required power to overcome this pressure loss is

and inserting the values gives

Worked Example #7

Air enters a 50 ft long part of an air-conditioning duct made of smooth galvanized steel with a square cross-section of side 2 ft. The equivalent roughness of the duct is = 0.00015 ft. The air is pushed through the duct with a fan at a volume flow rate of 1,800 ft per minute. (a) Determine the pressure drop and head loss along this part of the duct. (b) Determine the power needed to overcome the pressure losses over this part of the duct. Hints: 1. Disregard all entrance effects. 2. Consult the Moody chart.

(a) We are dealing with a duct that has a square cross-section, so we first need to find the hydraulic diameter , which is

where is the cross-sectional area, and is the perimeter of the section. We have a square cross-section of side , so the equivalent hydraulic diameter is

We are given a hint to consult the Moody chart, so the flow through the pipe will likely be turbulent. The equation for the pressure loss through a pipe is

where is the friction factor, which is still to be determined. Its value depends on the Reynolds number and pipe roughness found in the Moody chart.

The average flow velocity can be found from the flow rate and cross-sectional area. The pipe area is

So, the average flow velocity is

To use the Moody chart, we need the corresponding Reynolds number, which is

where we find the density and viscosity of air at ISA conditions.

Because , the flow through the duct will be fully turbulent. We also need the relative pipe roughness, which is

Therefore, from the Moody chart, the friction factor is

Returning to the calculation of the pressure drop, then

which corresponds to a head loss of

(b) The power required to move the air through this part of the duct and overcome this pressure drop is

so

which is a minimal amount of power.

Worked Example #8

You are asked to design an irrigation system in which water is pumped through a long horizontal pipe made of smooth plastic tubing. The pipe is 150 m long and has a diameter of 10 cm. The pump output must provide a maximum volume flow rate of 0.09 m s. For satisfactory operation, the sprinklers along the entire length of the pipe must operate at 205.0 kPa or higher pressure. (a) Find the pressure drop along the length of the pipe. (b) Calculate the minimum pressure that needs to be produced at the pump. (c) What pressure capability for the pump (in terms of head) would you specify? Notes: Assume turbulent flow. The density of water = 1,000 kg m and the viscosity of water kg m s.

(a) The volume flow rate, , is given by

The average flow velocity can be calculated using

Calculating the Reynolds number gives

For a smooth pipe the roughness factor, mm and . Obtaining the friction factor from the Moody chart gives . The pressure drop along the pipe can be calculated using

and substituting values gives

(b) The pressure at the pump located at point 1 is then the required pressure at the last sprinkler plus the pressure drop

Substituting values in the above equation (notice that both and are expressed in kPa) gives

(c) The corresponding head loss can be determined using

So, we would likely need to specify a pump with a head capability of at least 122, which would be a reasonably large pump.

Worked Example #9

Cooling water to a wind tunnel heat exchanger flows at a rate of 0.075 ms through an asphalted cast-iron pipe 30 m long and 15 cm in diameter. Assume for water that its density = 1,000 kg m and its kinematic viscosity = 10ms, and the pipe material has an equivalent roughness = 0.12 mm. Hint: Consult the Moody Chart. (a) Using the basic principles of pipe flows, show how to calculate the pressure loss and head loss (i.e., frictional losses) along the length of the pipe. (b) Determine the pumping power required.

(a) The general equation for the pressure drop is

where is the friction factor, which depends on the Reynolds number and pipe roughness. The average flow velocity can be found from the flow rate and pipe area. The pipe area is

so the average flow velocity is

To use the Moody chart, we need the Reynolds number, which is

We also need the relative roughness, which is

Therefore, from the Moody chart, it can be found that

Returning to the calculation of the pressure drop, then

The corresponding head loss can be determined using

(b) The pumping power required to overcome this pressure drop is

where the volume flow rate is

Therefore

Worked Example #10

Consider a pump pushing water steadily through a section of cast iron pipe 200 ft long, as shown in the figure below. The rectangular cross-section of the pipe has a height 3 in and a width 6 in. The internal surface roughness is 0.006 inches. The volumetric flow rate through the pipe is 0.75 ft s. Assume that the density of the water is 1.940 slug ft and its viscosity is 2.1 slug ft s. Also, assume that entrance effects can be ignored.

1. Determine the hydraulic diameter of the pipe.
2. Determine the average flow velocity through the pipe.
3. Determine the Reynolds number of the flow in the pipe.
4. Is the flow in the pipe laminar or turbulent? Explain.
5. Determine the pipe’s friction factor .
6. Determine the pressure drop and head loss through the pipe.

1. The cross-sectional area of the rectangular pipe is

so the hydraulic diameter is

2. The average flow velocity in the pipe is determined from the volumetric flow rate and the cross-sectional area, i.e.,

3. The Reynolds number based on the hydraulic diameter is

4. Because the Reynolds number is greater than 2,000, the flow in the pipe will be turbulent.

5. The relative surface roughness is

Using the Moody chart, the friction factor for this Reynolds number and relative roughness is about 0.023. Minor errors in reading the chart are acceptable.

6. The pressure drop over a pipe length of 200 ft is

Inserting the known values gives

Worked Example #11

RP-1 rocket fuel will be transferred from one tank to another using a pump with a flow rate of = 52.1 ft per minute. The fuel passes through a 4-inch diameter steel pipe with a length, , of 40 ft. The pipe has a roughness, = 0.002 inches. Assume the density of RP-1 is 1.746 slug ft and its viscosity is slug fts. How much pumping power is required to move the RP-1 along the pipe?

The average flow velocity in the pipe is

The Reynolds number based on the diameter is

This result means the pipe flow will be turbulent, so we must refer to the Moody chart to find the friction factor. The relative surface roughness is

Using the Moody chart, the friction factor is about 0.035. Some minor errors in reading the chart are acceptable. The pressure drop over the 40 ft length of pipe is

and inserting the values gives

The corresponding pumping power required is

Worked Example #12

A liquid rocket fuel system utilizes a high-strength nickel alloy pipe with a diameter of cm to transport liquid oxygen (LOX) from the tank to the engine. The LOX has a density kg m and a viscosity Pa s. The mass flow rate of the LOX is kg s. Assume that the average roughness of the pipe is 0.04 mm. How much pumping power is required to move the LOX along the pipe?

The average flow velocity in the pipe is

The Reynolds number of this flow (based on pipe diameter) is

This Reynolds number is significantly greater than 2,000, so the flow is in the turbulent regime. The relative roughness is

Referring to the Moody chart for = 3,270,000 and = 0.001, then 0.02.
The pressure drop along the pipe is

Substituting the numerical values gives

Therefore, the power required to pump this flow is