58 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 m/s 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 needed to move the oil.
(a) The average flow velocity is calculated from the volume flow rate, i.e.,
so
(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 200 m length of smooth pipe of 15 mm 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 crosssectional 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 (socalled Couette flow), 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 (where the density and viscosity are = 888 kg/m and = 0.800 kg/m/s, respectively) is flowing steadily through a 6 cm diameter 40 m long pipe. The pressure at the pipe inlet and outlet are 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), so 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, and the expectation of laminar flow is verified, confirming the correct equation 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 airconditioning system of an aircraft is flowing through a smooth 0.12inch 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 units of “head of inches of water,” what is the minimum pump head would you need?
The pressure losses associated with the pipe flow are determined using the equation
where is the DarcyWelsbach 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, and this loss 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
In 1865, Van Syckle revolutionized petroleum oil delivery by using an oil pipeline. His oil company transported petroleum oil over a fivestatute milelong wrought iron pipe with a 2inch diameter. If the company delivered 40 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 = 0.045 mm for a wrought iron 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 larger 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
Worked Example #6
Air enters a 50 ft long part of an airconditioning duct made of smooth galvanized steel with a square crosssection 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/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 crosssection, so we first need to find the hydraulic diameter , which is
where is the crosssectional area, and is the perimeter of the section. We have a square crosssection 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, depends on the Reynolds number and pipe roughness, and its value will come from the Moody chart.
The average flow velocity can be found from the flow rate and crosssectional 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
(b) The power required to move the air through this part of the duct and overcome this pressure drop is
so
Worked Example #7
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 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 dynamic 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 . So, 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 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 121.3~, which would be a reasonably large pump.
Worked Example #8
Cooling water to a wind tunnel heat exchanger flows at a rate of 0.075 m/s through an asphalted castiron pipe 30 m long and 15 cm in diameter. Assume for water that its density = 1,000 kg/m and its kinematic viscosity = 10m/s, 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 #9
Consider a pump pushing water steadily through a section of cast iron pipe 200 ft long, as shown in the figure below. The rectangular crosssection 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.

 Determine the hydraulic diameter of the pipe.
 Determine the average flow velocity through the pipe.
 Determine the Reynolds number of the flow in the pipe.
 Is the flow in the pipe laminar or turbulent? Explain.
 Determine the friction factor of the pipe.
 Determine the pressure drop and head loss through the pipe.
1. The crosssectional 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 crosssectional 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. Some minor error in reading the chart is acceptable.
6. The pressure drop over a pipe length of 200 ft is
and inserting the known values gives
The corresponding head loss is