Worked Examples: Internal Flows

J. Gordon Leishman

83 Worked Examples: Internal Flows

Many of these worked examples have been fielded as homework problems or exam questions.

Worked Example #1

Oil with a density $\varrho_{\rm oil} = 900.0$ kg m ${^{-3}}$ and kinematic viscosity $\nu_{\rm oil}$ = $10^{-5}$ m ${^{2}}$ s $^{-1}$ flows at 0.2 m ${^3}$ s $^{-1}$ through a 500 m length of 300 mm diameter cast iron pipe. The average roughness of the pipe’s surface is $\epsilon$ = 0.26 mm. Calculate: (a) The average flow velocity in the pipe. (b) Determine the Reynolds number and explain whether the flow is laminar or turbulent. (c) The pressure drop and head loss along the pipe. (d) The minimum power required by the pump to move the oil.

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

$V_{\rm av} = \frac{Q}{A}$

so the average flow velocity is given by

$V_{\rm av} = \frac{4Q}{\pi d^2} = \frac{4 \times 0.2}{\pi \times 0.3^2} = 2.83~\mbox{m s$^{-1}$}$

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

$Re_d = \frac{\varrho_{\rm oil} \, V_{\rm av} \, d}{\mu_{\rm oil} } =\frac{V_{\rm av} \, d}{\nu_{\rm oil}} = \frac{2.83\times 0.3 }{10^{-5}} = 84,900$

The pipe flow is turbulent because the Reynolds number exceeds 2,000, so we use the Moody chart to determine the friction factor ${\scriptstyle{f}}$ .

(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

$\frac{\epsilon}{d} = \frac{0.26}{300} = 0.00087$

From the Moody chart for a Reynolds number of 84,900 and a relative roughness of 0.00087 (using interpolation), we have $\scriptstyle f$ $\approx 0.022$ . Therefore, the pressure loss over the length of the pipe is

$\Delta p = \frac{1}{2} \varrho_{\rm oil} V_{\rm av}^2 f \left( \frac{L}{d}\right)$

and inserting the values gives

$\Delta p = 0.5\times 900.0\times 2.83^2 \times 0.022\times \left( \frac{500.0}{0.3}\right) = 132.15~\mbox{kPa}$

The corresponding head loss over this pipe is

$h = \frac{\Delta p}{\varrho_{\rm oil} \, g} = \frac{132.15\times 10^3}{900\times 9.81} = 14.97~\mbox{m}$

(d) The minimum hydraulic power delivered to the fluid by an ideal pump is

$P_{\rm hyd} = Q \, \Delta p = 0.2 \times 132.15 \times 10^3 = 26.43~\mbox{kW}$

For a real pump with efficiency $\eta_p$ , the required input power would be

$P_{\rm input} = \frac{P_{\rm hyd}}{\eta_p}$

Worked Example #2

You are tasked with designing the fuel delivery system. The system requires a flow through a smooth pipe 200 m long and 15 mm in diameter. The required fuel flow rate is 125 kg hr $^{-1}$ . The fluid properties of the fuel are given as: $\varrho =$ 800 kg m ${^{-3}}$ and $\mu =$ 0.00164 kg m $^{-1}$ s $^{-1}$ . All entrance effects can be disregarded. (a) What is the pressure drop along 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

$A_c = \frac{\pi D^2}{4} = \frac{\pi \times 0.015^2}{4} = 0.0001767~\mbox{m${^{2}}$}$

The mass flow rate $\overbigdot{m}$ is given as 125 kg hr $^{-1}$ = 0.0347~kg s $^{-1}$ , i.e.,

$\overbigdot{m} = \varrho \, A_c V_{\rm av} = 0.0347~\mbox{kg s$^{-1}$}$

So, the average flow velocity in the pipe is

$V_{\rm av} = \frac{\overbigdot{m}}{\varrho \, A_c} = \frac{0.0347}{800.0 \times 0.0001767} = 0.246~\mbox{m s$^{-1}$}$

The Reynolds number based on pipe diameter is

$Re = \frac{\varrho V_{\rm av} \, D}{\mu} = \frac{ 800.0 \times 0.246 \times 0.015}{0.00164} = 1,797$

Notice that this Reynolds number is in the laminar regime, so the friction factor $\scriptstyle f$ is given by

$f = \frac{64}{Re} = \frac{64}{1,797} = 0.0356$

The pressure drop $\Delta p$ is given by

$\Delta p = \frac{1}{2} \varrho V_{\rm av}^2 \, f \left( \frac{L}{d}\right)$

and inserting the values gives

$\Delta p = \frac{1}{2} \times 800.0 \times 0.246^2 \times 0.0356 \times \left( \frac{200.0}{0.015} \right) = 11,457.4~\mbox{Pa} = 11.46~\mbox{kPa}$

(b) The equivalent head loss ${h}$ will be

$h = \frac{\Delta p}{\varrho \, g} = \frac{11,457.4}{800.0 \times 9.81} = 1.46~\mbox{m}$

(c) The pumping power $P$ can be determined from

$P = \frac{\overbigdot{m} \, \Delta p}{\varrho} = \frac{0.0347 \times 11,457.4}{800.0} = 0.497~\mbox{W}$

Worked Example #3

Oil at 20 $^{\circ}$ C, with density $\varrho$ = 888.0 kg m ${^{-3}}$ and viscosity $\mu$ = 0.800 kg m $^{-1}$ s $^{-1}$ , respectively, is flowing steadily through a 6 cm diameter pipe that is 40 m long. The pressures at the pipe inlet and outlet are 745 kPa and 97 kPa, respectively. Determine the oil flow rate through the pipe.

This pressure drop is a direct consequence of viscous effects. The fluid is oil (dense and thick), so we expect a very low Reynolds number (i.e., laminar) flow. One of the first goals in pipe flow problems is to determine the Reynolds number to verify the laminar assumption or to show that the flow is turbulent ( $Re$ > 2,000). Based on the information provided, the only unknown is $V_{\rm av}$ . From laminar pipe flow theory, we have that

$\Delta p = p_1 - p_2 = \frac{8 \mu \, L \, V_{\rm av}}{R^2} = \frac{32 \, \mu \, L \, V_{\rm av}}{D^2}$

Notice that for a given pipe length, 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

$V_{\rm av} = \frac{\left( p_1 - p_2 \right) D^2}{32 \, \mu \, L} = \frac{\left(745.0 - 97.0 \right) \times10^3 \times (0.06)^2}{32 \times 0.8 \times 40} = 2.28~\mbox{m s$^{-1}$}$

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

$Re = \frac{\varrho V_{\rm av} \, D}{\mu} = \frac{888.0 \times 2.28 \times 0.06}{0.80} = 151.8$

It is very low, well below 2,000, and the expected laminar-flow condition is verified, confirming that the laminar pressure-drop equation used above is applicable. The volume flow rate through the pipe will be

$Q = A \, V_{\rm av} = \frac{\pi D^2}{4} V_{\rm av} = \frac{\pi \left( p_1 - p_2 \right) D^4}{128 \, \mu \, L}$

Substituting the known values gives

$Q = \frac{\pi \left( p_1 - p_2 \right) D^4}{128 \, \mu \, L} = \frac{\pi \left(745.0 - 97.0 \right) \times10^3 \times (0.06)^4}{128 \times 0.8 \times 40.0} = 0.00644~\mbox{m${^3}$ s$^{-1}$}$

The corresponding mass flow rate of oil will be

$\overbigdot{m} = \varrho \, Q = 888.0 \times 0.00644 = 5.72~\mbox{kg s$^{-1}$}$

Worked Example #4

A coolant-type fluid in an aircraft air-conditioning system flows steadily through a smooth, 0.12-inch-diameter, 30-ft-long horizontal pipe at an average speed of 3 ft s $^{-1}$ . The fluid has a temperature of 40 $^{\circ}$ F, $\varrho$ = 1.93 slug ft ${^{-3}}$ , and $\mu$ = 3.326 $\times$ 10 $^{-5}$ slug ft $^{-1}$ s $^{-1}$ . 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

$\Delta p_L = \frac{1}{2}\varrho V_{\rm av}^2 \, f \left( \frac{L}{D} \right)$

where ${\scriptstyle{f}}$ is the Darcy-Welsbach friction factor. The friction factor ${\scriptstyle{f}}$ depends on the flow Reynolds number and the pipe material’s roughness. Thus, pressure loss in a pipe also depends on the Reynolds number and the pipe’s roughness. Viscous frictional effects cause this pressure loss, which is irrecoverable and appears as heat.

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

$Re = \frac{\varrho V_{\rm av} \, D}{\mu} = \frac{1.93 \times 3.0 \times 0.12/12}{3.326 \times 10^{-5}} = 1,740 < 2,300, \mbox{i.e., laminar flow}$

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

$\Delta p_L = \frac{1}{2}\varrho V_{\rm av}^2 \, f \left( \frac{L}{D} \right)$

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

$f = \frac{64}{Re}$

Therefore

$\Delta p_L = \frac{1}{2}\varrho V_{\rm av}^2 \frac{64}{Re} \left( \frac{L}{D} \right) = 32\varrho V_{\rm av}^2 \left( \frac{L}{D} \right) \frac{\mu}{\varrho V_{\rm av}D} = \frac{32\mu V_{\rm av} L}{D^2}$

and so

$\Delta p_L = \frac{32 \times (3.326 \times 10^{-5}) \times 3.0 \times 30.0}{(0.01)^2} = 957.89 \mbox{~lb ft$^{-2}$}$

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

$P = Q \, \Delta p_L$

where the volume flow rate is

$Q = A V_{\rm av} = \frac{\pi}{4}D^2V_{\rm av} = \frac{\pi(0.01)^2 \times 3}{4} = 2.356 \times 10^{-4}~\mbox{ft${^3}$ s$^{-1}$}$

Therefore

$P = 2.356 \times 10^{-4} \times 957.89 = 0.2257~\mbox{ft-lb s$^{-1}$}$

(d) The head loss expressed as inches of water can be determined using

$h_L = \frac{\Delta p_L}{\varrho_{\rm water}g} = \frac{957.89}{1.94 \times 32.17} = 15.37~\mbox{ft} = 184.4~\mbox{inches~of~H}_2\text{O}$

So, we would likely need to specify a pump with a head capability of at least 185 $\mbox{in H}_2\text{O}$ .

Worked Example #5

Write MATLAB code to calculate the friction factor using the Colebrook-White equation and plot it versus the Reynolds number.

The following MATLAB code evaluates the Darcy friction factor for turbulent pipe flow using the Colebrook-White equation and plots it as a function of the Reynolds number for a specified relative roughness.

clear; clc;

% Reynolds number range
Re = logspace(4, 7, 100);

% Pipe diameter and roughness
D = 0.10;          % Pipe diameter, m
epsilon = 0.00005; % Pipe roughness, m

% Calculate friction factor
f = frictionFactor(Re, D, epsilon);

% Plot the result
figure;
semilogx(Re, f, 'LineWidth', 1.5);
grid on;
xlabel('Reynolds number, Re');
ylabel('Darcy friction factor, f');
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
        f_new = 1 / ...
            (-2 * log10((epsilon / D) / 3.7 + 2.51 / (Re(i) * sqrt(f_current))))^2;
        if abs(f_new - f_current) < error
            f(i) = f_new;
            break;
        end
        f_current = f_new;
    end
    if f(i) == 0
        f(i) = f_current;
        warning('Colebrook iteration did not converge for Re = %.3g', Re(i));
    end
end
end

Worked Example #6

In 1865, Samuel Van Syckel helped revolutionize petroleum delivery by using an iron pipeline. Consider a hypothetical pipeline based loosely on this historical development. Assume that petroleum oil is transported through a five-statute-mile-long pipe with a 4-inch diameter at a rate of 200 barrels per hour. What is the minimum power (in horsepower) required to pump the oil along the full length of the pipe? Assume $\varrho_{\rm oil}$ = 1.6046 slug ft ${^{-3}}$ and $\mu_{\rm oil}$ = $8.4627 \times 10^{-5}$ slug ft $^{-1}$ s $^{-1}$ . The pipe’s roughness is $\epsilon$ = 0.09 mm. Assume 1 barrel = 5.615 ft ${^3}$ .

We must determine the pressure loss, $\Delta p_L$ , 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.,

$Re = \frac{\varrho_{\rm oil} V_{\rm av} \, D}{\mu_{\rm oil}}$

From the above equation, the average flow velocity can be calculated from the oil’s flow rate. The flow rate of the oil is

$Q = V_{\rm av} \, \rm{Area} = V_{\rm av} \, \frac{\pi}{4} \, D^2 = 200.0 \text{ barrels/hr} = \frac{200\times 5.615}{3,600} = 0.312~\mbox{ft${^{3}}$ s$^{-1}$}$

Therefore,

$V_{\rm av} = \frac{4Q}{\pi D^2} = \frac{4 \times 0.312}{\pi (4/12)^2} = 3.58\mbox{ ft s$^{-1}$}$

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

$Re = \frac{\varrho_{\rm oil} V_{\rm av} \, D}{\mu_{\rm oil}} = \frac{(1.6046 \times 3.58 \times 4/12)}{8.4627 \times 10^{-5}} = 22,626.6$

The Reynolds number exceeds 2,000. Hence, the flow is turbulent. For a turbulent flow, the pressure loss can be determined using the equation

$\Delta p_L = \frac{1}{2}\varrho_{\rm oil}V_{\rm av}^2 \, f \left( \frac{L}{D} \right)$

Here, the friction factor ${\scriptstyle{f}}$ is a function of relative roughness ${\epsilon/D}$ and Reynolds number. For the wrought iron pipe, the relative roughness is

$\frac{\epsilon}{D} = \frac{0.09}{25.4 \times 4} = 0.00089$

Consulting the Moody chart for $\epsilon/D = 0.00089$ and $Re = 22,626$ , we find the friction factor ${\scriptstyle{f}} \ approx 0.0254$ . The pressure drop in the pipe can now be calculated, i.e.,

$\Delta p_L = \frac{1}{2}\varrho_{\rm oil}\, V_{\rm av}^2 \, f \left( \frac{L}{d}\right)$

and inserting the values gives

$\Delta p_L = \frac{1}{2}\times 1.6046 \times (3.58)^2 \times 0.0254 \left(\frac{5.0 \times 5,280.0}{(4.0/12)}\right) = 2.069\times 10^4~\mbox{lb ft$^{-2}$}$

The required power to overcome this pressure loss is

$\mbox{Power} = \Delta p_L \, Q$

and inserting the values gives

$\mbox{Power} = \left(2.069\times 10^4\right) (0.312) = 6.46\times 10^3~\mbox{ft-lb s$^{-1}$} = \dfrac{6.46\times 10^3}{550} = 11.7~\mbox{hp}$

Worked Example #7

Air enters a 50 ft-long section of an air-conditioning duct made of smooth, galvanized steel with a square cross-section, each side measuring 2 ft. The equivalent roughness of the duct is $\epsilon$ = 0.00015 ft. The air is pushed through the duct with a fan at a volume flow rate of 1,800 ft ${^3}$ per minute. (a) Determine the pressure drop and head loss along this part of the duct. (b) Determine the power required to overcome the pressure losses in this section 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 $D_h$ , which is

$D_h = \frac{4 A_c}{p}$

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

$D_h = \frac{4 A_c}{p} = \frac{4 a^2}{4a} = a = 2~\mbox{~ft}$

We are advised to consult the Moody chart, indicating that the flow through the pipe will likely be turbulent. The equation for the pressure loss through a pipe is

$\Delta p_L = \frac{1}{2} \varrho V_{\rm av}^2 f \left( \frac{L}{D_h} \right)$

where ${\scriptstyle{f}}$ is the friction factor, which is still to be determined. Its value depends on the Reynolds number and pipe roughness, as shown in the Moody chart.

The average flow velocity, $V_{\rm av}$ , can be determined from the flow rate and the cross-sectional area. The pipe area $A_c$ is

$A_c = a^2 = 2^2 = 4.0~\mbox{ft${^{2}}$}$

So, the average flow velocity is

$V_{\rm av} = \frac{Q}{A_c} = \frac{1,800}{60 \times 4.0} = 7.5~\mbox{ft s$^{-1}$}$

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

$Re = \frac{\varrho V_{\rm av} \, D_h}{\mu} = \frac{0.002378 \times 7.5 \times 2}{3.737 \times 10^{-7}} = 95,451$

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

Because $Re$ is well above the usual transition value for internal flow, the flow through the duct will be turbulent. We also need the relative pipe roughness, which is

$\frac{\epsilon}{D_h} = \frac{0.00015}{2} = 0.000075 = 7.5 \times 10^{-5}$

Therefore, from the Moody chart, the friction factor is

$f \approx 0.018$

Returning to the calculation of the pressure drop, then

$\begin{eqnarray*} \Delta p & = & \frac{1}{2} \varrho V_{\rm av}^2 \, f \left( \frac{L}{D_h} \right) \\ & = & \frac{1}{2} \times 0.002378 \times 7.5^2 \times 0.018 \times \left( \frac{50.0}{2.0} \right) \\ & = & 0.03~\mbox{lb ft$^{-2}$} \end{eqnarray*}$

which corresponds to a head loss of

$h = \frac{\Delta p}{\varrho \, g} = \frac{0.03}{0.002378 \times 32.17} = 0.39~\mbox{ft}$

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

$P = Q \, \Delta p_L$

so

$P = 1,800/60.0 \times 0.03 = 0.9~\mbox{ft-lb s$^{-1}$}$

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 to a sprinkler manifold located at the downstream end. The pipe is 150 m long and has a 10 cm diameter. The required volume flow rate through the pipe is 0.09 m ${^3}$ s $^{-1}$ . For satisfactory operation, the pressure at the downstream sprinkler manifold must be at least 205.0 kPa. (a) Find the pressure drop along the length of the pipe. (b) Calculate the minimum required pump pressure, $p_1$ . (c) What pressure capability (in terms of head) would you specify for the pump? Notes: Assume turbulent flow. The density of water $\varrho_{w}$ = 1,000 kg m ${^{-3}}$ and the viscosity of water $\mu_{w} = 1 \times 10^{-3}$ kg m $^{-1}$ s $^{-1}$ .

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

$Q = V_{\rm av} \, A$

The average flow velocity can be calculated using

$V_{\rm av} = \frac{Q}{A} = \frac{Q}{\displaystyle{\frac{\pi D^2}{4}}} = \frac{4Q}{\pi D^2}$

$V_{\rm av} = \frac{4 \times 0.09}{\pi (0.1)^2} = 11.46~\mbox{m s$^{-1}$}$

Calculating the Reynolds number gives

$Re = \frac{\varrho V_{\rm av} \, D}{\mu} = \frac{1000 \times 11.46 \times 0.1}{0.001} = 1.1 \times 10^6$

For the smooth plastic tubing, the absolute roughness is very small; using $\epsilon = 0.0015$ mm gives

$\frac{\epsilon}{D} = \frac{0.0015}{100} = 1.5 \times 10^{-5}$

Obtaining the Darcy friction factor from the Moody chart gives ${\scriptstyle{f}} \approx 0.0117$ . The pressure drop along the pipe can be calculated using

$\Delta p = p_1 - p_2 = \frac{1}{2} \varrho V_{\rm av}^2 f \left( \frac{L}{d}\right)$

And substituting values gives

$\Delta p = \bigg(\frac{1000 \times 11.46^2}{2}\bigg) \times 0.0117 \bigg(\frac{150}{0.1}\bigg) = 1.152 \times 10^6~\mbox{Pa} \approx 1,152~\mbox{kPa}$

(b) The pressure at the pump located at point 1 is then the required pressure at the last sprinkler $p_2$ plus the pressure drop

$p_1 = p_2 + \Delta p$

Substituting values in the above equation (notice that both $p_2$ and $\Delta p$ are expressed in kPa) gives

$p_1 = 205.0 + 1,152 = 1,357~\mbox{kPa}$

(c) The corresponding pump head capability can be determined using

$h_p = \frac{p_1}{\varrho_{\rm H_2O} \, g} = \frac{1,357 \times 10^3}{1,000 \times 9.81} = 138.3~\mbox{m~of~H}_2\text{O}$

So, we would likely need a pump with a head of at least 139 ${\mbox{m~of~H}_2\text{O}}$ , which is a reasonably large pump.

Worked Example #9

Cooling water flows at a rate of 0.075 m ${^{3}}$ s $^{-1}$ through a 30 m-long, 15 cm-diameter asphalted cast-iron pipe. Assume for water that its density $\varrho$ = 1,000 kg m ${^{-3}}$ and its kinematic viscosity $\nu$ = 10 $^{-6}$ m ${^{2}}$ s $^{-1}$ , and the pipe material has an equivalent roughness $\epsilon$ = 0.12 mm. Hint: Consult the Moody Chart. (a) Using the basic principles of pipe flow, 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

$\Delta p = \frac{1}{2} \varrho V_{\rm av}^2 \, f \left( \frac{L}{D} \right)$

where ${\scriptstyle{f}}$ is the friction factor, which depends on the Reynolds number and pipe roughness. The average flow velocity, $V_{\rm av}$ , can be determined from the flow rate and the pipe area. The pipe area $A_p$ is

$A_p = \pi \left( \frac{D^2}{4} \right) = \pi \left( \frac{0.15^2}{4} \right) = 0.0177~\mbox{m${^{2}}$}$

so the average flow velocity is

$V_{\rm av} = \frac{Q}{A_p} = \frac{0.075}{0.0177} = 4.24~\mbox{m s$^{-1}$}$

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

$Re = \frac{V_{\rm av} \, D}{\nu} = \frac{4.24 \times 0.15}{10^{-6}} = 6.36\times 10^5$

We also need the relative roughness, which is

$\frac{\epsilon}{D} = \frac{0.00012}{0.15} = 0.0008.$

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

$f \approx 0.019$

Returning to the calculation of the pressure drop, then

$\begin{eqnarray*} \Delta p & = & \frac{1}{2} \varrho V_{\rm av}^2 \, f \left( \frac{L}{D} \right) \\ & = & \frac{1}{2} \times 1,000 \times 4.24^2 \times 0.019 \times \frac{30}{0.15} \\ & = & 34.157 \times 10^3~\mbox{Pa} = 34.157~\mbox{kPa} \end{eqnarray*}$

The corresponding head loss $h_L$ can be determined using

$h_L = \frac{\Delta p}{\varrho_{w} \, g} = \frac{34.157 \times 10^3}{1,000 \times 9.81} = 3.48~\mbox{m}$

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

$P = Q \, \Delta p$

where the volume flow rate is

$Q = 0.075~\mbox{m${^{3}}$ s$^{-1}$}$

Therefore

$P = 0.075 \times 34.157 \times 10^3 = 2.562 \times 10^3~\mbox{W} = 2.56~\mbox{kW}$

Worked Example #10

Consider a pump that steadily discharges water through a 200-ft-long section of cast-iron pipe, as shown in the figure below. The pipe has a rectangular cross-section with height $h = 3$ in and width $w = 6$ in. The internal surface roughness is $\epsilon =$ 0.006 inches. The volumetric flow rate through the pipe is 0.75 ft ${^3}$ s $^{-1}$ . Assume that the density of the water $\varrho_{\rm w}$ is 1.940 slug ft ${^{-3}}$ and its viscosity $\mu$ is 2.1 $\times 10^{-5}$ slug ft $^{-1}$ s $^{-1}$ . Also, assume that entrance effects can be ignored.

1. Determine the hydraulic diameter $D_h$ of the pipe.
2. Determine the average flow velocity $V_{\rm av}$ through the pipe.
3. Determine the Reynolds number $Re_{D_h}$ of the flow in the pipe.
4. Is the flow in the pipe laminar or turbulent? Explain.
5. Determine the pipe’s friction factor ${\scriptstyle{f}}$ .
6. Determine the pressure drop $\Delta p$ and head loss $\Delta h$ through the pipe.

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

$A = w \times h = (3.0/12.0)(6.0/12.0) = 0.125~\mbox{ft${^{2}}$}$

so the hydraulic diameter is

$D_h = \frac{4 \times \mbox{Area}}{\mbox{Perimeter}} = \frac{4A}{2 (w + h) } = \frac{2A}{(w + h) } = \frac{2.0 \times 0.125}{(3.0/12.0 + 6.0/12.0)} = 0.333~\mbox{ft}$

2. The average flow velocity $V_{\rm av}$ in the pipe is determined from the volumetric flow rate and the cross-sectional area, i.e.,

$V_{\rm av} = \frac{Q}{A} = \frac{0.75}{0.125} = 6.0~\mbox{ft s$^{-1}$}$

3. The Reynolds number based on the hydraulic diameter is

$Re_{D_h} = \frac{\varrho_{\rm w} \, V_{\rm av} \, D_h}{\mu} = \frac{1.940 \times 6.0 \times 0.333}{2.1\times 10^{-5}} = 184,549 \approx 185,000$

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

5. The relative surface roughness is

$\frac{\epsilon}{D_h} = \frac{0.006/12.0}{0.333} = 0.00150$

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

$\Delta p = \frac{1}{2} \varrho_w \, V_{\rm av}^2 \, f \left(\frac{L}{D_h}\right)$

Inserting the known values gives

$\Delta p = 0.5 \times 1.94 \times 6.0^2 \times 0.023 \left( \frac{200.0}{0.333} \right) = 482.38~\mbox{lb ft$^{-2}$}$

The corresponding head loss is

$\Delta h = \frac{\Delta p}{\varrho_w \, g} = \frac{482.38}{1.94 \times 32.17} = 7.73~\mbox{ft}$

Worked Example #11

RP-1 rocket fuel will be transferred from one tank to another using a pump with a flow rate of $Q$ = 52.1 ft ${^3}$ per minute. The fuel passes through a 4-inch-diameter, 40-ft-long steel pipe. The pipe has a roughness, $\epsilon$ = 0.002 inches. Assume the density of RP-1 is 1.746 slug ft ${^{-3}}$ and its viscosity is $7.4 \times 10^{-4}$ slug ft $^{-1}$ s $^{-1}$ . How much pumping power is required to move the RP-1 along the pipe?

The average flow velocity $V_{\rm av}$ in the pipe is

$V_{\rm av} = \frac{Q}{A} = \frac{4 Q}{\pi \, d^2 } = \frac{4 \times (52.1/60)}{\pi (4.0/12.0)^2} = 9.95~\mbox{ft s$^{-1}$}$

The Reynolds number based on the diameter is

$Re_{d} = \frac{\varrho \, V_{\rm av} \, d }{\mu} = \frac{1.746 \times 9.95 \times 4.0/12.0}{7.4\times10^{-4}} = 7,825 \approx 8,000$

This result indicates turbulent pipe flow; therefore, we must use the Moody chart to determine the friction factor. The relative surface roughness is

$\frac{\epsilon}{d} = \frac{0.002 }{4.0} = 0.0005$

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

$\Delta p = \frac{1}{2} \varrho \, V_{\rm av}^2 \, f \left(\frac{L}{d}\right)$

and inserting the values gives

$\Delta p = 0.5 \times 1.746 \times 9.95^2 \times 0.035 \left( \frac{40.0}{4.0/12.0} \right) = 363.0~\mbox{lb ft$^{-2}$}$

The corresponding pumping power required is

$P = Q \, \Delta p = (52.1/60) \times 363.0 = 315.2~\mbox{ft-lb s$^{-1}$}$

Converting to horsepower gives

$P = \frac{315.2}{550} = 0.57~\mbox{hp}$

Worked Example #12

A liquid rocket fuel system uses a high-strength nickel alloy pipe with a diameter of $D = 4.0$ cm and a length of $L = 5.0$ m to transport liquid oxygen (LOX) from the tank to the engine. The LOX has a density $\varrho = 1,140$ kg m ${^{-3}}$ and a viscosity $\mu = 0.0001947$ Pa s. The mass flow rate of the LOX is $\overbigdot{m} = 20.0$ kg s $^{-1}$ . Assume the pipe’s average roughness, $\epsilon$ , 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

$V_{\rm av} = \frac{\overbigdot{m}}{\varrho \, A_c} = \frac{4 \overbigdot{m}}{\varrho \pi D^2 } = \frac{4 \times 20.0}{1,140.0 \times \pi \times 0.04^2 } = 13.96~\mbox{m s$^{-1}$}$

The Reynolds number of this flow, based on pipe diameter, is

$Re = \frac{\varrho \, V_{\rm av} \, D}{\mu} = \frac{ 1,140 \times 13.96 \times 0.04}{0.0001947} \approx 3.27 \times 10^6$

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

$\frac{\epsilon}{D} = \frac{0.04}{40.0} = 0.001$

Referring to the Moody chart for $Re = 3.27 \times 10^6$ and $\epsilon/D = 0.001$ , then $f \approx 0.02$ . The pressure drop along the pipe is

$\Delta p = \frac{1}{2} \varrho V_{\rm av}^2 \, f \left( \frac{L}{D} \right)$

Substituting the numerical values gives

$\Delta p = \frac{1}{2} \times 1,140.0 \times 13.96^2 \times 0.02 \left( \frac{5.0}{0.04} \right) = 2.78 \times 10^5~\mbox{Pa}$

Therefore, the power required to pump this flow is

$P = Q \, \Delta p = \left( \frac{\overbigdot{m}}{\varrho} \right) \Delta p = \left( \frac{20.0}{1,140.0} \right) \times 2.78 \times 10^5 \approx 4.87~\mbox{kW}$

Worked Example #13

A 1,000 ft long steel pipe has an inside diameter of 15 inches and a wall thickness of 1.5 inches. The pipe carries water from a reservoir, and a valve is located downstream. When the valve is fully open, the flow rate through the pipe is 23.3 ft ${^3}$ /s. If the valve closes completely in 0.15 seconds, determine the water-hammer pressure in the pipe. Assume $E_b$ = 3.0 $\times 10^5$ lb/in ${^{2}}$ , and $E_p$ = 2.8 $\times 10^7$ lb/in ${^{2}}$ , and $K$ = 1.0. The density of water is 1.94 slugs/ft ${^{3}}$ .

The relevant equation for the water hammer pressure in an elastic pipe is

$\Delta p = \varrho \, a \, V_0$

where $a$ is the speed of the pressure wave and $V_0$ is the initial average flow velocity. The average velocity through the pipe is related to the flow rate, $Q$ , by

$V_0 = \frac{Q}{A} = \frac{4 Q}{\pi D^2} = \frac{4 \times 23.3}{\pi (15.0/12.0)^2} = 18.99~\mbox{ft/s}$

The effective bulk modulus is given by

$\frac{1}{E_c} = \frac{1}{E_b} + \frac{D K}{E_p e}$

Using $D=15.0$ in and $e=1.5$ in gives

$\frac{1}{E_c} = \frac{1}{3.0 \times 10^5} + \frac{15.0 \times 1.0}{2.8 \times 10^7 \times 1.5} = 3.69 \times 10^{-6}$

so that

$E_c = 2.71 \times 10^5~\mbox{lb/in${^2}$}$

Therefore, the wave propagation speed is

$a = \sqrt{ \frac{E_c}{\varrho} } = \sqrt{ \frac{144 \times 2.71 \times 10^5}{1.94} } = 4,484~\mbox{ft/s}$

and the wave reflection time is

$t_t = \frac{2 L}{a} = \frac{2 \times 1,000}{4,484} = 0.45~\mbox{s}$

With a valve closure time of 0.15 s, $t_c < t_t$ , indicating a sudden closure. Therefore, the water hammer change in pressure will be

$\Delta p = \varrho \, V_0 \, a = 1.94 \times 18.99 \times 4,484 = 1.65 \times 10^5~\mbox{lb/ft${^{2}}$}$

or

$\Delta p = \frac{1.65 \times 10^5}{144} = 1,146~\mbox{lb/in${^{2}}$}$

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Introduction to Aerospace Flight Vehicles Copyright © 2022–2026 by J. Gordon Leishman is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, except where otherwise noted.

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