Worked Examples: Fluid Flows

J. Gordon Leishman

doi:https://doi.org/10.15394/eaglepub.2022.1066.9

51 Worked Examples: Fluid Flows

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

Worked Example #1

Two pipes of diameters $d_1$ and $d_2$ converge to form a single pipe of diameter $d$ . If a liquid flows with a velocity of $V_1$ and $V_2$ in the two pipes, respectively, what will be the flow velocity $V_3$ in the third pipe? Hint: Assume steady, one-dimensional, incompressible flow.

In this case, two pipes merge to produce one pipe. Conserving mass and using average flow properties, then

$\overbigdot{m} = \mbox{constant} = \varrho_1 V_1 A_1 + \varrho_2 V_2 A_2 = \varrho_3 V_3 A_3$

If the fluid is a liquid then the assumption that $\varrho$ = $\varrho_1$ = $\varrho_2$ = $\varrho_3$ = constant is justified so

$\overbigdot{Q} = \mbox{constant} = V_1 A_1 + V_2 A_2 = V_3 A_3$

Therefore, the flow velocity in the third pipe is

$V_3 = \frac{V_1 A_1 + V_2 A_2}{A_3} = \frac{V_1 (\pi d_1^2/4) + V_2 (\pi d_2^2/4)}{ (\pi d^2/4)} = \frac{V_1 d_1^2 + V_2 d_2^2}{d^2}$

Worked Example #2

Consider the steady flow of a particular gas through a horizontal, converging pipe with an inlet diameter $d_1$ of 0.22 m and an outlet diameter $d_2$ of 0.16 m. The density of the gas is known to change from $\varrho_1 = 0.91$ kg/m $^3$ at the inlet to $\varrho_2 = 0.83$ kg/m $^3$ at the outlet. If the inlet flow velocity of the gas $V_1$ is 5.1 m/s, what is its exit velocity $V_2$ ? Hint: Assume one-dimensional flow.

Conserving mass and using average flow properties, we have

$\overbigdot{m} = \mbox{constant} = \varrho_1 V_1 A_1 = \varrho_2 V_2 A_2$

where in this case, the density must be retained as a variable. In terms of pipe diameters, then

$\varrho_1 V_1 d_1^2 = \varrho_2 V_2 d_2^2$

and solving for $V_2$ gives

$V_2 = \frac{\varrho_1 V_1 d_1^2}{\varrho_2 d_2^2}$

Substituting values gives

$V_2 = \frac{0.91 \times 5.1 \times 0.22^2}{0.83 \times 0.16^2} = 10.57~\mbox{m s$^{-1}$}$

Worked Example #3

Air flows steadily through a horizontal pipe of a constant circular cross-sectional area. The air temperature is constant at $T_{\rm air} = 26^{\circ}$ C along the flow direction. At point A, the static pressure is $p_A = 3 \times 10^5$ Pa, and the mean velocity is $V_A = 110$ m/s. At point B, the static pressure is $p_B = 2 \times 10^5$ Pa. What is the average flow velocity at B? Assume $R$ = 287.057 J kg $^{-1}$ K $^{-1}$ .

The requirement for flow velocities suggests the use of the continuity equation, which can be written as

$\varrho_A V_A A_A = \varrho_B V_B A_B$

Notice that the density of the air is not given, but the pressures and the temperature $T_{\rm air} = T_A = T_B$ are, so using the equation of state gives

$\varrho_A = \frac{p_A}{R \, T_A} = \frac{3 \times 10^5}{287.057 \times (26.0 + 273.15)} = 3.494~\mbox{kg/m$^3$}$

Likewise

$\varrho_B = \frac{p_B}{R \, T_B} = \frac{2 \times 10^5}{287.057 \times (26.0 + 273.15)} = 2.329~\mbox{kg/m$^3$}$

The use of the continuity equation gives

$V_B = \left( \frac{\varrho_A}{\varrho_B} \right) \left( \frac{A_A}{A_B} \right) V_A = \left( \frac{3.494}{2.329} \right) 110.0 = 165.04~\mbox{m/s}$

noting that

$\left( \frac{A_A}{A_B} \right) = 1$

Worked Example #4

A gas of density 1.7 kg/m $^3$ flows through a pipe with a volume flow rate of 0.12 m $^3$ /s. The pipe has an inlet section diameter of $d_1$ of 21 cm and an outlet section diameter $d_2$ of 9 cm. Assuming an ideal incompressible fluid flow, calculate the flow velocities at the pipe’s inlet and outlet, i.e., $V_1$ and $V_2$ , respectively.

From the information given, it is reasonable to assume a one-dimensional, steady, incompressible, inviscid flow. The flow rates, flow velocities, and Mach numbers are low enough that compressibility effects can be neglected. Let the inlet be condition 1 and the outlet condition 2. The continuity equation can be used to relate the inlet and outlet conditions, i.e.,

$\overbigdot{m} = \varrho A_1 V_1 = \varrho A_2 V_2 = \mbox{constant}$

or because the flow is assumed incompressible, then just

$\overbigdot{Q} = A_1 V_1 = A_2 V_2 = \mbox{constant} = 0.12~\mbox{m$^3$/s}$

We see that

$V_1 = \frac{\overbigdot{Q} }{A_1}$

and

$V_2 = \frac{\overbigdot{Q} }{A_2}$

We can calculate $A_1$ , i.e.,

$A_1 = \frac{\pi d_1^2}{4} = \frac{\pi \, 0.21^2}{4} = 0.0346~\mbox{m$^2$}$

and also $A_2$ , i.e.,

$A_2 = \frac{\pi d_2^2}{4} = \frac{\pi \, 0.09^2}{4} = 0.00636~\mbox{m$^2$}$

so that

$V_1 = \frac{\overbigdot{Q} }{A_1} = \frac{0.12}{0.0346} = 3.47~\mbox{m/s}$

and

$V_2 = \frac{\overbigdot{Q} }{A_2} = \frac{0.12}{0.00636} = 18.87~\mbox{m/s}$

We have confirmed that the flow velocities are very low, much lower than a Mach number of 0.3, so the initial assumption of incompressible flow was justified.

Worked Example #5

A flowing liquid enters a pipe of diameter $d_1$ with a velocity $V_1$ . What will the velocity $V_2$ of the liquid be at the exit if the diameter of the pipe progressively reduces to $d_2 = 0.6d$ ? Hint: Assume one-dimensional flow.

Conserving mass and using the continuity equation with average flow properties (i.e., one-dimensional flow) gives

$\oiint_S \vec{V} \bigcdot d\vec{S} = \overbigdot{m} = \mbox{constant} = \varrho_1 V_1 A_1 = \varrho_2 V_2 A_2$

Because the fluid is a liquid, the assumption that $\varrho$ = constant is justified. So the continuity equation reduces to

$\overbigdot{Q} = \mbox{constant}$

In this case

$\overbigdot{Q} = V_1 \left( \frac{\pi d^2}{4} \right) = V_2 \left(\pi \frac{(0.6d)^2}{4} \right) = V_2 \left(\frac{0.36 \pi d^2}{4} \right) = 0.36 V_2 \left(\frac{\pi d^2}{4} \right)$

where $V_2$ is the exit velocity. Rearranging gives

$V_2 = \frac{V_1}{0.36} = 2.78 V_1$

Worked Example #6

Consider the flow-turning block shown below. A circular jet of water with diameter $D_j =$ 10 cm at a velocity $V_j = 13$ ms $^{-1}$ enters the block. The block turns the water back through 180 degrees and then exits through an orifice with an area $A_e$ of 120 cm $^2$ . First, find the velocity $V_e$ of the water exiting the block. Second, determine the force $F_x$ required to hold the block in place. Hint: Assume one-dimensional flow.

First, apply the conservation of mass to the inlet and outlet flow from the control volume, i.e., for a steady flow.

$\oiint_S \varrho \, \vec{V} \bigcdot d\vec{S} = 0$

and in one-dimensional form, then

$-\varrho V_j A_j + \varrho V_e A_e = 0$

or

$\varrho V_j A_j = \varrho V_e A_e = \overbigdot{m}$

The fluid is water so $\varrho =$ constant and so

$V_j A_j = V_e A_e$

Rearranging to solve for the exit velocity $V_e$ gives

$V_e = V_j \left( \frac{A_j }{A_e} \right) = \left( \frac{ \displaystyle{\frac{\pi}{4} } D_j^2}{A_e} \right) V_j$

Substituting the known values gives

$V_e = \left( \frac{ \displaystyle{\frac{\pi}{4} } D_j^2}{A_e} \right) V_j = \left( \frac{ \displaystyle{\frac{\pi}{4} } 0.1^2}{0.012} \right) 13.0 = 8.51\mbox{m s$^{-1}$}$

Second, from the principle of conservation of momentum the

$\vec{F} = \oiint_S (\varrho \, \vec{V} \bigcdot d\vec{S}) \vec{V}$

where $\vec{F}$ on the fluid from the block. In this case, then

$\vec{F} = \overbigdot{m} (-V_j ) - \overbigdot{m} V_e = -\overbigdot{m} \left( V_e + V_j \right)$

the minus reminding that the force on the fluid is to the left, i.e., in the negative $x$ direction. The mass flow rate is conserved so

$\overbigdot{m} = \varrho V_j A_j =\varrho V_e A_e = 1,000.0 \times 8.51 \times 0.012 = 102.12~\mbox{kg/s}$

where the density of water has been assumed to be 1,000 kg/m $^3$ , and so

$\vec{F} = \overbigdot{m} \left( V_e + V_j \right) = 102.12 \left( 8.51 + 13.0\right) = 2,196.6~\mbox{N}$

and the reaction force on the block is to the right, i.e., in the positive $x$ direction. Therefore, the force to restrain the block will be 2196.6 N toward the left.

Worked Example #7

An incompressible gas flow in a pipe is passed through a venturimeter. The diameter of the pipe is $D_2$ = 4.1 inches and the diameter of the throat of the venturimeter $D_1$ = 2.2 inches. The gas flow in the main pipe has a velocity $V_2$ = 14.3 ft/s. If the static pressure difference between the inlet and the throat of the venturi is measured using a U-tube water manometer, then calculate the height $h$ between the two columns. Assume 1-dimensional flow and no losses. Also assume that $\varrho_{\rm H_2O} = 1.94$ slug/ft $^3$ and $\varrho_{\rm gas} = 0.0023$ slug/ft $^3$ .

Assuming steady, 1-dimensional flow applying the continuity equation between inlet and the throat gives

$\varrho_1 \, A_1\, V_1 = \varrho_2 \, A_2 \, V_2$

and because $\varrho_1 = \varrho_2 = \varrho =$ constant in this case (the flow is stated to be incompressible) then

$A_1 \, V_1 = A_2 \, V_2$

Therefore, solving for $V_1$ gives

$V_1 = \frac{A_2 \, V_2}{A_1} = \frac{\left(\displaystyle{\frac{\pi}{4}}\right)D_2^2 \, V_2}{\left(\displaystyle{\frac{\pi}{4}}\right)D_1^2} = \frac{\left(\displaystyle{\frac{4.1}{12}}\right)^2 \times 14.3 }{\left(\displaystyle{\frac{2.2}{12}} \right)^2} = \frac{4.1^2 \times 14.3 }{2.2^2} = 49.67 \mbox{ ft/s}$

The application of Bernoulli’s equation gives

$p_1 + \frac{1}{2} \, \varrho_{\rm gas}\, V_1^2 + \varrho_{\rm gas} \, g \, z_1 = p_2 + \frac{1}{2} \, \varrho_{\rm gas} \, V_2^2 + \varrho_{\rm gas} \, g \, z_2$

There is no change in vertical height of the mean flow in this case, so $z_1 = z_2 = 0$ and

$p_1 + \frac{1}{2} \, \varrho_{\rm gas}\, V_1^2 = p_2 + \frac{1}{2} \, \varrho_{\rm gas} \, V_2^2$

Rearranging for the pressure difference $p_2 - p_1$ gives

$p_2 - p_1 = \frac{1}{2}\varrho_{\rm gas}\left(V_1^2 - V_2^2\right)$

The U-tube manometer will measure this static pressure difference $p_2 - p_1$ , so using the hydrostatic equation the pressure difference is

$p_2 - p_1 = \varrho_{\rm H_2O} \, g \, h$

making sure the density of water is used. Therefore, equating the latter two equations gives

$\varrho_{\rm H_2O} \, g \, h = \frac{1}{2}\varrho_{\rm gas}\left(V_1^2 - V_2^2\right)$

and solving for $h$ leads to

$h = \frac{\varrho_{\rm gas}}{2 \, \varrho_{\rm H_2O} \, g}\left(V_1^2 - V_2^2\right) = \frac{0.0023}{2 \times 1.94 \times 32.17}\left( 49.67^2 - 14.3^2\right) = 0.0417~\mbox{ft}$

Worked Example #8

A rain barrel with a circular cross-section allows the water to leave at the bottom through a drain valve. The discharge velocity $V_d$ of the flow from the drain varies with the height $h$ of the water level above the drain according to $V_d = \sqrt{ 2 g h}$ . If the initial height of the water level above the drain is 0.5 m, how long will it take to empty the tank? The density of water can be assumed to be 1,000 kg m $^{-3}$ .

The volume flow rate $\overbigdot{Q}$ from the tank through the drain can be expressed as

$\overbigdot{Q} = a V_d$

where $a$ is the area of the jet of water from the drain and

$V_d = \sqrt{ 2 g h}$

Notice that the flow rate depends on the instantaneous height of the water in the tank. This result is actually called Torricelli’s theorem, and it is a special case of using the Bernoulli equation for a fluid that flows out of an orifice based on the height of the fluid above the orifice.

The volume of water $\cal{V}$ in the tank is for any height $h$ is

${\cal{V}} = A h$

where $A$ is the area of the tank. In a small time interval $dt$ then the decrease in the volume $\cal{V}$ of water in the tank will be

$\frac{d \cal{V}}{dt} = - A \frac{dh}{dt}$

the minus sign denoting that the value of $h$ decreases as the tank is emptied. We are told that the initial height $h = h_0$ of the water in the tank is 0.5 m. Notice that the units in the foregoing equation are in terms of a flow rate.

Now, by continuity considerations (conservation of mass) this flow rate must be equal to the flow rate of fluid coming out of the drain, i.e.

$\frac{d \cal{V}}{dt} = \overbigdot{Q}$

so

$- A \frac{dh}{dt} = a V_d = a \sqrt{ 2 g h}$

Rearranging gives

$\frac{dh}{dt} = - \frac{a}{A} \sqrt{ 2 g h}$

This is an ordinary differential equation relating the height of the fluid level to time. We now need to solve this equation subject to the boundary conditions given in this question. Separating the variables and integrating gives

$t = \int_{0}^{t} dt = - \frac{A}{a} \int_{h_0}^0 \frac{dh}{\sqrt{ 2 g h}} = - \frac{A}{a \sqrt{ 2 g}} \int_{h_0}^0 \frac{dh}{\sqrt{h}} = \frac{A}{a \sqrt{ 2 g}} \int_0^{h_0} \frac{dh}{\sqrt{h}}$

where the limits of integration are noted, i.e., at time $t = 0$ then $h = h_0$ and when $h = 0$ the tank is considered to be empty. Performing the integration gives

$t = \frac{2A}{a \sqrt{ 2 g}} \sqrt{h_0}$

We are given the diameter of the tank $D$ and the outlet diameter $d$ , so in terms of ratios then

$\frac{A}{a} = \frac{D^2}{d^2}$

so

$t = \frac{2D^2}{d^2 \sqrt{ 2 g}} \sqrt{h_0}$

Substituting the known values gives

$t = \frac{2D^2}{d^2 \sqrt{ 2 g}} \sqrt{h_0} = \frac{2 \times 1.0^2}{0.1^2 \sqrt{ 2 \times 9.81}} \sqrt{0.5} = 31.93~\mbox{secs}$

Worked Example #9

A water jet with velocity $V$ that exits from a nozzle strikes a vertical plate. The exit area of the nozzle is $A$ . Find an expression for the horizontal force on the plate. If the volume flow rate through the nozzle is 0.5 m $^{3}$ s $^{-1}$ and the diameter of the jet is 5 cm, then find the force on the plate in Newtons. The density of water can be assumed to be 1,000 kg m $^{-3}$ .

The force on the plate is obviously related to the force of the fluid and so to the exit velocity from the nozzle. Using the momentum equation in its one-dimensional form, the the force on the plate from the water jet is just the time rate of change in horizontal momentum, i.e.,

$F = \overbigdot{m} V - 0$

The use of the continuity equation gives

$\varrho V A = \overbigdot{m}$

and because we are dealing with water then it is incompressible, so

$V A = \overbigdot{Q} = 0.5~\mbox{m$^{3}$ s$^{-1}$}$

The value of $V$ is

$V = \frac{\overbigdot{Q}}{A} = \frac{0.5}{\pi d^2/4} = \frac{0.5}{\pi (0.05)^2/4} = 254.65~\mbox{m s$^{-1}$}$

Therefore, the force on the plate is

$F = \overbigdot{m} V = \left( \varrho \overbigdot{Q} \right) V = 1,000 \times 0.5 \times 254.65 =127.33~\mbox{kN}$

Worked Example #10

Water flows into a container through both of pipes 1 and 3 and simultaneously exits through pipe 2. The flow rates are adjusted such that the height $h$ of water in the container stays constant. The volume flow rate $\overbigdot{Q}_3 = 0.015$ m $^3$ /s and also $V_{1} = 3.4$ m/s, $D_1 = 0.05$ m and $D_2 = 0.075$ m. From this information, determine the value of $V_2$ . Make any assumptions that may be justified.

If the level $h$ stays constant then the net mass (or volume) flow the system is a constant in that the mass of water coming in per unit time is equal to the mass (or volume) of water leaving per unit time. In terms of the problem parameters then

$\overbigdot{Q} = \overbigdot{Q}_1 + \overbigdot{Q}_3 = \overbigdot{Q}_2$

where $\overbigdot{Q}_3 = 0.015$ ~m $^3$ /s. Remember that

$\overbigdot{Q} = A V$

so

$V = \frac{\overbigdot{Q}}{A}$

and so

$A_1 V_{1} + \overbigdot{Q}_3 = A_2 V_2$

and solving for $V_2$ , which is what we are asked to find, then

$V_2 = \frac{A_1 V_{1} + \overbigdot{Q}_3}{A_2}$

We are given the diameters $D_1 = 0.05$ m and $D_2 = 0.075$ m so

$A_1 = \frac{\pi D_1^2}{4} = \frac{\pi \times 0.05^2}{4} = 0.00196~\mbox{m$^2$}$

and

$A_2 = \frac{\pi D_2^2}{4} = \frac{\pi \times 0.075^2}{4} = 0.00442~\mbox{m$^2$}$

so

$V_2 = \frac{A_1 V_{1} + \overbigdot{Q}_3}{A_2} = \frac{ 0.00196 \times 3.4 + 0.015}{0.00442} = 4.901~\mbox{m/s}$

Worked Example #11

A rigid tank of volume $\volume$ has air pumped into it at a constant mass flow rate. Find the density and pressure in the tank as a function of time if the initial density and pressure are $\varrho_0$ and $p_0$ , respectively. You may assume that the process is isothermal. Make any other assumptions that can be justified.

This is an unsteady flow problem because a mass of gas is pumped into a fixed volume and for mass conservation then the density of the gas must increase in time. The general form of the continuity equation is

$\frac{\partial}{\partial t}\iiint \varrho d \volume + \oiint_S \varrho \, \vec{V} \bigcdot d\vec{S} = 0$

In this case

$\oiint_S \varrho \, \vec{V} \bigcdot d\vec{S} = -\overbigdot{m}_{\rm in}$

also

$\frac{\partial}{\partial t}\iiint \varrho d \volume = \volume \frac{d \varrho}{dt}$

We will assume uniform mixing and so the density inside the volume uniform so we have

$\volume \frac{d \varrho}{dt} = \overbigdot{m}_{\rm in}$

Integrating using separation of variables gives

$\int_{\varrho_0}^{\varrho} d \varrho = \frac{ \overbigdot{m}_{\rm in}}{\volume} \int_{t_0}^{t} dt$

so

$\varrho = \varrho_0 + \frac{ \overbigdot{m}_{\rm in}}{\volume} \left( t - t_0 \right)$

which shows that the density of the gas increases linearly with time. The corresponding pressure can be obtained from the equation of state, i.e., $p = \varrho R T$ . If the process is isothermal then $T = T_0$ and so

$p = \varrho_0 R T_0 + \frac{ R T_0 \overbigdot{m}_{\rm in}}{\volume} \left( t - t_0 \right) = p_0 + \frac{R T_0 \overbigdot{m}_{\rm in}}{\volume} \left( t - t_0 \right)$

Worked Example #12

Starting from the most general form of the continuity equation identify and explain each of the terms. Then show, and explain carefully, how both of these general equations can be simplified for application to problems involving: (a) steady, compressible flows, (b) incompressible flows, (c) inviscid, one-dimensional, steady flows.

The general form of the continuity equation is

$\frac{\partial}{\partial t}\iiint_{\cal{V}} \varrho d {\cal{V}} + \oiint_S \varrho \, \vec{V} \bigcdot d\vec{S} = 0$

The first term is the time rate of change of the mass of fluid inside the control volume and the second term in the net mass flow rate of fluid out of the control volume. Conservation of mass requires that the net sum be zero.

(a) Steady, compressible flow. In this case $\partial/\partial t \equiv 0$ so that

$\oiint_S \varrho \, \vec{V} \bigcdot d\vec{S} = 0$

which is a statement that an equal mass of fluid coming into a control volume per unit time will also leave the control volume in the same time.(b) Steady, incompressible flow. In this case, $\varrho$ = constant so that we can write

$\oiint_S \vec{V} \bigcdot d\vec{S} = 0$

which is a statement the at the volume flow rate is conserved, i.e., an equal volume of fluid coming into a control volume per unit time will also leave the control volume in the same time.(c) One-dimensional, steady flow. In this case we must again retain the density in the integral but if the flow is one-dimensional then the flow properties can change only in one dimension, so for a fixed control volume with an inlet and and outlet plane defined accordingly, by conserving mass we would end up with

$\oiint_S \vec{V} \bigcdot d\vec{S} = -\varrho_1 V_1 A_1 + \varrho_2 V_2 A_2 = 0$

or

$\varrho_1 V_1 A_1 = \varrho_2 V_2 A_2 = \overbigdot{m} = \mbox{constant}$

This is a common form of the continuity equation and is very useful for problem solving when average flow properties are desired.

Worked Example #13

Starting from the most general form of the momentum equation, identify the meaning of each of the terms and show how the equation can then be simplified for application to problems involving: (a) Steady flows, (b) Steady, incompressible flows, (c) Steady, inviscid flow. (d) Unsteady flows with no body forces. In each case, explain the physical meaning of the terms in the respective equations.

The most general form of the momentum equation in integral form is

$\iiint_{\cal{V}} \varrho \vec{f} _b d{\cal{V}} -\oiint_S p dS + \vec{F}_{\mu} = \frac{\partial}{\partial t}\iiint_{\cal{V}} \varrho \, \vec{V} d {\cal{V}} + \oiint_S (\varrho \, \vec{V} \bigcdot d\vec{S}) \vec{V}$

In words then: Body Forces + Pressure Forces + Viscous Forces = Time rate of change of momentum inside ${\cal V}$ from unsteadiness + Net flow of momentum out of ${\cal V}$ per unit time. It is a vector equation, so three scalar equations rolled into one. This equation is very general and will apply to any type of flow.(a) For steady flows then $\partial/\partial t \equiv 0$ and the reduced form of the momentum equation will be

$\iiint_{\cal{V}} \varrho \vec{f}_b d{\cal{V}} - \oiint_S p dS + \vec{F}_{\mu} = \oiint_S (\varrho \, \vec{V} \bigcdot d\vec{S}) \vec{V}$

(b) For steady, incompressible flows then $\partial/\partial t \equiv 0$ and $\varrho =$ constant so

$\varrho \iiint_{\cal{V}} \vec{f}_b d{\cal{V}} - \oiint_S p dS + \vec{F}_{\mu} = \varrho \oiint_S (\vec{V} \bigcdot d\vec{S}) \vec{V}$

(c) For steady, inviscid flow then $\partial/\partial t \equiv 0$ and $\vec{F}_{\mu} = 0$ so

$\iiint_{\cal{V}} \varrho \vec{f}_b d{\cal{V}} - \oiint_S p dS = \oiint_S \varrho (\vec{V} \bigcdot d\vec{S}) \vec{V}$

(d) For unsteady flows with no body forces then $\vec{f} _b$ = 0 so

$-\oiint_S p dS + \vec{F}_{\mu} = \frac{\partial}{\partial t}\iiint_{\cal{V}} \varrho \, \vec{V} d {\cal{V}} + \oiint_S (\varrho \, \vec{V} \bigcdot d\vec{S}) \vec{V}$

Worked Example #14

Starting from the most general form of the energy equation, identify the meaning of each of the terms and show how the equation can then be simplified for application to problems involving: (a) Steady flows, (b) Steady, inviscid flow. (c) Steady flows with no body forces, heat or work addition. In each case, explain the physical meaning of the terms in the respective equations.

The most general form of the energy equation is

$\begin{eqnarray*} \iiint_{\cal{V}} \overbigdot{q} \varrho d{\cal{V}} + \overbigdot{Q}_{\mu} -\oiint_{S} (p d\vec{S}) \bigcdot \vec{V} + \iiint_{\cal{V}} ( \varrho \vec{f}_b d{\cal{V}}) \bigcdot \vec{V} + \overbigdot{W}_{\mu} + \overbigdot{W}_{\rm mech} = && \nonumber \\ \frac{\partial}{\partial t} \iiint_{\cal{V}} \varrho \left( e + \frac{V^2}{2} + g z\right) d{\cal{V}} + \oiint_{S} (\varrho \, \vec{V} \bigcdot d\vec{S}) \left( e + \frac{V^2}{2} + g z\right) \end{eqnarray*}$

which is a statement of energy conservation for a fluid. While this is a very general equation, there are obviously many practical problems in which we can simplify it, i.e., by making certain justifiable assumptions, which commensurately makes the mathematics and solution process easier.

(a) In the case of a steady flow then $\partial/\partial t \equiv 0$ so

$\begin{eqnarray*} \iiint_{\cal{V}} \overbigdot{q} \varrho d{\cal{V}} + \overbigdot{Q}_{\mu} -\oiint_{S} (p d\vec{S}) \bigcdot \vec{V} + \iiint_{\cal{V}} ( \varrho \vec{f}_b d{\cal{V}}) \bigcdot \vec{V} + \overbigdot{W}_{\mu} + \overbigdot{W}_{\rm mech} = && \nonumber \\ \oiint_{S} (\varrho \, \vec{V} \bigcdot d\vec{S}) \left( e + \frac{V^2}{2} + g z\right) \end{eqnarray*}$

(b) In the case of a steady, inviscid flow, then $\partial/\partial t \equiv 0$ and all of the terms that have their origin in the viscosity of the fluid can be neglected, i.e.,

$\begin{eqnarray*} \iiint_{\cal{V}} \overbigdot{q} \varrho d{\cal{V}} -\oiint_{S} (p d\vec{S}) \bigcdot \vec{V} + \iiint_{\cal{V}} ( \varrho \vec{f}_b d{\cal{V}}) \bigcdot \vec{V} + \overbigdot{W}_{\rm mech} = && \nonumber \\ \oiint_{S} (\varrho \, \vec{V} \bigcdot d\vec{S}) \left( e + \frac{V^2}{2} + g z\right) \end{eqnarray*}$

(c) In the case of a steady flow with no body forces, heat or work addition, then

$\begin{equation*} -\oiint_{S} (p d\vec{S}) \bigcdot \vec{V} = \oiint_{S} (\varrho \, \vec{V} \bigcdot d\vec{S}) \left( e + \frac{V^2}{2} + g z\right) \end{equation*}$

Worked Example #15

The airflow into a wind tunnel test section first passes through an upstream contraction section and then exhausts through a diffuser. Show by using the conservation equations that the flow velocity in the test section can be determined (calculated) by using:

(a) The difference between the total pressure in the opening (mouth) of the contraction section and the static pressure at the inlet to the test section.

(b) The difference in static pressure between the mouth of the contraction section and the static pressure at the inlet to the test section.

In each case, derive an equation relating the flow velocity in the test section to the pressure drop as well the cross-sectional areas of the inlet section and the test section.

The working part of a wind tunnel is essentially a venturi. The airflow at the settling chamber (i.e., the mouth of the venturi) of area $A_1$ can be assumed to have a flow velocity $V_1$ with static pressure $p_1$ . The wind tunnel cross-section then contracts to a smaller area $A_2$ at the test section where the velocity has increased to $V_2$ and the static pressure has decreased to $p_2$ . Notice that the velocity must increase if continuity (conservation of mass) is to be satisfied. The test section is where a model (such as a wing or complete airplane model) is tested, and measurements of the aerodynamic characteristics are made. The flow then passes downstream into a diverging duct called a diffuser.

First approach: From the continuity equation, the flow velocity in the test section will be

$\begin{equation*} V_2 = \left( \frac{A_1}{A_2} \right) V_1 \end{equation*}$

The pressures at the inlet (i.e., settling chamber) and the test section of wind tunnel can be related to the flow velocities by means of the Bernoulli’s equation, i.e.,

$\begin{equation*} p_1 + \frac{1}{2} \varrho V_1^2 = p_2 + \frac{1}{2} \varrho V_2^2 = p_T \end{equation*}$

where $p_T$ is the total pressure. This means that

$\begin{equation*} V_{2} = \sqrt{ \frac{2\left(p_T - p_{2}\right)}{\varrho} } \end{equation*}$

Therefore, the foregoing equation can be used to determine the velocity in the test section from a measurement of total pressure measured in the contraction section (using a Pitot probe) and the static pressure at the test section (using static pressure taps). However, the density of the air must be known and this is determined be measuring the temperature and pressure of the air in the flow and then calculating the density using the equation of state. This technique is used in many low-speed wind tunnels to measure the flow speed in the test section.

Second approach: From the continuity equation, the flow velocity in the test section will be

$\begin{equation*} V_2 = \left( \frac{A_1}{A_2} \right) V_1 \end{equation*}$

The static pressures at the inlet (i.e., settling chamber) and the test section of the wind tunnel can be related to the flow velocities by means of the Bernoulli’s equation, i.e.,

$\begin{equation*} p_1 + \frac{1}{2} \varrho V_1^2 = p_2 + \frac{1}{2} \varrho V_2^2 \end{equation*}$

The velocity in the test section can be related to the static pressure drop across sections 1 and 2, i.e., the pressure drop $p_2 - p_1$ between the contraction section and the test section. From the Bernoulli equation we have

$\begin{equation*} V_2^2 = \frac{2}{\varrho} \left( p_2 - p_1 \right) + V_1^2 \end{equation*}$

This means that

$\begin{equation*} V_2^2 = \frac{2}{\varrho} \left( p_1 - p_2 \right) + \left( \frac{A_2}{A_1} \right)^2 V_2^2 \end{equation*}$

Solving for $V_2$ (the flow velocity in the test section) we obtain

$\begin{equation*} V_{2} = \sqrt{ \frac{2\left(p_{1} - p_{2}\right)}{\varrho\left[ 1 - \left(\displaystyle {\frac{A_{2}}{A_{1}} }\right)^{2} \right] } } </div> <div>\end{equation*}$

In the second case, the area ratio $A_2/A_1$ must be known, but this is easily measured and is fixed for a given wind tunnel. Again, the density of the air must be known and this is determined by measuring the temperature (using a thermocouple) and pressure (using a pressure transducer) of the air in the flow, and then calculating the density using the equation of state. This technique is also used in low-speed wind tunnels to determine the flow speed in the test section, although it is not as common an approach because, in practice, it is found to be a little less accurate than the first method. The reason is that the static pressure drop $p_1 - p_2$ is much smaller than the difference between $p_T - p_2$ , and so it is more difficult to measure accurately using a pressure transducer.

Worked Example #16

A horizontal water jet with diameter $d$ = 4 cm impacts on a vertical plate at a velocity of 10 m/s. Draw an appropriate sketch and control volume to analyze this problem. By using the conservation laws of fluid dynamics, find the force on the plate, $F$ . Make a one-dimensional, steady flow assumption.

The mass flow rate in the jet is

$\overbigdot{m_j} = \varrho V_j A_j$

where $V_j$ = 10 m/s and $A_j = \pi d_j^2/4 = \pi (0.04)^2/4$ = 0.00126 m $^2$ .

The horizontal jet hits the plate and is brought to rest in the horizontal direction. To do this, the force on the fluid has to be equal to the time rate of change of momentum in that direction and the force on the plate is in the opposite direction, i.e.,

$F = \overbigdot{m_j} V_j = (\varrho V_j A_j) V_j = \varrho A_j V_j^2$

and substituting numbers gives

$F = \varrho A_j V_j^2 = (1,000) (0.00126) 10^2 = 125.66~\mbox{N}$

Worked Example #17

An incompressible gas flow in a exhaust system encounters a change in the size of the exhaust pipe as it transitions to the muffler. The exit diameter of the muffler pipe is $D_2$ = 4.1 inches and the diameter of the inlet header pipe is $D_1$ = 2.2 inches. The exhaust gas is discharged at a speed of $V_2$ = 14.3 ft/s. If the static pressure difference between the header and the muffler is measured using a U-tube water manometer, then calculate the height $h$ between the two columns. Assume 1-dimensional flow and no losses. Also assume that $\varrho_{\rm H_2O} = 1.94$ slug/ft $^3$ and $\varrho_{\rm gas} = 0.0023$ slug/ft $^3$ .

Assuming steady, 1-dimensional flow applying the continuity equation between inlet (section 1) and the exit (section 2) gives

$\varrho_1 \, A_1\, V_1 = \varrho_2 \, A_2 \, V_2$

and because $\varrho_1 = \varrho_2 = \varrho =$ constant in this case (the flow is stated to be incompressible) then

$A_1 \, V_1 = A_2 \, V_2$

Therefore, solving for the inlet velocity $V_1$ gives

$V_1 = \frac{A_2 \, V_2}{A_1} = \frac{\left(\displaystyle{\frac{\pi}{4}}\right)D_2^2 \, V_2}{\left(\displaystyle{\frac{\pi}{4}}\right)D_1^2} = \frac{\left(\displaystyle{\frac{4.1}{12}}\right)^2 \times 14.3 }{\left(\displaystyle{\frac{2.2}{12}} \right)^2} = \frac{4.1^2 \times 14.3 }{2.2^2} = 49.67 \mbox{ ft/s}$

The application of Bernoulli’s equation gives

$p_1 + \frac{1}{2} \, \varrho_{\rm gas}\, V_1^2 + \varrho_{\rm gas} \, g \, z_1 = p_2 + \frac{1}{2} \, \varrho_{\rm gas} \, V_2^2 + \varrho_{\rm gas} \, g \, z_2$

There is no change in vertical height of the mean flow in this case, so $z_1 = z_2 = 0$ and

$p_1 + \frac{1}{2} \, \varrho_{\rm gas}\, V_1^2 = p_2 + \frac{1}{2} \, \varrho_{\rm gas} \, V_2^2$

Rearranging for the pressure difference $p_2 - p_1$ gives

$p_2 - p_1 = \frac{1}{2}\varrho_{\rm gas}\left(V_1^2 - V_2^2\right)$

The U-tube manometer, as shown, will measure this static pressure difference $p_2 - p_1$ , so using the hydrostatic equation the pressure difference is

$p_2 - p_1 = \varrho_{\rm H_2O} \, g \, h$

making sure we use the density of water here. Therefore, equating the latter two equations gives

$\varrho_{\rm H_2O} \, g \, h = \frac{1}{2}\varrho_{\rm gas}\left(V_1^2 - V_2^2\right)$

and solving for $h$ leads to

$h = \frac{\varrho_{\rm gas}}{2 \, \varrho_{\rm H_2O} \, g}\left(V_1^2 - V_2^2\right) = \frac{0.0023}{2 \times 1.94 \times 32.17}\left( 49.67^2 - 14.3^2\right) = 0.0417~\mbox{ft} = 0.5~\mbox{inches}$

Worked Example #18

Gas is flowing at speed $V_2$ through a horizontal section of pipe whose cross-sectional area is $A_2 = 0.0700$ m $^2$ . The gas has a density of $\varrho = 1.34$ kg/m $^3$ . A venturi meter with a cross-sectional area of $A_1 = 0.0500$ m $^2$ and has been substituted for a section of the larger pipe. The pressure difference between the two sections is $p_2 - p_1 = 120.1$ Pa. Find: (a) Find the speed $V_2$ of the gas in the original pipe. (b) Find the speed $V_1$ in the throat of the venturi. (c) Find the volume flow rate $\overbigdot{Q}$ .

(a) We are told the gas has a density of 1.34 kg/m $^3$ implying constant density, so the Bernoulli equation can be used, i.e.,

$p_1 + \frac{1}{2} \varrho V_1^2 + \varrho g z_1 = p_2 + \frac{1}{2} \varrho V_2^2 + \varrho g z_2$

We can assume from the figure that both pressure gages are at the same height so $z_1 = z_2$ so the Bernoulli equation is now

$p_1 + \frac{1}{2} \varrho V_1^2 = p_2 + \frac{1}{2} \varrho V_2^2$

The continuity equation (constant density gas) is

$V_1 A_1 = V_2 A_2$

so

$V_1 = V_2 \left( \frac{ A_2}{A_1} \right)$

The Bernoulli equation becomes

$p_1- p_2 = \frac{1}{2} \varrho \left( V_2^2 - V_1^2 \right) = \frac{1}{2} \varrho V_2^2 \left( 1 - \left( \frac{A_2}{A_1} \right)^2 \right)$

We are given that $A_1 = 0.05$ m $^2$ and $A_2 = 0.07$ m $^2$ so $V_2$ becomes

$V_2 = \sqrt{ \frac{ 2(p_1 - p_2) }{\varrho \left( \left( 1- \displaystyle{\frac{A_2}{A_1}} \right)^2 \right) }} = \sqrt{ \frac{ 2 \times (-120.1)}{1.34 \left( 1 -\left( \displaystyle{\frac{0.07}{0.05}} \right)^2 \right) }} = 13.67~\mbox{m/s}$

(b) Using the continuity equation then

$V_1= V_2 \left( \frac{ A_2}{A_1} \right) = 13.67 \left( \frac{ 0.07}{0.05} \right) = 19.14~\mbox{m/s}$

(c) The volume flow rate $\overbigdot{Q}$ is

$\overbigdot{Q} = A_1 V_1 (= A_2 V_2)$

so we get

$\overbigdot{Q} =A_1 V_1 = 0.05 \times 19.14 = 0.957~\mbox{m$^3$/s}$

Worked Example #19

Water ( $\varrho$ = 1,000 kg/m $^3$ ) is flowing in a hose with a velocity of 1.0 m/s and a pressure of 300 kPa. At the nozzle the pressure decreases to atmospheric pressure (101.3 kPa). There is no change in height. (a) Use the Bernoulli equation to calculate the velocity of the water exiting the nozzle. (b) The hose ejects the water stream through the nozzle of final diameter 3 cm. The stream is then directed onto a stationary wall. Determine the force on the wall from the stream.

(a) We are told to use of the Bernoulli equation ( $\varrho =$ constant), which is

$p_1 + \frac{1}{2} \varrho V_1^2 = p_2 + \frac{1}{2} \varrho V_2^2$

Rearranging to solve for $V_2$ gives

$V_2 = \sqrt{ \frac{2 (p_1 - p_2)}{\varrho} + V_1^2} = \sqrt{ \frac{2 (300.0 - 101.3) \times 10^3}{1,000} + 1.0^2} = 19.96~\mbox{m/s}$

(b) The need for a force suggests the use of the momentum equation. The momentum of the jet before the wall is $\overbigdot{m} V_2$ where

$\overbigdot{m} = \varrho A_2 V_2$

When the jet hits the wall is in brought to rest in the horizontal direction (it will be deflected in all other directions)

$\overbigdot{m} = \varrho A_2 V_2^2 - 0 = F$

where $F$ is the force. Solving for $F$ gives

$F = \varrho A_2 V_2^2 = \varrho \left( \frac{\pi d^2}{4} \right) V_2^2 = 1,000 \left( \frac{\pi \, 0.03^2}{4} \right) 19.96^2 = 281.6~\mbox{N}$

Worked Example #20

Water flows through a converging nozzle. If the pressure at point A at the intake to the nozzle is 19.7 lb/in $^2$ , then determine the height $h$ shown on the U-tube manometer. Notes: Assume atmospheric pressure $p_a$ is 14.7 lb/in $^2$ . The density of water can be assumed to be 1.93 slugs/ft $^3$ , and the specific gravity of mercury is 13.6.

Let the height from point A to point B be $h_1$ . Let $h_2$ be the height from the bottom of the U-tube to the lowest point of the mercury. Considering the left leg of the U-tube, then the pressure at the lowest point in the U-tube will be

$p = p_A + \varrho_{\rm H_2O} \, g h_1 + \varrho_{\rm H_2O} \, g h + \varrho_{\rm Hg} \, g h_2$

For the right leg, then the pressure at the same point will be

$p = p_a + \varrho_{\rm Hg} \, g h + \varrho_{\rm Hg} \, g h_2$

Equating these latter two equations gives

$p_A + \varrho_{\rm H_2O} \, g h_1 + \varrho_{\rm H_2O} \, g h = p_a + \varrho_{\rm Hg} \, g h$

and rearranging gives

$\left( \varrho_{\rm Hg} - \varrho_{\rm H_2O} \right) g h = \left( p_A - p_a \right) + \varrho_{\rm H_2O} \, g h_1$

and solving for $h$ gives

$h = \frac{\left( p_A - p_a \right) + \varrho_{\rm H_2O} \, g h_1}{\left( \varrho_{\rm Hg} - \varrho_{\rm H_2O} \right) g}$

Now we can substitute the information given. Notice that $p_A - p_a$ = 5 lb/in $^2$ = (5 $\times$ 144) lb/ft $^2$ = 720.0 lb/ft $^2$ . Therefore,

$h = \frac{720.0 + 1.93 \times 32.17 \times (24.0/12) }{(13.6 - 1) \times 1.93 \times 32.17} = 1.079~\mbox{ft} = 12.94~\mbox{inches}$

Worked Example #21

Water flows through a pipe of circular cross-section with a volume flow rate of 14.2 ft $^3$ /s encounters a change in the area and height of the pipe as it passes through an elbow-type of coupling. The exit area of the pipe is $A_2$ = 2.4 ft $^2$ and the area of the inlet is $A_1$ = 4.8 ft $^2$ . Assume 1-dimensional flow and no losses with $\varrho_{\rm H_2O} = 1.94$ slug/ft $^3$ . (a) Determine the flow velocities at the entrance ( $V_1$ ) and the exit ( $V_2$ ) in units of ft/s. (b) If $z_1$ = 6 inches and $z_2$ = 25 inches, what is the static pressure difference between the inlet and outlet, i.e., the value of $p_1 - p_2$ , in units of lb/in $^2$ .

(a) The fluid is water so it is incompressible. The application of the continuity equation in one-dimensional form gives

$V_1 \, A_ 1 = A_ 2 V_2 = \overbigdot{Q} = \mbox{constant}$

Therefore, the inlet flow velocity $V_1$ is

$V_1 = \frac{\overbigdot{Q}}{A_1} = \frac{14.2}{4.8} = 2.96~\mbox{ft/s}$

and for the outlet flow velocity $V_2$ then we know that

$V_1 \, A_ 1 = A_ 2 V_2$

so

$V_2 = \left( \frac{A_2}{A_1} \right) V_1 = \left( \frac{4.8}{2.4} \right) V_1 = 2 \, V_1 = 5.92~\mbox{ft/s}$

It will also be noted that

$V_2 = \frac{\overbigdot{Q}}{A_2} = \frac{14.2}{2.4} = 5.92~\mbox{ft/s}$

(b) The Bernoulli equation is

$p + \frac{1}{2} \varrho V^2 + \varrho g z = \mbox{constant}$

so in this case

$p_1 + \frac{1}{2} \varrho_{\rm H_2O} \, V_1^2 + \varrho_{\rm H_2O} \, g \, z_1 = p_2 + \frac{1}{2} \varrho_{\rm H_2O} \, V_2^2 + \varrho_{\rm H_2O} \, g \, z_2$

We are asked for the pressure difference $p_1 - p_2$ so using the Bernoulli equation gives

$p_1 - p_2 = \frac{1}{2} \varrho_{\rm H_2O} \left( V_2^2 - V_1^2 \right) + \varrho_{\rm H_2O} \, g \left( z_2 - z_1 \right)$

and substituting the known values gives

$p_1 - p_2 = \frac{1}{2} \times 1.94 \left( 5.92^2 - 2.96^2 \right) + 1.94 \times 32.17 \times \left( 25.0 - 6.0 \right)/12$

noting to convert the hydrostatic head (last term) from units of inches to feet to keep the entire equation in consistent base units of lb, ft, and seconds (s). Evaluating the arithmetic gives

$p_1 - p_2 = 124.31~\mbox{lb/ft$^2$} = 141.31/144 ~\mbox{lb/in$^2$} = 0.863~\mbox{lb/in$^2$}$

noting that 144 lb/in $^2$ is equal to 1 lb/ft $^2$ .

Worked Example #22

The 6 cm diameter water jet as shown in the figure below strikes a flat plate containing a hole of 4 cm in diameter. Part of the jet passes through the hole and part is deflected radially symmetrically away from the plate. Assume the density of water is 1,000 kg m $^{-3}$ . (a) Calculate the mass flow rate of the jet toward the plate; (b) Calculate the mass flow rate of the part of the jet that is deflected; (c) Determine the magnitude and direction of the force required to hold the plate stationary.

(a) Assumptions: Incompressible, steady, no body force, inviscid, and one-dimensional flow. The mass flow rate of the jet towards the plate is

$\overbigdot{m}_1 = \varrho V_1 A_1$

and inserting the values gives

$\overbigdot{m}_1 = 998.23\times 25\times(\pi/4)\times(0.06)^2 = 70.56~\mbox{kg/s}$

(b) Conservation of mass in the one-dimensional form gives

$\varrho V_1 (-A_1) + \varrho V_2 A_2 + \varrho V_3 A_3 + \varrho (-V_4) (-A_4) = 0$

and so

$-70.56 + (998.23 \times 25 \times \pi/4 \times 0.04^2) + \overbigdot{m}_3 + \overbigdot{m}_4 = 0$

The velocities $V_3 = V_4$ , so the mass flow at section 3 and section 4 are equal so

$-70.56 + 31.36 + 2\overbigdot{m}_3 = 0$

so the mass flow rate of the part of the jet that is deflected is

$$2\overbigdot{m}_3$ = 39.2~\mbox{kg/s}$

and

$$\overbigdot{m}_3$ = 19.6~\mbox{kg/s}$

(c) The system is subjected to atmospheric pressure and the net pressure acting on the control volume is zero, Therefore, from conservation of momentum

$\vec{F_x} = \varrho (V_1 (-A_1))V_1 + \varrho (V_2 A_2)V_2$

and inserting the numbers gives

$F_x = (-70.56\times 25) + (31.36 \times 25) = -980.01~\mbox{N}$

Therefore, the force applied to the plate is

$R_x = 980.01 N$

and so to hold the plate stationary, a force of -980.01 N is required, i.e., to the left.

Worked Example #23

Two plates, set 4 cm apart, are separated by a bed of oil. The bottom plate is stationary but the top plate moves at a speed of 3.5 m/s. Calculate the shear stress in the oil. Assume that the coefficient of dynamic viscosity of the oil is 583.95 Pa s. Note: Pa s is an SI unit of viscosity, and is equivalent to Newton-second per square meter (N s m $^{-2}$ ), which is sometimes referred to as the Poiseuille (Pl).

It is necessary to calculate the shear stress in a fluid, $\tau$ , based on the formation of a velocity gradient. The use of Newton’s formula of viscosity gives

$\tau = \mu \left( \frac{du}{dy} \right)$

Newtonian flow will give a constant uniform velocity gradient in this case between the upper and lower plates will be

$\frac{du}{dy} = \frac{3.4}{0.04} = 87.5 \mbox{ s$^{-1}$}$

This means that the shear stress per unit depth in the fluid will be

$\tau = \mu \left( \frac{du}{dy} \right) = 51.1 \mbox{ kPa}$

Worked Example #24

Water ( $\varrho_w$ = 1.94 slug ft $^{-3}$ ) flows at a mass flow rate of $\overbigdot{m}$ = 6.0 slug s $^{-1}$ through a nozzle with an entrance diameter $d_e = 10.0$ in and and exit diameter of $d_j = 4.0$ . The water hits an angled bracket, as shown in the figure below, such that exactly half of the mass flow rate of the stream deflects to the left and half deflects to the right. The diameter of the deflected jets, $d_d$ , is 1.3 in. Assume that the water all stays in a horizontal plane and one-dimensional flow.

Calculate the following:

The velocity of the water entering and exiting the nozzle, $V_e$ and $V_j$ .
The change in pressure $\Delta p$ between the nozzle entrance and its exit.
The velocity $V_d$ of the water in the deflected jets.
The magnitude and direction of the force on the bracket.

1. The velocity of the water entering and exiting the nozzle can be determined using conservation of mass (continuity equation), i.e.,

$\overbigdot{m} = \varrho_w \, A_e \, V_e = \varrho_w \, A_j \, V_j$

The entrance area $A_e$ is

$A_e = \frac{\pi d_e^2}{4} = \frac{\pi (10.0/12.0)^2}{4} = 0.5454~\mbox{ft$^2$}$

The entrance velocity is

$V_e = \frac{\overbigdot{m}}{ \varrho_w \, A_j}$

and substituting the values gives

$V_e = \frac{6.0}{1.94 \times 0.5454} = 5.67~\mbox{ft s$^{-1}$}$

The exit jet area $A_j$ is

$A_j = \frac{\pi d^2}{4} = \frac{\pi (4.0/12.)^2}{4} = 0.0873~\mbox{ft$^2$}$

The exit jet velocity is

$V_j = \frac{\overbigdot{m}}{ \varrho_w \, A_j}$

and substituting the values gives

$V_j = \frac{6.0}{1.94 \times 0.0873} = 35.43~\mbox{ft s$^{-1}$}$

2. The change in pressure $\Delta p$ between the nozzle inlet and its exit can be found using the Bernoulli equation, i.e.,

$p + \frac{1}{2} \varrho V^2 = \mbox{constant}$

In this case then

$p_e + \frac{1}{2} \varrho V_e^2 = p_j + \frac{1}{2} \varrho V_j^2$

We are asked for the change in pressure so

$p_e - p_j = \Delta p = \frac{1}{2} \varrho \left( V_j^2 - V_e^2 \right)$

and substituting the values gives

$\Delta p = 0.5 \times 1.94 \times \left( 35.43^2 - 5.67^2 \right) = 1,1186.8~\mbox{lb ft$^{-2}$}$

3. The velocity $V_2$ of the water in the deflected jet can be found using continuity. In this case, half of the mass flow is deflected in each direction so

$\frac{\overbigdot{m}}{2} = \varrho_w \, A_d \, V_d$

The deflected jet area $A_j$ is

$A_d = \frac{\pi d_d^2}{4} = \frac{\pi (1.3/12.0)^2}{4} = 0.00922~\mbox{ft$^2$}$

The deflected jet velocity is

$V_d = \frac{\overbigdot{m}}{ 2 \varrho_w \, A_d}$

and substituting the values gives

$V_d = \frac{6.0}{2 \times 1.94 \times 0.00922} = 167.72~\mbox{ft s$^{-1}$}$

4. The force of the plate from the water can be determined from the time rate of change of momentum of the fluid, a suitable control volume being shown below. To this end, the rate of momentum of the water coming onto the deflector in the downstream (horizontal) direction is $\overbigdot{m} V_j$ . Likewise, the rate of deflected momentum coming out away from the defector is $-\overbigdot{m} V_d \, \cos 45^{\circ}$ . Notice that the $\cos 45^{\circ}$ term give the correct horizontal component of momentum.

Therefore, the force on the plate will be the time rate of the net momentum so that

$F = \overbigdot{m} V_j - (- \overbigdot{m} V_d \, \cos (45^{\circ}) = \overbigdot{m} V_j + \overbigdot{m} V_d \, \cos (45^{\circ}) = \overbigdot{m} \left( V_j + V_d \, \cos (45^{\circ} \right)$

Inserting the numbers gives

$F = \overbigdot{m} \left( V_j + V_d \, \cos 45^{\circ} \right) = 6.0 \times \left(35.43 +167.72 \times 0.707 \right) = 924.16~\mbox{lb}$

which will be applied by the water in the downstream direction away from the nozzle

Worked Example #25

Consider the steady flow of a liquid through an elbow-shaped converging axisymmetric pipe joint with an inlet area $A_1$ and outlet area $A_2$ , as shown in the figure below. The volumetric flow rate is $\overbigdot{Q}$ . The elbow angle is $\theta$ in a horizontal plane. Assume that the flow inside the pipe is one-dimensional. Use the continuity and momentum equations to set up the procedure to find the resultant force on the pipe. Note: You do not have to find the force; just show how to use and set up the relevant equations.

The flow is steady, so $\partial/\partial t \equiv 0$ , and the flow can be assumed as inviscid, so $\vec{F}_{\mu} = 0$ . The fluid is a liquid (which has a relatively high density), but there is no change in elevation from one side of the nozzle to the other. Therefore, gravitational effects can be neglected.

The general form of the momentum equation is

$\iiint_{\cal{V}} \varrho \vec{F} _b d{\cal{V}} -\oiint_S p \, d\vec{S} + \vec{F}_{\mu} = \frac{\partial}{\partial t}\iiint_{\cal{V}} \varrho \, \vec{V} d {\cal{V}} + \oiint_S (\varrho \, \vec{V} \bigcdot d\vec{S}) \vec{V}$

and the reduced form for the assumptions made in this question leads to

$- \oiint_S p \, d\vec{S} = \oiint_S (\varrho \, \vec{V} \bigcdot d\vec{S}) \vec{V}$

The pressure integral on the left-hand side can be written as

$\oiint_S p \, d\vec{S} = \iint_{A_1} p \, d\vec{S} + \iint_{A_2} p \, d\vec{S} + \iint_{\rm Walls} p \, d\vec{S}$

In this case, the force of the walls on the fluid, say $\vec{F}$ (to change its momentum), needs to be determined, i.e., this is the unknown. There is also an equal and opposite force on the nozzle, say $\vec{R}$ , which ultimately is the force to be determined.

We are told to assume one-dimensional over the cross-section of the nozzle, but the flow moves in two directions in this case, $x$ and $y$ . Therefore, there will be two scalar momentum equations in the $x$ and $y$ directions, respectively. In general form, then the momentum equation is

$-\oiint_S p \, d\vec{S} = - \vec{F} - \iint_{A_1} p \, d\vec{S} - \iint_{A_2} p \, d\vec{S} = \oiint_S (\varrho \, \vec{V} \bigcdot d\vec{S}) \vec{V}$

With the one-dimensional assumption, then for the $x$ direction, then

$-F_{x} - p_1 A_1 + p_2 A_2 \cos \theta = (\varrho_2 A_2 V_2) V_2 \cos \theta - \varrho_1 A_1 V_1 ^2$

so that

$F_{x} = (p_2 A_2 \cos \theta - p_1 A_1) + \varrho_2 A_2 V_2^2 \cos \theta - \varrho_1 A_1 V_1^2$

For the momentum equation in the $y$ direction, then

$-F_{y} - 0 - p_2 A_2 \sin \theta = (\varrho_2 A_2 V_2) V_2 \sin \theta - 0$

so that

$F_{y} = p_2 A_2 \sin \theta + \varrho_2 A_2 V_2^2 \sin \theta$

We are not asked to proceed any further from this point, just set up the equations to be solved, and in summary these are

$F_{x} = (p_2 A_2 \cos \theta - p_1 A_1) + \varrho_2 A_2 V_2^2 \cos \theta - \varrho_1 A_1 V_1^2$

and

$F_{y} = p_2 A_2 \sin \theta + \varrho_2 A_2 V_2^2 \sin \theta$

We also have the continuity equation to round out the other two equations, i.e.,

$\overbigdot{Q} = A_1 V_1 = A_2 V_2$

License

Icon for the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License

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

Digital Object Identifier (DOI)

https://doi.org/https://doi.org/10.15394/eaglepub.2022.1066.9