A decorative fountain uses a nozzle that is press-ﬁtted to the end of a high pressure water line as shown in Figure 7.17A. Find the force that must be exerted on the nozzle by the press ﬁtting to keep it in place. Assume uniform velocities and pressures at the inlet and exit of the nozzle.

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Accepted Answer

Figure 7.17A is a sketch of the physical arrangement. We will assume steady ﬂow. The nozzle is the object of interest, and the force applied by the ﬁtting to the nozzle is an external force F_E on the nozzle. Since we are asked to ﬁnd this external force, we begin as recommended by writing a vector force balance on the nozzle as shown in Figure 7.17B. The forces acting on the nozzle include the forces due to the water inside and air outside, the force of gravity, and the external force F_E exerted on the nozzle by the press ﬁt. The force balance in vector form is

F_water + F_air + F_gravity + F_E = 0

Thus the external force applied by the ﬁtting to the nozzle is given by

F_E= −F_water − F_air − F_gravity (A)

To ﬁnd the external force, we must ﬁnd the forces applied to the nozzle by gravity, the water, and the air. The force of gravity on the nozzle is F_gravity = −W_nozzlek. The force due to the water is found by considering the surface force applied by the water to the interior wetted surface of the nozzle. This surface force is given by integrating the stress vector over this surface, thus we can write

[latex]F_{water}=\int_{nozzle}^{}{\sum{dS}} =\int_{nozzle}^{}{(−pn+\pmb{ au} )dS } [/latex]

Since we are not given information about the pressure and shear stress distributions on the nozzle, this integral cannot be evaluated directly. Thus we will ﬁnd this force by a momentum balance on the CV shown in Figure 7.17C. Note that by placing a decal surface adjacent to the wetted surface of the nozzle, we can use the principle of action–reaction to show that the unknown force F_water is equal and opposite to the reaction force R on the decal surface. Thus the force of the water on the nozzle can be found from a momentum balance using the chosen CV.

Now consider the force of the air at atmospheric pressure on the outside of the nozzle. The air is at rest, so the force is given by the integral [latex]F_{air}=\int_{S_{air}}^{}{\sum{dS}} =\int_{S_{air}}^{}{−p_An \ dS } [/latex], where the integral is taken over the exterior surface S_air of the nozzle in contact with the air. We can use the following trick to evaluate this integral over a curved surface. Consider the surface force due to atmospheric pressure on the surface S shown in Figure 7.17D. This surfaced uplicates the exterior surface of the nozzle, but with the inlet and exit openings closed. Since a constant pressure acts everywhere on this closed surface, the integral [latex]\int_{S}^{}{−p_An \ dS}[/latex] over the entire surface must be zero. Thus we can write

[latex]\int_{S}^{}{−p_An \ dS} =\int_{S_{air}}^{}{−p_An \ dS }+\int_{I}^{}{−p_An \ dS}+\int_{\Pi}^{}{−p_An \ dS}=0[/latex]

Solving for the desired integral of atmospheric pressure over the exterior surface, which is equal to F_air we have

[latex]F_{air}=\int_{I}^{}{−p_An \ dS}-\int_{\Pi}^{}{−p_An \ dS}[/latex]

If the nozzle wall is thin, the open surfaces I and II are virtually identical to the inlet and exit CV surfaces, and the desired force of atmospheric pressure over the exterior surface is given by

[latex]F_{air}=-\int_{inlet}^{}{−p_An \ dS}-\int_{exit}^{}{−p_An \ dS}[/latex] (B)

This methodology for evaluating the force of atmospheric pressure on part of the surface of an object should be noted for future use, since it allows us to calculate the surface force due to atmospheric pressure on any surface with openings by considering the standard integral of atmospheric pressure over each opening and changing the sign in front of each integral.

Upon inserting the preceding results into the force balance (A), we see that the external force supplied by the press ﬁtting is given by

[latex]F_{E}=-\int_{nozzle}^{}{\sum{dS} } +\int_{inlet}^{}{−p_An \ dS}+\int_{exit}^{}{−p_An \ dS}+W_{nozzle}k[/latex] (C)

To evaluate the integral over the nozzle interior surface in (C), we use the CV with a decal surface adjacent to the interior wall of the nozzle as shown in Figure 7.17C. Applying Eq. 7.19, we have

[latex]\int_{CS}^{}{(ρu)(u • n)dS}=\int_{CV}^{}{ρf dV}+\int_{CS}^{}{\sum{dS} }[/latex]

The momentum transport terms on the inlet and exit are evaluated by inspection. On the inlet we have u = V₁k, n = −k, and u • n = −V₁, so that [latex]\int_{inlet}^{}{} [/latex] (ρu)(u • n)dS = (ρV₁k)(−V₁)A₁. On the exit, u = V₂k, n = k, and u • n = V₂, so that [latex]\int_{exit}^{}{} [/latex] (ρu)(u • n)dS = (ρV₂k)(V₂)A₂. The body force integral is simply the weight of the ﬂuid in the CV or −W_CVk. The stress terms are evaluated by noting that the stress on the inlet and exit is due to the pressure. Thus we ﬁnd

[latex]\int_{CS}^{}{\sum{dS} } =\int_{inlet}^{}{−p_1n \ dS}+\int_{exit}^{}{−p_2n \ dS}+\int_{decal}^{}{\sum{dS} }[/latex]

Although the stress on the decal surface is unknown, this integral is the reaction force R. By the principle of action–reaction, we know that F_water = − R. Thus in this problem we can write [latex]\int_{decal}^{}{\sum{dS} }[/latex] = R = −F_water = − [latex]\int_{nozzle}^{}{\sum{dS} }[/latex]. The completed momentum balance is then

[latex]ρV^2 _2 A_2k−ρV^2 _1 A_1k=−W_{CV}k+\int_{inlet}^{}{−p_1n \ dS}+\int_{exit}^{}{−p_2n \ dS}-\int_{nozzle}^{}{\sum{dS} }[/latex] (D)

We now have two equations, (C) and (D), to determine the external force applied to the nozzle by the ﬁtting. Rearranging (D) we have

[latex]-\int_{nozzle}^{}{\sum{dS} }= \left(ρV^2 _2 A_2k−ρV^2 _1 A_1j ight) +W_{CV}k+\int_{inlet}^{}{p_1n \ dS}+\int_{exit}^{}{p_2n \ dS}[/latex]

Inserting this into (C) and combining the pressure integrals gives

[latex]F_E= \left(ρV^2 _2 A_2k−ρV^2 _1 A_1k ight)+\int_{inlet}^{}{(p_1-p_A)n \ dS}+\int_{exit}^{}{(p_2-p_A)n \ dS}+(W_{CV}+W_{nozzle})k[/latex]

We see that the effect of the air at atmospheric pressure on the outside of the nozzle is equivalent to using gage pressure at the open surfaces of the control volume, i.e., the inlet and exit. Since the pressure is uniform on these surfaces, these integrals give [latex]\int_{inlet}^{}{} [/latex] (p₁ − p_A)n dS = (p₁ − p_A)(−k)A₁ and [latex]\int_{exit}^{}{} [/latex] (p₂ − p_A)n dS = (p₂ − p_A)(k)A₂. Thus the external force on the nozzle is given by

[latex]F_E =\left(p_{2_{gage} } +ρV^2 _2 ight) A_2k−\left(p_{1_{gage} } +ρV^2 _1 ight) A_1k+(W_{CV} +W_{nozzle} )k[/latex] (E)

This is the answer to the problem. To see whether it is reasonable, we can imagine the situation in which the nozzle is plugged so that no water is ﬂowing and a hydrostatic pressure [latex]p_{1_{gage} } [/latex] acts at the nozzle inlet, as shown in Figure 7.17E. The force of the water on the inside of the nozzle is found by using the indirect method for a curved surface as explained in Section 5.5.3. The result is

F_water = − [latex]\left(p_{1_{gage} } ight)[/latex] (−k)A₁ −W_CVk = [latex]p_{1_{gage}}[/latex] A₁k−W_CVk

which is the inlet gage pressure acting on the projected area of the nozzle less the expected small weight contribution, which accounts for the fact that the static pressure on the inside wetted surface of the nozzle is slightly less than [latex]p_{1_{gage} } [/latex] at the higher elevations. From the force balance (A), the required external force with the nozzle plugged is thus F_E= − [latex]\left(p_{1_{gage} } ight)[/latex] A₁k + (W_CV + W_nozzle)k. By putting V₁ = V₂ = 0 and A₂ = 0 into (E), we get exactly the same result. In most cases the weight of the ﬂuid in the CV and weight of the nozzle are negligible, so the external force required with the nozzle plugged is usually approximated as

F_E= -[latex]\left(p_{1_{gage} } ight)[/latex] A₁k (F)

As expected, the press ﬁtting must resist the tendency of the hydrostatic pressure to blow the nozzle off the pipe.

When the water is moving, the jet of liquid exiting the nozzle is at atmospheric pressure. Neglecting the effects of gravity, we have from (E) [latex]F_E = ρV^2 _2 A_2k−(p_{1_{gage} }+ρV^2 _1 )A_1k[/latex]. Amass balance reveals that [latex]\dot M = ρA_1V_1 = ρA_2V_2[/latex], so we can write this as

[latex]F_E =−\left(p_{1_{gage} }A_1 ight)k+\dot M(V_2 −V_1)k[/latex] (G)

Once again we see that the press ﬁtting must create a frictional force that resists the tendency of the nozzle to be blown off the pipe. With the water moving, both pressure and shear stress act on the nozzle, but now the pressure distribution inside the nozzle is far from hydrostatic. If the nozzle inlet pressure [latex]p_{1_{gage} } [/latex] is the same in both cases, which is true for a reasonably small nozzle opening, comparing (G) and (F) shows that the force required to keep the nozzle on the pipe is smaller with the nozzle open than when the nozzle is plugged. Does this agree with your intuition?

We can also observe that the completed momentum balance (D) allows us to write the unknown force of the water on the nozzle as

[latex]F_{water}=\int_{nozzle}^{}{\sum{dS} } =-\left(ρV^2 _2 A_2j−ρV^2 _1 A_1j ight)-W_{CV}k+\int_{inlet}^{}{-p_1n \ dS}+\int_{exit}^{}{-p_2n \ dS}[/latex]

Evaluating the pressure terms and neglecting the weight of water in the control volume, we have

[latex]F_{water} =\left(p_1 +ρV^2 _1 ight)A_1k−\left(p_A +ρV_2 ^2 ight)A_2k [/latex] (H)

This example problem can be solved more quickly and efficiently by using a mixed control volume that contains both ﬂuid and solid, as shown in Figure 7.17F.The CV surrounds the entire nozzle and has a decal surface along the press ﬁt, where the external force on the nozzle inside the CV occurs. Writing the momentum balance for this CV yields

[latex]\int_{inlet}^{}{(ρu)(u • n)dS}+\int_{exit}^{}{(ρu)(u • n)dS}=\int_{CV}^{}{ ho fdV}+\int_{inlet}^{}{\sum{dS}}+\int_{exit}^{}{\sum{dS}}+\int_{decal}^{}{\sum{dS}}+\int_{air}^{}{\sum{dS}}[/latex]

Where the integral of the surface force over the decal surface adjacent to the press ﬁt deﬁnes F_E, the external force acting on the nozzle inside the CV. Solving for this force, we have

[latex]F_E=\int_{decal}^{}{\sum{dS}}=\int_{inlet}^{}{(ρu)(u • n)dS}+\int_{exit}^{}{ (ρu)(u • n)dS}=\int_{CV}^{}{ ho fdV}-\int_{inlet}^{}{\sum{dS}}-\int_{exit}^{}{\sum{dS}}-\int_{air}^{}{\sum{dS}}[/latex]

Using the procedures outlined earlier to evaluate each term on the right-hand side yields

F_E = (ρV₁k)(−V₁)A₁ +(ρV₂k)(V₂)A₂ +(W_CV + W_nozzle)k − p₁A₁k + p₂A₂k − (−p_AA₁k + p_AA₂k)

Introducing gage pressures shows that this is identical to (E), the preceding result. Note that the body force term in the momentum balance for a mixed CV must account for the weight of the nozzle and the ﬂuid inside the CV, since both the nozzle and its ﬂuid contents are inside the mixed CV. In this example, using a mixed CV to ﬁnd the external force is much quicker and easier than using a CV ﬁlled only with ﬂuid. However, if the problem had asked for the force of the water on the nozzle, this mixed CV would not work because it does not have a decal surface adjacent to the wetted surface of the nozzle.

Question 7.9: A decorative fountain uses a nozzle that is press-ﬁtted to t...

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