Net Force on a Flange
Water flows at a rate of 18.5 gal/min through a flanged faucet with a partially closed gate valve spigot (Fig. 6–27). The inner diameter of the pipe at the location of the flange is 0.780 in (= 0.0650 ft), and the pressure at that location is measured to be 13.0 psig. The total weight of the faucet assembly plus the water within it is 12.8 lbf. Calculate the net force on the flange.
SOLUTION Water flow through a flanged faucet is considered. The net force acting on the flange is to be calculated.
Assumptions 1 The flow is steady and incompressible. 2 The flow at the inlet and at the outlet is turbulent and fully developed so that the momentum-flux correction factor is about 1.03. 3 The pipe diameter at the outlet of the faucet is the same as that at the flange.
Properties The density of water at room temperature is 62.3 lbm/ft³.
Analysis We choose the faucet and its immediate surroundings as the control volume, as shown in Fig. 6–27 along with all the forces acting on it. These forces include the weight of the water and the weight of the faucet assembly, the gage pressure force at the inlet to the control volume, and the net force of the flange on the control volume, which we call \vec{F}_{R}. We use gage pressure for convenience since the gage pressure on the rest of the control surface is zero (atmospheric pressure). Note that the pressure through the outlet of the control volume is also atmospheric since we are assuming incompressible flow; hence, the gage pressure is also zero through the outlet.
We now apply the control volume conservation laws. Conservation of mass is trivial here since there is only one inlet and one outlet; namely, the mass flow rate into the control volume is equal to the mass flow rate out of the control volume. Also, the outflow and inflow average velocities are identical since the inner diameter is constant and the water is incompressible, and are determined to be
V_{2}=V_{1}=V=\frac{\dot{V}}{A_{c}}=\frac{\dot{V}}{\pi D^{2} / 4}=\frac{18.5 gal / min }{\pi(0.065 ft )^{2} / 4}\left(\frac{0.1337 ft ^{3}}{1 gal }\right)\left(\frac{1 min }{60 s }\right)=12.42 ft / s
Also,
\dot{m}=\rho \dot{V}=\left(62.3 lbm / ft ^{3}\right)(18.5 gal / min )\left(\frac{0.1337 ft ^{3}}{1 gal }\right)\left(\frac{1 min }{60 s }\right)=2.568 lbm / s
Next we apply the momentum equation for steady flow,
\sum \vec{F}=\sum_{\text {out }} \beta \dot{m} \vec{V}-\sum_{\text {in }} \beta \dot{m} \vec{V} (1)
We let the x– and z-components of the force acting on the flange be F_{R x} \text { and } F_{R z} \text {, } and assume them to be in the positive directions. The magnitude of the velocity in the x-direction is +V_{1} at the inlet, but zero at the outlet. The magnitude of the velocity in the z-direction is zero at the inlet, but -V_{2} at the outlet. Also, the weight of the faucet assembly and the water within it acts in the –z-direction as a body force. No pressure or viscous forces act on the chosen (wise) control volume in the z-direction.
The components of Eq. 1 along the x– and z-directions become
\begin{aligned}&F_{R x}+P_{1, \text { gage }} A_{1}=0-\dot{m}\left(+V_{1}\right) \\&F_{R z}-W_{\text {faucet }}-W_{\text {water }}=\dot{m}\left(-V_{2}\right)-0\end{aligned}
Solving for F_{R x} \text { and } F_{R z} \text {, } and substituting the given values,
\begin{aligned}F_{R x} &=-\dot{m} V_{1}-P_{1, \text { gage }} A_{1} \\&=-(2.568 lbm / s )(12.42 ft / s )\left(\frac{1 lbf }{32.2 lbm \cdot ft / s ^{2}}\right)-\left(13 lbf / in ^{2}\right) \frac{\pi(0.780 in )^{2}}{4} \\&=-7.20 lbf\end{aligned}
\begin{aligned}F_{R z} &=-\dot{m} V_{2}+W_{\text {faucet }+\text { water }} \\&=-(2.568 lbm / s )(12.42 ft / s )\left(\frac{1 lbf }{32.2 lbm \cdot ft / s ^{2}}\right)+12.8 lbf =11.8 lbf\end{aligned}
Then the net force of the flange on the control volume is expressed in vector form as
\vec{F}_{R}=F_{R x} \vec{i}+F_{R z} \vec{k}=-7.20 \vec{i}+11.8 \vec{k} lbf
From Newton’s third law, the force the faucet assembly exerts on the flange is the negative of \vec{F}_{R},
\vec{F}_{\text {faucet on flange }}=-\vec{F}_{R}=7.20 \vec{i}-11.8 \vec{k} \text { lbf }
Discussion The faucet assembly pulls to the right and down; this agrees with our intuition. Namely, the water exerts a high pressure at the inlet, but the outlet pressure is atmospheric. In addition, the momentum of the water at the inlet in the x-direction is lost in the turn, causing an additional force to the right on the pipe walls. The faucet assembly weighs much more than the momentum effect of the water, so we expect the force to be downward. Note that labeling forces such as “faucet on flange” clarifies the direction of the force.