Faucet at the Top Floor!

Use the continuity equation and Bernoulli’s Law to solve for the velocity of the faucet with a \(2 \times 2\) system of equations.

The continuity equation states:

\begin{equation*}
v_{\text{pipe}}A_{\text{pipe}}=v_{\text{faucet}}A_{\text{faucet}},
\end{equation*}

where the area are transverse circular areas, then \(A = \pi r^2\) which can allows us to relate the velocities. For this problem, Bernoulli’s equation can be written as:

\begin{equation*}
P_{\text{pipe}}+\frac{1}{2}\rho v^2_{\text{pipe}}+\rho g z_{\text{pipe}}=P_{\text{faucet}}+\frac{1}{2}\rho v^2_{\text{faucet}}+\rho g z_{\text{faucet}},
\end{equation*}

where using the relation of the velocities, and solving for \(v_{\text{faucet}}\) after a lot of algebra we get:

\begin{equation*}
v_{\text{faucet}}=\sqrt{\frac{2(P_{\text{pipe}}-P_0-\rho gh)}{\rho\left(1-\frac{ r_{\text{faucet}}^4}{r_{\text{pipe}}^4}\right) }}.
\end{equation*}

With numerical values:

\begin{equation*}
v_{\text{faucet}} \approx 4.18 \, \text{m/s}.
\end{equation*}

For a more detailed explanation of any of these steps, click on “Detailed Solution”.

We need to find the speed and pressure of the water on the top floor. First, notice that the pressure \(P\) when water pours out of a facet is the same pressure as the surroundings, thus the pressure at the faucet is the atmospheric pressure \(P_0=101325\,\text{Pa}\).

To solve for the speed we must use the equations of steady flow: the conservation of mass equation and the Bernoulli equation. The conservation of mass equations states that at any point in a flow, the amount of mass \(m\) per unit time \(t\) is always the same. That is

\begin{equation}
\label{dmdt}
\frac{dm}{dt}=\text{constant}.
\end{equation}

Writing the mass \(m\) in terms of the fluid’s density \(\rho\) and its volume \(V\) as

\begin{equation}
m=\rho V,
\end{equation}

we can rewrite equation \eqref{dmdt} as

\begin{equation}
\label{flux}
\rho\frac{dV}{dt}=\text{constant}.
\end{equation}

If the density of the fluid is constant along all the flow, then the quantity that must be a constant is the flux

\begin{equation}
\label{flux2}
\frac{dV}{dt}=\text{constant}.
\end{equation}

Thus, the flux at the inlet of the pipe must be the same at the faucet when fully open, that is

\begin{equation}
\label{flux3}
\frac{dV}{dt}\Big|_{\text{pipe}}=\frac{dV}{dt}\Big|_{\text{faucet}}.
\end{equation}

For a steady flow, the flux can be calculated at any point as the product of the velocity of the fluid \(v\) by the transverse area \(A\), namely

\begin{equation}
\label{va}
\frac{dV}{dt}=vA.
\end{equation}

Using equation \eqref{va} into both sides of equation \eqref{flux3}, we obtain

\begin{equation}
\label{va2}
v_{\text{pipe}}A_{\text{pipe}}=v_{\text{faucet}}A_{\text{faucet}},
\end{equation}

where \(v_{\text{pipe}}\) and \(v_{\text{faucet}}\) is the velocity of water at the beginning of the pipes and at the faucet respectively. The terms \(A_{\text{pipe}}\) and \(A_{\text{faucet}}\) are the transverse areas of the pipes and faucets respectively. Because both transverse areas are circular, they can be calculated according to the expression

\begin{equation}
\label{area}
A=\pi r^2,
\end{equation}

where \(r\) is the radius. Using equation \eqref{area} into equation \eqref{va2}, we obtain

\begin{equation}
v_{\text{pipe}}\pi r_{\text{pipe}}^2=v_{\text{faucet}}\pi r_{\text{faucet}}^2,
\end{equation}

where we can solve for \(v_{\text{pipe}}\) to get

\begin{equation}
v_{\text{pipe}}=\frac{v_{\text{faucet}}\pi r_{\text{faucet}}^2}{\pi r_{\text{pipe}}^2},
\end{equation}

which after simplification becomes

\begin{equation}
\label{vfaucet}
v_{\text{pipe}}=\frac{v_{\text{faucet}} r_{\text{faucet}}^2}{ r_{\text{pipe}}^2}.
\end{equation}

The relation for velocities given in equation \eqref{vfaucet} will be useful to solve in the Bernoulli equation. The Bernoulli equation applies to steady and irrotational and incompressible flows, such as the one that we have. The equation reads that for each point in a streamline

\begin{equation}
\label{bernoulli}
P+\frac{1}{2}\rho v^{2}+\rho g z=\text{constant},
\end{equation}

where \(P\) is the absolute pressure in the fluid, \(v\) the velocity of the fluid along the streamline, \(z\) the height of the point in the streamline, and \(g\) the gravitational acceleration.

We can then write Bernoulli’s equation for two points in the same streamline, that is when the flow enters the pipes and when the flow pours out of the faucet. Then using equation \eqref{bernoulli} we get

\begin{equation}
\label{ber2}
P_{\text{pipe}}+\frac{1}{2}\rho v^2_{\text{pipe}}+\rho g z_{\text{pipe}}=P_{\text{faucet}}+\frac{1}{2}\rho v^2_{\text{faucet}}+\rho g z_{\text{faucet}}.
\end{equation}

The pressure \(P_{\text{pipe}}\) is given numerically by the problem while the pressure \(P_{\text{faucet}}\) is the atmospheric pressure \(P_0\) due to the water pouring out to the exterior. From figure 1, we see that from our coordinate system, with origin at the floor, \(z_{\text{pipe}}=0\) and \(z_{\text{faucet}}\) is the height of the building \(h\).

Figure 1: Water flowing through the pipe from the street level up to the open faucet. The arrow indicates the flow direction. The difference along the Z-axis between the street level and the faucet is also shown.

Using the results above to simplify equation \eqref{ber2}, we get

\begin{equation}
\label{ber3}
P_{\text{pipe}}+\frac{1}{2}\rho v_{\text{pipe}}^2=P_0+\frac{1}{2}\rho v_{\text{faucet}}^2+\rho g h,
\end{equation}

where we can use the result of equation \eqref{vfaucet} to write

\begin{equation}
\label{ber4}
P_{\text{pipe}}+\frac{1}{2}\rho \left(\frac{v_{\text{faucet}} r_{\text{faucet}}^2}{r_{\text{pipe}}^2}\right)^2=P_0+\frac{1}{2}\rho v_{\text{faucet}}^2 +\rho g h,
\end{equation}

where the only unknown is \(v_{\text{faucet}}\). Taking all the terms involving \(v_{\text{faucet}}\) to the right side and all the other terms to the left side

\begin{equation}
P_{\text{pipe}}-P_0-\rho gh=\frac{1}{2}\rho v_{\text{faucet}}^2-\frac{1}{2}\rho \left(\frac{v_{\text{faucet}} r_{\text{faucet}}^2}{r_{\text{pipe}}^2}\right)^2,
\end{equation}

which after evaluating the parenthesis explicitly becomes

\begin{equation}
P_{\text{pipe}}-P_0-\rho gh=\frac{1}{2}\rho v_{\text{faucet}}^2-\frac{1}{2}\rho \frac{v_{\text{faucet}}^2 r_{\text{faucet}}^4}{r_{\text{pipe}}^4}.
\end{equation}

After factorizing the terms on the right side, we can write

\begin{equation}
P_{\text{pipe}}-P_0-\rho gh=\frac{1}{2}\rho v_{\text{faucet}}^2\left(1-\frac{ r_{\text{faucet}}^4}{r_{\text{pipe}}^4}\right).
\end{equation}

Solving for \(v_{\text{faucet}}\) in the equation above, we obtain

\begin{equation}
v_{\text{faucet}}=\sqrt{\frac{2(P_{\text{pipe}}-P_0-\rho gh)}{\rho\left(1-\frac{ r_{\text{faucet}}^4}{r_{\text{pipe}}^4}\right) }}.
\end{equation}

Using the numerical values in SI units, we obtain

\begin{equation}
v_{\text{faucet}}=\sqrt{\frac{2(6\times 10^{5}\,\text{Pa}-101325\,\text{Pa}-(1000\,\text{kg/m}^2)(9.81\,\text{m/s}^2)(50\,\text{m}))}{1000\,\text{kg/m}^3\left(1-\frac{(2\,\text{cm})^4}{(4\,\text{cm})^4}\right)}},
\end{equation}

\begin{equation}
v_{\text{faucet}}\approx 4.18 \,\text{m/s}.
\end{equation}

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