A U-shaped tube that is open at one end holds water and mercury, as seen in the figure.
a) What is the absolute pressure at point A?
b) What is the absolute pressure at point B?
c) What is the air pressure at point C?
a) Use the hydrostatic pressure formula.
b) Consider what the pressure would be at the same height.
c) Same hint as part (a).
a) The hydrostatic pressure is:
\begin{equation*}
P_A = P_0 + \rho_w g H,
\end{equation*}
[mepr-show rules=”4409″ unauth=”both”]
where with numerical values:
\begin{equation*}
P_A = 105245 \, \text{Pa}.
\end{equation*}
b) Since \(P_A = P_B\), then:
\begin{equation*}
P_B = 105245 \, \text{Pa}.
\end{equation*}
c) For point C we have:
\begin{equation*}
P_B = P_C + \rho_{\text{Hg}} g h,
\end{equation*}
where solving for \(P_C\) and with numerical values:
\begin{equation*}
P_C = 102579.4 \, \text{Pa}.
\end{equation*}
For a more detailed explanation of any of these steps, click on “Detailed Solution”.
[/mepr-show]
Figure 1: U-shaped tube with water (blue) and mercury (dark grey) at equilibrium. The tube is closed on the left and open on the right.
a) In order to find the absolute pressure in point A, we should use the formula for the hydrostatic pressure while considering the contributions from both the atmosphere and the column of water above point A.
[mepr-show rules=”4409″ unauth=”both”]
The region of the u-tube where point A is located is open. Hence, point A is subjected to both the atmospheric pressure and the pressure exerted by the water. In this case, the hydrostatic pressure is given by
\begin{equation}
\label{EQ:hydro}
P_A = P_0 + \rho_w g H,
\end{equation}
where \(P_0\) is the atmospheric pressure, \(\rho_w\) is the density of water, and \(H\) is the height of the column of water above point A, as shown in figure 1.
Inserting numerical values gives
\begin{equation}
P_A = \left(101325 \ \text{Pa}\right) + 1000 \frac{\ \text{Kg}}{\ \text{m}^3} \left(9.8 \frac{\ \text{m}}{s^2}\right) \left(0.4 \ \text{m}\right),
\end{equation}
which is the same as
\begin{equation}
\label{EQ:npa}
P_A = 101325 + 3920 \ \text{Pa}
= 105245 \ \text{Pa} = 1.039 \, \text{atm}.
\end{equation}
b) We now need to find the absolute pressure at point B. As shown in figure 1, points A and B are at the same vertical position. We can use this observation to find the pressure at point B as follows: Point A and B are located along the same horizontal line on a static fluid. Hence, their pressures must be the same (if the pressures at point A and B were not the same, there will be a flow from the point of larger pressure towards the point of lower pressure). We can write this as
\begin{equation}
P_B = P_A,
\end{equation}
where \(P_A\) is given by eq. \eqref{EQ:hydro}.
Finally, substituting the numerical value from eq. \eqref{EQ:npa} yields
\begin{equation}
\label{EQ:nbp}
P_B = 105245 \, \text{Pa} = 1.039 \, \text{atm}.
\end{equation}
c) The absolute pressure at point C is produced by both the column of mercury (Hg) and the air in the region where point C is. Therefore, we can find the pressure at point C by writing an expression for \(P_B\) that includes both contributions, and solving it for \(P_C\).
Following this reasoning, the pressure at point B can be written as
\begin{equation}
P_B = P_C + \rho_{\text{Hg}} g h,
\end{equation}
where \(P_C\) is the air pressure at point C, and the second term is the pressure produced by the column of mercury of height \(h\). Solving for \(P_C\) gives
\begin{equation}
P_C = P_B – \rho_{Hg} g h,
\end{equation}
and we know the value of \(P_B\) from part (b). After inserting numerical values we obtain
\begin{equation}
P_C = 105245 \ \text{Pa} \, – \left(13600 \ \frac{\text{Kg}}{\text{m}^3}\right)\left(9.8 \ \frac{\text{m}}{\text{s}^2}\right)\left(0.02 \ \text{m}\right),
\end{equation}
which is equivalent to
\begin{equation}
P_C = 105245 \ \text{Pa} \, – 2665.6 \ \text{Pa}
= 102579.4 \ \text{Pa} = 1.012 \ \text{atm}.
\end{equation}
[/mepr-show]
Leave A Comment