Question

The container shown in (figure 1) is filled with oil. It is open to the atmosphere on the left.

What is the pressure at point A?

Pa=                           kPa

What is the pressure difference between points A and B?

Pb-Pa=                          kPa

What is the pressure difference between points A and C?

Pc-Pa=                           kPa

 

Answer 1

Concepts and reason

The concept used to solve this problem is pressure in fluid. First find the pressure at point A by subtracting the final height and height at the point A. Again, find the pressure difference between the points \(\mathrm{A}\) and \(\mathrm{B}\) by subtracting the height at point \(\mathrm{A}\) and height at point \(B\). Finally, find the pressure difference between the points \(\mathrm{A}\) and \(\mathrm{C}\) by subtracting the height at point \(\mathrm{A}\) and height at point \(C\)

Fundamentals

The pressure is defined as the force divided by per unit area. Here, the oil is filled in a partially closed container. A force applied in one area can result in a greater force in another area. Due to the force of gravity the pressure exerted by a fluid at equilibrium at a given point within the fluid, is called hydrostatic pressure. The expression for the liquid pressure is, \(p=\rho g h\)

Here, \(p\) is the pressure, \(\rho\) is the density of the oil, \(g\) is the gravitational constant, and \(h\) is the height.

 

(1) The expression for the liquid pressure is, \(p=\rho g h \ldots \ldots\)

Here the height of the pressure for point \(\mathrm{A}\) is, The expression for the height of the pressure is, \(h=h_{f}-h_{i}\)

Here, \(h\) is the height of the pressure, \(h_{f}\) is the final height and \(h_{i}\) is the height at the point \(\mathrm{A}\). From the diagram the final height is \(100 \mathrm{cmand}\) the height at the point \(\mathrm{A}\) is \(50 \mathrm{~cm}\).

$$ \begin{array}{c} h=100 \mathrm{~cm}\left(\frac{1 \mathrm{~m}}{10^{2} \mathrm{~cm}}\right)-50 \mathrm{~cm}\left(\frac{1 \mathrm{~m}}{10^{2} \mathrm{~cm}}\right) \\ =100 \times 10^{-2} \mathrm{~m}-50 \times 10^{-2} \mathrm{~m} \\ =0.5 \mathrm{~m} \end{array} $$

Substitute \(900 \mathrm{~kg} / \mathrm{m}^{3}\) for \(\rho, 9.8 \mathrm{~m} / \mathrm{s}^{2}\) for \(g\) and \(0.5 \mathrm{~m}\) for \(h\) in equation (1)

\(p=\left(900 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)(0.5 \mathrm{~m})\)

$$ =4410 \mathrm{pa}\left(\frac{10^{-3} \mathrm{kpa}}{\mathrm{pa}}\right) $$

\(=4.410 \mathrm{kpa}\)

The absolute pressure \(\mathrm{P}\) is, \(P=P_{0}+p\)

Here, \(P_{0}\) is the atomic pressure. Substitute 4.410 kpa for \(p\) and \(101.325 \mathrm{kPa}\) for \(P_{0}\).

$$ \begin{array}{c} P=101.325 \mathrm{kPa}+4.410 \mathrm{kpa} \\ =105.7 \mathrm{kPa} \end{array} $$

Part 1 The pressure at the point \(\mathrm{A}\) is \(105.7 \mathrm{kPa}\).

The Pressure of a liquid depends on the density of the liquid, acceleration due to gravity, and the distance. This can be viewed from the expression of the pressure: \(p=\rho g h\).

 

(2) The expression for the liquid pressure is, \(p=\rho g h \ldots \ldots(1)\)

Here the height of the pressure difference between the point \(\mathrm{A}\) and \(\mathrm{B}\) is, The expression for the height of the pressure is, \(h=h_{A}-h_{B}\)

Here, \(h_{A}\) is the height at the point \(\mathrm{A}\) and \(h_{B}\) is the height at the point \(\mathrm{B}\). From the diagram the height at the point Ais \(50 \mathrm{cmand}\) the height at the point \(\mathrm{B}\) is \(0 \mathrm{~cm}\).

$$ \begin{array}{c} h=50 \mathrm{~cm}\left(\frac{1 \mathrm{~m}}{10^{2} \mathrm{~cm}}\right)-0 \mathrm{~cm} \\ =50 \times 10^{-2} \mathrm{~m} \\ =0.5 \mathrm{~m} \end{array} $$

Substitute \(900 \mathrm{~kg} / \mathrm{m}^{3}\) for \(\rho, 9.8 \mathrm{~m} / \mathrm{s}^{2}\) for \(g\) and \(0.5 \mathrm{~m}\) for \(h\) in equation (1)

$$ \begin{array}{c} p=\left(900 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)(0.5 \mathrm{~m}) \\ =4410 \mathrm{pa} \\ =4.410 \mathrm{kpa} \end{array} $$

Part 2 The pressure difference between the points \(\mathrm{A}\) and \(\mathrm{B}\) is \(4.410 \mathrm{kpa}\).

The atmospheric pressure that acts on the surface of a liquid is the total pressure acting on a liquid. The Pressure of a liquid depends on the density of the liquid, acceleration due to gravity, and the distance.

 

(3) The expression for the liquid pressure is, \(p=\rho g h \ldots \ldots(1)\)

Here the height of the pressure difference between the point \(\mathrm{A}\) and \(\mathrm{C}\) is, The expression for the height of the pressure is, \(h=h_{A}-h_{C}\)

Here, \(h_{A}\) is the height at the point \(\mathrm{A}\) and \(h_{C}\) is the height at the point \(\mathrm{C}\). From the diagram the height at the point \(\mathrm{A}\) is \(50 \mathrm{cmand}\) the height at the point \(\mathrm{C}\) is \(0 \mathrm{~cm}\).

$$ \begin{array}{c} h=50 \mathrm{~cm}\left(\frac{1 \mathrm{~m}}{10^{2} \mathrm{~cm}}\right)-0 \mathrm{~cm} \\ =50 \times 10^{-2} \mathrm{~m} \\ =0.5 \mathrm{~m} \end{array} $$

Substitute \(900 \mathrm{~kg} / \mathrm{m}^{3}\) for \(\rho, 9.8 \mathrm{~m} / \mathrm{s}^{2}\) for \(g\) and \(0.5 \mathrm{~m}\) for \(h\) in equation (1)

$$ \begin{array}{c} p=\left(900 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)(0.5 \mathrm{~m}) \\ =4410 \mathrm{pa} \\ =4.410 \mathrm{kpa} \end{array} $$

Part 3 The pressure difference between the points \(\mathrm{A}\) and \(\mathrm{C}\) is \(4.410 \mathrm{kpa}\).

The pressure won't depend on the amount of liquid present. Pressure is same in all directions of the fluid for given points.

 

Part 1

The pressure at the point \(\mathrm{A}\) is \(105.7 \mathrm{kPa}\).

Part 2

The pressure difference between the points \(\mathrm{A}\) and \(\mathrm{B}\) is \(4.410 \mathrm{kpa}\).

Part 3

The pressure difference between the points \(A\) and \(C\) is \(4.410 \mathrm{kpa}\).