The problem describes a Venturi meter used to measure the velocity of a fluid flow. The Venturi meter consists of a large tube with cross-sectional area $A = 48 \text{ cm}^2$ and a smaller tube with cross-sectional area $a = 12 \text{ cm}^2$. A manometer is connected to both the larger and smaller tubes, and the difference in the mercury levels in the manometer is $h = 10 \text{ cm}$. The density of mercury is $13600 \text{ kg/m}^3$, and the density of water is $1000 \text{ kg/m}^3$. We need to find: i. An expression for the pressure at point A. ii. An expression for the pressure at point B. iii. An expression for the pressure difference $P_A - P_B$. iv. A relationship between the velocities $V_x$ and $V_y$. v. The velocity $V_x$ at point x.

1. Problem Description

The problem describes a Venturi meter used to measure the velocity of a fluid flow. The Venturi meter consists of a large tube with cross-sectional area $A = 48 \text{ cm}^2$ and a smaller tube with cross-sectional area $a = 12 \text{ cm}^2$. A manometer is connected to both the larger and smaller tubes, and the difference in the mercury levels in the manometer is $h = 10 \text{ cm}$. The density of mercury is $13600 \text{ kg/m}^3$, and the density of water is $1000 \text{ kg/m}^3$. We need to find:

i. An expression for the pressure at point A.

ii. An expression for the pressure at point B.

iii. An expression for the pressure difference $P_A - P_B$.

iv. A relationship between the velocities $V_x$ and $V_y$.

v. The velocity $V_x$ at point x.

2. Solution Steps

i. Pressure at point A:

Using Bernoulli's equation at point A, we can express the pressure $P_A$ as:

$P_A + \frac{1}{2} \rho V_x^2 + \rho g h_1$

Where:

$P_A$ is the pressure at point A.

$\rho$ is the density of the water.

$V_x$ is the velocity of the water at point A.

$g$ is the acceleration due to gravity.

$h_1$ is the height of the water column at point A.

ii. Pressure at point B:

Using Bernoulli's equation at point B, we can express the pressure $P_B$ as:

$P_B + \frac{1}{2} \rho V_y^2 + \rho g h_1$

Where:

$P_B$ is the pressure at point B.

$\rho$ is the density of the water.

$V_y$ is the velocity of the water at point B.

$g$ is the acceleration due to gravity.

$h_1$ is the height of the water column at point B.

However, it is important to consider the pressure due to the mercury.

Since the pressure at the bottom of the manometer must be equal on both sides, we can write:

$P_A + \rho_{water} g h = P_B + \rho_{mercury} g h$

$P_B = P_A + \rho_{water} g h - \rho_{mercury} g h$

iii. Pressure difference $P_A - P_B$:

$P_A - P_B = \frac{1}{2} \rho (V_y^2 - V_x^2)$

From the manometer reading:

$P_A - P_B = g h (\rho_{mercury} - \rho_{water})$

iv. Relationship between $V_x$ and $V_y$:

Using the equation of continuity:

$A V_x = a V_y$

$V_y = \frac{A}{a} V_x = \frac{48}{12} V_x = 4 V_x$

v. Velocity $V_x$:

From (iii), we know that $P_A - P_B = g h (\rho_{mercury} - \rho_{water})$.

Also, $P_A - P_B = \frac{1}{2} \rho (V_y^2 - V_x^2)$.

Therefore, $g h (\rho_{mercury} - \rho_{water}) = \frac{1}{2} \rho (V_y^2 - V_x^2)$

Substituting $V_y = 4 V_x$, we have:

$g h (\rho_{mercury} - \rho_{water}) = \frac{1}{2} \rho ((4 V_x)^2 - V_x^2) = \frac{1}{2} \rho (16 V_x^2 - V_x^2) = \frac{1}{2} \rho (15 V_x^2)$

$V_x^2 = \frac{2 g h (\rho_{mercury} - \rho_{water})}{15 \rho}$

$V_x = \sqrt{\frac{2 g h (\rho_{mercury} - \rho_{water})}{15 \rho}}$

$V_x = \sqrt{\frac{2 (9.8) (0.1) (13600 - 1000)}{15 (1000)}} = \sqrt{\frac{2 (9.8) (0.1) (12600)}{15000}} = \sqrt{\frac{24696}{15000}} = \sqrt{1.6464} \approx 1.283 \text{ m/s}$

3. Final Answer

i. $P_A + \frac{1}{2} \rho V_x^2 + \rho g h_1$

ii. $P_A + \rho_{water} g h - \rho_{mercury} g h$

iii. $P_A - P_B = g h (\rho_{mercury} - \rho_{water})$

iv. $V_y = 4 V_x$

v. $V_x = \sqrt{\frac{2 g h (\rho_{mercury} - \rho_{water})}{15 \rho}} \approx 1.283 \text{ m/s}$