A cylindrical tank has small holes drilled vertically along its side, as shown in the diagram. The tank is filled with water, and the water jets coming out of the holes travel different distances horizontally. The goal is to analyze this phenomenon using Bernoulli's theorem. We are given the following: atmospheric pressure is $7 \, \text{cmHg}$, the density of water is $\rho \, \text{kgm}^{-3}$, $h = 25 \, \text{cm}$, and $H = 100 \, \text{cm}$. We need to: (a) State the conditions under which Bernoulli's equation is valid. (b) Write Bernoulli's equation and identify its terms. Show that it is dimensionally correct. (c) Obtain an expression for the velocity $V_A$ of the water coming out of hole A using Bernoulli's theorem. (d) Obtain an expression for the horizontal distance $x$ the water jet from hole A travels before hitting the ground, in terms of $H$, $h$, and $g$. (e) Obtain an expression for the horizontal distance $x_C$ the water jet from hole C travels before hitting the ground, in terms of $H$, $h$, and $g$. (f) Given $h = 25 \, \text{cm}$, $H = 100 \, \text{cm}$, $\rho = 1000 \, \text{kgm}^{-3}$, and $g = 10 \, \text{ms}^{-2}$, find the velocity $A$ at $C$. (g) Find the distance the water jet from hole B in the middle of the tank travels before hitting the ground. (h) Sketch a graph showing how the horizontal distance of the water jet varies with depth.

1. Problem Description

A cylindrical tank has small holes drilled vertically along its side, as shown in the diagram. The tank is filled with water, and the water jets coming out of the holes travel different distances horizontally. The goal is to analyze this phenomenon using Bernoulli's theorem. We are given the following: atmospheric pressure is $7 \, \text{cmHg}$, the density of water is $\rho \, \text{kgm}^{-3}$, $h = 25 \, \text{cm}$, and $H = 100 \, \text{cm}$. We need to:

(a) State the conditions under which Bernoulli's equation is valid.

(b) Write Bernoulli's equation and identify its terms. Show that it is dimensionally correct.

(c) Obtain an expression for the velocity $V_A$ of the water coming out of hole A using Bernoulli's theorem.

(d) Obtain an expression for the horizontal distance $x$ the water jet from hole A travels before hitting the ground, in terms of $H$, $h$, and $g$.

(e) Obtain an expression for the horizontal distance $x_C$ the water jet from hole C travels before hitting the ground, in terms of $H$, $h$, and $g$.

(f) Given $h = 25 \, \text{cm}$, $H = 100 \, \text{cm}$, $\rho = 1000 \, \text{kgm}^{-3}$, and $g = 10 \, \text{ms}^{-2}$, find the velocity $A$ at $C$.

(g) Find the distance the water jet from hole B in the middle of the tank travels before hitting the ground.

(h) Sketch a graph showing how the horizontal distance of the water jet varies with depth.

2. Solution Steps

(a) Conditions for Bernoulli's Equation:

Bernoulli's equation is valid under the following conditions:

1. The fluid is incompressible (density is constant).

2. The fluid is non-viscous (no internal friction).

3. The flow is steady (velocity at a point does not change with time).

4. The flow is along a streamline.

(b) Bernoulli's Equation:

$P + \frac{1}{2} \rho v^2 + \rho g h = \text{constant}$

where:

- $P$ is the pressure.

- $\rho$ is the density of the fluid.

- $v$ is the velocity of the fluid.

- $g$ is the acceleration due to gravity.

- $h$ is the height above a reference point.

Dimensional analysis:

$[P] = \text{ML}^{-1}\text{T}^{-2}$

$[\frac{1}{2} \rho v^2] = \text{ML}^{-3} (\text{LT}^{-1})^2 = \text{ML}^{-1}\text{T}^{-2}$

$[\rho g h] = \text{ML}^{-3} \text{LT}^{-2} \text{L} = \text{ML}^{-1}\text{T}^{-2}$

All terms have the same dimensions, so the equation is dimensionally correct.

(c) Velocity at Hole A ($V_A$):

Let's apply Bernoulli's equation at the surface of the water in the tank (point 1) and at hole A (point 2). We take the reference level to be at the height of hole A.

$P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2$

$P_1 = P_{atm}$, $v_1 \approx 0$ (since the tank is large), $h_1 = H$ (height of the water surface above hole A)

$P_2 = P_{atm}$, $v_2 = V_A$, $h_2 = 0$

$P_{atm} + 0 + \rho g H = P_{atm} + \frac{1}{2}\rho V_A^2 + 0$

$\rho g H = \frac{1}{2}\rho V_A^2$

$V_A^2 = 2gH$

$V_A = \sqrt{2gH}$

(d) Horizontal Distance for Hole A ($x$):

The water jet from hole A is launched horizontally with velocity $V_A = \sqrt{2gH}$. The time it takes to fall a distance $h$ to the ground is given by:

$h = \frac{1}{2} g t^2$

$t = \sqrt{\frac{2h}{g}}$

The horizontal distance $x$ is given by:

$x = V_A t = \sqrt{2gH} \sqrt{\frac{2h}{g}} = \sqrt{4Hh} = 2\sqrt{Hh}$

(e) Horizontal Distance for Hole C ($x_C$):

Let the depth of hole C be $h_c$ from the surface of the water. Then $H-h_c$ is the height of C from the bottom. So we have to change h by $H-h_c$. We know that we use the formula from d where h is $h =H-h_c$.

$v_c=\sqrt{2gh_c}$, $t=\sqrt{\frac{2(H-h_c)}{g}}$

$x_c=\sqrt{2gh_c}\sqrt{\frac{2(H-h_c)}{g}} = \sqrt{4h_c(H-h_c)}= 2\sqrt{h_c(H-h_c)}$

(f) Given $h = 25 \, \text{cm} = 0.25 \, \text{m}$ and $H = 100 \, \text{cm} = 1 \, \text{m}$, we want to find the velocity at $C$. Let's assume that hole C is at the bottom of the tank.

From question (e), hole C distance $x= 2\sqrt{h(H-h)}$. So, $x= 2\sqrt{0.25(1-0.25)} = 2\sqrt{0.25*0.75} =2\sqrt{0.1875} = 0.866 m$

$V= \sqrt{2gh} = \sqrt{2*10*0.25} = \sqrt{5} = 2.236 \frac{m}{s}$

(g) Hole B is in the middle of the tank. So, the depth of hole B from the surface of the water is $h_B = H/2 = 1/2 = 0.5 \, \text{m}$. The height of hole B from the ground is also $H/2 = 0.5 \, \text{m}$.

$x_B = 2\sqrt{h_B(H-h_B)} = 2\sqrt{0.5(1-0.5)} = 2\sqrt{0.5(0.5)} = 2\sqrt{0.25} = 2(0.5) = 1 \, \text{m}$

(h) Sketch of the graph:

The horizontal distance $x$ is given by $x = 2\sqrt{h(H-h)}$. This is a parabola. The maximum distance occurs when $h = H/2$, and the distance is zero when $h=0$ or $h=H$.

3. Final Answer

(a) The conditions under which Bernoulli's equation is valid are: incompressible fluid, non-viscous fluid, steady flow, and flow along a streamline.

(b) $P + \frac{1}{2} \rho v^2 + \rho g h = \text{constant}$, where $P$ is pressure, $\rho$ is density, $v$ is velocity, $g$ is acceleration due to gravity, and $h$ is height. The equation is dimensionally correct.

(c) $V_A = \sqrt{2gH}$

(d) $x = 2\sqrt{Hh}$

(e) $x_c = 2\sqrt{h_c(H-h_c)}$, where $h_c$ is depth of C from the surface.

(f) $v=2.236 m/s$

(g) $x_B = 1 \, \text{m}$

(h) The graph of $x$ vs. $h$ is a parabola.