A fluid with density $d$ flows through a pipe. I. The cross-sectional area at point x is $A_1 = 48 \text{ cm}^2$. The fluid is water with density $d_2 = 1000 \text{ kgm}^{-3}$. The cross-sectional area at point y is $A_2 = 12 \text{ cm}^2$. The velocity of the fluid at point y is $V_2 = 24 \text{ ms}^{-1}$. Find the velocity $V_1$ at point x. II. The pressure at point x is $P_x = 3 \times 10^5 \text{ Nm}^{-2}$. Find the pressure $P_y$ at point y.

1. Problem Description

A fluid with density $d$ flows through a pipe.

I. The cross-sectional area at point x is $A_1 = 48 \text{ cm}^2$. The fluid is water with density $d_2 = 1000 \text{ kgm}^{-3}$. The cross-sectional area at point y is $A_2 = 12 \text{ cm}^2$. The velocity of the fluid at point y is $V_2 = 24 \text{ ms}^{-1}$. Find the velocity $V_1$ at point x.

II. The pressure at point x is $P_x = 3 \times 10^5 \text{ Nm}^{-2}$. Find the pressure $P_y$ at point y.

2. Solution Steps

I.

According to the continuity equation:

$A_1V_1 = A_2V_2$

$V_1 = \frac{A_2V_2}{A_1}$

$V_1 = \frac{12 \text{ cm}^2 \times 24 \text{ ms}^{-1}}{48 \text{ cm}^2}$

$V_1 = \frac{24}{4} \text{ ms}^{-1}$

$V_1 = 6 \text{ ms}^{-1}$

II.

Using Bernoulli's equation:

$P_1 + \frac{1}{2} \rho V_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho V_2^2 + \rho g h_2$

Since the pipe is horizontal, $h_1 = h_2$. Thus the $\rho g h$ terms cancel out. Also, we are given the density of the fluid as $d_2$, and we computed $V_1$ above.

$P_x + \frac{1}{2} d_2 V_1^2 = P_y + \frac{1}{2} d_2 V_2^2$

$P_y = P_x + \frac{1}{2} d_2 V_1^2 - \frac{1}{2} d_2 V_2^2$

$P_y = P_x + \frac{1}{2} d_2 (V_1^2 - V_2^2)$

$P_y = 3 \times 10^5 \text{ Nm}^{-2} + \frac{1}{2} \times 1000 \text{ kgm}^{-3} \times ((6 \text{ ms}^{-1})^2 - (24 \text{ ms}^{-1})^2)$

$P_y = 3 \times 10^5 + 500 (36 - 576)$

$P_y = 3 \times 10^5 + 500 (-540)$

$P_y = 3 \times 10^5 - 270000$

$P_y = 300000 - 270000$

$P_y = 30000 \text{ Nm}^{-2}$

$P_y = 3 \times 10^4 \text{ Nm}^{-2}$

3. Final Answer

I. $V_1 = 6 \text{ ms}^{-1}$

II. $P_y = 3 \times 10^4 \text{ Nm}^{-2}$