Water enters a pipe with a velocity of $1.5 \, m/s$ and a diameter of $0.1 \, m$. The diameter at the exit is $0.4 \, m$. We need to find the mass flow rate at the entrance and the velocity at the exit. The density of water is given as $\rho = 1000 \, kg/m^3$.

1. Problem Description

Water enters a pipe with a velocity of $1.5 \, m/s$ and a diameter of $0.1 \, m$. The diameter at the exit is $0.4 \, m$. We need to find the mass flow rate at the entrance and the velocity at the exit. The density of water is given as $\rho = 1000 \, kg/m^3$.

2. Solution Steps

First, we calculate the area at the entrance:

$A_1 = \pi r_1^2 = \pi (\frac{d_1}{2})^2 = \pi (\frac{0.1}{2})^2 = \pi (0.05)^2 = 0.0025 \pi \, m^2$

The mass flow rate at the entrance is given by:

$\dot{m} = \rho A_1 v_1$

where $\rho$ is the density, $A_1$ is the area at the entrance, and $v_1$ is the velocity at the entrance.

$\dot{m} = (1000 \, kg/m^3) (0.0025 \pi \, m^2) (1.5 \, m/s) = 3.75 \pi \, kg/s \approx 11.78 \, kg/s$

Next, we use the principle of mass conservation:

$\dot{m}_1 = \dot{m}_2 = \dot{m}$

$\rho A_1 v_1 = \rho A_2 v_2$

Since the density is constant, we have:

$A_1 v_1 = A_2 v_2$

We need to find the velocity at the exit, $v_2$.

$v_2 = \frac{A_1 v_1}{A_2}$

$A_2 = \pi r_2^2 = \pi (\frac{d_2}{2})^2 = \pi (\frac{0.4}{2})^2 = \pi (0.2)^2 = 0.04 \pi \, m^2$

Therefore,

$v_2 = \frac{(0.0025 \pi \, m^2)(1.5 \, m/s)}{0.04 \pi \, m^2} = \frac{0.0025 \times 1.5}{0.04} \, m/s = \frac{0.00375}{0.04} \, m/s = 0.09375 \, m/s$

3. Final Answer

The mass flow rate at the entrance is $3.75 \pi \, kg/s \approx 11.78 \, kg/s$.

The velocity at the exit is $0.09375 \, m/s$.