A steam of high pressure with mass $m = 8000 \, \text{kg}$ enters a steam network pipe with velocity $v_1 = 90 \, \text{m/s}$. At the outlet, the velocity is reduced to $v_2 = 80 \, \text{m/s}$. Find the kinetic energy of the steam at the inlet and outlet, and the change in kinetic energy.

1. Problem Description

A steam of high pressure with mass $m = 8000 \, \text{kg}$ enters a steam network pipe with velocity $v_1 = 90 \, \text{m/s}$. At the outlet, the velocity is reduced to $v_2 = 80 \, \text{m/s}$. Find the kinetic energy of the steam at the inlet and outlet, and the change in kinetic energy.

2. Solution Steps

First, we need to find the kinetic energy at the inlet. The formula for kinetic energy is:

$KE = \frac{1}{2}mv^2$

At the inlet, we have:

$KE_1 = \frac{1}{2}mv_1^2$

$KE_1 = \frac{1}{2}(8000 \, \text{kg})(90 \, \text{m/s})^2$

$KE_1 = 4000 \, \text{kg} \times 8100 \, \text{m}^2/\text{s}^2$

$KE_1 = 32400000 \, \text{J}$

$KE_1 = 32.4 \times 10^6 \, \text{J}$

Next, we need to find the kinetic energy at the outlet.

$KE_2 = \frac{1}{2}mv_2^2$

$KE_2 = \frac{1}{2}(8000 \, \text{kg})(80 \, \text{m/s})^2$

$KE_2 = 4000 \, \text{kg} \times 6400 \, \text{m}^2/\text{s}^2$

$KE_2 = 25600000 \, \text{J}$

$KE_2 = 25.6 \times 10^6 \, \text{J}$

Now, we calculate the change in kinetic energy:

$\Delta KE = KE_2 - KE_1$

$\Delta KE = 25600000 \, \text{J} - 32400000 \, \text{J}$

$\Delta KE = -6800000 \, \text{J}$

$\Delta KE = -6.8 \times 10^6 \, \text{J}$

3. Final Answer

The kinetic energy at the inlet is $32.4 \times 10^6 \, \text{J}$.

The kinetic energy at the outlet is $25.6 \times 10^6 \, \text{J}$.

The change in kinetic energy is $-6.8 \times 10^6 \, \text{J}$.