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.

Applied MathematicsPhysicsKinetic EnergyEnergy Conservation
2025/5/10

1. Problem Description

A steam of high pressure with mass m=8000kgm = 8000 \, \text{kg} enters a steam network pipe with velocity v1=90m/sv_1 = 90 \, \text{m/s}. At the outlet, the velocity is reduced to v2=80m/sv_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=12mv2KE = \frac{1}{2}mv^2
At the inlet, we have:
KE1=12mv12KE_1 = \frac{1}{2}mv_1^2
KE1=12(8000kg)(90m/s)2KE_1 = \frac{1}{2}(8000 \, \text{kg})(90 \, \text{m/s})^2
KE1=4000kg×8100m2/s2KE_1 = 4000 \, \text{kg} \times 8100 \, \text{m}^2/\text{s}^2
KE1=32400000JKE_1 = 32400000 \, \text{J}
KE1=32.4×106JKE_1 = 32.4 \times 10^6 \, \text{J}
Next, we need to find the kinetic energy at the outlet.
KE2=12mv22KE_2 = \frac{1}{2}mv_2^2
KE2=12(8000kg)(80m/s)2KE_2 = \frac{1}{2}(8000 \, \text{kg})(80 \, \text{m/s})^2
KE2=4000kg×6400m2/s2KE_2 = 4000 \, \text{kg} \times 6400 \, \text{m}^2/\text{s}^2
KE2=25600000JKE_2 = 25600000 \, \text{J}
KE2=25.6×106JKE_2 = 25.6 \times 10^6 \, \text{J}
Now, we calculate the change in kinetic energy:
ΔKE=KE2KE1\Delta KE = KE_2 - KE_1
ΔKE=25600000J32400000J\Delta KE = 25600000 \, \text{J} - 32400000 \, \text{J}
ΔKE=6800000J\Delta KE = -6800000 \, \text{J}
ΔKE=6.8×106J\Delta KE = -6.8 \times 10^6 \, \text{J}

3. Final Answer

The kinetic energy at the inlet is 32.4×106J32.4 \times 10^6 \, \text{J}.
The kinetic energy at the outlet is 25.6×106J25.6 \times 10^6 \, \text{J}.
The change in kinetic energy is 6.8×106J-6.8 \times 10^6 \, \text{J}.

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