What is the increase in mechanical energy if pressure of water is increased from 100 kPa to 900 kPa by a pump? Express your answer in kJ/kg.

The mechanical energy of a fluid is increased by a pump through an increase in pressure. The increase in mechanical energy per unit mass (also known as specific energy) can be calculated using the formula:

ΔE = ΔP / ρ

where:

• ΔE is the increase in specific energy,
• ΔP is the increase in pressure, and
• ρ is the density of the fluid.

1. We know that the increase in pressure (ΔP) is from 100 kPa to 900 kPa, so ΔP = 900 kPa - 100 kPa = 800 kPa. However, we need to convert this to a standard unit of pressure, which is pascals (Pa). 1 kPa is equal to 1000 Pa, so ΔP = 800 kPa * 1000 = 800,000 Pa.

2. The density of water (ρ) is approximately 1000 kg/m³.

3. Now, we can substitute these values into the formula to find ΔE:

   • ΔE = 800,000 Pa / 1000 kg/m³ = 800 J/kg

4. However, the question asks for the answer in kJ/kg, so we need to convert J/kg to kJ/kg. 1 J/kg is equal to 0.001 kJ/kg, so ΔE = 800 J/kg * 0.001 = 0.8 kJ/kg.

So, the increase in mechanical energy if the pressure of water is increased from 100 kPa to 900 kPa by a pump is 0.8 kJ/kg.