What are the correct SI units for hydraulic power defined as P_h = ΔP × Q?

What are the correct SI units for hydraulic power defined as P_h = ΔP × Q?

• A. Newtons.
• B. Watts.
• C. Joules.
• D. Cubic meters.

Answer: (B)

Hydraulic power is the rate at which energy is transferred by the fluid, and it’s defined as the product of the pressure difference and the volumetric flow rate. The pressure difference ΔP is measured in pascals (Pa), which are newtons per square meter. The flow rate Q is in cubic meters per second (m^3/s). Multiplying them gives Pa × m^3/s. Since Pa is N/m^2, this becomes N·m/s, which is energy per unit time, i.e., joules per second, which is watts. So the SI unit is watts. To put it another way, 1 W = 1 J/s, so if you have a pressure drop of, say, 100 kPa with a flow rate of 0.01 m^3/s, the hydraulic power is 100,000 Pa × 0.01 m^3/s = 1000 N·m/s = 1000 W.

Hydraulic power is the rate at which energy is transferred by the fluid, and it’s defined as the product of the pressure difference and the volumetric flow rate. The pressure difference ΔP is measured in pascals (Pa), which are newtons per square meter. The flow rate Q is in cubic meters per second (m^3/s). Multiplying them gives Pa × m^3/s. Since Pa is N/m^2, this becomes N·m/s, which is energy per unit time, i.e., joules per second, which is watts. So the SI unit is watts.

To put it another way, 1 W = 1 J/s, so if you have a pressure drop of, say, 100 kPa with a flow rate of 0.01 m^3/s, the hydraulic power is 100,000 Pa × 0.01 m^3/s = 1000 N·m/s = 1000 W.