Input interpretation
water hammer pressure with fast closure
Equation
ΔP = ρ v_s δv | ΔP | pressure difference ρ | fluid density v_s | speed of sound δv | change in fluid velocity (assuming an incompressible fluid)
Input values
fluid density | 1000 kg/m^3 (kilograms per cubic meter) speed of sound | 343.2 m/s (meters per second) change in fluid velocity | 1 m/s (meter per second)
Results
pressure difference | 343.2 kPa (kilopascals) = 49.78 psi (pounds-force per square inch) = 7168 lbf/ft^2 (pounds-force per square foot) = 0.3432 MPa (megapascals) = 343200 Pa (pascals) = 3.387 atm (atmospheres) = 3432 mbar (millibars) = 2574 mmHg (millimeters of mercury)
Possible intermediate steps
Calculate the pressure difference using the following information: known variables | | ρ | fluid density | 1000 kg/m^3 v_s | speed of sound | 343.2 m/s δv | change in fluid velocity | 1 m/s Convert known variables into appropriate units using the following: 1 kg/m^3 = 1000 g/m^3: known variables | | ρ | fluid density | 1×10^6 g/m^3 v_s | speed of sound | 343.2 m/s δv | change in fluid velocity | 1 m/s The relevant equation that relates pressure difference (ΔP), fluid density (ρ), speed of sound (v_s), and change in fluid velocity (δv) is: ΔP = ρ v_s δv Substitute known variables into the equation: known variables | | ρ | fluid density | 1×10^6 g/m^3 v_s | speed of sound | 343.2 m/s δv | change in fluid velocity | 1 m/s | : ΔP = 1×10^6 g/m^3×343.2 m/s×1 m/s Separate the numerical part, 1×10^6×343.2×1, from the unit part, g/m^3×m/s×m/s = g/(m s^2): ΔP = 1×10^6×343.2×1 g/(m s^2) Evaluate 1×10^6×343.2×1: ΔP = 3.432×10^8 g/(m s^2) Convert 3.432×10^8 g/(m s^2) into kPa (kilopascals) using the following: 1 g/(m s^2) = 1×10^-6 kPa: Answer: | | ΔP = 343.2 kPa