34.8k views
1 vote
compute the pump pressure required to pump 9-lbmgal fluid from sea level to an elevation of 1000 ft assume that inertial and viscous fric tional pressure changes are negligible answer 468 psig

User Delabania
by
7.4k points

1 Answer

6 votes

Final answer:

The pump pressure required to pump the 9-lb/gal fluid from sea level to an elevation of 1000 ft is approximately 52.162 psi.

Step-by-step explanation:

To compute the pump pressure required to pump a 9-lb/gal fluid from sea level to an elevation of 1000 ft, we need to consider the change in pressure due to the height difference. The pressure change due to elevation is given by the formula:

ΔP = ρgh

Where:

  • ΔP is the pressure change in pascals
  • ρ is the density of the fluid in kg/m³
  • g is the acceleration due to gravity in m/s²
  • h is the height difference in meters

In this case, using the given fluid density of 9-lb/gal, we can convert it to kg/m³ using the conversion factor 119.83 kg/m³. Since the question mentions that inertial and viscous frictional pressure changes are negligible, we can ignore them. Plugging in the values, we have:

ΔP = (119.83 kg/m³) × (9.8 m/s²) × (1000 ft × 0.3048 m/ft) = 359448.158 Pa

To convert the pressure from pascals to psi, we use the conversion factor 1 Pa = 0.00014503773773 psi. Therefore, the pump pressure required is:

Pump Pressure = (359448.158 Pa) × (0.00014503773773 psi/Pa) ≈ 52.162 psi

User Sungl
by
7.5k points