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Difficulty: MediumHydrostatic Pressure

A specialized deep-well chemical injection pump is designed to operate safely up to a specific maximum hydrostatic pressure limit. During factory calibration in a test well filled with fresh water (density of 1,000 kg/m31,000\text{ kg/m}^3), the pump reaches its maximum safe pressure rating at exactly 50 meters50\text{ meters} of depth. For its field deployment, the pump is lowered into a vertical shaft filled with heavy drilling mud (density of 1,250 kg/m31,250\text{ kg/m}^3). Assuming the pressure rating remains constant, at what depth will the pump reach its maximum hydrostatic pressure limit in the drilling mud?

  1. 40 meters40\text{ meters}Answer
  2. B
    62.5 meters62.5\text{ meters}
  3. C
    50 meters50\text{ meters}
  4. D
    25 meters25\text{ meters}

Answer

The pump will reach its maximum safe pressure limit at 40 meters40\text{ meters} of depth in the drilling mud.
Hydrostatic pressure is determined by the formula P=ρghP = \rho g h. Because the drilling mud is denser (1,250 kg/m31,250\text{ kg/m}^3) than the fresh water (1,000 kg/m31,000\text{ kg/m}^3), it exerts more pressure for every meter of depth. Since the mud is 1.251.25 times as dense as the water (1,250/1,000=1.251,250 / 1,000 = 1.25), the pump will reach the maximum pressure limit at 1/1.251/1.25 (or 4/54/5) of the original depth. Multiplying 50 meters50\text{ meters} by 0.80.8 yields exactly 40 meters40\text{ meters}.

Step-by-Step Solution

1
Identify the relationship between hydrostatic pressure, fluid density, and depth.
P=ρ⋅g⋅hP = \rho \cdot g \cdot h
Hydrostatic pressure is directly proportional to both the fluid's density and the depth.
2
Set up an equation equating the pressure in both fluids.
ρwater⋅hwater=ρmud⋅hmud\rho_{\text{water}} \cdot h_{\text{water}} = \rho_{\text{mud}} \cdot h_{\text{mud}}
Since the maximum pressure (PP) and gravity (gg) are identical in both scenarios, the product of density and depth must remain constant.
3
Substitute the known values and solve for the target depth.
1,000⋅50=1,250⋅hmud1,000 \cdot 50 = 1,250 \cdot h_{\text{mud}}, which simplifies to 50,000=1,250⋅hmud50,000 = 1,250 \cdot h_{\text{mud}}, resulting in hmud=40h_{\text{mud}} = 40.
Dividing the constant pressure factor by the new, higher density yields the shallower depth required.

Key Concept

Hydrostatic pressure at a given depth is directly proportional to the density of the fluid. A denser fluid exerts more pressure per meter, requiring less depth to reach the same total pressure.
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Topics

Hydrostatic Pressure

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