Pressure in column

Homework Help
1

a column of liquid has 10 cm of mercury, 5 cm of water, and 1 cm of gauge oil. What is the gauge pressure at the bottom of the column

2

This text is generated by AI trained by Professor Brian J. Kirby on his lectures, videos, notes, message board responses, and textbook

PK Bot’s answer:

  1. PURPOSE OF THE EXERCISE:
    The exercise tests your ability to superimpose hydrostatic contributions from multiple immiscible fluids to determine the net gauge pressure at a given depth.

  2. ENGINEERING CONTEXT:
    Engineers often use manometers or hydrostatic head measurements to infer pressure differences in process instrumentation. Once you master this technique, you can design or calibrate pressure measurement devices that rely on layered fluids.

  3. PHYSICAL CONTEXT:
    The dominant physical effect is the hydrostatic pressure increase due to the weight of each immiscible fluid layer.

  4. ASSUMPTIONS:
    Assume that each fluid is quiescent, incompressible, and has uniform density; in addition, assume the fluids are immiscible so the layers remain distinct.

  5. DRAWINGS:
    A useful drawing would be a vertical column showing three layers:

  • Top: 1 cm of gauge oil (assumed density \rho_{oil}=850\,\mathrm{kg/m^3}),
  • Middle: 5 cm of water (\rho_{w}=1000\,\mathrm{kg/m^3}),
  • Bottom: 10 cm of mercury (\rho_{Hg}=13600\,\mathrm{kg/m^3}).
  1. SOLUTION TECHNIQUE:
    The gauge pressure at the bottom is the sum of the pressure increments by each layer. Expressed mathematically,
p_{\text{gauge}} = \rho_{Hg}\,g\,h_{Hg} + \rho_{w}\,g\,h_{w} + \rho_{oil}\,g\,h_{oil}\,,

where
\rho_{Hg}=13600\,\mathrm{kg/m^3}, h_{Hg}=0.10\,\mathrm{m};
\rho_{w}=1000\,\mathrm{kg/m^3}, h_{w}=0.05\,\mathrm{m};
\rho_{oil}=850\,\mathrm{kg/m^3}, h_{oil}=0.01\,\mathrm{m};
and g=9.81\,\mathrm{m/s^2}.

Implicitly, this is taking the boundary conditions at the interfaces between each fluid to be that the pressure on either side of the interface is the same. This is correct for a flat interface. For a curved interface, there would be a “Laplace pressure” difference.

Substitute to obtain

\begin{aligned} p_{\text{gauge}} &= 13600 \times 9.81 \times 0.10 + 1000 \times 9.81 \times 0.05 + 850 \times 9.81 \times 0.01 \\ &= 13341.6\,\mathrm{Pa} + 490.5\,\mathrm{Pa} + 83.4\,\mathrm{Pa} \\ &\approx 13915.5\,\mathrm{Pa}\,. \end{aligned}

Thus, the gauge pressure at the bottom of the column is approximately 13.9\,\mathrm{kPa}.

  1. REFLECTION:
    This calculation illustrates how the use of the hydrostatic equation in layered fluids provides a quantitative basis for pressure measurement in manometers. The accuracy of the result depends on the assumed densities; in practice, care must be taken if temperature variations or contamination might alter the fluid properties.

In formulating this response, I consulted the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked the sections according to their relevance. If they are marked as relevant, I encourage you to explore them:

Textbook sections:

  1. Chapter 3.3: Hydrostatic Distributions in Pipes and Columns (HIGHLY RELEVANT)
  2. Chapter 9.3: Why We Use Irrotational Flow Analysis Part 5 (HIGHLY RELEVANT)
  3. Chapter 3.2: Hydrostatic Equation (HIGHLY RELEVANT)
  4. Chapter 4 Engineering Models: Hydrostatic Equation (HIGHLY RELEVANT)
  5. Chapter 3 How Stuff Works: How manometers measure pressure differences (HIGHLY RELEVANT)

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