riaology
Unit 03 · Lesson 02 of 09

How does pressure change with depth?

Depth sets pressure; density sets its slope.

Explore the model. Derive the result. Check its limits.

What you’ll be able to do
  • Derive a small-column balance.
  • Integrate layer by layer.
  • Gauge is relative to the surface reference.
Before you touch the controls

Will pressure jump when the probe crosses a flat interface between two fluids?

An experiment you can change

Probe a two-layer liquid column

Predict the direction of change, move one control, then explain the response using the equation.

Two-layer column with probe at 2 m depth. Upper density 800 kg per cubic meter.Upper liquidWaterProbe: 17.658 kPa gaugeProbe a two-layer liquid column. Depth (m) versus Gauge pressure (kPa). Selected result 17.658 kPa. Pressure is continuous at the interface; the slope changes with density. Absolute pressure adds the local surface pressure.0013.7341.527.4683Gauge pressure (kPa)Depth (m)
pg=9.811000[800min⁡(2,1)+1000max⁡(2−1,0)]=17.658 kPap_g=\frac{9.81}{1000}\left[800\min(2,1)+1000\max(2-1,0)\right]=17.658\,\mathrm{kPa}

Pressure is continuous at the interface; the slope changes with density. Absolute pressure adds the local surface pressure.

Model assumptions and units

Static open tank; upper layer 1 m deep, lower layer water at 1000 kg/m³, g = 9.81 m/s². No surface-tension pressure jump at the flat interface.

dpdz=−ρg,p(h)−ps=g∑iρiΔhi\frac{dp}{dz}=-\rho g,\quad p(h)-p_s=g\sum_i\rho_i\Delta h_i
01

Derive a small-column balance.

Take a stationary fluid element with vertical coordinate z pointing upward. Upward bottom pressure force balances downward top pressure force and weight. Dividing by element volume and taking the limit gives the pressure gradient.

p(z)A−p(z+dz)A−ρgA dz=0⟹dpdz=−ρgp(z)A-p(z+dz)A-\rho gA\,dz=0\quad\Longrightarrow\quad\frac{dp}{dz}=-\rho g
02

Integrate layer by layer.

For a constant-density layer, moving down by depth h increases pressure by ρgh\rho gh. In a layered column, add each layer’s contribution. The flat interface pressure is continuous when surface-tension curvature effects are absent.

p−ps=g∑iρiΔhip-p_s=g\sum_i\rho_i\Delta h_i
03

Gauge is relative to the surface reference.

An open surface is at local atmospheric pressure. Gauge pressure subtracts that reference; absolute pressure adds it back. Equal depth gives equal pressure only in the appropriate connected static fluid, not across arbitrary separate fluids.

pabs=patm+pgp_{abs}=p_{atm}+p_g
04

Pressure loads a submerged wall.

A vertical rectangular wall under a single liquid has a triangular gauge-pressure distribution. Integrating pressure over area gives the resultant force, which acts below the centroid because pressure is greater at larger depth.

F=∫Apg dA=12ρgbH2,hCP=2H3F=\int_A p_g\,dA=\frac12\rho gbH^2,\quad h_{CP}=\frac{2H}{3}
A worked example

Two meters below a two-layer surface.

  1. The first meter is oil at 800 kg/m3800\,\mathrm{kg/m^3} and the next meter is water at 1000 kg/m31000\,\mathrm{kg/m^3}.
  2. Gauge pressure is 9.81[800(1)+1000(1)]=17658 Pa=17.658 kPa9.81[800(1)+1000(1)]=17658\,\mathrm{Pa}=17.658\,\mathrm{kPa}.
  3. If the atmosphere is 100 kPa, absolute pressure is 117.658 kPa117.658\,\mathrm{kPa}. The oil–water interface pressure is 7.848 kPa7.848\,\mathrm{kPa} gauge.
Make the idea your own

Calculate. Compare. Test the assumptions.

Work out your answer, check it, then explore the visual solution. The answer checker uses a 1.5% rounding tolerance; for model-limit questions it checks the displayed model result.

1 · Calculate

Cross the interface

Find gauge pressure 2 m below an open surface: first meter oil at 800 kg/m³, second meter water at 1000 kg/m³.

Hint

Add the pressure rise through each layer.

Explore the visual solution
Two-layer column with probe at 2 m depth. Upper density 800 kg per cubic meter.Upper liquidWaterProbe: 17.658 kPa gaugeProbe a two-layer liquid column. Depth (m) versus Gauge pressure (kPa). Selected result 17.658 kPa. Pressure is continuous at the interface; the slope changes with density. Absolute pressure adds the local surface pressure.0013.7341.527.4683Gauge pressure (kPa)Depth (m)
pg=9.811000[800min⁡(2,1)+1000max⁡(2−1,0)]=17.658 kPap_g=\frac{9.81}{1000}\left[800\min(2,1)+1000\max(2-1,0)\right]=17.658\,\mathrm{kPa}

Pressure is continuous at the interface; the slope changes with density. Absolute pressure adds the local surface pressure.

Model assumptions and units

Static open tank; upper layer 1 m deep, lower layer water at 1000 kg/m³, g = 9.81 m/s². No surface-tension pressure jump at the flat interface.

dpdz=−ρg,p(h)−ps=g∑iρiΔhi\frac{dp}{dz}=-\rho g,\quad p(h)-p_s=g\sum_i\rho_i\Delta h_i

Controls explore variations. The steps below retain the original question’s values.

Static open tank; upper layer 1 m deep, lower layer water at 1000 kg/m³, g = 9.81 m/s². No surface-tension pressure jump at the flat interface.

dpdz=−ρg,p(h)−ps=g∑iρiΔhi\frac{dp}{dz}=-\rho g,\quad p(h)-p_s=g\sum_i\rho_i\Delta h_i
2 · Compare

Stay in the upper liquid

Find gauge pressure 0.50 m below the surface in upper fluid of density 700 kg/m³.

Hint

The lower water contributes nothing above the interface.

Explore the visual solution
Two-layer column with probe at 0.5 m depth. Upper density 700 kg per cubic meter.Upper liquidWaterProbe: 3.433 kPa gaugeProbe a two-layer liquid column. Depth (m) versus Gauge pressure (kPa). Selected result 3.4335 kPa. Pressure is continuous at the interface; the slope changes with density. Absolute pressure adds the local surface pressure.0013.2431.526.4873Gauge pressure (kPa)Depth (m)
pg=9.811000[700min⁡(0.5,1)+1000max⁡(0.5−1,0)]=3.4335 kPap_g=\frac{9.81}{1000}\left[700\min(0.5,1)+1000\max(0.5-1,0)\right]=3.4335\,\mathrm{kPa}

Pressure is continuous at the interface; the slope changes with density. Absolute pressure adds the local surface pressure.

Model assumptions and units

Static open tank; upper layer 1 m deep, lower layer water at 1000 kg/m³, g = 9.81 m/s². No surface-tension pressure jump at the flat interface.

dpdz=−ρg,p(h)−ps=g∑iρiΔhi\frac{dp}{dz}=-\rho g,\quad p(h)-p_s=g\sum_i\rho_i\Delta h_i

Controls explore variations. The steps below retain the original question’s values.

Static open tank; upper layer 1 m deep, lower layer water at 1000 kg/m³, g = 9.81 m/s². No surface-tension pressure jump at the flat interface.

dpdz=−ρg,p(h)−ps=g∑iρiΔhi\frac{dp}{dz}=-\rho g,\quad p(h)-p_s=g\sum_i\rho_i\Delta h_i
3 · Challenge the model

Compare equal depths

At total depth 3 m, change upper-layer density to 1000 kg/m³. Find gauge pressure and explain the shape of the pressure curve.

Hint

Both layers now have the same density.

Explore the visual solution
Two-layer column with probe at 3 m depth. Upper density 1000 kg per cubic meter.Upper liquidWaterProbe: 29.43 kPa gaugeProbe a two-layer liquid column. Depth (m) versus Gauge pressure (kPa). Selected result 29.43 kPa. Pressure is continuous at the interface; the slope changes with density. Absolute pressure adds the local surface pressure.0014.7151.529.433Gauge pressure (kPa)Depth (m)
pg=9.811000[1000min⁡(3,1)+1000max⁡(3−1,0)]=29.43 kPap_g=\frac{9.81}{1000}\left[1000\min(3,1)+1000\max(3-1,0)\right]=29.43\,\mathrm{kPa}

Pressure is continuous at the interface; the slope changes with density. Absolute pressure adds the local surface pressure.

Model assumptions and units

Static open tank; upper layer 1 m deep, lower layer water at 1000 kg/m³, g = 9.81 m/s². No surface-tension pressure jump at the flat interface.

dpdz=−ρg,p(h)−ps=g∑iρiΔhi\frac{dp}{dz}=-\rho g,\quad p(h)-p_s=g\sum_i\rho_i\Delta h_i

Controls explore variations. The steps below retain the original question’s values.

Static open tank; upper layer 1 m deep, lower layer water at 1000 kg/m³, g = 9.81 m/s². No surface-tension pressure jump at the flat interface.

dpdz=−ρg,p(h)−ps=g∑iρiΔhi\frac{dp}{dz}=-\rho g,\quad p(h)-p_s=g\sum_i\rho_i\Delta h_i
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