Theory — Pressure and Flow in Fluids

Pressure Increases with Depth

In a fluid at rest, the pressure at a point comes from the weight of everything above it: the column of fluid plus the atmosphere pressing on the surface. The deeper you go, the more fluid sits above you, so the pressure rises in direct proportion to depth.

Pressure at Depth P = P_atm + ρ · g · h

P_atm = atmospheric pressure (≈ 101.3 kPa at sea level)
ρ = fluid density (kg/m³), g = gravity (m/s²), h = depth (m)
The fluid part of the pressure, ρgh, is the gauge pressure

Why Doubling Density Does Not Double the Reading

The pressure gauge reads the total pressure, P_atm + ρgh. Only the ρgh part depends on the fluid. If you double the density, you double ρgh, but the atmospheric part stays the same — so the total reading goes up, but it does not double. Turn the atmosphere off (place the tank in vacuum) and P_atm = 0, so the reading is just ρgh; then doubling the density really does double the reading.

Pascal's Principle

Pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid. In a connected (enclosed) tank, pressing down on the fluid on one side raises the pressure everywhere by the same amount, so gauges at the same height read the same value — this is the basis of the hydraulic press.

Continuity — Flow Through a Pipe

For an incompressible fluid, the volume flow rate is the same everywhere along a pipe. Where the pipe is narrow, the area A is small, so the speed v must be large to carry the same flow; where it is wide, the fluid slows down.

Continuity Equation Q = A · v = constant  →  A₁v₁ = A₂v₂
Narrow section → faster flow; wide section → slower flow

Bernoulli's Equation and the Water Tower

For an ideal fluid, the sum of pressure energy, kinetic energy, and gravitational energy per unit volume is constant along a streamline. For a tank open to the atmosphere, this gives Torricelli's result: fluid leaving an opening a depth h below the surface comes out at v = √(2gh). A stream directed straight up would, in the absence of losses, rise back to the height of the water surface in the tank.

Bernoulli / Torricelli P + ½ρv² + ρgy = constant (along a streamline)
Exit speed from depth h: v = √(2 · g · h)
Maximum stream height ≈ water level in the tank (ideal fluid)

Deeper or denser

More fluid weight above → higher gauge pressure (ρgh grows).

Narrower pipe

Smaller area → faster flow, since A·v stays constant.

Lower water level

Smaller depth h → slower exit speed and a shorter stream.

QuantityEquationNotes
Pressure at depthP = P_atm + ρghgauge reads total pressure
Gauge (fluid) pressureρghdoubles if density doubles
Pascal's principleΔP transmitted equallyenclosed fluid, hydraulics
ContinuityA₁v₁ = A₂v₂flow rate constant
Exit speed (Torricelli)v = √(2gh)opening depth h below surface

Apparatus

The equipment a real fluid-pressure experiment uses. In the simulation these are modelled for you, but the readings correspond to what each instrument would measure.

holds the fluid
Fluid tank
Holds the liquid column whose pressure varies with depth.
pressure difference
U-tube manometer
Measures the pressure difference between two points.
measures pressure
Pressure gauge
Reads the absolute pressure at a chosen depth.
measures volume
Graduated cylinder
Measures fluid volume for the density and flow work.
measures density
Hydrometer
Measures the fluid density used in P = rho g h.
measures length
Meter rule
Measures the depth h below the surface.

Instructions — Running the Virtual Experiment

The simulation has four tabs — Pressure, Enclosed Tank (Pascal), Flow, and Water Tower — matching the parts of the worksheet. This is a measurement exercise: read each value from the simulation and write it down in your worksheet as you go. Do not just watch the readings change. Pressures are shown in kilopascals (kPa).

Part 1 — Pressure with depth (Pressure tab)
1
Open Simulation → Pressure with water selected. Drag the pressure gauge to a depth of 1.0 m and record the total pressure. Repeat at 1.5, 2.0, 2.5, and 3.0 m, recording each reading. Tabulate pressure against depth — your values should rise in direct proportion to depth.
2
Keep the gauge at 2.0 m. Switch the fluid from water (1000 kg/m³) to gasoline (700 kg/m³), then to honey (1400 kg/m³), and record the reading for each. Compare them: the reading changes with density, but it does not simply double when the density doubles, because the atmospheric part of the pressure stays the same.
3
With the gauge still at 2.0 m, record the reading with the atmosphere on, then uncheck it to place the tank in vacuum and record again. The difference between your two readings is atmospheric pressure; with the atmosphere off, the reading is just ρgh.
Part 2 — Pascal's principle (Enclosed Tank tab)
1
Open Enclosed Tank (Pascal). Place a gauge at the same height on each side and record both readings. Add the 250 kg weight on one side, then record both gauges again and note how much each one rose.
2
Add more weight and record the new readings. The two gauges rise by the same amount each time — confirm Pascal's principle from your recorded numbers, not just by watching.
Part 3 — Flow and continuity (Flow tab)
1
Open Flow. The flow rate is held constant. Drag the speed gauge to the wide section and record the area and the speed; move it to the narrow section and record the area and speed there. Check that area × speed gives the same flow rate at both points (A·v = constant).
2
Make the middle narrower, then wider, recording the speed at the gauge for each setting. Then toggle friction and switch the fluid to honey or gasoline, recording how the speed near the walls changes.
Part 4 — Water tower and Torricelli (Water Tower tab)
1
Open Water Tower. Set the water level to 3.0 m, open the spout, and record the exit speed and the range of the stream. Repeat at 2.0 m and 1.0 m, recording each pair. Confirm that your recorded exit speeds follow v = √(2gh).
2
Switch on the Hose so the stream points up. Record the maximum height the stream reaches and compare it with the water level in the tank.

Simulation — Fluid Pressure and Flow

Fluid Pressure and Flow Virtual LabDrag the gauge to read pressure at depth
pressure gauge
Drag the gauge depth with the slider. Reading in kPa.

Tank & fluid

Pressure gauge
Fluid density ρ1000 kg/m³
Depth h2.0 m
Atmosphereon
Gauge (fluid) ρgh19.6 kPa
Total reading P120.9 kPa
P = P_atm + ρgh.
Enclosed U-tank · gauges at equal height on both sides.

Enclosed tank — add weight

Weight is placed on the LEFT piston (area 0.10 m²). Pascal: the added pressure appears on both sides.
Gauge readings (same height)
Added weight (left)0 kg
Added pressure ΔP0.0 kPa
Left gauge120.9 kPa
Right gauge120.9 kPa
With no weight, both gauges read the same.
speed gauge
Flow rate held constant. Speed gauge reads local speed.

Pipe & fluid

Flux meter / speed gauge
Flow rate Q2.00 m³/s
Area at gauge1.00 m²
Speed at gauge2.00 m/s
Q = A·v stays constant along the pipe.
water stream
Exit speed v = √(2gh). Hose stream rises toward the water level.

Water tower

Water tower
Depth h3.0 m
Exit speed v = √(2gh)7.67 m/s
Range of stream— m
Open the spout and watch the trajectory.

Team Questions

Question 1. Water (ρ = 1000 kg/m³) sits 2.0 m above a gauge, g = 9.8 m/s². What is the gauge (fluid) pressure ρgh, in kPa? (Type just the number — e.g. 19.6)
Question 2. With the atmosphere ON (P_atm = 101.3 kPa), if you change the fluid from gasoline (700) to honey (1400) at the same depth, does the total gauge reading double? (Answer "yes" or "no")
Question 3. With the atmosphere OFF (vacuum), does the reading double when the density doubles from 700 to 1400? (Answer "yes" or "no")
Question 4. In a pipe carrying a constant flow rate, the area is halved in the narrow section. What happens to the fluid's speed there — does it halve, stay the same, or double? (One word)
Question 5. A water tower has its surface 3.0 m above the opening. What is the exit speed of the water? Use v = √(2gh), g = 9.8. (Type just the number in m/s — e.g. 7.67)
Question 6. When a hose on the water tower points straight up and leakage is matched, roughly how high does the stream rise compared with the water level in the tank? (Answer "lower", "about the same", or "higher")
Question 7 — Challenge. In an enclosed tank, you press down on the fluid on the left side and the pressure on the right side rises by the same amount. What principle does this demonstrate? (One or two words)

Example Lab Report

Sample report demonstrating the expected format and level of detail. Use as a guide if your instructor asks for a full report; otherwise complete the worksheet.

Fluid Pressure and Flow

Physics | Section: [Your Section] | Date: [Date]

Lab Members: [Names of all members present]

Purpose

To investigate how pressure in a fluid depends on depth, density, gravity, and atmospheric pressure; to confirm that an enclosed fluid transmits pressure equally (Pascal's principle); to observe the continuity relation A·v = constant for flow in a pipe; and to study the stream from a water tower using Torricelli's and Bernoulli's relations.

Theory

Pressure at depth is P = P_atm + ρgh. The gauge reads the total pressure, so the fluid contribution ρgh doubles when the density doubles, but the total reading does not (the atmospheric part is unchanged). With the atmosphere off, the reading is ρgh alone and doubles with density. An enclosed fluid transmits an applied pressure equally everywhere. For flow, A₁v₁ = A₂v₂. For a tank open to air, the exit speed at depth h is v = √(2gh).

P = P_atm + ρgh · A₁v₁ = A₂v₂ · v = √(2gh)

Calculations — Sample values

Gauge pressure (water, h = 2 m): ρgh = 1000 × 9.8 × 2 = 19 600 Pa = 19.6 kPa; total = 19.6 + 101.3 = 120.9 kPa

Density effect (vacuum): gasoline ρgh = 700 × 9.8 × 2 = 13.7 kPa; honey = 1400 × 9.8 × 2 = 27.4 kPa — exactly doubled

Gravity effect: at g = 12, ρgh = 1000 × 12 × 2 = 24.0 kPa (up from 19.6 kPa); the gauge reading rises in proportion to g

Water tower (h = 3 m): v = √(2 × 9.8 × 3) = 7.67 m/s

Results Table — Pressure (h = 2 m, g = 9.8 m/s²)

Fluidρ (kg/m³)ρgh (kPa)Atmosphere on (kPa)Atmosphere off (kPa)
Gasoline70013.7115.013.7
Honey140027.4128.827.4

With the atmosphere on, the reading rose but did not double; with the atmosphere off, doubling the density doubled the reading.

Discussion

The pressure rose with depth and density as predicted by ρgh. With the atmosphere on, doubling the density did not double the reading because the gauge also includes the fixed atmospheric pressure; in vacuum the reading was just ρgh and doubled exactly. Increasing gravity raised the gauge pressure proportionally. In the enclosed tank, adding weight on one side raised the pressure equally on both gauges, confirming Pascal's principle. In the pipe, narrowing the middle increased the speed while the flow rate stayed constant (A·v = constant), and the water-tower stream slowed and fell shorter as the level dropped, consistent with v = √(2gh) and Bernoulli's equation.

Conclusion

Pressure in a fluid follows P = P_atm + ρgh; the gauge (fluid) part scales with density and gravity, while the atmospheric part is fixed. Enclosed fluids obey Pascal's principle, flowing fluids obey continuity, and the water-tower stream obeys Torricelli's/Bernoulli's relations. Experimental readings agreed with theory throughout.

Practice Questions

Show all work and include units. Use P_atm = 101.3 kPa and g = 9.8 m/s² unless told otherwise.

Question 1
Find the gauge pressure 5.0 m below the surface of a freshwater lake (ρ = 1000 kg/m³). Then find the absolute (total) pressure at that depth.
Hint: gauge = ρgh; absolute = P_atm + ρgh.
Question 2
A gauge sits 2.0 m deep in gasoline (700 kg/m³). Compute the reading with the atmosphere on and in vacuum. Explain why only one of them doubles when you switch to honey (1400 kg/m³).
Hint: total = P_atm + ρgh; the atmospheric part does not scale with density.
Question 3
Water flows at 3.0 m/s through a pipe of cross-section 0.020 m². The pipe narrows to 0.005 m². Find the flow rate and the speed in the narrow section.
Hint: Q = A₁v₁; then v₂ = Q/A₂.
Question 4
A water tower's surface is 4.0 m above a small opening. Find the exit speed of the water. If the level falls to 1.0 m, what is the new exit speed?
Hint: v = √(2gh) for each depth.
Question 5
Explain, using Pascal's principle, how a hydraulic lift can raise a heavy car with a small applied force. Why does the pressure (not the force) stay the same across the two pistons?
Hint: P is transmitted equally; F = PA, so a larger piston area gives a larger force.
Question 6 — Challenge
Using Bernoulli's equation between the water surface (speed ≈ 0, open to air) and a stream at its maximum height, show that an ideal vertical stream rises to the height of the water surface. State which real-world effect prevents it from quite reaching that height.
Hint: equate ρg·y_surface to ½ρv² then to ρg·y_max; friction/air resistance and turbulence cause losses.