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# Pressure in Fluids

## Pressure is force per unit area and increases with depth in an incompressible fluid.

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Practice Pressure in Fluids
Progress
Estimated7 minsto complete
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Pressure in Fluids

Students will learn about pressure and solving pressure problems in the context of fluids.

### Key Equations

P=FAP=PoρghPressure is force per unit areaPressure in an incompressible fluid as a function of depth\begin{align*} P = \frac{F}{A} && \text{Pressure is force per unit area}\\ P = P_o - \rho gh && \text{Pressure in an incompressible fluid as a function of depth}\\ \end{align*}

Guidance
• The pressure of a fluid is a measure of the forces exerted by a large number of molecules when they collide and bounce off its boundary. The unit of pressure is the Pascal (Pa).
• In a fluid at rest, pressure increases linearly with depth – this is due to the weight of the water above it.
• Pascal’s Principle reminds us that, for a fluid of uniform pressure, the force exerted on a small area in contact with the fluid will be smaller than the force exerted on a large area. Thus, a small force applied to a small area in a fluid can create a large force on a larger area. This is the principle behind hydraulic machinery.
• Liquids obey a continuity equation which is based on the fact that liquids are very difficult to compress. This means that the total volume of a fluid will remain constant in most situations. Imagine trying to compress a filled water balloon!

#### Example 1

A weather balloon is ascending through the atmosphere. If the density of air is 1.2 kg/m3 and atmospheric pressure at sea level is 101.3 kPa, then what is the pressure on the balloon at (a) 100 m above the ground, (b) 500 m above the ground, and (c) 1000 m above the ground?

##### Solution

For all parts of these problems, we'll be using the equation for pressure given above where the atmospheric pressure at sea level is Po.

(a):

PPP=Poρairgh=101.3kPa1.2kg/m39.8m/s2100m=100.12kPa\begin{align*} P&=P_o-\rho_{air}gh\\ P&=101.3\;\text{kPa} - 1.2\;\text{kg/m}^3 * 9.8\;\text{m/s}^2 * 100\;\text{m}\\ P&=100.12\;\text{kPa}\\ \end{align*}

(b):

PP=101.3kPa1.2kg/m39.8m/s2500m=95.4kPa\begin{align*} P&=101.3\;\text{kPa} - 1.2\;\text{kg/m}^3 * 9.8\;\text{m/s}^2 * 500\;\text{m}\\ P&=95.4\;\text{kPa}\\ \end{align*}

(c):

PP=101.3kPa1.2kg/m39.8m/s21000m=89.5kPa\begin{align*} P&=101.3\;\text{kPa} - 1.2\;\text{kg/m}^3 * 9.8\;\text{m/s}^2 * 1000\;\text{m}\\ P&=89.5\;\text{kPa}\\ \end{align*}

### Time for Practice

1. A 1500kg\begin{align*}1500 \;\mathrm{kg}\end{align*} car is being lifted by a hydraulic jack attached to a flat plate. Underneath the plate is a pipe with radius 24cm\begin{align*}24 \;\mathrm{cm}\end{align*}.
1. If there is no net force on the car, calculate the pressure in the pipe.
2. The other end of the pipe has a radius of 2.00cm\begin{align*}2.00\;\mathrm{cm}\end{align*}. How much force must be exerted at this end?
3. To generate an upward acceleration for the car of 1.0m/s2\begin{align*}1.0 \;\mathrm{m/s}^2\end{align*}, how much force must be applied to the small end of the pipe?
2. A SCUBA diver descends deep into the ocean. Calculate the water pressure at each of the following depths.
1. 15m.\begin{align*}15 \;\mathrm{m}.\end{align*}
2. 50m.\begin{align*}50 \;\mathrm{m}.\end{align*}
3. 100m.\begin{align*}100 \;\mathrm{m}.\end{align*}
3. Ouch! You stepped on my foot! That is, you put a force of 550N\begin{align*}550 \;\mathrm{N}\end{align*} in an area of 9cm2\begin{align*}9 \;\mathrm{cm}^2\end{align*} on the tops of my feet!
1. What was the pressure on my feet?
2. What is the ratio of this pressure to atmospheric pressure?

1. a. 83,000Pa\begin{align*}83,000 \;\mathrm{Pa}\end{align*} b. 104N\begin{align*}104 \;\mathrm{N}\end{align*} c. 110N\begin{align*}110 \;\mathrm{N}\end{align*}
2. a. 248kPa\begin{align*}248 \;\mathrm{kPa}\end{align*} b. 591kPa\begin{align*}591 \;\mathrm{kPa}\end{align*} c. 1081kPa\begin{align*}1081 \;\mathrm{kPa}\end{align*}
3. a. 611kPa\begin{align*}611 \;\mathrm{kPa}\end{align*} b. 6atm\begin{align*}6 \;\mathrm{atm}\end{align*}