
A fluid is a form of matter that easily flows and changes shape in response to an applied force. Forces in fluids (gases and liquids) act through pressure. Pressure is force per unit area and has units of pascals (N/m^{2}) and pounds per square inch (psi or lbs/in^{2}). Weight creates hydrostatic pressure, including atmospheric pressure and pressure underwater. Energy conservation along a streamline leads to Bernoulli’s equation, which relates pressure, volume, and height in a moving fluid.

fluid, pressure, compressible, incompressible, Boyle’s law, Charles’s law, ideal gas, ideal gas law, density, Bernoulli’s equation, streamline


  
  $$\rho gh+{\scriptscriptstyle \frac{1}{2}}\rho {v}^{2}+P=\text{constant}$$


 Review problems and questions 

 A cubic box contains gas at a gauge pressure of 400,000 N/m^{2} (400,000 Pa) and measures 0.5 m × 0.5 m × 0.5 m. What force does the gas inside the box exert on any single side of the box?

Answer: 100,000 N The direction of the force will always be outward. The area of any one side of the box is given by$$A=\left(0.5\text{m}\right)\times \left(0.5\text{m}\right)=0.25{\text{m}}^{2}$$Since P = F/A, the force on any one side of the box will be$$\begin{array}{l}F=P\times A=(400,000\text{Pa)}\left(0.25{\text{m}}^{\text{2}}\right)=100,000\text{N}\end{array}$$

 Put the following pressures in order from lowest to highest.
 1 atmosphere
 1 psi
 1 N/m^{2}
 100 psi
 10,000 Pa

c. 1 N/m^{2} = 1 Pa b. 1 psi = 6,895 Pa e. 10,000 Pa a. 1 atmosphere = 101,325 Pa d. 100 psi = 689,500 Pa

 A submarine must be able to dive to a depth of 800 m. What is the absolute pressure at this depth? How does this compare to atmospheric pressure of 14.7 psi?

Answer: The absolute pressure is 7.9×10^{6} Pa, which is much larger than one atmosphere. The increase in pressure due to the water comes from the hydrostatic pressure equation, where the density of water (assumed freshwater here) is 1,000 kg/m^{3}:$$P=\rho gh=(\mathrm{1,000}{\text{kg/m}}^{3})(9.8{\text{m/s}}^{2})(800\text{m})=7.8\times {10}^{6}\text{Pa}$$To this you must add the pressure of one atmosphere (for the air above the water), which is 1.01×10^{5} Pa = 0.1×10^{6} Pa, for a total pressure of 7.9×10^{6} Pa. This is much larger (by a factor of 78!) than one atmosphere, which is 14.7 psi or 101,325 Pa.

 A cylindrical container full of liquid mercury is 80 cm high and has a radius of 10 cm. The density of mercury is 13,580 kg/m^{3}.
 What is the volume of the container?
 What is the mass of the mercury?
 What pressure does the container exert on the floor? Assume that the mass of the container is negligibly small compared to the mass of the mercury.

Answer: 0.025 m^{3}
 340 kg
 106,500 Pa

 Calculate the net force on a 3 m square window when a wind outside blows parallel to the window at 20 m/s. Assume that the quantity ρgh + ½ρv^{2} + P is the same inside and outside but that a sudden wind gust allows the outside pressure to drop compared to the inside pressure where v = 0. Express your answer in newtons and pounds. Do you think this is a significant problem for building engineers? Why? (Assume the density of air to be 1.29 kg/m^{3}.)

Answer: 2,300 N = 520 lb In Bernoulli’s equation, all of the quantities in ρgh are the same on both sides of the window, because it is the same air on both sides and at the same height above ground. So it is changes in ½ρv^{2} that result in changes of P to keep (ρgh + ½ρv^{2} + P) constant on both sides of the window. Set P equal to ½ρv^{2} and calculate:$$P={\scriptscriptstyle \frac{1}{2}}\rho {v}^{2}={\scriptscriptstyle \frac{1}{2}}(1.29{\text{kg/m}}^{3}){\left(20\text{m/s}\right)}^{2}=258{\text{N/m}}^{2}$$The window has dimensions of (3 m)×(3 m) = 9 m^{2}, so the total force on the window (F = P×A) is nine times 258 N/m^{2} or 2,300 N. There are 4.45 N/lb, so 2,300 N is equivalent to 520 lb of force.
This is a lot of force, equivalent to a 230 kg person (or, more likely, two 115 kg people) standing on the window glass! In hurricane winds, this can cause window glass to shatter, so it is an important design constraint for engineers.

 A certain quantity of gas has an absolute pressure of 250,000 Pa at 20ºC. What would the pressure be if the temperature increased to 1,000ºC while the volume remained constant?

Answer: 1.1×10^{6} Pa The most important first step of solving pressure and temperature problems is to convert the temperature to kelvins! The initial temperature is 293 K and the final temperature is 1,273 K. From the ideal gas law, pressure changes are proportional to temperature changes. Since the temperature increased by a factor of (1,273/293) = 4.3, the pressure increased by a factor of 4.3 from 250,000 Pa to 1.1×10^{6} Pa.
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