# Examples – Fluid Properties & Hydrostatics

These worked examples have been fielded as homework problems or exam questions

Worked Example #1

An illusionist floated up to nearly 25,000 feet by hanging from 52 helium-filled balloons. He then detached himself from the balloons, skydived several thousand feet, and parachuted to the ground. Assuming the illusionist weighed 90 kg, then determine the volume required for each of his balloons at the point of takeoff. You may assume that the density of the ambient air at takeoff is 1.19 kg m and that the density of helium at the same conditions is 0.1778 kg m.

The physics involved in this problem is the principle of buoyancy, for which Archimedes theorem applies. We will ignore the weight the ropes and the composition of the balloon material itself, which is practically negligible. Archimedes principle says that for neutral buoyancy then the upthrust, , must be equal to weight of the illusionist, , plus the weight of the helium, , in the balloons, so

The volume of air displaced by each balloon is , which is the value to be determined. Let be the number of balloons. The weight of air displaced by the balloons will be

The weight of the helium inside the balloons is

Therefore,

or

Solving for gives

Using the information given leads to the volume of each balloon, i.e.,

which seems reasonable based on the photos we have of the event.

Worked Example #2

The hydrostatic principle of buoyancy is applicable here. Using Archimedes’s principle then the upforce (lift), , on the airship will be equal to the net weight of the fluid displaced less the weight of the gas inside the envelope, i.e.,

where is the volume of the gas envelope.

The gross weight of the airship will be the sum of the empty weight, , plus the fuel weight, , plus the payload, , i.e.,

For vertical force equilibrium at takeoff then so

and rearranging for the payload weight gives

Using the numerical values supplied gives

Therefore, the maximum payload at the point of takeoff would be 4,491 lb.

Worked Example #3

with = 287 and = 2077 which are both in units of J Kg K. Neglect the weight of the balloon itself.

The physics of floatation centers around the use of Archimedes’ principle, which says that the upthrust on an object (in this case the balloon) is equal to the weight of fluid displaced by that object. To be able to “float” then the upthrust produced by the balloon must be equal to the weight of the mass plus the weight of the helium inside the balloon. We are told to neglect the weight of the balloon itself. The governing equation to solve, therefore, is

or

To find the upthrust produced by the balloon, we need to find the weight of air it displaces. We also need to find the weight of the helium inside the balloon.

where is the volume. Also we have that

so the governing equation is

We are given as an altitude of = 1,000 m, as well as two equations we can use to calculate the temperature and pressure at that altitude. It is reasonable to assume that the temperature of the helium in the balloon will be the same as that of the ambient air. We can then calculate the density of both air and of helium using the equation of state, and so proceed to find the needed volume of the balloon to obtain the required buoyancy or upthrust.

At 1,000 m then the temperature of the air and the helium will be

The pressure of the ambient air will be

and the pressure inside the balloon will be

Using the equation of state then density of the air will be

and the density of helium will be

We are told that the mass of the block is 10 kg so its weight is

Therefore, the needed volume of the balloon will be

Worked Example #4

The engineering form of the equation of state is

where is pressure, is density, is absolute temperature, and is the gas constant. In this form of the equation of state, then the value of is gas-specific (i.e., its value depends on the gas). Rearranging for the density, which is the most common use of the equation of state, gives

Worked Example #5

**SI Units:**During measurements in a wind tunnel, the pressure and temperature of the air in SI units are found to be 102.3 kPa and 15.7C, respectively. Calculate the density of the air in the wind tunnel. Assume MSL ISA conditions.

In this case, the absolute temperature is 15.7 + 273.15 = 288.85 K. The gas constant for air in SI units is 287.057 J kg K so the density of the air will be

**USC units:**During measurements in a wind tunnel, the pressure and temperature of the air in USC units is 14.61 pounds per square inch (psi) at a temperature of 71.1F. Calculate the density of the air in the tunnel. Assume MSL ISA conditions.

In this case, the absolute temperature is 71.1F + 459.67 = 530.77 R. The pressure is given in terms of pounds per square inch (psi) so to convert to base units of pounds per square foot (psf or lb/ft) then it is necessary to multiply by 144. The gas constant for air in USC units is 1716.59 ft lb slugR so the density of the air will be

Worked Example #6

Because pressure and temperature are involved in this problem, the equation of state is useful, i.e.,

Therefore, we can write that

We are told that the volume of the tire is the same before (condition 1) and after landing (condition 2), so this means that the density of the gas in the tires (usually nitrogen) remains the same (constant volume process) and so the pressure ratio is

We are given that (= 40F) and (165F) so that

and remembering to convert to absolute temperature units, i.e., Rankine in this case. We are also given the pressure in units of lb/in, but we do not need to convert to lb/ft because we are dealing with ratios and so the conversion factor cancels out. Therefore, the pressure in the tires after landing will be

Worked Example #7

The relevant equation is the hydrostatic equation for a constant density fluid, i.e., constant, where is the vertical height. In this case, the working fluid is water, so 1,000 kg/m.

We are told that the manometer is inclined at an angle of 30 to the horizontal, so the relationship between the length of the measured displacement of the levels of water in the tube, , and the vertical height is .

Therefore, the difference in the applied pressure will be

and substituting the given values leads to

Worked Example #8

The governing hydrostatic equation for the pressure in the fluid subject to a uniform external acceleration is

So for a uniformly accelerating fluid in static equilibrium in the – plane then

The minus sign on the acceleration terms appearing because these are inertial terms, i.e., in the opposite direction to the accelerations. The relevant subset of equations in this particular two-dimensional case are

Integrating these equations and solving for the pressure leads to

The shape of the free surface is obtained by letting = constant. Therefore, the shape of the free surface becomes a linear function of , i.e.,

and the angle that the free surface makes with the horizontal will be

Worked Example #9

A cylindrical container of radius filled is with water and then is rotated at an angular velocity of about a vertical axis to obtain “solid-body rotation” of the container and the water. Find an equation that describes the shape of the free surface of the water. Hints: 1. The body force in this case is a centrifugal force, so the body force per unit mass is a centripetal acceleration. 2. Even though this is a three-dimensional problem, it is axisymmetric so it can be solved in two dimensions.

We can assume axi-symmetry about the axis of spin, so we will define in the radial direction and in the vertical direction. Using the basic equations for a pressure field (with a body force) then the relevant subset of equations in this particular case are

The centripetal acceleration (i.e., body force per unit mass) of a fluid element in this case will be

so we have

Integrating the first differential equation with respect to gives

Integrating the second differential equation with respect to gives

Therefore, the solution for the pressure is

for which the pressure will be constant on the free surface of the water, i.e., the equation of the free surface will be

which is a parabola. It can be shown with some further effort that compared to the original free surface (i.e., when the container is not rotating), the free surface dips down at the center of the spin axis by an amount equal to the amount the surface rises at the rim of the container at by where

where the radius. Notice that the density of the fluid does not appear in this equation. Finally, the equation of the free surface is

where is an initial height of the free-surface of the non-rotating case. A graph showing the shape of the free surface for increasing values of rotational rate is given below – you should try to plot up these results by yourself such as by using MatLab.

Worked Example #10

The relevant equation is the hydrostatic equation for a constant density fluid, i.e., constant, where is the vertical height. In this case, the working fluid is kerosine with 820 kg/m. The problem can clearly be assumed to be one-dimensional, but the fuel tank is subjected to 5 times the normal acceleration under gravity. Therefore, the excess pressure (over and above any internal pressure) on the bottom of the tank will be

Worked Example #11

A U-tube manometer is connected to the inlet and outlet of a water pump as shown below; the left side is connected to the inlet at pressure and right side to the outlet at pressure . Assuming that the inlet and outlet conditions are at the same elevation, then determine the pressure increase produced by the pump. Note: Remember that specific gravity SG = . The density of water can be assumed to be 1.93 slugs/ft.

The pressure at point “a” will be equal to the pressure produced by each arm of the U-tube manometer. If we look at the left side arm then we can write

Similarly, for the right side arm then

Therefore, the difference in pressure becomes

and substituting the numerical values gives

Worked Example #12

A manometer contains two liquids of specific weight (Fluid 1 = water) and (Fluid 2 = mercury). Note: Specific weight . Determine the equation relating the pressure in the spherical reservoir, i.e., , in terms of , , , , and .

or in terms of specific weight

and correspondingly for the right leg

Therefore,

Solving for in terms of , , , , and as asked for in the question then

We are then given that is atmospheric pressure = = 14.7 lb/in = 2116.8 lb/ft. Also, we can calculate and , i.e.,

and

Therefore, solving for the value of in this case gives

Worked Example #13

To measure small pressure changes, a liquid manometer of the type shown in the figure below is frequently used. One leg of the manometer is inclined at an angle and the differential reading of the two heights of the liquid is measured along the inclined tube. The two pipes (which run into the screen) contain gases, the gas in pipe A being of density , and the gas in pipe B being of density . The density of the liquid in the tube is . Determine the equation for the difference in pressure between A and B in terms of the length .

Remember that the pressures at two points in the same fluid at the same vertical height are equal. In this case we will look at point (1) on the left leg of the manometer. Using the principles of hydrostatics then pressure there, , will be

Looking to the other leg of the manometer then the pressure at the same vertical height will be

remembering that it is the vertical height that matters here. Therefore, equating the two pressures at the same level gives

or that

which is what we are asked for. But we can take this answer a step further because if A and B are gasses then their density is very low compared to the density of a liquid, i.e., and and the difference in heights is small so that

Therefore, to a good approximation if A and B are gasses then

or

Worked Example #14

where the – plane is horizontal and is measured positive upwards. Find the equation that describes the pressure field and so the isobars, i.e., lines of constant pressure).

We are given two equations for the particular pressure field in stagnant flow, i.e.,

and

Also, in the direction we will have

Integrating the first of the three equations gives

and integrating the second gives

and integrating the third gives

Therefore, by inspection we see that and and so that the pressure field will be described by

which are cylindrical isobars. This latter result can then be verified using partial differentiation, which will obtain the initial equations given in this problem.

Worked Example #15

The physics of the problem says that for neutral buoyancy: The weight of hot air inside the balloon + the weight of the balloon itself = the upthrust according to Archimedes’ Principle caused by the weight of displaced cool air, i.e., there is no net vertical force on the balloon so that

The volume of air displaced by the balloon is = 7,000 m. Therefore, the weight of cool air displaced by the balloon is then

which is equal to the upward buoyancy force according to Archimedes’ Principle. The mass of the balloon is 500 kg so its weight, , is 500 = 500 9.81 = 4905.0 N. Notice is the density of the displaced air, in this case at a height of 1,100 m, which is given as 1.102 kg m.

Therefore, for neutral buoyancy (i.e., if the net vertical force is zero) the weight of the hot air in the balloon must be

This means that the corresponding density of the hot air inside the balloon must be

which, of course, is less than the density of the surrounding cool air, as expected.

We are asked to determine the corresponding temperature of the hot air, for which we can use the equation of state, i.e.,

We are told to assume that the pressure inside the balloon is the same as that outside the balloon so

and rearranging gives the ratio of the temperatures as

We are told that the outside air temperature at 1,100 m is 10.5 C or 283.65K. Therefore, the temperature of the hot air needed for neutral buoyancy will be

so a differential balloon internal temperature of about 20 C at this altitude and temperature conditions.

Worked Example #16

The volume of the tank will be

To find the weight of helium we need its density so use the equation of state and so

We are told to use ISA MSL temperature conditions so = 15C = (273.15 + 15)K = 288.15K, which we can assume as the same temperature as the helium . Therefore

The weight of helium in the tank, therefore, will be

Worked Example #17

Consider a circular cylinder of initial volume = 5 liters whose height = 16 cm. It contains helium gas at ISA MSL pressure and temperature conditions. A weightless (and frictionless) piston seals the top end of the cylinder. A mass is then placed on the piston, which has the effect of progressively pushing the piston down into the cylinder and finally compressing the helium gas to half of its original volume in an isothermal process.

(a) Determine the mass of helium gas (in kg) in the cylinder. (b) What is the new pressure of the helium gas in the cylinder after it is compressed? (c) How large a mass (in kg) must be placed on the top of the piston to compress the gas? Assume that = 2,077.0 J Kg K.

The volume is given as 5 liters, so = 5,000 cm = 0.005 m. The conditions are specified as MSL ISA, so from the formula sheet then pressure = 1.0132510 Pa and temperature = 15C = 288.15K.

We see that volume, pressure and temperature are involved in this question, so the use of the equation of state is appropriate, which can be written in this case as

where is the mass of gas.

(a) Assume that state 1 refers to the uncompressed state and state 2 refers to the fully compressed state. We can calculate the mass of the gas (in kg) using

This mass of gas is fixed because it is all contained in the volume below the piston.

(b) We have for the initial pressure at state 1

and for the fully compressed pressure at state 2

In this case we are told that is an isothermal process so so we can write for a given mass of gas that

Therefore, we have that

(c) In this part, we need to find the force on the top of the piston to fully compress the helium to half its volume. Force is a pressure difference between the chamber and ambient times an area, so the area of the piston, using the symbol , is

The force, , required to fully compress the gas, therefore, will be

and the mass that needs to be placed on top of the piston will be

Worked Example #18

Consider a large concrete dam that keeps sea water contained in a dock. The dam has dimensions of = 20 m wide (in the direction perpendicular to the screen), and is = 40 m high and = 30 m thick at its base. The density of sea-water can be assumed to vary linearly with depth according to the formula , where = 1,028 kg m at = 0 and = 0.0026 kg m.

(a) Find an expression for the hydrostatic water pressure acting on the wall of the dam below point A at some general depth .

(b) What is the value of the maximum water pressure (in kPa) at the foot of the dam at point B?

(a) Starting from the usual form of the hydrostatic equation

then we have in this case the specific differential equation

where the specified density variation has been included and recognizing that is measured downward, i.e., in the opposite direction to the normal upward definition of .

(b) Separating the variables and integrating downward from the water surface to a general depth gives

which gives for the hydrostatic pressure from the water as

When , which is at a the foot of the dam, the previous result gives

and substituting the numbers gives

Worked Example #19

where is the altitude in feet. You may also assume that the mass of air in the balloon does not change during the ascent.

(a) Develop the buoyancy equation(s) when the balloon is in static equilibrium.

(b) When the balloon leaves the ground, the air inside is heated to 210F and the pressure is equal to the MSL atmospheric pressure. Determine the maximum potential altitude the balloon can achieve when the temperature of the air remains at 210F.

(a) The forces acting on the balloon include buoyancy force, weight of balloon structure and the weight of hot air inside. When the balloon reaches the maximum altitude, the forces are in static equilibrium. We have

(b) At the initial state, the density of air inside of the balloon is

Assume the ballon is a sphere so its volume is

and the weight of air inside is

By using the condition of static equilibrium, when the balloon achieves the maximum altitude then the air density is

or the density ratio is

Substitute this density into MSL ISA model for the density and solving for gives the maximum potential altitude as

Worked Example #20

A 6-inch diameter piston is located within a cylinder and connected to an inclined U-tube manometer. The fluid in the cylinder and the manometer is oil with a density = 1.83 slug ft. When a weight is placed on the top of the piston, the fluid level in the inclined manometer tube moves 6 inches along the tube from point 1 to 2. What is the value of the weight ? Assume that the change in vertical position of the piston is negligible compared to the movement in oil along the tube and the entire system is open to the atmosphere.

Both sides of the manometer are equally exposed to atmospheric pressure , so it will be apparent there will be no differential effect of on the height of the manometer column, which is initially at level 1.

When the weight is placed on the top of the piston, the excess pressure on the oil from the weight (i.e., pressure = weight/area) causes the column to rise to level 2. For pressure equilibrium with the weight applied, and recognizing that the pressure at the same height in the same fluid is equal, then

where is the cross sectional area of the piston. Remember that the it is the vertical height of the fluid levels that matters in manometry, hence the term to account for the inclined leg in this case. The effects of the atmospheric pressure obviously cancel out in the preceding equation so

and rearranging gives

The diameter of the piston is = 6-inches = 0.5 ft, so the area of the piston is

Therefore, the value of the weight to be placed on top of the piston to get the fluid to move along the inclined leg the required 6-inches is

Worked Example #21

A fuel tank resides inside a spacecraft and is a part of a specialized passive pressure relief valve for use in an emergency. The hinged gate opens only if subjected to an excess acceleration during a normal launch. To accomplish this, the gate is kept closed during a normal launch by a preload torque , exerted by a coil spring.

If the length of the door (normal to screen) is , determine the minimum required preload torque () in terms of the parameters given. When the maximum acceleration is achieved, the spacecraft is in space and the outside pressure is 0 atm. Calculate the minimum preload torque if = 101,325 Pa, = 45 cm, = 125 cm, = 820 kg/m, = 80 cm, and = 5.

The pressure, the pressure force, and the moment of the pressure force about the hinge must be determined. This is a hydrostatics problem and the pressure in the liquid varies linearly with depth. Using the hydrostatic equation

where is measured downward from the surface of the liquid and is the translational acceleration load factor. The pressure variation on the door will produce a force . Considering a small strip of the door of height and width then the incremental force on that strip is

and so the net force on the door is obtained by integration, i.e.,

To find the moment then the distance (arm) of the pressure force about the hinge must be determined, i.e., by inspection of the figure then the arm is . Therefore, the net moment about the hinge is

which must now be solved. Expanding out gives

Integrating gives

Therefore

and simplifying gives

and so the moment is

Substituting the values given in this case leads to

P_0 = 101325; %Pa

h_1 = 45./100; %m

h_2 = 125./100; %m

rho = 820; %kg m-3

w = 80./100; %m

n = 5.0; % load factor

g = 9.81; %m s-2

M = (0.5.*P_0.*w.*((h_2-h_1).^2))+ …

((1./6).*rho.*n.*g.*w.*((2.*(h_2.^3))-(3.*h_1.*(h_2.^2))+(h_1.^3))); %Nm

Worked Example #22

A manometer containing water ( = 1,000 kg m) and mercury (SG = 13.6) is connected to a tank containing air at an internal pressure , the other end being open to atmospheric pressure of 101 Pa.

- Using hydrostatic principles, derive an expression for the pressure in terms of .
- Derive a corresponding expression for the pressure in the tank in terms of .
- Obtain a final expression relating to .
- If = 25 cm, = 50 cm, = 60 cm, and = 40 cm, then determine the internal pressure in the tank.
- If the temperature of the air in the tank is measured to be 30C, determine its corresponding density.

1. Let be the density of mercury. Using the principles of hydrostatics then the pressure will be

2. Continuing to work around the leg in the manometer (and now the fluid is water with density ) then

and rearranging gives

3. Equating the two previous results from parts (a) and (b) gives

and rearranging gives

or in final form as

where is the specific gravity of mercury, which is 13.6.

4. Now we can substitute in the known numerical values into the prior equation to give

5. The temperature of the air in the tank is given as 30C, so in units of Kelvin then

The density of the air can be found from the equation of state, i.e.,

Worked Example #23

A manometer containing water ( = 1,000 kg m) and Galinstan (SG = 6.44) is connected to a cylindrical water tank with a piston on top, as shown in the figure below. The diameter of the tank is 0.45 m. A mass = 6 kg is placed on top of the piston; the piston can be assumed as weightless. Both sides of the system are open to an atmospheric pressure = 101.3 kPa. The height values are: = 15.0 cm, = 50.0 cm, = 65.0 cm, and = 95.0 cm.

Part 1: Determine an equation for and evaluate the pressure (in kPa) at:

- at point A (just below the piston).
- at point B.
- at point C.
- at point D.

Part 2: Answer the following:

- What is the relationship (as an equation) between the value of the mass and the height of the Galinstan column?
- For the values of the parameters as given, what will be the value of (in units of cm)?

Part 1:

1. at point A (just below the piston): Pressure is defined as force/area so the pressure at point A will be

where is the weight ( and is the cross-sectional area of the cylinder. Therefore,

and inserting the values gives

2. at point B: We are going DOWN in the water with an increase in depth so the pressure at point B will be

Inserting the values gives

3. at point C: We are going UP in the water with a decrease in depth so the pressure at point C will be

Inserting the values gives

4. at point D: We are going DOWN in the water with an increase in depth so the pressure at point D will be

Inserting the values gives

Part 2:

1. In this problem we just continue the foregoing process until we get to point E, i.e.,

and noting at the is the density of Galistan and . Collecting terms and also noting that the effects of will balance out on each side of the problem (piston side and open end of U-tube) then

so

which establishes the relationship between and (the only unknown). Rearranging gives

and by cancelling out the acceleration under gravity “” term gives

so the value of is

Notice that the grouping has dimensions of length, as it should if the equation is to be dimensionally homogeneous. Also, the ratio of the densities is the specific gravity of the Galinstan = 6.44.

2. The final step in this question is just to insert the numerical values that are known to us. Therefore, the value of is

so

Worked Example #24

1. Determine the mass of helium gas in the cylinder.

2. What is the new pressure of the helium gas after it is fully compressed.

3. What will be the value of the mass placed on the top of the piston to compress the gas to these conditions?

4. If the helium cools back to ambient conditions, what will be the value of ?

where is the mass of gas contained inside the cylinder.1. State 1 refers to the uncompressed state and State 2 refers to the initially compressed state. We can calculate the mass of the gas (in kg) using

Notice that helium is really quite light. This mass of helium gas is fixed because it is all contained in the volume below the piston. Therefore, the corresponding density of the helium at State 1 is

2. We have for the initial pressure at State 1

and for the compressed pressure at State 2

where = 288.15 + 23.3 = 311.45K. The density is

and so the corresponding pressure is

3. We now need to find the force on the top of the piston to compress the helium. Force is a pressure times an area, so the area of the piston is

The weight, , on the piston that is required to fully compress the gas to State 2 will be

and the corresponding mass that needs to be placed on top of the piston will be

4. If the helium gas now cools down back to ambient temperature = 15C = 288.15K at State 3, then the weight will push the piston down a little further to . In this case

where the final compressed volume is just . Therefore, is given by

Worked Example #25

1. Determine the air density in the tank.

2. Determine the oil pressure in the pipe.

3. For a decrease in temperature in the air tank from 80C to 40C and assuming that the pressure in the oil pipe remains constant then determine:

(i) The new pressure in the air tank.

(ii) The changes in the level of the manometer fluid at the oil/Gordium interface and the manometer fluid at the air/Gordium interface.

1. We see that pressure, density, and temperature are all involved in this question, so the use of the equation of state is appropriate, i.e.,

Rearranging for the density gives

In this case we are given the pressure and temperature in the compressed air tank so

2. To find the pressure in the oil pipe over on the other leg of the manometer then we need to use hydrostatic principles. Starting at the oil pipe then adding a hydrostatic pressure increment as we go down and subtracting hydrostatic pressure increment as we go up then

also remembering that we are told to ignore the (negligible) hydrostatic effects of the air column inside the tank and the air volume in the manometer. Rearranging gives the pressure in the oil pipe as

Inserting the values of the known quantities gives

noting that and that . Performing the arithmetic gives

3. For a decrease in temperature in the air tank from 80C to 40C then:

(i) The new pressure in the air tank will be

Let and we note that tank contains the same mass of gas so the density is unchanged, therefore the new pressure is

(ii) Notice that the left-side leg of the manometer has a larger bore that the right-side leg, in the ratio 3:1. This means that if the level in the left-side leg drops by some amount, say then it is necessary to determine the change in the fluid length along the right-side leg as well as the corresponding vertical change. The volume of the Gordium is fixed so the relationship is

so . Because we need the vertical height change of the Gordium on the right-side leg, then

Returning back to the equation for the hydrostatic balance then

In this case, the changes in the heights of the Gordium for a change in air pressure will manifest as

which can be rewritten more simply in terms of the SG, i.e.,

Substituting the numerical values gives

so

The changes in the level of the manometer fluid at the oil/Gordium interface will drop 19 cm and the level of the manometer fluid at the air/Gordium interface will increase 1.1 m. Of course, it is noted that the change in the height of the Gordium along the length of the inclined tube is very sensitive to the change in pressure in the air tank.