# 61 Worked Examples: Fluid Properties & Hydrostatics

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

Worked Example #1

Convert the following temperature values. Write down the appropriate formulas and show all steps in your calculations. Do not use Google or similar to “convert” these temperature values–use the formulas.

(a) C to units of F.

The conversion is

so with C, then

(b) C to units of K.

The conversion is

so with C, then

(c) F to units of R.

The conversion is

so with F, then

(d) R to units of K.

The conversion is

so with R, then

(e) C to units of R.

The conversion is

so with C, then

Worked Example #2

The air in the test section of a wind tunnel has a measured temperature of 82F and a pressure of 101.1 kPa. Determine the density of the air in both SI and USC units.

SI units:

A temperature of 82F converted to base units of Kelvin is

The equation of state is

where is 286.9 J kgK in the SI system. Therefore,

USC units:

A temperature of 82F converted to base units of Rankine is

The equation of state is

where is 1716.59 ft lb slugR in the USC system. A pressure of 101.1 kPa (N/m) is the same as 2111.52 lb ft. Therefore,

Notice: Raw values (e.g., measurements) in both SI and USC units are common in practice. However, it is critically important that the calculations be conducted in a consistent set of units, i.e., entirely in SI units or entirely in USC units.

Worked Example #3

A bed of SAE 10 oil separates two plates that are 20 mm apart. The bottom plate is stationary, but a force applied to the top plate is such that it moves at a steady speed of 25 mm/s. It is found that the velocity profile of the oil between the plates can be approximated by

Calculate the shear stress in the oil as a function of (per unit depth or length of the plate) and use MATLAB to plot your results on a graph. Plot also the velocity profile itself. Assume that the dynamic viscosity of the oil is = 0.084 kg m s.

The shear stress can be calculated using Newton’s law of viscosity, i.e.,

This problem only involves one dimension () so

We are told that

so

Therefore, the shear stress for this velocity gradient is

Some MATLAB code is given below:

clc
figure
axis([0.0 25 0.0 200])
y = linspace(0.0,20,100); %set the range of values
u = 11.822 * y.^0.25; %velocity profile
tau = 0.248 * y.^-0.75; %shear stress
plot(u,y);hold on
xlabel(‘u (mm/s)’)
ylabel(‘y (mm)’)
clc
figure
plot(tau,y);hold on
xlabel(‘tau (N/mm/unit depth)’)
ylabel(‘y (mm)’)

The results are easily plotted, being sure to pay attention to the units, which would be per unit depth. Notice that the shear stress is high near the surface (wall), where the velocity gradient is higher and drops off quickly when moving away from the wall, typical of a boundary layer flow.

Worked Example #4

Suppose a measurement in the air gives a temperature of 102F, calculate the dynamic viscosity coefficient. What happens to the viscosity of air as its temperature increases, and why?

Sutherland’s Law can be expressed as

where R, R and slugs s ft. The absolute temperature is

Inserting the values gives

Bonding between the molecules in a gas is relatively low compared to a liquid. Therefore, the intermolecular momentum transfer between the molecules increases more with increasing temperature, which manifests as an increase in viscosity.

Worked Example #5

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, determine the volume required for each balloon at the takeoff point. 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 buoyancy principle, which Archimedes’ theorem applies. We will ignore the weight of 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 the weight of the illusionist, , plus the weight of the helium, , in the balloons, so

The air volume displaced by each balloon 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.

Note: We should be careful about using the term “weight” because we inevitably use units of kg in commonly used SI units. However, a kg is a unit mass, and the unit of weight is a force (Newtons), so in engineering calculations, we need to be careful and much more precise and use Newtons for force (i.e., mass times acceleration under gravity).

Worked Example #6

An airship has a gas envelope with a 296,520 ft volumetric capacity; helium is used to fill it. If the airship has an empty weight of 14,188 lb and a maximum fuel load of 960 lb, what would be the maximum payload the airship could carry? Assume MSL ISA standard atmospheric conditions. The density of helium can be assumed to be 0.0003192 slugs/ft.

The hydrostatic principle of buoyancy is applicable here. Using Archimedes’ principle, then the upforce (lift), , on the airship will be equal to the net weight of the fluid displaced minus 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 takeoff point would be 4,491 lb.

Worked Example #7

For a class science experiment, the students want to see if they can lift a 10 kg block off the ground using a helium-filled balloon. Determine the balloon volume required for the block to float at = 1,000 m under standard ISA conditions. The helium pressure inside the balloon is 5% greater than the local ambient atmospheric pressure. You can assume that ISA properties are modeled using

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 using Archimedes’ principle, which says that the upthrust on an object (in this case, the balloon) is equal to the weight of the fluid displaced by that object. To ” float,” the upthrust produced by the balloon must equal 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 the air it displaces and the weight of the helium inside the balloon.

where is the volume. Also, we have that

So, the governing equation is

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

At 1,000 m, 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 , the 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 #8

Write down the engineering form of the equation of state and identify the terms. In what format do we most commonly use the equation of state?

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, 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 #9

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.

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 are 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.

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), 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 #10

The tires of the main landing gear of an airplane are at a temperature of 40F and pressure of 40 lb/in just before landing. After landing and braking, the temperature of the tires increases to 165F. Determine the new pressure in the tires. Hint: Assume that the volume of the tire stays the same.

Because pressure and temperature are involved in this problem, the equation of state is valid, 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

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, so the conversion factor cancels out. Therefore, the pressure in the tires after landing will be

Worked Example #11

A well-type manometer filled with water has a leg inclined at an angle of 30 to the horizontal. When pressure is applied on the well side, the water moves along the leg side from level 1 to level 2 by a distance equal to 25.25 cm. Find the pressure difference that has been applied.

The relevant equation is the hydrostatic equation for a constant density fluid, i.e., constant, where is the vertical height. The working fluid is water, so 1,000 kg/m. Remember that the liquid level in the well does not change for a well-type manometer, at least not significantly enough to affect the pressures.

We are told that the manometer is inclined at an angle of 30 to the horizontal. Hence, the relationship between the length of the measured displacement of the water levels 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 #12

A tank that is vented to the atmosphere is partly filled with fuel. The tank is subject to external vertical and lateral accelerations, and , respectively. Find an expression for the slope (and the angle ) of the free surface of the fuel in the tank in terms of the accelerations.

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 appears 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 #13

A cylindrical container of radius is filled 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 centripetal acceleration. 2. Even though this is a three-dimensional problem, it is axisymmetric so that it can be solved in two dimensions.

We can assume axisymmetry about the spin axis to define in the radial direction and in the vertical direction. Using the basic equations for a pressure field (with a body force), the relevant subset of equations in this particular case are

The centripetal acceleration (i.e., body force per unit mass) of a fluid element 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 is 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 the initial height of the free surface of the non-rotating case. Below is a graph showing the free surface’s shape for increasing rotational rate values.

Worked Example #14

The first stage of a 20 m tall cylindrical rocket fuel tank is filled with liquid RP-1 fuel. This stage experiences a maximum vertical acceleration of 5 g during the launch. Calculate the maximum excess pressure on the bottom of the tank during launch. Assume that the density of RP-1 is 820 kg/m.

The relevant equation is the hydrostatic equation for a constant density fluid, i.e., constant, where is the vertical height. The working fluid is kerosine with 820 kg/m. The problem can 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 #15

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 the right side to the outlet at pressure . Assuming that the inlet and outlet conditions are at the same elevation, determine the pressure increase produced by the pump. Remember that specific gravity SG = . The density of water can be assumed to be 1.93 slugs/ft.

The pressure at point “a” will equal the pressure produced by each arm of the U-tube manometer. Looking 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 #16

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 .

If the open end of the manometer in the above figure is now exposed to the atmosphere and = 5.2 inches and = 18.6 inches, then find the pressure . Assume atmospheric pressure is 14.7 lb/in. The density of water can be assumed to be 1.93 slugs/ft, and the specific gravity (SG) of mercury is 13.6. Use = 32.17 ft/s. Note: SG = .

Use the foundational principle that the hydrostatic pressure at the same height in the same fluid will be constant. Using the hydrostatic equation for the left leg

or in terms of specific weight

and correspondingly for the right leg

Therefore,

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

In the second part of the question, we are then given that is atmospheric pressure = = 14.7 lb/in = 2,116.8 lb/ft. Also, we can calculate and , i.e.,

and

Therefore, solving for the value of gives

Worked Example #17

A liquid manometer, shown in the figure below, is frequently used to measure small pressure changes. 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. We will look first at point (1) on the left leg of the manometer. Using the principles of hydrostatics, then pressure there, , will be

Looking at the other leg of the manometer, 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 #18

In a particular industrial application, the three-dimensional pressure field in a stagnant fluid is known to behave according to

where the plane is horizontal and is measured positive upwards. Find the equation that describes the pressure field and 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

Finally, 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 be verified using partial differentiation, which will obtain the initial equation for this problem.

Worked Example #19

A hot air balloon becomes neutrally buoyant at an altitude of 1,100 m where the outside air temperature is 10.5C. With all its cargo, the balloon has a mass of 500 kg (excluding the air in the balloon), displacing a volume of 7,000 m. Assume that the pressure inside the balloon is the same as that outside. Determine the air temperature inside the balloon to give it neutral buoyancy. Hint: At 1,100 m, the air density can be considered 1.102 kg m.

The physics of the problem says that for neutral buoyancy, then 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. There is no net vertical force on the balloon, so

The air volume 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 at the 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 #20

A spherical holding tank with a diameter 2.7 m stores gaseous helium for a blimp under a pressure of 250 kPa. Determine the total weight of the gas inside the tank. Assume that = 2077.1 J kg K and ISA MSL temperature conditions prevail.

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 #21

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 progressively pushes the piston down into the cylinder and compresses the helium gas to half 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 must a mass (in kg) be placed on the top of the piston to compress the gas? Assume that = 2,077.0 J Kg K.

The volume is 5 liters, so = 5,000 cm = 0.005 m. The conditions are specified as MSL ISA, so = 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 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 gas mass 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

We are told that it 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 compress the gas fully, therefore, will be

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

Worked Example #22

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 seawater 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 dam wall 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

gives the differential equation

where the specified density variation has been included. Notice that this equation recognizes that is measured downward, i.e., in the opposite direction to the standard upward definition of .

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

which gives the hydrostatic pressure from the water as

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

and substituting the numbers gives

Worked Example #23

After the Great Flood, the human race built the Tower of Babel to reach Heaven. A hot-air balloon was made to lift heavy materials from the ground to the tower. Consider a hot-air balloon as a sphere with an inflated diameter ft. The system’s total weight (including the fabric, basket, and payload, but without air inside) is 1,000 lb. You may assume that the local air density in the ISA relative to MSL is given by

where is the altitude in feet. Assume that the balloon’s air mass 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, the weight of the 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 the air inside the balloon is

Assume the balloon 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, the air density is

or the density ratio is

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

Worked Example #24

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 an 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 points 1 to 2. What is the value of the weight ? Assume that the change in the 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 vertical height of the fluid levels matters in manometry. Hence, the term accounts for the inclined leg. The effects of the atmospheric pressure 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 #25

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

If the door length (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; the liquid’s pressure 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 , the distance (arm) of the pressure force about the hinge must be determined, i.e., by inspection of the figure, 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 leads to

MATLAB code check
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 #26

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.

1. Using hydrostatic principles, derive an expression for the pressure in terms of .
2. Derive a corresponding expression for the pressure in the tank in terms of .
3. Obtain a final expression relating to .
4. If = 25 cm, = 50 cm, = 60 cm, and = 40 cm, then determine the internal pressure in the tank.
5. If the air temperature 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 the given 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 #27

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 to be 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:

1. at point A (just below the piston).
2. at point B.
3. at point C.
4. at point D.

1. What is the relationship (as an equation) between the value of the mass and the height of the Galinstan column?
2. 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 continue the preceding 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 canceling out the acceleration under gravity, the “” term gives

So the value of is

Notice that the grouping has length dimensions, 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 to insert the numerical values given to us. Therefore, the value of is

so

Worked Example #28

Consider a cylinder of circular cross-section containing helium gas, initially at MSL ISA pressure and temperature conditions. A frictionless piston seals the top end of the cylinder, as shown in the figure below. The initial volume of helium is 5.2 liters, and the bottom of the piston is at a height 16.2 cm. A mass is then placed on the piston, which pushes the piston down into the cylinder to 8.3 cm and compresses the helium gas inside. During this process, the temperature of the gas also increases by 23.3C. Assume that for helium = 2,077 J Kg K.

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 ?

The volume is 5.2 liters, so = 5,200 cm = 0.0052 m. The conditions are specified as MSL ISA, so the pressure Nm = 1.01325 Pa, and temperature = 15C = 288.15K. We see that volume, pressure, density, and temperature are all involved in this question, so the use of the equation of state is appropriate, i.e.,

where is the mass of gas 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 relatively 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 #29

An oil pipeline and a rigid air tank containing compressed air at an initial pressure of 1.169 MPa and temperature of 80C are connected by a manometer filled with Gordium, as shown in the figure below. Assume that the hydrostatic effects of the air column inside the tank and the air volume in the manometer are negligible. The specific gravity for the oil is 2.68 and 11.8 for Gordium. Assume that the water density is 1,000 kg m.

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

We are given the pressure and temperature in the compressed air tank, so

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

Also, remember 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 given 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 the tank contains the same gas mass, so the density is unchanged. Therefore, the new pressure is

(ii) Notice that the left-side leg of the manometer has a larger bore than 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 to the equation for the hydrostatic balance, then

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 manometer fluid at the air/Gordium interface will increase by 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.

Worked Example #30

Consider the manometer system below, which taps off and connects two pipes containing fluids A and B, respectively. Fluid A has a specific gravity (S.G.) of 0.9, and fluid B has an S.G. of 3.7. Assume that the density of water is 1.940 slugs/ft.

1. Determine an expression for the pressure at point B.
2. Determine an expression for the pressure at point C.
3. Determine an expression for the pressure at point D.
4. Write down a final expression for the value of .
5. If the pressure were to decrease, what would happen to the level at point C?
6. If = 10 inches, = 13 inches, = 5 inches, = 15 inches, and = 45, then calculate the pressure difference in units of lb/in.

1. The pressure at point B will be

where is the density of fluid A.

2. The pressure at point C will be

3. Because points C and D are at the same height connected by the same fluid, then the pressure at point D is the same as at point C, i.e.,

4. For the final part of the manometer, we are given a diagonal length in fluid B rather than a vertical height. However, the vertical height between points D and E is just , so that

where is the density of fluid B, and

Therefore, the final expression for is

and collecting terms gives

5. If the pressure were to decrease, then the level at point C would increase, so the level would increase, and the value of would decrease.

6. In the final part, we are asked to substitute some known heights for the various levels and calculate the value of the pressure difference . We are given the specific gravity of each fluid, so

Substituting the known values gives

Finally, after doing the arithmetic, then the required pressure difference is

Worked Example #31

The viscosity of an oil is to be determined using a falling ball viscometer, as shown in the figure below. A heavy steel ball of diameter is dropped into the oil and slowly descends at a steady velocity , the time taken for the ball to pass between two reference points being recorded. The forces acting on the ball will be its weight, buoyancy, and fluid drag. The fluid drag force, , is given by

where is the viscosity (coefficient of dynamic viscosity), which is to be determined. The steel ball has a diameter of 2.1 mm and a density of 7,750 kg m. The oil has a density of 940.2 kg m. What is the oil’s viscosity if the steel ball falls through a distance of 30 cm in 12.6 s?

The balance of forces in equilibrium is such that the weight of the ball, , less any buoyancy force, , is equal to the viscous drag on the ball, . The weight will be density times volume times acceleration under gravity, i.e.,

where is the density of the steel ball and . The (upward) buoyancy force on the ball (Archimedes’s principle) will be equal to the weight of oil displaced by the ball, i.e.,

Therefore, the equilibrium equation is

The drag force on a sphere of radius moving through a fluid of high viscosity at low speed (this is called a creeping flow) is given by Stokes’s law, i.e.,

Therefore, in equilibrium, then

and rearranging to solve for gives

The value of can be found using the measured time over the fixed distance, i.e.,

Therefore, introducing all of the numerical values gives

Worked Example #32

The density of an irregularly shaped solid object needs to be determined. The object’s shape is such that it cannot be determined geometrically. Given a container of water, a thin wire, and a load cell to measure force, explain how you would measure the density of this object. Explain all the steps you would take and use equations to support your approach.

This question is about buoyancy and Archimedes’ principle, which states that the buoyancy force on a body immersed in a fluid equals the weight of the fluid displaced. In this case, you should tie the body to the wire, weigh it in the air, and then weigh it when immersed in the water. The difference, then, is the buoyancy force.

The weight of the body measured on the scale in air is

where is the volume of the body. The buoyancy force, according to Archimedes’ principle, is

The effective weight of the body measured on the scale when immersed in water is

Therefore,

from which the volume of the body is

The density of water, , is known. And because then

Therefore, the object’s density can be determined using the two readings on the load cell, and , and the known density of water.