Worked Examples: Fluid Properties & Hydrostatics

J. Gordon Leishman

doi:https://doi.org/10.15394/eaglepub.2022.1066.9

49 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. Be sure to write down the appropriate formulas you use and show all steps in your calculations. Do not use Google or similar to “convert” these temperature values.

(a) $12 ^{\circ}$ C to units of $^{\circ}$ F.

The conversion is

$Y \, ^{\circ}\mbox{F} = \frac{9}{5} \, X \, ^{\circ}\mbox{C} + 32$

so in this case with $X = 12.0 ^{\circ}$ C, then

$Y \, ^{\circ}\mbox{F} = \frac{9}{5} \, 12.0 \ ^{\circ} \mbox{C} + 32 = 53.6 \, ^{\circ}\mbox{F}$

(b) $67 ^{\circ}$ C to units of $^{\circ}$ K.

The conversion is

$Y \, ^{\circ}\mbox{K} = X \, ^{\circ}\mbox{C} + 273.15$

so in this case with $X = 67.0 ^{\circ}$ C, then

$Y \, ^{\circ}\mbox{K} = 67.0 \, ^{\circ}\mbox{C} + 273.15 = 340.15 \, ^{\circ}\mbox{K}$

(c) $-14 ^{\circ}$ F to units of $^{\circ}$ R.

The conversion is

$Y \, ^{\circ}\mbox{R} = X \, ^{\circ}\mbox{F }+ 459.67$

so in this case with $X = -14.0 ^{\circ}$ F, then

$Y \, ^{\circ}\mbox{R} = -14.0 \, ^{\circ}\mbox{F} + 459.67 = 445.67 \, ^{\circ}\mbox{R}$

(d) $386^{\circ}$ R to units of $^{\circ}$ K.

The conversion is

$Y \, ^{\circ}\mbox{K} = \frac{5}{9} \, X \, ^{\circ}\mbox{R}$

so in this case with $X = 386.0^{\circ}$ R, then

$Y \, ^{\circ}\mbox{K} = \frac{5}{9} \, 386.0 \, ^{\circ}\mbox{R} = 214.44 \, ^{\circ}\mbox{K}$

(e) $-49^{\circ}$ C to units of $^{\circ}$ R.

The conversion is

$Y \, ^{\circ}\mbox{R} = \left( \frac{9}{5} \, X \, ^{\circ}\mbox{C } + 32 \right) + 459.67 \, ^{\circ}\mbox{R}$

so in this case with $X = -49.0^{\circ}$ C, then

$Y \, ^{\circ}\mbox{R} = \left( \frac{9}{5} (-49.0) \, ^{\circ}\mbox{C } + 32 \right) + 459.67 \, ^{\circ}\mbox{R} = 403.47 \, ^{\circ}\mbox{R}$

Worked Example #2

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

SI units:

A temperature of 82 $^{\circ}$ F converted to base units of Kelvin is

$82^{\circ}\mbox{F} = \frac{5}{9} \times ( 82 - 32) = 27.78^{\circ}\mbox{C} = 300.93^{\circ}\mbox{K}$

The equation of state is

$\varrho = \frac{p}{R \ T}$

where $R$ is 286.9 J kg $^{-1}$ K $^{-1}$ in the SI system. Therefore,

$\varrho = \frac{p}{R T} = \frac{101.1 \times 10^3}{286.9 \times 300.93} = 1.171~\mbox{kg m$^{-3}$}$

USC units:

A temperature of 82 $^{\circ}$ F converted to base units of Rankine is

$82^{\circ}\mbox{F} = 541.67^{\circ}\mbox{R}$

The equation of state is

$\varrho = \frac{p}{R \ T}$

where $R$ is 1716.59 ft lb slug $^{-1}$ R $^{-1}$ in the USC system. A pressure of 101.1 kPa (N/m $^2$ ) is the same as 2111.52 lb ft $^{-2}$ . Therefore,

$\varrho = \frac{p}{R T} = \frac{2111.52}{1716.59 \times 541.67} = 0.002271~\mbox{slugs ft$^{-3}$}$

Notice: Raw values (e.g., measurements) in both SI and USC units is 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 $P$ 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

$u (y) = 11.822 \ y^{1/4}$

Calculate the shear stress $\tau$ in the oil as a function of $y$ (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 $\mu$ = 0.084 kg m $^{-1}$ s $^{-1}$ .

The shear stress $\tau$ can be calculated using Newton’s law of viscosity, i.e.,\tau = \mu \left( \frac{\partial u}{\partial y} \right)
\]
In this case, the problem only involves one dimension ( $y$ ) so

$\tau = \mu \left( \frac{d u}{d y} \right)$

We are told that

$u (y) = 11.822 \ y^{1/4}$

so

$\frac{d u}{d y} = 11.822 \left( \frac{1}{4} \right) y^{-3/4} = 2.956 y^{-3/4}$

Therefore, the shear stress for this velocity gradient is

$\tau = \mu \frac{d u}{d y} = 2.956 \times 0.084 \, y^{-3/4} = \frac{0.248}{y^{3/4}}$

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

If a measurement in air gives a temperature of 102 $^{\circ}$ F, calculate the coefficient of dynamic viscosity. What happens to the viscosity of air as its temperature increases, and why?

Sutherland’s Law can be expressed as

$\mu = \mu_{\rm ref} \left( \frac{T}{T_{\rm ref}} \right)^{1.5} \left( \frac{T_{\rm ref} + S}{T+S} \right)$

where $T_{\rm ref} = 518.67^{\circ}$ R, $S = 198.72^{\circ}$ R and $\mu_{\rm ref} = 3.63 \times 10^{-7}$ slugs s $^{-1}$ ft $^{-1}$ . In this case, then the absolute temperature is

$102 \ ^{\circ}\mbox{F} \equiv 561.67 \ ^{\circ}\mbox{R}$

Inserting the values gives

$\mu = 3.63 \times 10^{-7} \left( \frac{561.67}{518.67} \right)^{1.5} \left( \frac{518.67 + 198.72}{561.67 + 198.72} \right) = 3.859 \times 10^{-7} \mbox{ slugs s$^{-1}$ ft$^{-1}$}$

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 $^{-3}$ and that the density of helium at the same conditions is 0.1778 kg m $^{-3}$ .

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, $F_u$ , must be equal to the weight of the illusionist, $W$ , plus the weight of the helium, $W_{\rm He}$ , in the balloons, so

$F_u - W - W_{\rm He} = 0$

The air volume displaced by each balloon is $\cal{V}$ , which is the value to be determined. Let $N_b$ be the number of balloons. The weight of air displaced by the balloons will be

$N_b {\cal{V}} \varrho_{\rm air} g = F_u$

The weight of the helium inside the balloons is

$N_b {\cal{V}} \varrho_{\rm He} g = W_{\rm He}$

Therefore,

$F_u - W - W_{\rm He} = N_b {\cal{V}} \varrho_{\rm air} g - W - N_b {\cal{V}} \varrho_{\rm He} g = 0$

or

$N_b {\cal{V}} \left( \varrho_{\rm air} - \varrho_{\rm He} \right) g = W$

Solving for ${\cal{V}}$ gives

${\cal{V}} = \frac{W}{N_b \left( \varrho_{\rm air} - \varrho_{\rm He} \right) g }$

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

${\cal{V}} = \frac{90.0 g}{N_b ( 1.19 - 0.1778 ) g } = \frac{90}{52 ( 1.19 - 0.1778) } = 1.71~\mbox{m$^3$}$

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

Note: We should be careful about the use of the term “weight” because in commonly used SI units, then we inevitably use units of kg. 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 $^3$ volumetric capacity; the gas used is helium. If the airship has an empty weight of 14,188 lb and has maximum fuel load of 960 lb, what would be the maximum payload that the airship could carry? Assume MSL ISA standard atmospheric conditions. The density of helium can be assumed to be 0.0003192 slugs/ft $^3$ .

The hydrostatic principle of buoyancy is applicable here. Using Archimedes’ principle then the upforce (lift), $L$ , 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.,

$L = \left( \varrho_{\rm air} - \varrho_{\rm He} \right) g {\cal{V}}$

where ${\cal{V}}$ is the volume of the gas envelope.

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

$W = W_e + W_f + W_p$

For vertical force equilibrium at takeoff then $L = W$ so

$W_e + W_f + W_p = \left( \varrho_{\rm air} - \varrho_{\rm He} \right) g {\cal{V}}$

and rearranging for the payload weight gives

$W_p = \left( \varrho_{\rm air} - \varrho_{\rm He} \right) g {\cal{V}} - W_e - W_f$

Using the numerical values supplied gives

$W_p = \left( 0.002378 - 0.0003192 \right) \times 32.17 \times 296,520 - 14,188.0 - 960.0 = 4,491 \mbox{lb}$

Therefore, the maximum payload at the point of takeoff 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 volume of the balloon required for the block to float at a height of $h$ = 1,000 m under standard ISA conditions. The pressure of the helium inside the balloon is 5% greater than the local ambient atmospheric pressure. You can assume that ISA properties are modeled using

$T_{\rm air} = 288.15 \left( 1 - 0.00002256 \, h \right) \mbox{ in units of K}$

$p_{\rm air} = 101325 \left( 1 - 0.00002256 \, h \right) ^{5.25} \mbox{ in units of Pa}$

with $R_{\rm air}$ = 287 and $R_{\rm He}$ = 2077 which are both in units of J Kg $^{-1}$ K $^{-1}$ . 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

$\mbox{Upthrust~on~balloon} = \rm{Weight~of~block} + \rm{Weight~of~helium~inside~balloon}$

or

$\mbox{Upthrust~on~balloon} - \rm{Weight~of~helium~inside~balloon} = \rm{Weight~of~block}$

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.

$\mbox{Upthrust~on~balloon} = \varrho_{\rm air} {{\cal{V}}} g$

where ${\cal{V}}$ is the volume. Also we have that

$\mbox{Weight~of~helium~inside~balloon} = \varrho_{\rm He} {\cal{V}} g$

so the governing equation is

$\left( \varrho_{\rm air} - \varrho_{\rm He} \right) {\cal{V}} g = W$

We are given as an altitude of $h$ = 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

$T_{\rm air} = T_{\rm He} = 288.15 \left( 1 - 0.00002256 \, h \right) = 288.15 \left( 1 - 0.00002256 \times 1000 \right) = 281.65~\mbox{K}$

The pressure of the ambient air will be

$p_{\rm air} = 101325 \left( 1 - 0.00002256 \, \times 1000 \right) ^{5.25} = 89885.5~\mbox{Pa}$

and the pressure inside the balloon will be

$p_{\rm He} = 1.05 \times p_{\rm air} = 1.05 \times 89885.5 = 94379.8~\mbox{Pa}$

Using the equation of state $p = \varrho R T$ then density of the air will be

$\varrho_{\rm air} = \frac{ p_{\rm air} }{R_{\rm air} T_{\rm air} } = \frac{89885.5}{287 \times 281.65} = 1.112~\mbox{kg m$^{-3}$}$

and the density of helium will be

$\varrho_{\rm He} = \frac{ p_{\rm He} }{R_{\rm He} T_{\rm He} } = \frac{ p_{\rm He} }{R_{\rm He} T_{\rm air} } = \frac{94379.8}{2077 \times 281.65} = 0.161~\mbox{kg m$^{-3}$}$

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

$W = M \, g = 10 \times 9.81 = 98.1 \mbox{N}$

Therefore, the needed volume of the balloon will be

${\cal{V}} = \frac{W}{\left( \varrho_{\rm air} - \varrho_{\rm He} \right) g} = \frac{98.1}{(1.112-0.161) \times 9.81} = 10.52~\mbox{m$^{3}$}$

Worked Example #8

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

The engineering form of the equation of state is

$p = \varrho R T$

where $p$ is pressure, $\varrho$ is density, $T$ is absolute temperature, and $R$ is the gas constant. In this form of the equation of state, then the value of $R$ 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

$\varrho = \frac{p}{R T}$

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 15.7 $^{\circ}$ C, 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 $R$ for air in SI units is 287.057 J kg $^{-1}$ K $^{-1}$ so the density of the air will be

$\varrho = \frac{p}{R T} = \frac{102.3 \times 10^3}{287.057 \times 288.85} = 1.2304~\mbox{kg m$^{-3}$}$

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 71.1 $^{\circ}$ F. Calculate the density of the air in the tunnel. Assume MSL ISA conditions.

In this case, the absolute temperature is 71.1 $^{\circ}$ F + 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 $^{2}$ ) then it is necessary to multiply by 144. The gas constant for air in USC units is 1716.59 ft lb slug $^{-1}$ R $^{-1}$ so the density of the air will be

$\varrho = \frac{p}{R T} = \frac{14.61 \times 144.0 }{1716.59 \times 530.77} = 0.002309~\mbox{slugs ft$^{-3}$}$

Worked Example #10

The tires of the main landing gear of an airplane are at a temperature of 40 $^{\circ}$ F and pressure of 40 lb/in $^2$ just before landing. After landing and braking, the temperature of the tires increase to 165 $^{\circ}$ F. 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 useful, i.e.,

$p = \varrho R T$

Therefore, we can write that

$\frac{p_2}{p_1} = \left( \frac{\varrho_2}{\varrho_1} \right) \frac{T_2}{T_1}$

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

$\frac{p_2}{p_1} = \frac{T_2}{T_1}$

We are given that $T_1$ (= 40 $^{\circ}$ F) and $T_2$ (165 $^{\circ}$ F) so that

$\frac{p_2}{p_1} = \frac{165+459.67}{40+459.67} = \frac{624.67}{499.67} = 1.25$

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 $^2$ , but we do not need to convert to lb/ft $^2$ because we are dealing with ratios and so the conversion factor cancels out. Therefore, the pressure in the tires after landing will be

$p_2 = 1.25 p_1 = 1.25 \times 40.0 = 50.0 ~\mbox{lb/in$^2$}$

Worked Example #11

A well-type manometer filled with water has a leg inclined at an angle of 30 $^{\circ}$ to the horizontal. When a pressure $p_a$ is applied on the well side then the water moves along the leg side from level 1 to level 2 by a distance $\Delta l$ equal to 25.25 cm. Find the difference in pressure that has been applied.

The relevant equation is the hydrostatic equation for a constant density fluid, i.e., $p + \varrho g h =$ constant, where $h$ is the vertical height. In this case, the working fluid is water, so $\varrho =$ 1,000 kg/m $^3$ . Remember that for a well-type manometer the level of the fluid in the well does not change, at least not significantly enough to affect the pressures.

We are told that the manometer is inclined at an angle of 30 $^{\circ}$ to the horizontal, so the relationship between the length of the measured displacement of the levels of water in the tube, $\Delta l$ , and the vertical height is $\Delta h = \Delta l \sin 30^{\circ}$ .

Therefore, the difference in the applied pressure will be

$\Delta p = \varrho g \Delta h = \varrho g \Delta l \sin 30^{\circ}$

and substituting the given values leads to

$\Delta p = 1,000 \times 9.81 \times 0.2525 \sin 30^{\circ} = 1.238~\mbox{kPa}$

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, $a_z$ and $a_x$ , respectively. Find an expression for the slope (and the angle $\alpha$ ) 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 $\vec{a} = a_x \vec{i} + a_y \vec{j} + a_z \vec{k}$ is

$\nabla p = \varrho \vec{a} - \varrho g \vec{k}$

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

$\begin{eqnarray*} \frac{\partial p}{\partial x} & = & \varrho (-a_x) \nonumber \\ \frac{\partial p}{\partial y} & = & \varrho (-a_y) \nonumber \\ \frac{\partial p}{\partial z} & = & \varrho ((-a_z) - g) \nonumber \end{eqnarray*}$

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

$\begin{eqnarray*} \frac{\partial p}{\partial x} & = & -\varrho a_x \nonumber \\ \frac{\partial p}{\partial z} & = & -\varrho (a_z + g) \nonumber \end{eqnarray*}$

Integrating these equations and solving for the pressure $p$ leads to

$p = -\varrho a_x x - \varrho (a_z + g ) z + \mbox{constant}$

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

$a_x x + (a_z + g) z = \mbox{constant}$

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

$\alpha = \tan^{-1} \left( \frac{-a_x}{a_z + g} \right)$

Worked Example #13

A cylindrical container of radius $R$ filled is with water and then is rotated at an angular velocity of $\Omega$ 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 axisymmetry about the axis of spin, so we will define $r$ in the radial direction and $z$ 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

$\begin{eqnarray*} \frac{\partial p}{\partial r} & = & \varrho a_r \nonumber \\ \frac{\partial p}{\partial z} & = & -\varrho g \nonumber \end{eqnarray*}$

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

$a_r = \omega^2 r$

so we have

$\begin{eqnarray*} \frac{\partial p}{\partial r} & = & \varrho \omega^2 r \nonumber \\ \frac{\partial p}{\partial z} & = & -\varrho g \nonumber \end{eqnarray*}$

Integrating the first differential equation with respect to $r$ gives

$p = \frac{1}{2} \varrho \omega^2 r^2 + C_1$

Integrating the second differential equation with respect to $z$ gives

$p = -\varrho g z + C_2$

Therefore, the solution for the pressure is

$p = \frac{1}{2} \varrho \omega^2 r^2 -\varrho g z + C_3$

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

$p = \frac{1}{2} \varrho \omega^2 r^2 - \varrho g z = \mbox{constant}$

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 $z = h/2$ equal to the amount the surface rises at the rim of the container at $r = R$ by $z = h/2$ where

$h = \frac{R^2 \omega^2}{2 g}$

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

$z = z_0 - \frac{\omega^2}{4 g} \left( R^2 - 2r^2 \right)$

where $z_0$ 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 #14

The first stage of a 20 m tall cylindrical fuel tank of a rocket is filled entirely with liquid RP-1 fuel. During the launch, this stage experiences a maximum vertical acceleration of 5g’s. Calculate the maximum excess pressure on the bottom of the tank during launch. Assume that the density of RP-1 is 820 kg/m $^3$ .

The relevant equation is the hydrostatic equation for a constant density fluid, i.e., $p + \varrho g h =$ constant, where $h$ is the vertical height. In this case, the working fluid is kerosine with $\varrho =$ 820 kg/m $^3$ . 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

$p = \varrho (5g) h = 820 \times 5 \times 9.81 \times 20 = 804.42 ~\mbox{k Pa}$

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 $p_{\rm in}$ and right side to the outlet at pressure $p_{\rm out}$ . 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 = $\varrho/\varrho_{H_2 0}$ . The density of water can be assumed to be 1.93 slugs/ft $^3$ .

Illustration showing the elevation of a U-tube manometer connected to an inlet and outlet water pump.

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

$p_a = \varrho_{\rm Hg} \, g h_1 + \varrho_{\rm Hg} \, g h_2 + \varrho_{\rm H_{2}O} \, g h_3 + p_{\rm in}$

Similarly, for the right side arm then

$p_a = \varrho_{\rm Hg} \, g h_1 + \varrho_{\rm H_{2}O} \, g h_2 + \varrho_{\rm H_{2}O} \, g h_3 + p_{\rm out}$

Therefore, the difference in pressure becomes

$\Delta p = p_{\rm out} - p_{\rm in} = \varrho_{\rm Hg} \, g h_2 - \varrho_{\rm H_{2}O} \, g h_2 = \left( \varrho_{\rm Hg} - \varrho_{\rm H_{2}O} \right) g h_2$

and substituting the numerical values gives

$\Delta p = \left( 13.6 - 1.0 \right) \times 1.93 \times 32.17 \times (6.0/12) = 391.16 \quad \mbox{lb/ft$^2$}$

Worked Example #16

A manometer contains two liquids of specific weight $\gamma_1$ (Fluid 1 = water) and $\gamma_2$ (Fluid 2 = mercury). Note: Specific weight $\gamma = \varrho g$ . Determine the equation relating the pressure in the spherical reservoir, i.e., $p_0$ , in terms of $p_2$ , $h_1$ , $h_2$ , $\gamma_1$ , and $\gamma_2$ .

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

$p_1 = p_0 + \varrho_1 g h_1$

or in terms of specific weight

$p_1 = p_0 + \gamma_1 h_1$

and correspondingly for the right leg

$p_1 = p_2 + \gamma_2 h_2$

Therefore,

$p_0 + \gamma_1 h_1 = p_2 + \gamma_2 h_2$

Solving for $p_0$ in terms of $p_2$ , $h_1$ , $h_2$ , $\gamma_1$ , and $\gamma_2$ as asked for in the question then

$p_0 = p_2 + \gamma_2 h_2 - \gamma_1 h_1$

If the open end of the manometer in the above figure is now exposed to the atmosphere and $h_1$ = 5.2 inches and $h_2$ = 18.6 inches, then find the pressure $p_0$ . Assume atmospheric pressure $p_a$ is 14.7 lb/in $^2$ . The density of water can be assumed to be 1.93 slugs/ft $^3$ , and the specific gravity (SG) of mercury is 13.6. Use $g$ = 32.17 ft/s $^2$ . Note: SG = $\varrho/\varrho_{H_2 0}$ .

We are then given that $p_2$ is atmospheric pressure = $p_a$ = 14.7 lb/in $^2$ = 2,116.8 lb/ft $^2$ . Also, we can calculate $\gamma_1$ and $\gamma_2$ , i.e.,

$\gamma_1 = 1.93 \times 32.17 = 62.09~\mbox{slugs/ft$^2$/s$^2$}$

and

$\gamma_2 = 13.6 \times 1.93 \times 32.17 = 844.40~\mbox{slugs/ft$^2$/s$^2$}$

Therefore, solving for the value of $p_0$ in this case gives

$\begin{eqnarray*} p_0 & = & p_2 + \gamma_2 h_2 - \gamma_1 h_1 \\ & = & 2,116.8 + (844.40 \times 18.6/12) - (62.09 \times 5.2/12) \\ & = & 3,398.71~\mbox{lb/ft$^2$} \\ & = & 23.6~\mbox{lb/in$^2$} \end{eqnarray*}$

Worked Example #17

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 $\theta$ and the differential reading of the two heights of the liquid $l_2$ is measured along the inclined tube. The two pipes (which run into the screen) contain gases, the gas in pipe A being of density $\varrho_A$ , and the gas in pipe B being of density $\varrho_B$ . The density of the liquid in the tube is $\varrho_L$ . Determine the equation for the difference in pressure between A and B in terms of the length $l_2$ .

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, $p_1$ , will be

$p_1 = p_A + \varrho_A \, g \, h_1$

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

$p_B + \varrho_B \, g \, h_3 + \varrho_L \, g \, l_2 \sin \theta$

remembering that it is the vertical height $l_2 \sin \theta$ that matters here. Therefore, equating the two pressures at the same level gives

$p_1 = p_A + \varrho_A g h_1 = p_B + \varrho_B g h_3 + \varrho_L g l_2 \sin \theta$

or that

$p_A - p_B = \varrho_L g l_2 \sin \theta + \left( \varrho_B g h_3 - \varrho_A g h_1 \right)$

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., $\varrho_A \ll \varrho_L$ and $\varrho_B \ll \varrho_L$ and the difference in heights $h_3 - h_1$ is small so that

$\left( \varrho_B g h_3 - \varrho_A g h_1 \right) \ll \varrho_L g l_2 \sin \theta$

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

$p_A - p_B = \varrho_L \, g \, l_2 \sin \theta$

or

$l_2 = \frac{p_A - p_B }{ \varrho_L \, g}$

Worked Example #18

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

$\frac{\partial p}{\partial x} = x$

$\frac{\partial p}{\partial y} = y$

where the $x$ – $y$ plane is horizontal and $z$ 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.,

$\frac{\partial p}{\partial x} = x$

and

$\frac{\partial p}{\partial y} = y$

Also, in the $z$ direction we will have

$\frac{\partial p}{\partial z} = -\varrho g$

Integrating the first of the three equations gives

$p = \frac{x^2}{2} + f_1(y) + f_2(z) + C_1$

and integrating the second gives

$p = \frac{y^2}{2} + f_3(x) + f_2(z) + C_2$

and integrating the third gives

$p = -\varrho g z + f_3(x) + f_1(y) + C_3$

Therefore, by inspection we see that $f_1(y) = y^2/2$ and $f_3(x) = x^2/2$ and $f_2(z) = -\varrho g z$ so that the pressure field will be described by

$p(x, y, z) = \frac{x^2}{2} + \frac{y^2}{2} -\varrho g z + C_4$

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 #19

A hot air balloon becomes neutrally buoyant at an altitude of 1,100 m where the outside air temperature is 10.5 $^{\circ}$ C. The balloon with all its cargo has mass of 500 kg (excluding the air in the balloon), and displaces a volume of 7,000 m $^3$ . Assume that the pressure inside the balloon is the same as that outside the balloon. Determine the temperature of the warm air inside the balloon to give it neutral buoyancy. Hint: At 1,100 m, the density of air can be considered to be 1.102 kg m $^{-3}$ .

The physics of the problem says that for neutral buoyancy: The weight of hot air inside the balloon $W_{\rm hot}$ + the weight of the balloon itself $W_b$ = the upthrust $F_u$ 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

$F_u - \left( W_b + W_{\rm hot} \right) = 0$

The volume of air displaced by the balloon is $\cal{V}$ = 7,000 m $^3$ . Therefore, the weight of cool air displaced by the balloon is then

$W_{\rm cool~air} = {\cal{V}} \, \varrho_{1,100} \, g = 7,000 \times 1.102 \times 9.81 = 75,674.3~\mbox{N}$

which is equal to the upward buoyancy force $F_u$ according to Archimedes’ Principle. The mass of the balloon is 500 kg so its weight, $W_b$ , is 500 $g$ = 500 $\times$ 9.81 = 4905.0 N. Notice $\varrho_{1,100}$ 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 $^{-3}$ .

Therefore, for neutral buoyancy (i.e., $F_u = W_b + W_{\rm hot}$ if the net vertical force is zero) the weight of the hot air in the balloon must be

$W_{\rm hot} = F_u - W_b = 75,674.3 - 4905.0= 70,769.3~\mbox{N}$

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

$\varrho_{\rm hot} = \frac{W_{\rm hot} }{g \cal{V}} = \frac{70,769.3}{9.81 \times 7,000} = 1.031~\mbox{kg m$^{-3}$}$

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.,

$p = \varrho R T$

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

$p = \varrho_{\rm cool} R T_{\rm cool} = \varrho_{\rm hot} R T_{\rm hot}$

and rearranging gives the ratio of the temperatures as

$\frac{T_{\rm hot}}{T_{\rm cool}} = \frac{\varrho_{\rm cool}}{\varrho_{\rm hot}} = \frac{1.102}{1.1031} = 1.069$

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

$T_{\rm hot} = 1.001 \, T_{\rm cool} = 1.001 \times 283.65 = 303.31~\mbox{$^{\circ}$K} = 30.16~\mbox{$^{\circ}$C}$

so a differential balloon internal temperature of about 20 $^{\circ}$ C at this altitude and temperature conditions.

Worked Example #20

A spherical holding tank with diameter $d =$ 2.7 m is used to store gaseous helium for a blimp under a pressure of 250 kPa. Determine the total weight of the gas inside the tank. Assume that $R_{\rm He}$ = 2077.1 J kg $^{-1}$ K $^{-1}$ and ISA MSL temperature conditions prevail.

The volume ${\cal{V}}$ of the tank will be

${\cal{V}} = \frac{4}{3} \pi \left(\frac{d}{2}\right)^3 = \frac{4}{3} \pi \left( \frac{2.7}{2} \right)^3 = 10.306~\mbox{m$^3$}$

To find the weight of helium we need its density so use the equation of state $p = \varrho R T$ and so

$\varrho = \frac{p}{R T}$

We are told to use ISA MSL temperature conditions so $T = T_0$ = 15 $^{\circ}$ C = (273.15 + 15) $^{\circ}$ K = 288.15 $^{\circ}$ K, which we can assume as the same temperature as the helium $T_{\rm He}$ . Therefore

$\varrho_{\rm He} = \frac{p_{\rm He} }{R_{\rm He} \ T_{\rm He} } = \frac{250 \times 10^3}{2077.1 \times 288.15} = 0.418~\mbox{kg m$^{-3}$}$

The weight $W_{\rm He}$ of helium in the tank, therefore, will be

$W_{\rm He} = {\cal{V}} \, \varrho_{\rm He} \, g = 10.306 \times 0.418 \times 9.81 = 42.23~\mbox{N}$

Worked Example #21

Consider a circular cylinder of initial volume $\cal{V}$ = 5 liters whose height $h$ = 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 $R_{\rm He}$ = 2,077.0 J Kg $^{-1}$ K $^{-1}$ .

The volume is given as 5 liters, so ${\cal{V}}$ = 5,000 cm $^3$ = 0.005 m $^3$ . The conditions are specified as MSL ISA, so from the formula sheet then pressure $p$ = 1.01325 $\times$ 10 $^{5}$ Pa and temperature $T$ = 15 $^{\circ}$ C = 288.15 $^{\circ}$ K.

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

$p = \varrho R_{\rm He} T = \left( \frac{m}{\cal{V}} \right) R_{\rm He} T$

where $m$ 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

$m = \frac{p_1 {{\cal{V}}_1} }{R_{\rm He} T_1} = \frac{ 1.01325 \times 10^{5} \times 0.005}{ 2,077.0 \times 288.15} = 8.47 \times 10^{-4}~\mbox{kg}$

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

$p_1 = \left( \frac{m}{{\cal{V}}_1} \right) R_{\rm He} T_1$

and for the fully compressed pressure at state 2

$p_2 = \left( \frac{m}{{\cal{V}}_2} \right) R_{\rm He} T_2$

In this case we are told that is an isothermal process so $T_1 = T_2$ so we can write for a given mass of gas that

$\frac{p_2}{p_1} = \frac{{\cal{V}}_1}{{\cal{V}}_2} = \frac{{\cal{V}}_1}{{\cal{V}}_1 /2} = \frac{2 {\cal{V}}_1}{{\cal{V}}_1} = 2$

Therefore, we have that

$p_2 = 2 p_1 = 2 \times 1.01325 \times 10^{5} = 2.0265 \times 10^{5}~\mbox{Nm$^{-2}$}$

(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 $A$ , is

$A = \frac{{\cal{V}}_1}{h} = \frac{0.005}{0.16} = 0.03125~\mbox{m$^2$}$

The force, $F_2$ , required to fully compress the gas, therefore, will be

$F_2 = \left(p_2-p_1\right) A = \left(2.0265-1.01325\right) \times 10^{5} \times 0.03125 = 3.223 \times 10^3~\mbox{N}$

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

$M = \frac{F_2}{g} = \frac{3.223 \times 10^3}{9.81} = 329~\mbox{kg}$

Worked Example #22

Consider a large concrete dam that keeps sea water contained in a dock. The dam has dimensions of $W$ = 20 m wide (in the direction perpendicular to the screen), and is $H$ = 40 m high and $B$ = 30 m thick at its base. The density of sea-water can be assumed to vary linearly with depth according to the formula $\varrho = \varrho_0+K \, h$ , where $\varrho_0$ = 1,028 kg m $^{-3}$ at $h$ = 0 and $K$ = 0.0026 kg m $^{-4}$ .

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

$\displaystyle{\frac{dp}{dz}} = -\varrho g$

then we have in this case the specific differential equation

$\displaystyle{\frac{dp}{dh}} = \varrho g = (\varrho_0+Kh) g$

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

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

$\int_{0}^{p}dp = \int_{0}^{h} (\varrho_0+Kh)g \, dh$

which gives for the hydrostatic pressure from the water as

$p = g \left( \varrho_0 h +\frac{1}{2} K h^2 \right)$

When $h = H$ , which is at a the foot of the dam, the previous result gives

$p_H = g \left( \varrho_0 H +\frac{1}{2} KH^2 \right)$

and substituting the numbers gives

$p = 9.81 \times \left( 1,028 \times 40 + 0.5 \times 0.0026 \times 40^2 \right) = 403.41~\mbox{kPa}$

Worked Example #23

After the Great Flood, the human race decided to build the Tower of Babel to reach Heaven. A hot-air balloon was built to lift heavy materials up to the tower from the ground. Consider such a hot-air balloon in the form of a sphere with an inflated diameter $d=50$ ft. The total weight of the system (including the fabric, basket and its payload, but without air inside) is 1,000 lb. You may assume that the local air density $\varrho$ in the ISA relative to MSL $\varrho_0$ is given by

$\frac{\varrho}{\varrho_0} = \left(1 - 6.876\times 10^{-6}h\right)^{4.252}$

where $h$ 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 210 $^\circ$ F and the pressure is equal to the MSL atmospheric pressure. Determine the maximum potential altitude $h_{\rm max}$ the balloon can achieve when the temperature of the air remains at 210 $^\circ$ F.

(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

$F_{\rm up} = W_{\rm air} + W_{\rm balloon}$

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

$\varrho_{in,0} = \frac{p_{\rm atm}}{RT} = \frac{2116.4}{1716.59 \times (459.67+210)} = 0. 001841~\mbox{slug/ft}^3$

Assume the ballon is a sphere so its volume $\mathcal{V}$ is

$\mathcal{V} = \frac{4}{3}\pi(d/2)^3 = 65,450~\mbox{ft}^3$

and the weight of air inside is

$W_{\rm air} = \varrho_{in,0} \, g \, \mathcal{V} = 0.001841 \times 65,450 \times 32.17= 3,876.4~\mbox{lb}$

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

$\varrho_{h_{\rm max}} = \frac{W_{\rm air} + W_{\rm balloon}}{g \mathcal{V} } = \frac{3,876.4+1,000}{32.17 \times 65,450} = 0.002316~\mbox{slug/ft}^3$

or the density ratio $\sigma$ is

$\sigma = \frac{\varrho_{h_{\rm max}} }{\varrho_0} = 0.9739$

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

$h_{\rm max} \approx 900 \mbox{ft}$

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 oil with a density $\varrho_{\rm oil}$ = 1.83 slug ft $^{-3}$ . When a weight ${\cal{W}}$ 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 ${\cal{W}}$ ? 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 $p_a$ , so it will be apparent there will be no differential effect of $p_a$ on the height of the manometer column, which is initially at level 1.

When the weight ${\cal{W}}$ 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

$p_a + \frac{ {\cal{W}} }{A_p} = p_a + \varrho_{\rm oil} \, g \, \left( 6.0/12.0 \right) \sin 30^{\circ}$

where $A_p$ 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 $\sin 30^{\circ}$ term to account for the inclined leg in this case. The effects of the atmospheric pressure obviously cancel out in the preceding equation so

$\frac{ {\cal{W}} }{A_p} = \varrho_{\rm oil} \, g \left( 0.5 \right) 0.5 = 0.25 \, \varrho_{\rm oil} \, g$

and rearranging gives

${\cal{W}} = 0.25 \, \varrho_{\rm oil} \, g\, A_p$

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

$A_p = \frac{\pi d^2}{4} = \frac{\pi 0.5^2}{4} = 0.1964~\mbox{ft$^2$}$

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

${\cal{W}} = 0.25 \, \varrho_{\rm oil} \, g\, A_p = 0.25 \times 1.83 \times 32.17 \times 0.1964 = 2.89~\mbox{lb}$

Worked Example #25

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 $n_{\rm max} g$ during a normal launch. To accomplish this, the gate is kept closed during a normal launch by a preload torque $M_h$ , exerted by a coil spring.

If the length of the door (normal to screen) is $w$ , determine the minimum required preload torque ( $M_h$ ) 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 $p_0$ = 101,325 Pa, $h_1$ = 45 cm, $h_2$ = 125 cm, $\varrho$ = 820 kg/m $^3$ , $w$ = 80 cm, and $n_{\rm max}$ = 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

$p = p_0 + \varrho n g h$

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

$dF = p w dh$

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

$F = \int_{h_1}^{h_2} p w dh = \int_{h_1}^{h_2} (p_0 + \varrho n g h) w dh$

To find the moment $M$ 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 $h - h_1$ . Therefore, the net moment about the hinge is

$M = \int_{h_1}^{h_2} (p_0 + \varrho n g h) (h - h_1) w dh$

which must now be solved. Expanding out gives

$M = \int_{h_1}^{h_2} \left( p_0 h - p_0 h_1 + \varrho n g h^2 - \varrho n g h h_1 \right) w dh$

Integrating gives

$M = w \left[ \frac{p_0 h^2}{2} - p_0 h_1 h + \frac{\varrho ng h^3}{3} - \frac{ \varrho n g h_1 h^2}{2} \right]_{h_1}^{h_2}$

Therefore

$M = \left[ \frac{p_0 w h_2^2}{2} - p_0 w h_1 h_2 + \frac{\varrho n g w h_2^3}{3} - \frac{ \varrho n g w h_1 h_2^2}{2} \right] - \left[ \frac{p_0 w h_1^2}{2} - p_0 w h_1^2 + \frac{\varrho n g w h_1^3}{3} - \frac{ \varrho n g w h_1^3}{2} \right]$

and simplifying gives

$M = \frac{1}{2} p_0 w \left( h_2^2 - h_1^2 \right) + p_0 w \left(h_1^2 - h_1 h_2 \right) + \frac{1}{3} \varrho n g w \left( h_2^3 - h_1^3 \right) + \frac{1}{2} \varrho n g w \left(h_1^3 - h_1 h_2^2 \right)$

and so the moment is

$M = \frac{1}{2} p_0 w \left( h_2 - h_1 \right)^2 + \frac{1}{6} \varrho n g w \left( 2 h_2^3 - 3h_1 h_2^2 + h_1^3 \right)$

Substituting the values given in this case leads to

$\begin{eqnarray*} M & = & \frac{1}{2} (101325) (0.8) \left( 1.25 - 0.45 \right)^2 \\ && + \frac{1}{6} (820) (5.0)(9.81) (0.8) \left( 2 (1.25)^3 - 3(0.45) 1.25^2 + 0.45^3 \right) \\ & = & 36.064~\mbox{kN \, m} \end{eqnarray*}$

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 ( $\varrho_w$ = 1,000 kg m $^{-3}$ ) and mercury (SG = 13.6) is connected to a tank containing air at an internal pressure $p$ , the other end being open to atmospheric pressure $p_{\rm atm}$ of 101 $\times 10^3$ Pa.

Using hydrostatic principles, derive an expression for the pressure $p_1$ in terms of $p_{\rm atm}$ .
Derive a corresponding expression for the pressure in the tank $p$ in terms of $p_1$ .
Obtain a final expression relating $p$ to $p_{\rm atm}$ .
If $h_1$ = 25 cm, $h_2$ = 50 cm, $h_3$ = 60 cm, and $h_4$ = 40 cm, then determine the internal pressure $p$ in the tank.
If the temperature of the air in the tank is measured to be 30 $^{\circ}$ C, determine its corresponding density.

1. Let $\varrho_{\rm Hg}$ be the density of mercury. Using the principles of hydrostatics then the pressure $p_1$ will be

$p_1 = p_{\rm atm} + \varrho_{\rm Hg} g h_4$

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

$p_1 - \varrho_w g h_3 + \varrho_w g h_2 - \varrho_w g h_1 = p$

and rearranging gives

$p_1 = p + \varrho_w g \left( h_3 - h_2 + h_1 \right)$

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

$p_{\rm atm} + \varrho_{\rm Hg} g h_4 = p + \varrho_w g \left( h_3 - h_2 + h_1 \right)$

and rearranging gives

$p = p_{\rm atm} + \varrho_{\rm Hg} g h_4 - \varrho_w g \left( h_3 - h_2 + h_1 \right)$

or in final form as

$p = p_{\rm atm} + \varrho_{w} g \left( SG_{\rm Hg} \, h_4 - h_3 + h_2 - h_1 \right)$

where $SG_{\rm Hg}$ is the specific gravity of mercury, which is 13.6.

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

$\begin{eqnarray*} p & = & p_{\rm atm} + \varrho_{w} g \left( SG_{\rm Hg} \, h_4 - h_3 + h_2 - h_1 \right) \\ & = & 101\times 10^3 + 1,000 \times 9.81 \left( 13.6 \times 0.4 - 0.6 + 0.5 - 0.25 \right) \\ & = & 150.9~\mbox{ kPa} \end{eqnarray*}$

5. The temperature $T$ of the air in the tank is given as 30 $^{\circ}$ C, so in units of Kelvin then

$T = 30 + 273.15 = 303.15 \mbox{K}$

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

$\varrho = \frac{p}{R \, T} = \frac{150.9 \times 10^3} {287.1 \times 303.15} = 1.734~\mbox{kg m$^{-3}$}$

Worked Example #27

A manometer containing water ( $\varrho_{w}$ = 1,000 kg m $^{-3}$ ) 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 $D$ is 0.45 m. A mass $M$ = 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 $p_{\rm atm}$ = 101.3 kPa. The height values are: $h_1$ = 15.0 cm, $h_2$ = 50.0 cm, $h_3$ = 65.0 cm, and $h_5$ = 95.0 cm.

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

$p_A$ at point A (just below the piston).
$p_B$ at point B.
$p_C$ at point C.
$p_D$ at point D.

Part 2: Answer the following:

What is the relationship (as an equation) between the value of the mass $M$ and the height $h_4$ of the Galinstan column?
For the values of the parameters as given, what will be the value of $h_4$ (in units of cm)?

Part 1:

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

$p_A = p_{\rm atm} + \frac{M \, g}{A} = p_{\rm atm} + \frac{W}{A}$

where $W$ is the weight ( $W = M g)$ and $A$ is the cross-sectional area of the cylinder. Therefore,

$p_A = p_{\rm atm} + \frac{W}{A} = p_{\rm atm} + \frac{4W}{\pi D^2}$

and inserting the values gives

$p_A = 101.3 \times 10^3 + \frac{ 4.0 \times 6.0 \times 9.81}{\pi \, 0.45^2} = 101.67~\mbox{kPa}$

2. $p_B$ at point B: We are going DOWN in the water with an increase in depth $h_1$ so the pressure at point B will be

$p_B = p_A + \varrho_w \, g \, h_1$

Inserting the values gives

$p_B = p_A + \varrho_w \, g \, h_1 = 101.67 \times 10^3 + 1,000 \times 9.81 \times 0.15 = 103.142~\mbox{kPa}$

3. $p_C$ at point C: We are going UP in the water with a decrease in depth $h_2$ so the pressure at point C will be

$p_C = p_B - \varrho_w \, g \, h_2$

Inserting the values gives

$p_C = p_B - \varrho_w \, g \, h_2 = 103.142 \times 10^3 - 1,000 \times 9.81 \times 0.50 = 98.237~\mbox{kPa}$

4. $p_D$ at point D: We are going DOWN in the water with an increase in depth $h_3$ so the pressure at point D will be

$p_D = p_C + \varrho_w \, g \, h_3$

Inserting the values gives

$p_D= p_C + \varrho_w \, g \, h_3 = 98.237 \times 10^3 + 1,000 \times 9.81 \times 0.65 =104.615~\mbox{kPa}$

Part 2:

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

$p_{\rm atm} + \frac{W}{A} + \varrho_w \, g \, h_1 - \varrho_w \, g \, h_2 + \varrho_w \, g \, h_3 + \varrho_g \, g \left( h_5 - h_3 \right) - \varrho_g \, g \, h_4 = P_E = p_{\rm atm}$

and noting at the $\varrho_g$ is the density of Galistan and $\varrho_g = 6.44 \varrho_w$ . Collecting terms and also noting that the effects of $p_{\rm atm}$ will balance out on each side of the problem (piston side and open end of U-tube) then

$\frac{W}{A} + \varrho_w \, g \left( h_1 - h_2 + h_3 \right) + \varrho_g \, g \left( h_5 - h_3 - h_4 \right) = 0$

so

$\frac{M \, g}{A} = \varrho_w \, g \left( h_2 - h_1 - h_3 \right) + \varrho_g \, g \left( h_3 - h_5 + h_4 \right)$

which establishes the relationship between $W$ and $h_4$ (the only unknown). Rearranging gives

$\varrho_g \, g \, h_4 = \varrho_w \, g \left( h_1 - h_2 + h_3 \right) + \varrho_g \, g \left( h_5 - h_3 \right) + \frac{M \, g}{A}$

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

$\varrho_g \, h_4 = \varrho_w \left( h_1 - h_2 + h_3 \right) + \varrho_g \left( h_5 - h_3 \right) + \frac{M}{A}$

so the value of $h_4$ is

$h_4 = \frac{\varrho_w}{\varrho_g} \, \left( h_1 - h_2 + h_3 \right) + \left( h_5 - h_3 \right) + \frac{M}{A \, \varrho_g}$

Notice that the grouping $M / (A \, \varrho_g)$ has dimensions of length, as it should if the equation is to be dimensionally homogeneous. Also, the ratio of the densities $\varrho_g/\varrho_w$ 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 given to us. Therefore, the value of $h_4$ is

$h_4 = \frac{\varrho_w}{\varrho_g} \, \left( h_1 - h_2 + h_3 \right) + \left( h_5 - h_3 \right) + \frac{M}{A \, \varrho_g}$

so

$h_4 = \frac{1.0}{6.44} \left( 0.15 - 0.50 + 0.65 \right) + \left( 0.95 - 0.65 \right) + \frac{ 4.0 \times 6.0 }{\pi \, 0.45^2 \times 6.44 \times 10^3} = 35.24~\mbox{cm}$

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 $h =$ 16.2 cm. A mass is then placed on the piston, which has the effect of pushing the piston down into the cylinder to $h =$ 8.3 cm and compressing the helium gas inside. During this process, the temperature of the gas also increases by 23.3 $^{\circ}$ C. Assume that for helium $R_{\rm He}$ = 2,077 J Kg $^{-1}$ K $^{-1}$ .

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 $h_3$ ?

The volume is given as 5.2 liters, so ${\cal{V}}$ = 5,200 cm $^3$ = 0.0052 m $^3$ . The conditions are specified as MSL ISA, so the pressure $p = 1.01325 \times 10^{5}$ Nm $^{-2}$ = 1.01325 $\times 10^{5}$ Pa, and temperature $T$ = 15 $^{\circ}$ C = 288.15 $^{\circ}$ K. 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.,

$p = \varrho R T = \left( \frac{m}{\cal{V}} \right) R T$

where $m$ 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

$m = \frac{p_1 {{\cal{V}}_1} }{R_{\rm He} \, T_1} = \frac{ 1.01325 \times 10^{5} \times 0.0052}{ 2,077.0 \times 288.15} = 8.804 \times 10^{-4}~\mbox{kg}$

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

$\varrho_1 = \frac{m}{{{\cal{V}}_1} } = \frac{8.804 \times 10^{-4}}{ 0.0052} = 0.1693~\mbox{kg m$^{-3}$}$

2. We have for the initial pressure $p_1$ at State 1

$p_1 = \varrho_1 \, R_{\rm He} \, T_1 = 1.01325 \times 10 ^{5}~\mbox{Pa}$

and for the compressed pressure $p_2$ at State 2

$p_2 = \varrho_2 \, R_{\rm He} \, T_2$

where $T_2$ = 288.15 + 23.3 = 311.45 $^{\circ}$ K. The density $\varrho_2$ is

$\varrho_2 = \varrho_1 \left( \frac{h_1}{h_2} \right) = 0.1693 \left( \frac{16.2}{8.3} \right) = 0.3304~\mbox{kg m$^{-3}$}$

and so the corresponding pressure $p_2$ is

$p_2 = \varrho_2 \, R_{\rm He} \, T_2 = 0.3304 \times 2,077.0 \times 311.45 = 2.1373 \times 10 ^{5}~\mbox{Pa}$

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

$A = \frac{{\cal{V}}}{h_1} = \frac{0.0052}{0.16} = 0.0321~\mbox{m$^2$}$

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

$W = p_2 \, A = 2.1373 \times 10^{5} \times 0.0321 = 6.861 \times 10^3~\mbox{N}$

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

$M = \frac{W}{g} = \frac{6.861 \times 10^3}{9.81} = 699.4~\mbox{kg}$

4. If the helium gas now cools down back to ambient temperature $T_1$ = 15 $^{\circ}$ C = 288.15 $^{\circ}$ K at State 3, then the weight will push the piston down a little further to $h_3$ . In this case

$p_3 = \frac{W}{A} = \varrho_3 \, R_{\rm He} \, T_1 = \left( \frac{m}{ A h_3} \right) R_{\rm He} T_1$

where the final compressed volume is just $A h_3$ . Therefore, $h_3$ is given by

$h_3 = \left( \frac{m}{W} \right) \, R_{\rm He} \, T_1 = \frac{8.804 \times 10^{-4}}{6.861 \times 10^3} \times 2,077.0 \times 288.15 = 0.0768~\mbox{m} = 7.68~\mbox{cm}$

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 80 $^{\circ}$ C are connected to each other 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 also that the density of water is $\varrho_w =$ 1,000 kg m $^{-3}$ .

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 80 $^{\circ}$ C to 40 $^{\circ}$ C 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 $\Delta h_1$ at the oil/Gordium interface and the manometer fluid $\Delta h_2$ 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.,

$p = \varrho R T$

Rearranging for the density gives

$\varrho = \frac{p}{R \, T}$

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

$\varrho_{\rm air} = \frac{p_{\rm air}}{R \, T_{\rm air}} = \frac{1.169 \times 10^6}{287.057 \times (273.15 + 80.0)} = 11.531~\mbox{kg m$^{-3}$}$

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

$p_{\rm oil} + \varrho_{\rm oil} \, g \, h_{\rm oil} + \varrho_{\rm Go} \, g \, h_{\rm Go} = p_{\rm air}$

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

$p_{\rm oil} = p_{\rm air} - \varrho_{\rm oil} \, g \, h_{\rm oil} - \varrho_{\rm Go} \, g \, h_{\rm Go}$

Inserting the values of the given quantities gives

$p_{\rm oil} = 1.169 \times 10^6 - (2.68 \times 10^3 \times 9.81 \times 0.75) - (11.8 \times 10^3 \times 9.81 \times 0.20)$

noting that $\varrho_{\rm oil} = {\rm SG}_{\rm oil} \, \varrho_w$ and that $\varrho_{\rm Go} = {\rm SG}_{\rm oil} \, \varrho_w$ . Performing the arithmetic gives

$p_{\rm oil} = 1.169 \times 10^6 -1.9718 \times 10^4 - 2.3152 \times 10^4 = 1.126 \times 10^6~\mbox{Pa} = 1.126~\mbox{MPa}$

3. For a decrease in temperature in the air tank from $T_1 =$ 80 $^{\circ}$ C to $T_1 =$ 40 $^{\circ}$ C then:

(i) The new pressure $p_2$ in the air tank will be

$p_2 = \varrho _{\rm air} \, R_{\rm air} \, T_2$

Let $p_{1} = p_{\rm air}$ and we note that tank contains the same mass of gas so the density is unchanged, therefore the new pressure is

$p_{2} = p_{1} \left( \frac{T_2}{T_1} \right) = 1.169 \times 10^6 \times \left( \frac{273.15 + 40.0}{273.15 + 80.0} \right) = 1.0366~\mbox{MPa}$

(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 $x$ 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

$x \left( \frac{\pi (3d)^2}{4} \right) = y \left( \frac{\pi d^2}{4} \right)$

so $y = 9 x$ . Because we need the vertical height change of the Gordium on the right-side leg, then

$y_{\rm vert} = 9 x \sin (40^{\circ}) = 5.785 x$

Returning back to the equation for the hydrostatic balance then

$p_{\rm oil} + \varrho_{\rm oil} \, g \, h_{\rm oil} + \varrho_{\rm Go} \, g \, h_{\rm Go} = p_{\rm air}$

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

$p_{\rm oil} + \varrho_{\rm oil} \, g \, ( h_{\rm oil} + x) + \varrho_{\rm Go} \, g \, ( h_{\rm Go} - 5.785 x) = p_{\rm air}$

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

${\rm SG}_{\rm oil} (h_{\rm oil} + x) + {\rm SG}_{\rm Go} \ ( h_{\rm Go} - 5.785 x) = \frac{p_{\rm air} - p_{\rm oil}}{\varrho_w \, g}$

Substituting the numerical values gives

$2.68 \times (0.75 + x) + 11.8 ( 0.20 - 5.785 x) = \frac{( 1.0366- 1.126) \times 10^6}{1,000 \times 9.81}$

so

$x = 0.19~\mbox{m}$

The changes in the level $\Delta h_1$ of the manometer fluid at the oil/Gordium interface will drop 19 cm and the level of the manometer fluid $\Delta h_2$ 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.

Worked Example #30

Consider the manometer system shown 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 a S.G. of 3.7. Assume that the density of water is 1.940 slugs/ft $^3$ .

Determine an expression for the pressure at point B.
Determine an expression for the pressure at point C.
Determine an expression for the pressure at point D.
Write down a final expression for the value of $p_A - p_E$ .
If the pressure $p_A$ was to decrease, what would happen to the level at point C?
If $h_1$ = 10 inches, $h_2$ = 13 inches, $h_3$ = 5 inches, $l_1$ = 15 inches, and $\theta$ = 45 $^{\circ}$ , then calculate the pressure difference $p_A - p_E$ in units of lb/in $^2$ .

1. The pressure at point B will be

$p_B = p_A - \varrho_A \, g \, h_1$

where $\varrho_A$ is the density of fluid A.

2. The pressure at point C will be

$p_C = p_B + \varrho_A \, g \, h_2 = p_A - \varrho_A \, g \, h_1 + \varrho_A \, g \, h_2$

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.,

$p_D = p_C = p_A - \varrho_A \, g \, h_1 + \varrho_A \, g \, h_2$

4. For the final part of the manometer, we are given a diagonal length $l_1$ in fluid B rather than a vertical height. However, the vertical height between points D and E is just $l_1 \sin \theta$ , so that

$p_E = p_D - \varrho_B \, g \, l_1 \sin \theta$

where $\varrho_B$ is the density of fluid B, and

$p_E = p_A - \varrho_A \, g \, h_1 + \varrho_A \, g \, h_2 - \varrho_B \, g \, l_1 \sin \theta$

Therefore, the final expression for $p_A - p_E$ is

$p_A - p_E = \varrho_A \, g \, h_1 - \varrho_A \, g \, h_2 + \varrho_B \, g \, l_1 \sin \theta$

and collecting terms gives

$p_A - p_E = \varrho_A \, g \left( h_1 - h_2 \right) + \varrho_B \, g \, l_1 \sin \theta$

5. If the pressure $p_A$ was to decrease, then the level at point C would move up and so the level $h_3$ will increase and the value of $h_2$ will 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 $p_A - p_E$ . We are given the specific gravity of each fluid so

$p_A - p_E = \varrho_w \, g \bigg( SG_A (h_1 - h_2) + SG_B \, l_1 \sin \theta \bigg)$

Substituting the known values gives

$p_A - p_E = 1.94 \times 32.17 \bigg( 0.9 \times (10.0 - 13.0)/12.0 + 3.7 \times (15.0/12.0) \sin 45^{\circ} \bigg)$

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

$\[ p_A - p_E = 190.061~\mbox{lb/ft$^2$} = 1.32~\mbox{lb/in$^2$}$

Worked Example #31

Consider the manometer system shown 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 a S.G. of 3.7. Assume that the density of water is 1.940 slugs/ft $^3$ . Notice that this problem is a variation of worked example #30.

Determine an expression for the pressure at point B.
Determine an expression for the pressure at point C.
Determine an expression for the pressure at point D.
Write down a final expression for the value of $p_A - p_E$ .
If the pressure $p_A$ was to decrease, what would happen to the level at point C?
If $h_1$ = 10 inches, $h_2$ = 13 inches, $h_3$ = 5 inches, $l_1$ = 15 inches, and $\theta$ = 45 $^{\circ}$ , then calculate the pressure difference $p_A - p_E$ in units of lb/in $^2$ .

1. The pressure at point B will be

$p_B = p_A + \varrho_A \, g \, l_1 \sin \theta$

where $\varrho_A$ is the density of fluid A. Notice that we are given a diagonal length $l_1$ in fluid A rather than a vertical height. However, the vertical height between points A and B is just $l_1 \sin \theta$ , as used in the equation.

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

$p_C = p_B = p_A +\varrho_A \, g \, l_1 \sin \theta$

3. The pressure at point D will be

$p_D = p_C - \varrho_B \, g \, h_1 = p_A +\varrho_A \, g \, l_1 \sin \theta - \varrho_B \, g \, h_1$

where $\varrho_B$ is the density of fluid B.

4. The pressure at point E will be

$p_E = p_D + \varrho_B \, g \, h_2 = p_A +\varrho_A \, g \, l_1 \sin \theta - \varrho_B \, g \, h_1 + \varrho_B \, g \, h_2$

Therefore, the final expression for $p_A - p_E$ is

$p_A - p_E = -\varrho_A \, g \, l_1 \sin \theta + \varrho_B \, g \, h_1 - \varrho_B \, g \, h_2$

and collecting the terms gives

$p_A - p_E = \varrho_B \, g \left( h_1 - h_2 \right) - \varrho_A \, g \, l_1 \sin \theta$

5. If the pressure $p_A$ was to decrease, then the level at point C would move down and so the level $h_3$ will decrease and the value of $h_2$ will increase.

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 $p_A - p_E$ . We are given the specific gravity of each fluid so

$p_A - p_E = \varrho_w \, g \bigg( SG_B (h_1 - h_2) - SG_A \, l_1 \sin \theta \bigg)$