Flow - MCAT Physical

Card 0 of 196

Question

Which of the following is/are assumptions of the Bernoulli equation?

I. The fluid is flowing turbulently

II. The volume of the fluid changes when pressure is applied

III. The fluid is frictionless

Answer

Bernoulli’s principle is an equation that states the relationship between the pressure and velocity of a fluid flowing through a pipe.

$P_1+\frac{1}{2}\rho v_1^2+\rho g h_1=P_2+\frac{1}{2}\rho v_2^2+\rho g h_2$

It operates on three main assumptions. First, the fluid has to have laminar flow. There are two types of flow: laminar and turbulent. Laminar flow is characterized by uniform and ordered flow, whereas turbulent flow is characterized by haphazard and irregular flow. For the Bernoulli equation, the fluid must flow uniformly (laminar). Second, the fluid must be incompressible. This means that the fluid must not change volume (should not be compressed) when an outside pressure is applied. Third, the fluid must experience no friction during flow. Recall that during fluid flow friction usually occurs between layers of fluid because the molecules from each layer interact with each other; however, for the Bernoulli equation to be valid this friction between layers of fluid must be negligible (must be frictionless).

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Question

A liquid of density $\small \rho$ enters a pipe at velocity $\small v_{1}$ and with pressure $\small P_{1}$ . The liquid then exits the pipe a height h above the starting point, at velocity $\small v_{2}=2v_{1}$ . What is the pressure, $\small P_{2}$ , at this exit point, in terms of $\small \rho$ , $\small v_{1}$ , $\small P_{1}$ , and h?

Answer

Use Bernoulli's equation: $\small P_{1}+\rho gh_{1}+\frac{1}{2}\rho v_{1}^{2} = P_{2}+\rho gh_{2}+\frac{1}{2}\rho v_{2}^{2}$

Plug in our given values: $\small P_{1}+0+\frac{1}{2}\rho v_{1}^{2}=P_{2}+\rho gh+\frac{1}{2}\rho (2v_{1})^{2}$

Rearrange to isolate $\small P_{2}$ : $\small P_{2}=P_{1}-\rho gh-\frac{3}{2}\rho v_{1}^{2}$

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Question

Which of the following is the gravitational equivalent in mechanical energy to Bernoulli's equation for fluid mechanics?

$P_1+\frac{1}{2}\rho v_1^2+\rho gh_1=P_2+\frac{1}{2}\rho v_2^2+\rho gh_2$

Answer

Bernoulli's equation states that:

$P_1+\frac{1}{2}\rho v_1^2+\rho gh_1=P_2+\frac{1}{2}\rho v_2^2+\rho gh_2$

Essentially, this equation notes that pressure, velocity, and height of a fluid during flow can be related by a constant term. Very similarly, kinetic energy and potential energy sum to a constant mechanical energy for static compounds.

$mgh_1+\frac{1}{2}mv_1^2=mgh_2+\frac{1}{2}mv_2^2$

Neglecting the term for pressure in Bernoulli's equation, there are direct correlations between kinematic and gravitational terms. Each term has an equivalent in both equations.

Kinematic terms: $\frac{1}{2}\rho v^2\ \text{and}\ \frac{1}{2}mv^2$

Gravitational terms: $\rho gh\ \text{and}\ mgh$

In Bernoulli's equation, kinetic energy is the kinematic equivalent and potential energy is the gravitational equivalent.

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Question

Use the following information to answer questions 1-6:

The circulatory system of humans is a closed system consisting of a pump that moves blood throughout the body through arteries, capillaries, and veins. The capillaries are small and thin, allowing blood to easily perfuse the organ systems. Being a closed system, we can model the human circulatory system like an electrical circuit, making modifications for the use of a fluid rather than electrons. The heart acts as the primary force for movement of the fluid, the fluid moves through arteries and veins, and resistance to blood flow occurs depending on perfusion rates.

To model the behavior of fluids in the circulatory system, we can modify Ohm’s law of V = IR to ∆P = FR where ∆P is the change in pressure (mmHg), F is the rate of flow (ml/min), and R is resistance to flow (mm Hg/ml/min). Resistance to fluid flow in a tube is described by Poiseuille’s law: R = 8hl/πr4 where l is the length of the tube, h is the viscosity of the fluid, and r is the radius of the tube. Viscosity of blood is higher than water due to the presence of blood cells such as erythrocytes, leukocytes, and thrombocytes.

The above equations hold true for smooth, laminar flow. Deviations occur, however, when turbulent flow is present. Turbulent flow can be described as nonlinear or tumultuous, with whirling, glugging or otherwise unpredictable flow rates. Turbulence can occur when the anatomy of the tube deviates, for example during sharp bends or compressions. We can also get turbulent flow when the velocity exceeds critical velocity vc, defined below.

vc = NRh/ρD

NR is Reynold’s constant, h is the viscosity of the fluid, ρ is the density of the fluid, and D is the diameter of the tube. The density of blood is measured to be 1060 kg/m3.

Another key feature of the circulatory system is that it is set up such that the organ systems act in parallel rather than in series. This allows the body to modify how much blood is flowing to each organ system, which would not be possible under a serial construction. This setup is represented in Figure 1.

Which of the following disorders would most likely cause an increase in turbulent flow?

I. Increased cardiac output

II. Anemia

III. Lung cancer

Answer

We can see from the equation that critical velocity depends on Reynolds number, viscosity, diameter of the tube, and density of the fluid. If we assume the heart is pumping normally and that the velocity of blood is constant, a decrease in the critical velocity would increase the chance that the normal velocity would exceed this critical velocity. This would result in turbulent flow.

Looking at the choices, anemia would cause a decrease in blood viscosity, resulting in a lower critical velocity, increasing the chance for turbulent flow. Increased cardiac output would increase the average velocity of the blood such that it may overcome the critical velocity, resulting in more turbulence. Lung cancer should not have a measurable effect on blood turbulence.

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Question

A researcher is observing a fluid flowing in a pipe. He measures the velocity of the flow at both ends of the pipe and wants to know the relative pressures each end. Which of the following equations will be most useful to him?

Answer

The question states that the researcher wants to know the relationship between velocity and pressure. Recall that the Bernoulli equation states this relationship for a fluid flowing through a pipe; therefore, the most appropriate principle for the researcher would be Bernoulli’s equation.

Poiseuille’s law and the continuity equation are both fluid mechanics principles, but neither of them state a relationship between velocity and pressure. Poiseuille’s law states the relationship between flow resistance and flow rate for a fluid flowing through a pipe, whereas the continuity equation states the relationship between velocity of the fluid and the area of the pipe. The Doppler effect is irrelevant to this question because it relates changes in frequency of a wave relative to the observer and the wave source; it is not a fluid mechanics principle.

Bernoulli equation: $P_1+\frac{1}{2}\rho v_1^2+\rho g h_1=P_2+\frac{1}{2}\rho v_2^2+\rho g h_2$

Poiseulle's law: $Q=\frac{\pi Pr^4}{8\eta l}$

Continuity equation:

Doppler effect: $f=f_o\frac{v\pm v_o}{v\pm v_s}$

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Question

A fluid is flowing through a pipe from left to right. If you increase the vertical height of the left end, what will happen to the pressure and velocity of the fluid at the left end?

Bernoulli’s equation:

$P_1 + \frac{1}{2}\rho v_1^2+\rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2+\rho gh_2$

Answer

To answer this question you need to know the Bernoulli equation:

$P_1 + \frac{1}{2}\rho v_1^2+\rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2+\rho gh_2$

In the equation, is pressure, $\rho$ is density, is velocity, is acceleration due to gravity, and is vertical height. Let’s assume that the left side of this equation is for the left end of the pipe and right side is for the right end of the pipe.

The question states that the vertical height, , is increased at the left end; therefore, the potential energy term on the left hand side of Bernoulli’s equation is increased ( $\rho gh_1$ ). Since the potential energy at the left end increased, the other term(s) on the left hand side of the equation must decrease so that the left hand side equals the right hand side.

This means that the sum of pressure and velocity must decrease; however, we do not have enough information to determine the kinds of changes that will occur in each individual term. It is possible that only one term decreases, while the other term stays constant, or it is possible that both terms decrease; therefore, without more information, we cannot determine the relative changes to pressure and velocity.

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Question

Due to plaque buildup, a small part of a patient’s aorta has a smaller radius than a regular, healthy aorta. Which of the following is true regarding the unhealthy and healthy aorta?

Bernoulli's equation:

$P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2$

Continuity equation:

Answer

The question states that a part of the unhealthy aorta has a smaller radius. This means that this portion of the aorta will have a smaller cross-sectional area. To solve this question we need to use both the continuity equation and the Bernoulli equation. The continuity equation is as follows:

and are area and velocity of fluid flow, respectively. This equation states that the product of area and velocity of fluid flow on one side of a pipe must equal the product of area and velocity of the fluid flow on the other side of the pipe. Since part of the unhealthy aorta has a smaller radius, it will have a smaller area. According to the continuity equation, the velocity of fluid flowing through the smaller part of the aorta will increase to compensate for the decrease in area; therefore, the velocity of the fluid flow will increase in the clogged region of the aorta.

The second equation we need to use is the Bernoulli equation:

$P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2$

Here, is pressure, $\rho$ is density, is velocity, is acceleration due to gravity, and is vertical height. Bernoulli’s equation states that an increase in velocity of fluid flow will decrease the pressure. This occurs because the pressure will decrease to compensate for the increase in velocity (to ensure that the left hand side of the equation equals the right hand side); therefore, the clogged region of the aorta will have a higher velocity and lower pressure.

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Question

If a pipe with flowing water has a cross-sectional area nine times greater at point 2 than at point 1, what would be the relation of flow speed at the two points?

Answer

Using the continuity equation we know that $A_{1}V_1= A_2V_2$ . The question tells us that the cross-sectional area at point 2 is nine times greater that at point 1 ().

Using the continuity equation we can make A1= 1 and A2 = 9.

$\frac{V_1}{V_2}=\frac{9}{1}$

Flow speed at point 1 is nine times that at point 2.

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Question

Use the following information to answer questions 1-6:

The circulatory system of humans is a closed system consisting of a pump that moves blood throughout the body through arteries, capillaries, and veins. The capillaries are small and thin, allowing blood to easily perfuse the organ systems. Being a closed system, we can model the human circulatory system like an electrical circuit, making modifications for the use of a fluid rather than electrons. The heart acts as the primary force for movement of the fluid, the fluid moves through arteries and veins, and resistance to blood flow occurs depending on perfusion rates.

To model the behavior of fluids in the circulatory system, we can modify Ohm’s law of V = IR to ∆P = FR where ∆P is the change in pressure (mmHg), F is the rate of flow (ml/min), and R is resistance to flow (mm Hg/ml/min). Resistance to fluid flow in a tube is described by Poiseuille’s law: R = 8hl/πr4 where l is the length of the tube, h is the viscosity of the fluid, and r is the radius of the tube. Viscosity of blood is higher than water due to the presence of blood cells such as erythrocytes, leukocytes, and thrombocytes.

The above equations hold true for smooth, laminar flow. Deviations occur, however, when turbulent flow is present. Turbulent flow can be described as nonlinear or tumultuous, with whirling, glugging or otherwise unpredictable flow rates. Turbulence can occur when the anatomy of the tube deviates, for example during sharp bends or compressions. We can also get turbulent flow when the velocity exceeds critical velocity vc, defined below.

vc = NRh/ρD

NR is Reynold’s constant, h is the viscosity of the fluid, ρ is the density of the fluid, and D is the diameter of the tube. The density of blood is measured to be 1060 kg/m3.

Another key feature of the circulatory system is that it is set up such that the organ systems act in parallel rather than in series. This allows the body to modify how much blood is flowing to each organ system, which would not be possible under a serial construction. This setup is represented in Figure 1.

Which of the following accurately demonstrates a relationship between critical velocity and resistance?

Answer

Viscosity (h) is common to both equations, so to find a relationship between resistance and critical velocity, we can use the resistance equation and solve for viscosity.

$R=\frac{8hl}{nr^4}$ and $v_c=\frac{N_rh}{\rho D}$

Rearranging the equation for resistance allows us to substitute for h in the critical velocity equation.

$h=\frac{Rnr^4}{8l}$

$v_c=\frac{RN_rnr^4}{8\rho Dl}$

This equation can be further reduced. Diameter is twice the radius, so D = 2r.

$v_c=\frac{RN_rnr^4}{8\rho l(2r)}=\frac{RN_rnr^3}{16\rho l}$

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Question

A liquid flows through a pipe with a diameter of 10cm at a velocity of 9cm/s. If the diameter of the pipe then decreases to 6cm, what is the new velocity of the liquid?

Answer

Rate of flow, A * v, must remain constant. Use the continuity equation, $A_{1}v_{1}=A_{2}v_{2}$ .

Solving the initial cross-sectional area yields: $A_{1}=\pi r^{2}=25\pi cm^{2}$ . The initial radius is 5cm.

Then find the final area of the pipe: $A_{2}=\pi r^{2}=9\pi cm^{2}$ . The final radius is 3cm.

Using these values in the continuity equation allows us to solve the final velocity.

$(25 \pi cm^{2})(9 cm/s) = (9\pi cm^{2})v_{2}$

$v_{2}= 25 cm/s$

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Question

If blood flows through the aorta with velocity, , with what velocity would blood flow through the capillaries in the body?

Answer

Just like the volume flow rate equation for fluids, the flow rate of blood through the body is equal to area times velocity.

$flow\ rate=Av$

The flow rate is a constant, so depending on the area that the blood is travelling through, the velocity is constantly changing; therefore the volume flow rate though the aorta is equal to the volume flow rate in the capillaries.

Because we can determine the area of the aorta and area of the capillaries, knowing the velocity through the aorta can give the velocity through the capillaries.

$v_c=\frac{A_av_a}{A_c}$

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Question

Which will produce the greatest increase in flow velocity through a tube?

Answer

The equation for volumetric flow rate is $\small Q=Av$ , where $\small A$ is the cross-sectional area of the tube and $\small v$ is the flow velocity. We can re-write this equation to solve for the velocity and include the tube radius.

$v=\frac{Q}{A}=\frac{Q}{\pi r^2}$

Halving the radius will reduce the tube area by a factor of four.

Volumetric flow rate is constant, thus, any reduction in area will cause a corresponding increase in velocity.

Halving the radius will thus quadruple the velocity, resulting in the greatest increase of the given options.

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Question

A pipe with a diameter of 4 centimeters is attached to a garden hose with a nozzle. If the velocity of flow in the pipe is $\small 2\frac{m}{s}$ , what is the velocity of the flow at the nozzle when it is adjusted to have a diameter of 8 millimeters?

Answer

Flow rate in a pipe must be constant in order to create linear flow. This flow rate is given by the product of the cross-sectional area and the velocity of the fluid.

The cross-sectional areas of the pipe and nozzle can be found using their radii. Note that you were given dimensions in terms of diameter, so be sure to divide by 2 to get the radius.

$A=\pi r^2$

$A_1=\pi (0.02m)^2=0.0004\pi m^2$

$A_2=\pi(0.004m)^2=0.000016\pi m^2$

Use these areas and the initial velocity to calculate the final velocity in the nozzle.

$(0.0004\pi m^2)(2\frac{m}{s})=(0.000016\pi m^2)v_2$

$v_2=\frac{(0.0004\pi m^2)(2\frac{m}{s})}{0.000016\pi m^2}$

$v_2=50\frac{m}{s}$

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Question

Blood travels through an artery at velocity . If a vasoconstricting chemical is consumed and the artery constricts to half the original diameter, what is the new velocity of the blood?

Answer

The continuity equation states that:

$A_{1}v_{1}=A_{2}v_{2}$

In other words, the volumetric flow rate stays constant throughout a pipe of varying diameter. If the diameter decreases (constricts), then the velocity must increase.

To establish the change in cross-sectional area, we need to find the area in terms of the diameter:

$A=\pi r^2=\pi(\frac{D}{2})^2=\frac{\pi D^2}{4}$

If the diameter is halved, the area is quartered.

$A_2=\frac{\pi(\frac{D}{2})^2}{4}=\frac{\pi(\frac{D^2}{4})}{4}=\frac{1}{4}\frac{\pi D^2}{4}$

$A_2=\frac{1}{4}A$

To keep the volumetric flow constant, the velocity would have to increase by a factor of 4.

$A_1v_1=(\frac{1}{4}A_1)v_2$

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Question

As water is traveling from a water tower, to somone's home, the pipes it travels in frequently change size.

Water is traveling at $5\frac{m}{s}$ in a tube with a diameter of . The tube gradually increases in size to a diameter of , and then gradually decreases to a diameter of . Neglecting any energy losses due to friction and pressure changes, what is the speed of the water when it reaches the tube diameter of ?

Answer

This problem covers the concept of continuity. As the tube diameter changes, the volumetric flow of water stays constant. Therefore, we can calculate the volumetric flow at the diameter of 0.5m, and use that to find the velocity of water at 1m.

$\dot{V} = vA_c$

Here, is the cross-sectional area of the pipe

$\dot{V} = v(\pi\frac{d^2}{4}) = (5\frac{m}{s})\pi\frac{(0.5m)^2}{4} = 0.9817 \frac{m^3}{s}$

Apply this flow rate to a tube diameter of 1m to find the velocity:

$0.9817\frac{m^3}{s} = v\pi\frac{(1m)^2}{4} = v\frac{\pi}{4}$

$v = 1.25 \frac{m}{s}$

Alternatively, this question can be solved by setting up a proportion.

$\dot{V}=v_1A_1=v_2A_2$

Rearranging for :

$v_2 = v_1\frac{A_1}{A_2} = v_1\frac{\pi\frac{d_1^2}{4}}{\pi\frac{d_2^2}{4}} = v_1\frac{d_1^2}{d_2^2}$

Plugging in our values:

$v_2 = (5\frac{m}{s}) \frac{(0.5m)^2}{(1m)^2} = 0.25(5\frac{m}{s}) = 1.25 \frac{m}{s}$

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Question

Diffusion can be defined as the net transfer of molecules down a gradient of differing concentrations. This is a passive and spontaneous process and relies on the random movement of molecules and Brownian motion. Diffusion is an important biological process, especially in the respiratory system where oxygen diffuses from alveoli, the basic unit of lung mechanics, to red blood cells in the capillaries.

Figure 1 depicts this process, showing an alveoli separated from neighboring cells by a capillary with red blood cells. The partial pressures of oxygen and carbon dioxide are given. One such equation used in determining gas exchange is Fick's law, given by:

ΔV = (Area/Thickness) · Dgas · (P1 – P2)

Where ΔV is flow rate and area and thickness refer to the permeable membrane through which the gas passes, in this case, the wall of the avlveoli. P1 and P2 refer to the partial pressures upstream and downstream, respectively. Further, Dgas, the diffusion constant of the gas, is defined as:

Dgas = Solubility / (Molecular Weight)^(1/2)

In higher altitudes, a decrease in which factors of Fick's law can change in order to achieve the same flow rate at lower altitudes.

Answer

It should be known that in higher altitudes, the partial pressure of oxygen falls. That is, the partial pressure in the alveoli will fall. Thus, flow rate will decrease, and we will need changes to increase flow rate.

By looking at Fick's equation, we can see that a decrease in thickness can help restore flow rate. Biologically speaking, this is less likely to happen, and more correctly, hemoglobin concentration and binding affinity to oxygen increases; however, this is extraneous information not needed for the MCAT.

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Question

Diffusion can be defined as the net transfer of molecules down a gradient of differing concentrations. This is a passive and spontaneous process and relies on the random movement of molecules and Brownian motion. Diffusion is an important biological process, especially in the respiratory system where oxygen diffuses from alveoli, the basic unit of lung mechanics, to red blood cells in the capillaries.

Figure 1 depicts this process, showing an alveoli separated from neighboring cells by a capillary with red blood cells. The partial pressures of oxygen and carbon dioxide are given. One such equation used in determining gas exchange is Fick's law, given by:

ΔV = (Area/Thickness) · Dgas · (P1 – P2)

Where ΔV is flow rate and area and thickness refer to the permeable membrane through which the gas passes, in this case, the wall of the avlveoli. P1 and P2 refer to the partial pressures upstream and downstream, respectively. Further, Dgas, the diffusion constant of the gas, is defined as:

Dgas = Solubility / (Molecular Weight)^(1/2)

Conceptually, if alveoli are considered to be perfectly spherical and assuming that its entire surface exchanges gas, which new relationship, introducing the variable, r, the radius of an alveoli, correctly describes Fick's Equation, assuming partial pressures remain constant?

Answer

The quickest approach to this equation is to see what variable in Fick's law might be affected by a change in the radius of an alveoli. Pressure is constant and can be ruled out. The diffusion constant would not be effected by radius (hence it being a constant).

The question stem mentions nothing about a change in thickness, therefore, we are left with area.

The surface area of a sphere is measured by 4πr2 and can be substituted. If you had difficulty remembering how to measure surface area, remember that the area of a circle is πr2 andlogically it follows that the surface area of an entire sphere must be greater due to it being three dimensional.

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Question

Diffusion can be defined as the net transfer of molecules down a gradient created by differing concentrations of the molecule in different locations. This is a passive, spontaneous process and relies on the random movement of molecules and Brownian motion. Diffusion is an important biological process, especially in the respiratory system where oxygen diffuses from alveoli, the basic units of lung mechanics, to red blood cells in the capillaries.

Figure 1 depicts this process, showing an alveolus separated from neighboring cells by a capillary with red blood cells. The partial pressures of oxygen and carbon dioxide are given. One equation used in determining gas exchange is Fick's law, given by:

$\Delta V=(\frac{\text{Area}}{\text{Thickness}})(D_{gas})(P_1-P_2)$

In this equation, $\Delta V$ is the flow rate. Area and thickness refer to the permeable membrane through which the gas passes_—_in this case, the wall of the alveolus. and refer to the partial pressures upstream and downstream, respectively. $D_{gas}$ , the diffusion constant of the gas, is defined as:

$D_{gas}=\frac{\text{Solubility}}{\sqrt\text{Molecular Weight}}$

At high altitudes, the partial pressure of oxygen quickly drops while that of carbon dioxide decreases by much less. Given the following table, at what elevation is the pressure gradient of oxygen equal to half that of carbon dioxide, assuming constant capillary partial pressures of oxygen and carbon dioxide of 50mmHg and 40mmHg, respectively?

Elevation (m) PO2 (mmHg) PCO2 (mmHg)
0 100 50
1000 80 40
2000 60 30
3000 40 20
4000 20 10
5000 10 0

Elevation (m)	PO2 (mmHg)	PCO2 (mmHg)
0	100	50
1000	80	40
2000	60	30
3000	40	20
4000	20	10
5000	10	0

Answer

This problem at first may seem straightforward, until you realize that the gradient (or difference in partial pressures of Fick's law) actually changes.

In normal circumstances, capillary pressure would be subtracted from arterial pressure; however, the body operates at a physiologic equilibrium, which essentially has a constant production of CO2 and demand for O2. At a certain point the capillary pressure will be greater than the arterial, and hence the gradient switches to Pcapillary – Parterial from the original Parterial – Pcapillary.

Simple arithmetic done alongside the table margins can quickly arrive to the correct answer. At an elevation of 3000 m, the gradients of oxygen (50 – 40 = 10 mmHg) and carbon dioxide (40 – 20 = 20 mmHg) differ by twofold. It is easy to be confused over a seemingly complicated table, when in reality, this question only asks for basic arithmetic.

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Question

A barrel of height 10m is filled with water to a height of 5m. If a hole is punctured 2m above the bottom of the barrel, what is the velocity of the water leaving the barrel?

Answer

To answer this question, we can use Toricelli’s law.

$v=\sqrt{2gh}$

In this case, h is the height of water above the hole. Since the barrel is filled to a height of 5m, and the hole is punctured 2m from the bottom of the barrel, the height of water is 3m.

$v=\sqrt{(2)(10\frac{m}{s^2})(3m)}=\sqrt{60}\frac{m}{s}$

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Question

What causes the lift experienced by the wing of an airplane in level flight?

Answer

Air density is considered to be uniform in Bernoulli's equation, which defines the air pressure on the upper and lower surfaces of an aircraft wing. Likewise, air humidity at any one place in the atmosphere is uniform over the space occupied by the wing. Central to the idea of producing lift is the ability of the wing to avoid turbulence, one reason for de-icing wings before winter takeoffs.

Air is considered to move in bulk during flight. This means that a volume of air that has to travel a longer distance than its twin volume (because the airplane wing split them apart) must speed up to arrive at the same place (the trailing edge of the wing) at the same time as its twin. Velocity is squared in Bernoulli's equation, meaning that small differences in flow velocity between the upper and lower wing surfaces translate to large pressure differences.

Attack angle is an important lift factor, but not in level flight.

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Flow - MCAT Physical

Which of the following is/are assumptions of the Bernoulli equation? I. The fluid is flowing turbulently II. The volume of the fluid changes when pressure is applied III. The fluid is frictionless

A liquid of density enters a pipe at velocity and with pressure . The liquid then exits the pipe a height h above the starting point, at velocity . What is the pressure, , at this exit point, in terms of , , , and h?

Use Bernoulli's equation: Plug in our given values: Rearrange to isolate :

Which of the following is the gravitational equivalent in mechanical energy to Bernoulli's equation for fluid mechanics?

A researcher is observing a fluid flowing in a pipe. He measures the velocity of the flow at both ends of the pipe and wants to know the relative pressures each end. Which of the following equations will be most useful to him?

A fluid is flowing through a pipe from left to right. If you increase the vertical height of the left end, what will happen to the pressure and velocity of the fluid at the left end? Bernoulli’s equation:

Due to plaque buildup, a small part of a patient’s aorta has a smaller radius than a regular, healthy aorta. Which of the following is true regarding the unhealthy and healthy aorta? Bernoulli's equation: Continuity equation:

If a pipe with flowing water has a cross-sectional area nine times greater at point 2 than at point 1, what would be the relation of flow speed at the two points?

Using the continuity equation we know that . The question tells us that the cross-sectional area at point 2 is nine times greater that at point 1 (). Using the continuity equation we can make A1= 1 and A2 = 9. Flow speed at point 1 is nine times that at point 2.

A liquid flows through a pipe with a diameter of 10cm at a velocity of 9cm/s. If the diameter of the pipe then decreases to 6cm, what is the new velocity of the liquid?

If blood flows through the aorta with velocity, , with what velocity would blood flow through the capillaries in the body?

Which will produce the greatest increase in flow velocity through a tube?

A pipe with a diameter of 4 centimeters is attached to a garden hose with a nozzle. If the velocity of flow in the pipe is , what is the velocity of the flow at the nozzle when it is adjusted to have a diameter of 8 millimeters?

Blood travels through an artery at velocity . If a vasoconstricting chemical is consumed and the artery constricts to half the original diameter, what is the new velocity of the blood?

A barrel of height 10m is filled with water to a height of 5m. If a hole is punctured 2m above the bottom of the barrel, what is the velocity of the water leaving the barrel?

To answer this question, we can use Toricelli’s law. In this case, h is the height of water above the hole. Since the barrel is filled to a height of 5m, and the hole is punctured 2m from the bottom of the barrel, the height of water is 3m.

What causes the lift experienced by the wing of an airplane in level flight?

Which of the following is/are assumptions of the Bernoulli equation?

I. The fluid is flowing turbulently

II. The volume of the fluid changes when pressure is applied

III. The fluid is frictionless

Which of the following is the gravitational equivalent in mechanical energy to Bernoulli's equation for fluid mechanics?

$P_1+\frac{1}{2}\rho v_1^2+\rho gh_1=P_2+\frac{1}{2}\rho v_2^2+\rho gh_2$

A fluid is flowing through a pipe from left to right. If you increase the vertical height of the left end, what will happen to the pressure and velocity of the fluid at the left end?

Bernoulli’s equation:

$P_1 + \frac{1}{2}\rho v_1^2+\rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2+\rho gh_2$

A pipe with a diameter of 4 centimeters is attached to a garden hose with a nozzle. If the velocity of flow in the pipe is $\small 2\frac{m}{s}$ , what is the velocity of the flow at the nozzle when it is adjusted to have a diameter of 8 millimeters?