MCAT Physical - Flow | Practice Hub

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:

Flow in the clogged part of the unhealthy aorta will have a higher velocity and lower pressure than in a healthy aorta

Flow in the clogged part of the unhealthy aorta will have a higher velocity and higher pressure than in a healthy aorta

Flow in the clogged part of the unhealthy aorta will have a lower velocity and lower pressure than in a healthy aorta

Flow in the clogged part of the unhealthy aorta will have a lower velocity and higher pressure than in a healthy aorta

Explanation

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.

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:

Flow in the clogged part of the unhealthy aorta will have a higher velocity and lower pressure than in a healthy aorta

Flow in the clogged part of the unhealthy aorta will have a higher velocity and higher pressure than in a healthy aorta

Flow in the clogged part of the unhealthy aorta will have a lower velocity and lower pressure than in a healthy aorta

Flow in the clogged part of the unhealthy aorta will have a lower velocity and higher pressure than in a healthy aorta

Explanation

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

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

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.

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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)

Under certain conditions, alveoli enlarge or constrict. How would either change affect ΔV?

ΔV increases if enlarged, decreases if constricted.

ΔV decreases if enlarged, increases if constricted.

ΔV decreases if enlarged, decreases if constricted.

Has no effect

Explanation

By Fick's law, we see that ΔV, or flow rate, is directly proportional to surface area. Thus, if alveoli enlarge, so will surface area, and therefore flow rate will also increase. Similarly, if alveoli constrict, so does the total surface area, and thus flow rate will also decrease.

This problem mainly tests to see if the test-taker can quickly see the relationship between ΔV and surface area. Also, intuitively, one might guess that the larger the surface area, the more easily and the more areas gas can diffuse.

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

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

Under certain conditions, alveoli enlarge or constrict. How would either change affect ΔV?

ΔV increases if enlarged, decreases if constricted.

ΔV decreases if enlarged, increases if constricted.

ΔV decreases if enlarged, decreases if constricted.

Has no effect

Explanation

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?

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

$\small P_{2}=P_{1}+\rho gh+2\rho v_{1}^{2}$

$\small P_{2}=P_{1}-\rho gh+4\rho v_{1}^{2}$

$\small P_{2}=P_{1}+\rho gh-\frac{3}{2}\rho v_{1}^{2}$

$\small P_{2}=P_{1}-\rho gh-\rho v_{1}^{2}$

Explanation

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

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?

Explanation

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.

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

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?

Explanation

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.

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

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

$\small P_{2}=P_{1}+\rho gh+2\rho v_{1}^{2}$

$\small P_{2}=P_{1}-\rho gh+4\rho v_{1}^{2}$

$\small P_{2}=P_{1}+\rho gh-\frac{3}{2}\rho v_{1}^{2}$

$\small P_{2}=P_{1}-\rho gh-\rho v_{1}^{2}$

Explanation

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

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 ?

$1.25 \frac{m}{s}$

$20\frac{m}{s}$

$0.56\frac{m}{s}$

$45\frac{m}{s}$

$17\frac{m}{s}$

Explanation

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:

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

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?

The flow speed at point 1 is nine times that at point 2

The flow speed at point 2 is nine times that at point 1

The flow speed at point 1 is three times that at point 2

The flow speed at point 2 is three times that at point 1

The flow speed relation will depend on the viscosity of the water

Explanation

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 ().