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Example Questions
Example Question #1 : Fluids
Consider the image shown below.
In the container on the left, there is a gas at an unknown pressure. The tubing connected to the container is open to the atmosphere, which has a pressure of . If the liquid contained within the tubing shows a pressure difference of
, as in the above image, what is the pressure of the gas in the container?
In this question, we're told that an unknown gas is held within a container. Connected to this container is a tube open to the atmosphere which also contains a liquid. Based on the image shown, the height difference of the liquid levels, and the atmospheric pressure, we're asked to determine the pressure of the unknown gas inside the container.
This is a fairly straight-forward problem that asks us about pressure. Because of the way the tubing is situated, the pressure from the atmosphere will push down on the right side of the tubing, while the pressure from the unknown gas in the container will push down on the left part of the tubing. Since the liquid level on the left is lower than on the right, we can infer that the gas in the container is "pushing" on the liquid more strongly than the atmosphere on the other side. Consequently, we know that the gas in the container has a pressure greater than atmospheric pressure.
The difference in height of the liquid levels allows us to quantitatively determine how much greater this pressure is. Since the height difference shows up as , we know that the gas in the container is
more than atmospheric pressure.
Example Question #1 : Fluids
An object is initially submerged below the surface of water. If this same object is then lowered to a depth of
below the surface of water, how does the absolute pressure on the object change?
The absolute pressure halves
The absolute pressure slightly decreases
The absolute pressure doubles
The absolute pressure slightly increases
The absolute pressure slightly increases
For this question, we need to understand the relationship between gauge pressure, ambient pressure, and absolute pressure.
The absolute pressure that an object experiences has two components; ambient and gauge pressure. The ambient pressure is an arbitrarily defined reference pressure that acts on the object. Generally, atmospheric pressure at sea level is used as the reference value, which is . The gauge pressure is measured relative to the ambient pressure that has been used as a reference point. Gauge pressure is dependent on the density of the fluid in which the object is submerged together with the distance under which the object is submerged in the fluid.
The equation for absolute pressure can be expressed as follows.
In the above expression, is one component of the gauge pressure and refers to the distance of the object from the surface of water. As the object is submerged deeper, it will experience a greater gauge pressure and thus a greater absolute pressure.
But what happens if were doubled from
to
, as stated in the question stem? This would certainly cause the gauge pressure to double, but would it cause the absolute pressure to double? The answer is no; there is still the atmospheric pressure to take into account, which maintains the same value in both cases. Because we are increasing the gauge pressure, we know that absolute pressure will increase, but not to the extent of doubling. Thus, we know that it will increase slightly.
Example Question #1 : Continuity Equation
Fluid flows into a pipe with a diameter of at a rate of
. If the other end of the pipe has a cross sectional area of
, what is the speed of the fluid as it exits the pipe?
This question tests the concept of the continuity equation which stipulates that at steady state the volume flowing into one end of a pipe must be exactly equal to the volume flowing out of the other end. That is, . Note that in this case, we are given the inlet diameter and not cross-sectional area. Thus, we must also find the inlet's cross-sectional area by using the formula for the area of a circle.
Example Question #1 : Continuity Equation
If the radius of a pipe with flowing liquid is decreased by a factor of , how does the velocity of the fluid need to change in order to maintain a constant flow rate?
Increase by a factor of
Decrease by a factor of
Increase by a factor of
Dcrease by a factor of
Increase by a factor of
To see how a change of radius affects the velocity of a moving fluid, we'll need to take the continuity equation into account.
Even though the radius is not shown in the above expression, remember that area is related to radius.
Or put another way:
When substituted into the continuity equation:
As the above proportionality shows, if the radius were to decrease by a factor of , this would cause the flow rate to decrease by a factor of
. To counteract this drop in flow rate, the velocity term would need to increase by a factor of
.
Example Question #1 : Fluids
The reason why a coconut floats in water is because __________.
the surface area of the coconut contacting the water is the same as the surface area of the water contacting the coconut
the density of the coconut is less than the density of the water it displaces
the density of the coconut is equal to the density of the water it displaces
the gravitational force acting on the coconut is less than the buoyant force
the gravitational force acting on the coconut is less than the buoyant force
The equation for the buoyant force is . The gravitational force is constant. In order for the coconut to float, the buoyant force must be greater than the gravitational force. If the gravitational force is greater than the buoyant force, then the object will sink.
Example Question #1 : Density, Specific Gravity, And Viscosity
A baseball has a mass of , but it weighs
when completely submerged in water. What is its volume assuming that the density of water is
?
We are given the mass of the baseball outside of the water. Using the weight equation with the gravitational constant being
and the mass being
, the weight of the baseball outside of the water is 4.905 N. (Be careful and convert the mass of the baseball from grams to kilograms since we are using SI units).
The buoyancy force is the difference of the weight of the baseball when it is in the air and when it is in the water. So subtract the two differences:
Now we use the buoyancy equation:
where
is the buoyancy force,
is the density of water,
is the volume of the baseball, and
is the gravitational constant. Plug in the known variables and solve for the volume.
and we get
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