Liquids are incompressible. Their density ρ = mass/volume is constant.
When a liquid flows through a pipe, conservation of mass leads to the
equation of continuity.
Consider the flow of a fluid through a pipe with varying cross sectional area A.
The volume V1 of liquid flowing into the pipe equals the volumeV2 flowing out of the pipe per unit time.
V1/Δt = V2/Δt, A1Δx1/Δt = A2Δx2/Δt, A1v1 = A2v2.
For the pipe we write the equation of continuity as A1v1 = A2v2, or Q = Av = constant. Q is called the volume flow rate.
In this session, you will try to verify the equation of continuity for water flowing out of the bottom of an elevated can through a small-diameter hose. You will also measure the fraction of the ordered energy that is lost because of friction. In a another experiment you will determine the density of a metal block using only a balance, by applying Archimedes’ principle.
- can and hose
- water containers, one with overflow spout
- metal blocks
- electronic balance
- meter stick
Open a Microsoft Word document to keep a log of your experimental procedures, results and discussions. This log will form the basis of your lab report. Address the points highlighted in blue. Answer all questions.
Each table will work as a group on this experiment.
One end of a rubber hose is attached to a can with a spout in the bottom. The can has an inside diameter of d1 = 9.8 cm and the rubber hose has an inside diameter of d2 = 0.8 cm. The other end of the hose is taped to a horizontal rod a certain distance below the can. While one member of your group plugs the hose with a finger, another member fills the can with water. Then you will allow the water to drain from the can through the hose into a catch pan. You will measure the speed with which the water level in the can drops, and the speed with which the water emerges from the hose.
Set up the apparatus on the floor, not on the table.
- The horizontal position of the lower end of the hose should be a few cm over the edge of the catch pan.
- Measure the height of ∆y of the center of the hose above the bottom of the catch pan.
- Measure the vertical distance h from the center of the hose to the center of the 5 cm long black tape inside the can.
- Record your measurements in table 1.
|can diameter (cm)||9.8|
|hose diameter (cm)||0.8|
|tape length (cm)||5|
Fill up the can with water to above the upper end of the black tape without allowing water to drain. Distribute tasks to be performed after you start draining the water.
- One member of your group will notify the others when the water level in the can reaches the upper end and when it reaches the lower end of the black tape.
- Several members will measure the time interval ∆t between these two events.
- Other members will mark the spots where the water hits the bottom of the pan for those two events. You will have to determine the average horizontal distance ∆x between the lower end of the hose and the spots where the water hits the pan.
Drain the water while making measurements and record your measurements in the table.
- Repeat the experiment to check for reproducibility.
- To find the speed v1 with which the water level drops calculate v1 = (5 cm)/∆t.
- To find the speed v2 with which the water emerges from the hose, use the formulas for projectile motion.
- For the vertical motion we have ∆y = ½gt2. We can solve for the time t it takes water to fall from the end of the hose to the bottom of the pan.
- For the horizontal motion we have ∆x = v2t. Inserting the time from above we have v2 = ∆x/(2∆y/g)1/2.
- The kinetic energy of a volume V of water moving with speed v = ½ρVv2
= ½mv2, since ρV = m.
The potential energy of the water changes as it moves. While all the water moves, the change in potential energy is the same as that of a volume V, which has been moved from the top of the can to the exit of the hose. The potential energy of the water in the rest of the can and hose is the same as the potential energy of the water that used to be there before the movement.
The change in the potential energy of a mass ρV = m of water coming out of the hose is mgh.
Here h is the distance from the center of the hose to the center of the 5 cm long black tape inside the can that you measured earlier.
If no ordered energy is converted into disordered energy we expect from energy conservation that the change in kinetic energy is equal to the change in potential energy.
½mv22 - ½mv12 = mgh or ½(v22 - v12) = gh.
But some of the ordered energy will be converted into disordered energy, so we expect ½(v22 - v12) < gh.
Dividing the change in kinetic energy by the change in potential energy we find the fraction R = (v22 - v12)/(2gh) of the potential energy that is converted to kinetic energy. If no ordered energy is lost, then R = 1. We expect it to be less than one, because friction is always present. But how much less?
Fill in table 2.
|v1 = 5/∆t (cm/s)|
|v2 = ∆x/(2∆y/980)1/2 (cm/s)|
|A1v1 = (πd12/4)v1 (cm3/s)|
|A2v2 = (πd22/4)v2 (cm3/s)|
|R = (v22 - v12)/(2*980*h)|
- Did you verify the equation of continuity? If not, explain what factors may be responsible for the difference in your two values of the volume flow rate? Does the difference seem reasonable considering those factors?
- What fraction of the potential energy of the water is converted into kinetic energy in this experiment? What fraction is converted into thermal energy? Does this result surprise you? If yes, pay attention to Poiseuille’s law in the next module.
Each group will perform their own experiment.
Each group has two metal blocks, a small metal can, a can with a spout, and a container to hold extra water. Each group will determine the density of the metal blocks using a balance only.
Record all your measurements in a table in your log.
|wblock||wempty small can||wsmall can with water||wdisplaced water||wbuoyant||Vwater||ρblock|
For each block:
Set up the balance. Turn it on and zero it.
Determine the weight of the metal block.
Note: The readout of the balance is in kg, the weight is mg.
Determine the weight of the small metal can.
Fill the can with the spout with water until it overflows. Catch the overflow in the small can and discard it.
Then lower the metal block into the water and catch the displaced water with the small can.
Determine the weight of the small metal can with the displaced water in it using the balance.
Calculate the weight of the displaced water by subtracting the weight of the can.
The weight of the displaced water should be equal to the buoyant force.
To get a more accurate measurement of the buoyant force, place the can with the spout onto the balance.
Make sure there is enough water in it so you can submerge the metal block, but not enough to cause it to overflow onto the balance when you do submerge it.
Zero the balance with the can on it.
Gently lower the block into the water. Make sure it is totally submerged but is not touching the walls or bottom of the can. The water is pushing up on the block with a force equal to wbuoyant. The scale now measures the reaction force with which the block is pushing down, which is equal in magnitude to the buoyant force by Newton's third law. Record wbuoyant.
Given the mass of the displaced water calculate its volume.
The density of water is ρ = m/V = 1000 kg/m3, so V = m/(1000 kg/m3).
The volume of the metal block is equal to the volume of the displaced water.
Calculate the density ρblock = mblock/Vwater.
Can you identify the materials your blocks are made of using your result and the table below and the appearance of the blocks?
- Are your experimental results consistent, i.e. is wdisplaced water equal to wbuoyant? If not, what do you think can account for the difference?
- What materials are your block is made of? Are your measured densities close to one of the densities in the table? How big is the difference, and does this seem reasonable given you experimental procedure?