Problems in Fluid Dynamics

Let's state all of the equations we have for treating fluids.

The solutions to the problems below can be found at the end of this page. As always, try all the problems before looking at the solutions. It's much easier to understand a solution put before you than to come up with the solution yourself. To develop the skills necessary to solve the problems yourself, you must spend the time doing it.

Problems

1. The main pipe of a city water supply system is 4" (10 cm) in diameter and carries water at a flow speed 5 m/s. There are 10 branch pipes connected to the main, each with a diameter 1" (2.5 cm) . What is the flow speed and flow rate in each of the branch pipes, assuming they are all equal? Will there be a pressure difference between the main pipe and the branches? Assume there is no elevation change.

2. Assume the pressure in the main supply line (coming from the pump) of your home is maintained at a constant value. Does the pressure just inside the branch pipe (which is smaller than the main pipe) to your upstairs shower change when the water is turned on or off? Think about flow speeds and look at Bernoulli's Eq and deduce the answer.

3. Here's a problem proposed in the first module. Assume the winds during a hurricane reach 100 mph. This is a speed of about 45 m/s. Let's assume you have the house closed up. The speed inside is close to zero. Take r of air to be about 1.3 kg/m³. What total force would act on the roof if it was 10 m by 10 m?

4. A standard 1" main supply water line runs horizontally. (That's a diameter of about 2.5 cm.) During peak use the main line must carry about one gallon per second. What pressure drop will be needed along a 10 meter length of the pipe? You will need to estimate the flow rate (one gallon is about 4 liters) and use h = .001 Pa^.s for water.

5. Here's a challenging, but very practical problem. If you can't figure it out on your own, team up with some friends.

   You wish to install an additional sink in the upstairs bathroom. You will have to run a 1/2 " line from the shower line along a horizontal distance of 10 m to reach the place where you wish to install the sink. The shower supply line is a 1/2 " line that runs down to the water pump, 5 meters below. You estimate the total length of the shower pipe to be 8 meters. To determine whether this will work or not, you remove the shower head and measure the maximum flow rate of the shower. It is about 2 gallons per minute. Before you start tearing the walls apart, will the flow rate at the new sink be acceptable?

Solutions

Don't look at the answers until you've tried the problems on your own!!

1. The flow in the main pipe must be equal to the total flow in the 10 branches (provided there are no leaks.) The flow rate in the main is Av = p(.10/2)² x 6 = .047 m³/s. Each branch will have 1/10 th of this, or a flow rate of .0047 m³/s. Applying the flow equation to each branch pipe we get .0047 m³/s = p(.025/2)²v and v = 8.9 m/s. Since the velocity is higher in the branch pipes, Bernoulli's Eq tells us that the pressure must be lower.

2. Look at B's Eq ... P₁ + ¹/₂ rv₁² + rgy₁ = P₂ + ¹/₂ rv₂² + rgy₂ and take position "1" to be at the main line and position "2" to be upstairs. When the shower is off, v is zero at both places and the difference in pressure is due to the difference in height y₂ and y₁. When the faucet is turned on, the speed in the main line (v₁) will be less than in the smaller shower line (v₂). The value of P₁ is held constant and the heights stay the same. Since the value of the ¹/₂rv₂² term will be greater than the value of the ¹/₂rv₁² term, then the value of P₂ will have to drop. This change in pressure as you turn faucets on and off creates the "water hammer" or pipe vibrations that you often hear.

3. The two points are outside at the top of the roof (1) and inside at the bottom of the roof (2). Therefore, y₁ and y₂ are essentially the same and cancel from the equation. (This gives us the special version of B's Eq. that we introduced in the Basic Structure module.) The pressure is greater inside where the velocity is smaller. The difference in pressure is P₂ - P₁ = DP = 1/2 x 1.3 x (45² - 0²) = 1300 Pa. The area of the roof is A = 10 m x 10 m = 100 m². The net lifting force would be F = PA = 1300 x 100 m = 1.3 x 10⁵ N, which is about 30,000 lb or 14 tons! You better have installed hurricane clips!

4. The flow rate of one gallon per second is about 4 liters/s = .004 m³/s. The radius of the pipe is 2.5 cm/ 2 = .013 m. Plugging into Poiseuille's Eq. we have

   Flow Rate = .004 = ^p / ₈ (^R4 / _h ) (^DP / _L ) =

   ^p / ₈ (^.0134 / _.001 ) (^DP / ₁₀ )

   Solving for DP we get DP = 3.6 x 10³ Pa = .52 psi. This is not a significant drop and should not cause problems. But note what happens if we were using a 1/2 " pipe. The only change would be that R is reduced by 2, so R⁴ decreases by 16. That means DP will have to increase by 16 to about 8.4 psi if we want to keep the flow rate the same. That is large enough to seriously effect the flow rate if the water pump can't maintain the additional pressure difference needed.

5. Foiled!! I'll only give hints here! If you want to see all the numbers involved, do the following: Determine the static head pressure that the pump must have to maintaine the 5 m vertical height of water. Determine the flow rate and then use P's Eq. to determine the pressure drop due to viscosity. The total pressure maintained by the pump must be the sum of these two pressures. Now you can find the flow rate for the sink, remembering that you must add the additional length of pipe.

   It is possible to arrive at the solution more easily. Think through what stays the same and what changes and you'll be able to get the answer with a simple ratio. Here's a potentially important follow-up question. What if the shower and sink are both running at the same time?