Fluid mechanics problem involving a hose and water coming out.

In summary, the conversation discusses a problem involving a conservation of mass for a steady state incompressible fluid. The initial approach of using V1*A1 = V2*A2 is deemed incorrect due to the varying flow of water through A1. The conversation then suggests using the total volume of water flowing in 1 second through A1 and equating it to the total volume flowing through A2. The correct equation is found to be (1/R)∫v(r)dr with limits 0 and R, = (4/3)Vavg. The conversation concludes by discussing the use of Bernoulli's equation and the assumption of no interaction with surrounding air to solve for diameter ratios.
  • #1
guitar24
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Homework Statement



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Homework Equations



I am having some trouble starting this problem. At first look I would say that a conservation of mass will give V1*A1=V2*A2 for steady state incompressible fluid. Then I would solve for A2 and relate the two diameters but I don't have the velocity distribution of the water stream. Any hints??
 
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  • #2
You're spot on to include conservation of volume, but V1*A1 = V2*A2 can't be right since the flow of water thru A1 varies with r. So think of the total volume of water flowing in 1 sec. thru A1 and equate that to the total volume of water flowing in 1 sec. thru A2.
 
  • #3
Thank you for your reply! could you elaborate a bit more on how I would do that?
The total volumetric flow rate in the tube is Vavg*Area of the tube. Multiply by one and you get the volume of water entering the cross section in one second. Then i would just divide that by the area of the water stream to get the volume of water flowing in one sec through the area of the water stream. Then just rearrange the equation to get Dout/Dhose. Is that correct?

Thank you!
 
  • #4
guitar24 said:
Thank you for your reply! could you elaborate a bit more on how I would do that?
The total volumetric flow rate in the tube is Vavg*Area of the tube. Multiply by one and you get the volume of water entering the cross section in one second. Then i would just divide that by the area of the water stream to get the volume of water flowing in one sec through the area of the water stream. Then just rearrange the equation to get Dout/Dhose. Is that correct?

Thank you!

I computed the average flow per unit time and found it to be not Vavg but (1/R)∫v(r)dr with limits 0 and R, = (4/3)Vavg which is somewhat weird.

Anyway, whatever average V is, you are right in using mass conservation flow to get the answer. But you still need to get v, the (uniform) velocity of the stream exiting at Dout. For that, go to Bernoulli.
 
  • #5
Given the assumption that there is no interaction with the surrounding air and thus no momentum loss, that suggests momentum is constant. Write integrals representing the momentum at the nozzle using the given velocity profile and one for the region downstream (utilizing conservation of mass flow) where the velocity is uniform. Equate them and solve for diameter ratios.
 

1. How does the diameter of the hose affect the velocity of water coming out?

The diameter of the hose directly affects the velocity of water coming out. According to Bernoulli's principle, as the diameter of the hose decreases, the velocity of the water increases. This is because the same volume of water must pass through a smaller area, resulting in an increase in velocity.

2. What is the relationship between the pressure of the water and the height of the water in the hose?

The pressure of the water in the hose is directly proportional to the height of the water in the hose. This is due to the hydrostatic pressure, which increases with depth. As the water level in the hose increases, so does the pressure exerted on the water inside.

3. How does the angle at which the hose is held affect the distance the water can reach?

The angle at which the hose is held can affect the distance the water can reach. This is because the distance the water travels is dependent on the initial velocity and angle of the water coming out of the hose. Holding the hose at a higher angle can result in a shorter distance, while holding it at a lower angle can result in a longer distance.

4. How does the density of the water affect the flow rate through the hose?

The density of the water has an inverse relationship with the flow rate through the hose. As the density of the water increases, the flow rate decreases. This is because the density of the water affects its viscosity, which is the measure of its resistance to flow. Higher viscosity results in a slower flow rate.

5. Can the flow rate through the hose be increased by increasing the pressure of the water?

Yes, the flow rate through the hose can be increased by increasing the pressure of the water. This is because the flow rate is directly proportional to the pressure difference between the two ends of the hose. By increasing the pressure at one end, the flow rate will increase as the water rushes to equalize the pressure difference.

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