Calculating Initial Velocity of Water After Nozzle Increase

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SUMMARY

The discussion centers on calculating the increase in velocity of water as it exits a nozzle, specifically noting that the velocity increases by a factor of approximately 6.25 times, not exactly 6 times. This conclusion is derived from the relationship between the diameters of the hose and the nozzle, as well as the principle of conservation of mass for incompressible fluids. The calculations involve determining the flow rate and the cross-sectional areas of both the hose and the nozzle, as illustrated in Example 5-1.

PREREQUISITES
  • Understanding of fluid dynamics principles, particularly the conservation of mass.
  • Familiarity with calculating flow rates using the formula Q = Area × Velocity.
  • Knowledge of cross-sectional area calculations for circular pipes.
  • Basic algebra for manipulating ratios and equations.
NEXT STEPS
  • Study the derivation of the continuity equation in fluid dynamics.
  • Learn how to calculate cross-sectional areas for different shapes, focusing on circular pipes.
  • Explore the implications of incompressible flow in various fluid systems.
  • Review practical applications of nozzle design in engineering contexts.
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Students in engineering or physics, fluid mechanics enthusiasts, and professionals involved in hydraulic systems or nozzle design will benefit from this discussion.

goldfish9776
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Homework Statement


how to know that the nozzle increases the velocity of water by 6 times? it's not given ... and no data for initial velocty given

Homework Equations

The Attempt at a Solution

 

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goldfish9776 said:

Homework Statement


how to know that the nozzle increases the velocity of water by 6 times? it's not given ... and no data for initial velocty given

Homework Equations

The Attempt at a Solution

Did you look at the start of the problem calculations given in Example 5-1 at the bottom of the first image?

That's the one where the hose is being used to fill a bucket of a certain size in a certain amount of time. You can work out the velocity of the flow from the information given in the example. The second image just sums up the calculations from the first image.
 
SteamKing said:
Did you look at the start of the problem calculations given in Example 5-1 at the bottom of the first image?

That's the one where the hose is being used to fill a bucket of a certain size in a certain amount of time. You can work out the velocity of the flow from the information given in the example. The second image just sums up the calculations from the first image.
yes, but i still can't understand the velocity increases by 6 times...
 
goldfish9776 said:
yes, but i still can't understand the velocity increases by 6 times...
What is the ratio between the hose diameter and the nozzle diameter?
What is the ratio of their cross-sectional areas?
Since the volume flow rate must be the same for both, what does that tell you about the ratio of the linear speeds?
 
goldfish9776 said:
yes, but i still can't understand the velocity increases by 6 times...
You've got a nozzle on the end of the hose. The water flowing through the hose is incompressible, so whatever amount goes in one end of the hose must come out the other end, in the same amount of time. You are also given the diameter of the hose and the diameter of the exit of the nozzle.

The flow rate through the hose Q = Area of the hose × velocity of the water. Since Qin = Qout and you can calculate Ain and Aout for the hose given the indicated sizes, then the only thing which remains is to calculate Vin and Vout. You are told how long it takes to fill a bucket with a certain volume of water, and this gives you Qin.
 
haruspex said:
What is the ratio between the hose diameter and the nozzle diameter?
What is the ratio of their cross-sectional areas?
Since the volume flow rate must be the same for both, what does that tell you about the ratio of the linear speeds?
2 /0.8 = 2.5
1 / (0.4^2) = 6.25
so , the ans is It's 6.25 times , not 6 times?
 
goldfish9776 said:
2 /0.8 = 2.5
1 / (0.4^2) = 6.25
so , the ans is It's 6.25 times , not 6 times?
Yes. I don't think they intended the 6 to be exact.
 
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