Calculating Initial Velocity of Water After Nozzle Increase

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The discussion centers on understanding how a nozzle increases the velocity of water, specifically questioning the claim that it increases by six times. Participants reference Example 5-1 to derive the velocity from the flow rate and the dimensions of the hose and nozzle. They highlight that the flow rate must remain constant, leading to a relationship between the cross-sectional areas and velocities of the hose and nozzle. Calculations reveal that the velocity actually increases by approximately 6.25 times rather than exactly six times. The conclusion suggests that the original figure of six was not intended to be precise.
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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