Velocity of a water jet exiting a water tank drain tube

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The discussion revolves around calculating the velocity of water exiting a tank drain tube, initially approached through pressure and impulse-momentum methods. The original poster (OP) faced challenges in their calculations and seeks assistance. Clarification is needed regarding the term "initial velocity," as it may refer to the exit point rather than the start of flow. The complexity of the problem increases without knowledge of Bernoulli's principle, suggesting that a more structured approach may be necessary. Overall, the OP's question lacks detail, making it difficult for others to provide effective help.
WhiteyM
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Homework Statement
Is given: h, H0, d (diameter of the tube), D (diameter of the cylinder), l (length of the tube), L (length of the water jet), ρ.

a) What is the initial velocity of the water jet exiting the tube based on time? (formula)
b) What is the length of the water jet exiting the tube based on time? (formula)

[CHECK ATTACHMENT]
Relevant Equations
p=F/S
p=ρgh
I originally thought about working out the force F with which it pushes the water thru the tube using pressure, and after using impulse and momentum to get the initial velocity, but sadly it wasn't successful. If anyone can help, please.
 

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WhiteyM said:
I originally thought about working out the force F with which it pushes the water thru the tube using pressure, and after using impulse and momentum to get the initial velocity, but sadly it wasn't successful. If anyone can help, please.
Do you expect us to guess in what way it was unsuccessful? Please post your attempt, as required by the forum rules.
 
The question did not make much sense to me until I realised that "initial" might not mean "when the flow starts". Rather, it might mean the velocity at the point of exit, rather than elsewhere in the stream, but at any time t after the flow starts.
 
If one is working from first principles without knowing Bernoulli's principle, this could be a daunting challenge.
 
The question seems a little open in its assumptions and a little closed in reply’s from the OP…
 
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My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...