Fluid dynamics: head losses and pressure

[AwfulPhysicist]

see I worked through 3a) and I was comparing answers, I saw that they omitted v2 (5m/s) their explanation was because it was "defined in the question" I'm not sure what that means 3)b I tried understanding what's going on but I am clueless unfortunately, how exactly did they get 12.1m ? Where is it from!

All equations and values are attached

 

[Astronuc]

see I worked through 3a) and I was comparing answers, I saw that they omitted v2 (5m/s) their explanation was because it was "defined in the question" I'm not sure what that means

"Defined in the question" means "as given in the problem statement", which in this case is the statement about "Neglect local losses and ignore the forces caused by momentum change."

3)b I tried understanding what's going on but I am clueless unfortunately, how exactly did they get 12.1m ? Where is it from!

All equations and values are attached

There is an energy loss and the equivalent 'head' would be given by equating that energy loss = ρgH. Note the calculation associated with the 12.1 m.

 

[Chestermiller]

As a fluid mechanics guy, I would never have done this problem this way. I would start out by getting the average fluid velocity in the needle:

$$v=18\frac{(5)^2}{(0.3)^2}=5000 mm/s =5 m/s$$
shear rate γ at needle wall = $8\frac{v}{d}=\frac{(8)(5000)}{0.3}=133333 s^{-1}$

shear stress τ at needle wall = μγ = (133333)(0.980x10^-3)=131 Pa

pressure drop Δp = $4\frac{L}{d}τ=4\frac{60}{0.3}131=104500Pa=104.5kPa$

pressure head $h = \frac{Δp}{ρg}=\frac{104500}{(800)(9.8)}=13.3 m$

For part b, I would have done the calculation in reverse, starting with the pressure drop, and working backwards to the flow rate.

Chet