Bernoulli equation exercise from Fanning and Moody

[williamcarter]

Homework Statement

I would really appreciate it if you could give me a hand with this exercise, not sure on what I've done.
Data:

Moody:

L=55*10^-3m
D=10^-1m
k=0.0002m

Homework Equations

$Re=\frac{D*u*ρ} {μ}$
$Re=\frac{4*m} {pi*D*μ}$
Relative roughness $ξ=\frac k D$ where k=rougness and D=diameter
f=fanning friction factor from Moddy chart as ξ ∩ Re
K=loss coefficient =$\frac{f*4L} {D}+∑Ki$
where Ki=other losses
$Δhloss=\frac{K*u^2} {2g}$
Bernoulli:$\frac {P1} {ρg}$+h1 +$\frac {u1^2} {2g}$ = $\frac {P2} {ρg}$+h2 +$\frac {u2^2} {2g} +Δhloss$

The Attempt at a Solution

i)Re=?
Q=0.05m^3/s
μ=1.2*10^-6m²/s

$Re=\frac{4*m} {pi*D*μ}$
where m=ρ*Q
m=10³*0.05
m=50 Kg/s
=>$Re=\frac{4*50} {pi*10^-1*1.2*10^-6}$
=>Re=530516477 turbulent flow

ii)f=?fanning friction
$ξ=\frac k D$ =$ξ=\frac {0.0002} {10^-1}$
ξ=2*10^-3
=> f=0.006 from ξ∩Re on Moody Chart

iii)Assumptions
A: hA=7m PA=? uA=0
B hB=0m(datum) PB=0(atomospheric) uB=0;

:$\frac {P1} {ρg}$+h1 +$\frac {u1^2} {2g}$ = $\frac {P2} {ρg}$+h2 +$\frac {u2^2} {2g} +Δhloss$

Pluggin in the assumptions this means:
$Δhloss=\frac {PA} {ρg} +hA$
=>$PA=(Δhloss-hA)*ρg$
Δhloss=K*u²/2*g
K=$\frac{f*4L} {D}+∑Ki$=$\frac{0.006*4*55*10^-3} {10^-1}+0.5+3*0.8+2*0.25+0.95$

$K=4.36$

Q=0.05 m^3/s=>$u=\frac {Q} {A}$ $u=\frac {0.05} {pi*(10^-1)^2/4}$
u=6.36m/s

$Δhloss=\frac {K*u^2} {2g}$=$Δhloss=\frac {4.36*6.36^2} {2*9.81}$

Δhloss=8.98m

$PA=(Δhloss-hA)*ρg$

$PA=(8.98-7)*10^3*9.81$

=>PA=19423.8 Pa

 

[Chestermiller]

Homework Statement

I would really appreciate it if you could give me a hand with this exercise, not sure on what I've done.
Data:
View attachment 104815
Moody:
View attachment 104816

L=55*10^-3m
D=10^-1m
k=0.0002m

Homework Equations

$Re=\frac{D*u*ρ} {μ}$
$Re=\frac{4*m} {pi*D*μ}$
Relative roughness $ξ=\frac k D$ where k=rougness and D=diameter
f=fanning friction factor from Moddy chart as ξ ∩ Re
K=loss coefficient =$\frac{f*4L} {D}+∑Ki$
where Ki=other losses
$Δhloss=\frac{K*u^2} {2g}$
Bernoulli:$\frac {P1} {ρg}$+h1 +$\frac {u1^2} {2g}$ = $\frac {P2} {ρg}$+h2 +$\frac {u2^2} {2g} +Δhloss$

The Attempt at a Solution

i)Re=?
Q=0.05m^3/s
μ=1.2*10^-6m²/s

$Re=\frac{4*m} {pi*D*μ}$
where m=ρ*Q
m=10³*0.05
m=50 Kg/s
=>$Re=\frac{4*50} {pi*10^-1*1.2*10^-6}$
=>Re=530516477 turbulent flow

The Reynolds number is calculated incorrectly. You are given the kinematic viscosity, not the dynamic viscosity.

 

[williamcarter]

The Reynolds number is calculated incorrectly. You are given the kinematic viscosity, not the dynamic viscosity.

$Re=\frac{D*u} {∂}$
where D=diameter u=velocity ∂=kinematic viscosity
u=Q/A=6.36m/s

$Re=\frac{10^{-1}*6.36} {1.2*10^{-6}}$

Re=530000=>Re=5.3*10⁵

$ξ=\frac{k} {D}=2*10^{-3}$=0.002

ξ∩Re=>f=0.006 from Moody Chart

 

[Chestermiller]

$Re=\frac{D*u} {∂}$
where D=diameter u=velocity ∂=kinematic viscosity
u=Q/A=6.36m/s

$Re=\frac{10^{-1}*6.36} {1.2*10^{-6}}$

Re=530000=>Re=5.3*10⁵

$ξ=\frac{k} {D}=2*10^{-3}$=0.002

ξ∩Re=>f=0.006 from Moody Chart

This is better. I haven't checked the rest of the analysis.

 

[williamcarter]

This is better. I haven't checked the rest of the analysis.

Thank you for your reply.
I would really appreciate it , if you can and have time , if you could check the rest of the exercise, as I am not sure if it is correct.
In both cases the f=0.006

 

[Chestermiller]

Thank you for your reply.
I would really appreciate it , if you can and have time , if you could check the rest of the exercise, as I am not sure if it is correct.
In both cases the f=0.006

It looks OK to me.