- #1
williamcarter
- 153
- 4
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
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