- #1
williamcarter
- 153
- 4
Homework Statement
Hello, I am struggling with this heat transfer and fluid flow problem.I would really appreciate if someone could have a look over what I have done on it.
I am not sure if what I have done is correct ,I would really appreciate if you could tell me if it is correct or wrong.And also I would really appreciate if you could correct what I did wrong.
Thank you in advance.
Question 2) image: http://imgur.com/aJCzNyB
Question 3) image http://imgur.com/wFl93SA
Table image: http://i.stack.imgur.com/aiO1J.jpg
Homework Equations
Q2.Data:
a)[/B]
layer of snow=x=20*10^-2 m
thermal conductivity(lambda)=0.3W/m*K
Tair(inside)=18 Degrees Celsius
Tinterface=0 Degrees Celsius
Roof area=9m^2
Polycarbonate sheet x'=25*10^-3 m
latent heat of ice=333.5 kj/kg
b)
A=100cm^2
Pin=100KPa
Pout=60KPa
u=15m/s
Q3 Data.
a)
L=30m
D=6*10^(-2)m
composite material x=2*10^(-2)m
lambda(thermal conduct)=0.4W/m*k
Ti (inside temp of insulation)=130 degrees celsius
Tair=10 degrees celsius
h=20W/m^2*k(convective heat transfer coefficient)
b)
D=10^-3m
Specific gravity=0.8=>Petrol ro(density)=800kg/m^3
Pvap=60KPa abs=>Pvap=101.3-60=41.3KPa gauge
The Attempt at a Solution
Q2 Solved
i)[/B]Q=delta T/Rtotal
delta T=(18-0)
Rtotal=1/UA=(1/hi+xsnow/lambda snow)*1/A=(1/25+0.2/03)*1/9=0.07851
=>Q=(18-0)/0.07851
=>Q=229.24W
ii)Q=? if ice melts at 1.8 kg/h
m=1.8kg/h=1.8kg/3600s=5*10^-4 kg/s
Q=m*hmelting (phase change)
Q=5*10^-4*333.5
Q=0.16675 Kw
iii)h=6W/m^2*k (convective heat transfer coefficient)
Toutside air=?
We know Qconv=hA*delta T or q=h*Delta T
Struggle on this one.
I stopped here , I am confused how to solve this one.
b)U bend pipe, force to keep in place
A=ct=100cm^2=0.01m^2
Pin=100Kpa
Pout=60Kpa
u=15m/s
F=?to hold pipe in place
Fm(momentum force)=m*u=ro*q*u=ro*A*u^2
Fmx=m*u=m*(u2-u1) if done on fluid
Fmx=m*u=m*(u1-u2) if done by fluid
Fmx=ro*A*u^2=10^3*0.01*15^2
Fmx=2250N(in X direction)
Fmy= - 2250N n(in Y direction)
Fpx=P1A1-P2A2
FPx=10^5Pa*10^-2 m^2
FPx=10^3 N(in X direction)
Fpy=-10^3 N(in Y direction)
Fx=Fmx+FPx=3250N
Fy=Fmy+Fpy=-3250N
Magnitude of force =sqrt(Fx^2+Fy^2)=sqrt(3250^2+3250^2)=4956N
Force to keep pipe in place=reaction force equal in magnitude but opposite in direction
=> Force to keep pipe in place Fr= -4956 N
Q3.solved
a)i) q(heat flux)=?
q=U*delta T
1/U=1/h+x/lambda
=>1/U=1/20+2*10^-2/0.4=>1/U=0.1=>U=10
q=U*delta T=10*(130-10)
q=1200 W/m^2*k
ii)if x'=2*x
=>q'=U' *delta T
1/U'=1/h+2x/lambda=1/20+4*10^-2/0.4=>U=6.6
=>q=800W/m^2
iii)increasing insulation layer is also increasing safety level by reducing q(heat flux)
b)
Vol=2L Petrol=2*10^-3 m^3
D=10^-3m
=>A=7.85*10^-5 m^2
Specific gravity=0.8
=>Petrol ro(density)=800kg/m^3
Pvap=60KPa abs
=>Pvap=101.3-60=41.3KPa gauge
We apply Bernoulli equation P1/ro*g+h1+u1^2/2*g=P2/ro*g+h2+u2^2/2*g
Taking points 1 and 2 as below
http://imgur.com/cDR2v51
Point 1 on the bottom of the tank
Point 2 discharged in atmosphere from siphon
time t=Volume/Volumetric flowrate
for point 1:
u1=0(we can neglect velocity due to difference in Area between point1 and 2)
h1=1m
P1=41.3*10^3Pa
for point2:
u2=?we will find it
P2=0(due to atmosphere)
h2=0(datum point)
Plug all those values in main Bernoulli equation
=> u2^2=(P1/ro*g+h1)*2g
u2=sqrt( (P1/ro*g +h1)*2g
u2=sqrt( (41.3*10^3/800*9.81+1)*2*9.81
=>u2=11.08 m/s
we know q(volumetric flowrate)= u*A
=>q=u2*A2(area of siphon)
q=11.08 m/s*7.85*10^-5 m^2
q=8.70*10^-4 m^3/s
but time t=V/q
t=2*10^-3 m^3/8.70*10^-4 m^3/s
=>time=2.29 seconds
ii) find maximum allowable h in order to maintain flow.
I struggle here.
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