- #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)

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)=?

a)i)

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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