1. The problem statement, all variables and given/known data 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 2. Relevant equations Q2.Data: a) 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 3. The attempt at a solution Q2 Solved i)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.