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Heat transfer and Fluid Flow Struggling Problems

  1. Aug 3, 2016 #1
    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.
     
    Last edited: Aug 3, 2016
  2. jcsd
  3. Aug 3, 2016 #2
    You showed that the heat flow through the roof layer is 229 W and that 167 W of this is used to bring about the phase change at the interface between the snow and the roof. So you are left with 62 W for the rate of heat flow from 0 C at the snow interface with the roof to the outside air. What outside air temperature is required to drive this heat flow?
    You kinda had the right idea here, but not quite. If you regard the total pipe bend as your control volume, there is no component of force in the y direction. For the x direction, if F is the force exerted by the bend on the fluid in the negative x direction, then $$P_{in}A-P_{out}A-F=-2\rho A v^2$$ where the right hand side is the rate of change in fluid momentum in the positive x direction.
    This problem is not done correctly because you haven't taken into account the cylindrical geometry. Within the insulation, the heat conduction equation should be $$q=-k\frac{dT}{dr}$$, and the heat balance equation should be $$\frac{d(qr)}{dr}=0$$ or equivalently$$qr=const$$ These equations need to be solved simultaneously to get the temperature profile and the heat flux. Then, this solution has to be properly merged with the convective heat transfer.

    In part iii, you should be comparing the ratio of the conductive resistance to the convective resistance.
    Bernoulli's equation is not applied correctly here. The vapor pressure of Petrol has nothing to do with the pressure at point 1. The pressure at point 1 is determined from the hydrostatic equation within the tank. Or you could apply the Bernoulli equation to the combination of a point at the very top of the tank (where the pressure is atmospheric) and point 2, where the pressure is again atmospheric. Either way, you will get the same answer.

    To answer part ii, you need to consider the absolute pressure at location h (at the top of the siphon) and compare this with the vapor pressure of the Petrol. If the pressure at this location is calculated to be less than the vapor pressure of the petrol, you will get cavitation (and the siphon will not work).
     
  4. Aug 3, 2016 #3
    Please, in the future, submit problems is separate threads. You have actually submitted 4 problems within this thread. This is not acceptable.
     
  5. Aug 4, 2016 #4
    for Question2) iii)
    I did delta Q=Q1-Q2
    where Q1=229.2 W from above subpoints of exercise
    Q2=166.75 W rate of heat transfer to melt ice
    => delta Q=62.49 W
    Q=h*A*delta T(convection) => delta T=Q/h*A =>(Tsnow-Toutside)=Q/hA
    =>To=Tsnow-Q/hA
    To=0-62.49/6*9
    To(outside air temp)= - 1.15 degrees Celsius

    for Question2 b) U-bend pipe
    F=force to keep pipe in place=Fm(momentum force)+Fp(pressure force)
    Fm=ro*A*u^2=10^3*10^-2*15^2=2250N
    Fp=P1A1-P2A2=(10^5Pa*10^-2)-(6*10^4Pa*10^-2)
    Fp=400N

    =>F=Fm+Fp
    F=2250+400
    F=2650 N( force to keep pipe in place)

    For question 3)
    i) I tried using Laplace equation and by integrating and putting boundary conditions I got:
    Temp profile T(x)=(T2-T1)*x/L+T1
    =>T(x)=(10-130)*2*10^-2/30 +130
    T(x)=129.92

    q(x)=-lambda/L *(T2-T1)
    q(x)= -0.4/30 *(10-130)
    q(x)=1.6 W/m^2 (heat flux)
    Now I do not know how to connect this to heat flux from convection q=h*delta T

    For question 3b)
    I do not know how to apply hydrostatic pressure
    for point 1(let's say is on the bottom of the tank) P1=ro*g*h1=800*9.81*1m=7848Pa ,if at point 2 h2=0(datum)

    However for point 3 let's say somewhere in fluid above, at 4m P3=ro*g*h3=800*9.81*4m
    P3=31392 Pa, which is incorrect because the deeper we go , the pressure should be bigger but in this case is opposite.

    I applied Bernoulli as above but I took the point 1 as at the top of the fluid of the tank.
    So for point 1: u1=0,P1=0, h1=4m;
    For point 2:u2=? P1=0; h2=0m(datum)
    => u2=sqrt(2*g*h1)
    u2=sqrt(2*9.81*4)
    u2=8.86 m/s
    Area of siphon=pi*D^2/4=pi*(10^-2)^2/4
    Asiphon=7.85*10^-5 m^2
    q=u2*Asiphon=8.86m/s*7.85*10^-5 m^2
    q=6.95*10^-4 m^3/s
    t(time)=V/q=2*10^-3 m^3/ 6.95*10^-4 m^3/s
    t=2.87 seconds

    For question 3 last part with maximum height
    I will calculate the hydrostatic Pressure in point 3 which is at top of siphon
    P=ro*g*h for Gauge Pressure
    P=Patm+ro*g*h to make Absolute Pressure
    =>P3=Patm + ro*g*h3=800 kg/m^3 *9.81 m/s^2 *4m
    P3=31392 Pa gauge
    P3=69933Pa=69.9 Kpa absolute > 60KPa vap absolute
    We apply again Bernoulli eq:
    for point 3: P3=31392 Pa gauge ; u3=0;h3=?
    for point 2: P2=0 ; u2=8.86 m/s (from above) h2=0;
    h3=u2^2/2*g -P3/ro*g
    h3=8.86^2/2*9.81 - 31392 /800*9.81
    h3=9.98*10^ -4 meters
     
    Last edited: Aug 4, 2016
  6. Aug 4, 2016 #5
    You omitted the (conductive) temperature difference across the layer of snow.
    I didn't check this result. Is it consistent with the equation I gave for the force exerted by the bend on the fluid in my previous post?
    You need to use the Laplace equation in cylindrical coordinates. Using the equation in cartesian coordinates is not correct.
    The two resistances are in series. $$(\Delta T)_A=\frac{Q}{R_A}$$$$(\Delta T)_B=\frac{Q}{R_B}$$
    $$(\Delta T)_{total}=(\Delta T)_A+(\Delta T)_B=\frac{Q}{R_A}+\frac{Q}{R_B}$$
    Then solve for Q.
    I don't understand what you did here.
    This is correct.
    This equation should have a minus sign.
    This equation should also have a minus sign:
    P=Patm-ro*g*h=100 kPa-ro*g*h
    In order to avoid cavitation, this pressure must be greater than the equilibrium vapor pressure of the liquid.

    EDIT: Actually, after further consideration, I realize that I left out a term in the equation for the pressure at the top of the bend. The correct application of the Bernoulli equation should be:
    $$P+\rho g h+\rho \frac{v^2}{2}=P_{atm}$$
    Therefore, $$P=P_{atm}-\rho g h-\rho \frac{v^2}{2}$$
     
    Last edited: Aug 4, 2016
  7. Aug 5, 2016 #6
    What exactly do you mean that I omitted conductive temp difference, and where exactly am I supposed to add that?
     
  8. Aug 5, 2016 #7
    There is a resistance to heat transfer through the layer of snow.
     
  9. Aug 5, 2016 #8
    Thank you for quick reply, but where am I supposed to plug in that temp difference of conduction?
     
  10. Aug 5, 2016 #9
    $$\frac{62.49}{9}=0.3\frac{(0-T)}{0.2}$$where T is the temperature at the outer surface of the snow.
     
  11. Aug 16, 2016 #10
    I know the first one is from Fourier for conduction
    I would really appreciate it if you could give me some more hints, I am stuck on this one.
    I knew that d/dr*(r*dT/dr)=0
     
  12. Aug 16, 2016 #11
    If Q is the total rate of heat flow through the insulation (in W), for a cylindrical surface of length L and arbitrary radial location r within the insulation, what is the heat flux at this radial location (in terms of Q, L, and r, but not T)?
     
  13. Aug 16, 2016 #12
    Q=q*A
    Q=q*2pi*r*L
    =>q=Q/2*pi*r*L
     
  14. Aug 16, 2016 #13
    Good. Now, in terms of the thermal conductivity k and the temperature gradient dT/dr, what is the heat flux q?
     
  15. Aug 16, 2016 #14
    Thank you for your answer.

    q= -k*dT/dr from Fourier's law of conduction
     
  16. Aug 16, 2016 #15
    You need to learn how to use LaTex. Look under INFO drop down menu and click on Help/How to, then click on LaTex Primer.
     
  17. Aug 16, 2016 #16
    Good. Now set them equal. What do you get?
     
  18. Aug 16, 2016 #17
    ##Q=-k* \frac {dT} {dr} *2\pi*r*L##
     
  19. Aug 16, 2016 #18
    Good. Now solve the ordinary linear differential equation for T as a function of r, subject to the boundary condition that ##T=T_i## at the inner boundary of the insulation, ##r=r_i##. (I feel like we've gone through all of this before).
     
  20. Aug 16, 2016 #19
    ##T(r)=-\frac Q {k*2*\pi*L}* ln|r|+c1##
     
  21. Aug 16, 2016 #20
    What happened to substituting the boundary condition?
     
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