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Homework Help: Fluid Flow/Heat Transfer question struggle

  1. Jul 31, 2016 #1
    1. The problem statement, all variables and given/known data
    I would really appreciate if someone could help me out with this exercise in fluid flow/heat transfer ,because I really struggling with it.Below I am attaching photo of problem+table with values. Thank you in advance !

    2. Relevant equations
    Question link:

    Table link:

    3. The attempt at a solution
    Struggling with it.
  2. jcsd
  3. Jul 31, 2016 #2


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    Do you know the density of water? Are you familiar with the continuity equation?
    Q =vA
    What do you know about Reynolds number?
  4. Jul 31, 2016 #3
    Hello, thank you for your quick reply.
    I know Reynolds number has 2 formulas
    #1 Re=Diameter*velocity*density/viscosity=D*u*ro/Mew
    #2Re=4*mass flowrate/Pi*Diameter*Viscosity=4*m/pi*D*Mew

    Reynolds number is dimensionless and can show us 3 types of flow:Laminar(Re<2100),Turbulent(2100<Re<10000) and Ideal(frictionless)

    Continuity eq:mass flowrate m=Density1*Velocity1*Area1=Density2*Velocity2*Area2
    For incompressible flow such as liquids and gasses at low Press, the density is the same, hence q which is vol flowrate has the formula

    Density is 1/specific volume, we have all the values in the above steam table , but specific volume is in mg/m^3 hence density will be 10^3/specific volume.
    First column in each field of the table is for liquid and 2nd for vapour.
  5. Jul 31, 2016 #4


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    Sounds like you have some understanding. Where exactly are you stuck? Can you find the density of water at point 1 and 2?
    Can you apply the continuity equation to find the velocities?
  6. Jul 31, 2016 #5
    Thank you again for your reply , really appreciated.
    we know mass flowrate m=720kg/h.However we need to convert it to kg/s
    The mass flowrate is constant m in =m out , no accumulation term,is at steady flow.
    Thus 720kg/h=720kg/3600s=0.2kg/s.

    At point 1 at T=10 deg Celsius we have :
    Density1=10^3/1.000 =1000kg/m^3
    Viscosity Mew1=1304*10^(-6)
    From continuity u1=mass flowrate/density1*Area1=0.2/10^3*0.01
    We have A1 ,we need D1;
    Hence D1=0.1128 m
    so Re=D*u*ro/mew

    So at point 1 we calculated:
    u1=0.02m/s and Re1=1730.06

    Same for point 2 at outlet where T2=80 degrees Celsius
  7. Jul 31, 2016 #6


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    That looks correct, good work.
    And what about the duty? Are you familiar with this equation?
    Q =m_dot * Cp * deltaT
  8. Jul 31, 2016 #7
    Thank you for your reply, I am glad I answered correctly.

    As for c)with Heat Duty required we are supposed to use the equation Q=mass flowrate*specific heat*delta T
    in our case heat duty transfer rate Q=m*c*(Tout-Tin)

    We will take the cp as the cp of (10+80) degrees /2 so we will look in table for cp of 45 degrees celsius
    As the value for cp is not there , and it is for values such as 44 and 46 degrees celsius we will have to linear interpolate.

    Hence cp at 45 deg Celsius=4179.5(I could have estimated it as 4180, but I wanted to be more precise)
    Furthermore it is not required to transform Celsius to Kelvin because the difference(delta T) would be the same , so I would leave it in Celsius.

    so we have everything as it follows:
    m=0.2 kg/s
    cp=4179.5 J/Kg*K
    Tout=T2=80 degrees Celsius
    Tin=T1=10 degrees Celsius

    We know Q=m*c*(Tout-Tin)
    hence Q=0.2*4179.5*(80-10)
    Q=58513 J (Which is the heat duty required to raise water from 10 degrees Celsius to 80 degrees Celsius)
  9. Jul 31, 2016 #8


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    That's right, except the 'per second' from the mass flow rate went missing. the answer should be joules per second (aka Watts).
    Any questions about the next questions?
  10. Jul 31, 2016 #9
    Thank you for your answer, that is correct it is in Watts(J/s), I do not know why I forgot about seconds,I did this probably because I was thinking of Heat which is in Joule and not Power which is Watt, and I forgot about seconds from the mass flowrate.
    As for d) I cannot completely understand what I am supposed to do
    Thank you in advance.
  11. Jul 31, 2016 #10
    Your steam tables give values for the enthalpy of liquid water at 80 C and 10 C. Why didn't you just use this?
  12. Jul 31, 2016 #11
    Hello, and thank you for your answer.
    I did the arithmetic mean between the 2 temperatures in order to get the mean cp (in my case at 45 degrees celsius) and use it in Q=m*c*delta T to get the duty as a power in Watts.The reason I did that was because I wanted to be consistent with the formula I was given in the course, however your path seems easier, cheers for that.

    I could also ,as you said use the enthalpies and calculate the Q as following:
    Q=m_dot*[(enthalpy h at 80 degrees Celsius of water)-(enthalpy h at 10 degrees Celsius of water)]

    This will be:Q=0.2kg/s*(335.0*10^3 J/Kg -42.03*10^3 J/Kg)
    =>Q=58594 Watt
  13. Jul 31, 2016 #12
    For part (d), what is the change in enthalpy per kg in going from saturated steam at 130C to saturated liquid water at 130C?
  14. Jul 31, 2016 #13
    Thank you for your reply!
    For part d)
    I would say Qdot on water side= Qdot on steam side.
    Q on water side was calculated at c) and it is 58594 Watt
    Q on steam side is : Q=m*lat heat ;but Qwater=Qsteam=58594(J/s)=58594 Watt
    and mdot=Qdot/latent heat
    From tables latent heat of steam at 130 degrees is 2174 kJ / Kg
    =>mdot=58594 Watt /2174*10^3 J/Kg
    mdot=0.0269 kg/s
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