Results 1 to 8 of 8

Math Help - Heat transfer problem,

  1. #1
    Super Member
    Joined
    Sep 2008
    Posts
    607

    Heat transfer problem,

    A heat exchanger produces steam by boiling water within vertical tubes, as shown in Figure 3.1

    i)Water is supplied at a rate of 2 kg s1 and an initial temperature of 80C. Calculate the rate of heat transfer required to produce saturated steam at 140C. What would the pressure in the top of the heat exchanger be (in kPa gauge)?

    ii)The water/steam passes through a bundle of 50 vertical tubes, each of 3 cm internal diameter, 4 cm outer diameter, and 4 m in length, made of stainless steel with thermal conductivity 19 W m1 K1. Calculate the overall resistance to heat transfer across the tube walls. Hence calculate the required temperature driving force to deliver the heat transfer rate calculated in part (i).
    I am really stuck, don't know what to do.

    Any help appreciated.
    Attached Thumbnails Attached Thumbnails Heat transfer problem,-untitled.png  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Sep 2008
    Posts
    607
    Okay I asked my teacher and he said I have to use this equation for the second part,

    thermal resistance for a single pipe =  \frac{ln(\frac{r_{o}}{r_{i}})}{2 \pi H \lambda}

     H = 4m
    so  r_{o} = 2 cm = 0.02 m

     r_{i} = 1.5cm = 0.015 m

     \lambda = 19 W m^{-1} K^{-1}

    pluggiing these numbers in I get  6.02\times 10^{-4}

    so the overall resistance, I am not sure but Can I just use the same formula as in electrical circuits, e.g  \frac{1}{R} = \frac{1}{R} + \frac{1}{R} ...

    and times that by 50, since there are 50 tubes
    so R =  \frac{1}{ 6.02 x 10^{-4}} \times 50

    Is this correct?

    I really need help, with the first part, since we are turning water into steam will I have to use Q = ml , for latent heat? or is it  Q = mc \delta T .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    1. For the first part, you'll need to need to use:

    Q = mc_{water}\theta

    That'll give you the energy required to raise the temperature of water from 80 to 100 C

    Then, use

    Q = mL_{vaporisation}

    That'll give you the energy required to boil the water.

    Q = mc_{steam}\theta

    Again to get the heat required to heat the steam at 100 to 140 C.

    Use m = 2 kg, this will give the heat required in 1 second.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    Quote Originally Posted by Tweety View Post
    Okay I asked my teacher and he said I have to use this equation for the second part,

    thermal resistance for a single pipe =  \frac{ln(\frac{r_{o}}{r_{i}})}{2 \pi H \lambda}

     H = 4m
    so  r_{o} = 2 cm = 0.02 m

     r_{i} = 1.5cm = 0.015 m

     \lambda = 19 W m^{-1} K^{-1}

    pluggiing these numbers in I get  6.02\times 10^{-4}

    so the overall resistance, I am not sure but Can I just use the same formula as in electrical circuits, e.g  \frac{1}{R} = \frac{1}{R} + \frac{1}{R} ...

    and times that by 50, since there are 50 tubes
    so R =  \frac{1}{ 6.02 x 10^{-4}} \times 50

    Is this correct?

    I really need help, with the first part, since we are turning water into steam will I have to use Q = ml , for latent heat? or is it  Q = mc \delta T .
    I don't know how to do this... but if what you wrote is correct, then this is not correct:

    R = \dfrac{1}{ 6.02 \times 10^{-4}} \times 50

    You should be having:

    \dfrac{1}{R} =  \dfrac{1}{ 6.02 \times 10^{-4}} \times 50

    So,

    R =  \dfrac{6.02 \times 10^{-4}}{50}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Sep 2008
    Posts
    607
    Quote Originally Posted by Unknown008 View Post
    1. For the first part, you'll need to need to use:

    Q = mc_{water}\theta

    That'll give you the energy required to raise the temperature of water from 80 to 100 C

    Then, use

    Q = mL_{vaporisation}

    That'll give you the energy required to boil the water.

    Q = mc_{steam}\theta

    Again to get the heat required to heat the steam at 100 to 140 C.

    Use m = 2 kg, this will give the heat required in 1 second.
    Thank you very much,

    so for the first one

     Q = 2 kg \times 4.198 KJ Kg^{-1} K^{-1} \times 100-80 C^{o}  = 167.92

    for the next one I looked up the latent heat of vaporization for water, in my steam table, at 100C and found it to be 2257 KJ/Kg

     Q = 2kg \times 2257 = 4514 KJ

     Q = 2kg \times 2.034 KJ Kg^{-1}  K^{-1} \times 140-100C^{o}  =  162.72

    Is this correct?

    So now do I just add all of my values up to get the rate of heat required to produce saturated steam at 140?

    Thanks so much
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    Yes, you're right

    For the pressure, I'm sorry but I don't know what you are supposed to be using. Personally, I would have use the ideal gas equation, but this might be seen mostly as chemistry...

    2 kg of water -> 111.1 moles of water.
    Assuming that steam behaves as an ideal gas, the pressure is given by:

    PV = nRT

    I would say that the volume is kept constant from liquid to gas, hence a volume of 2 litres, or 2 dm^3 = 0.002 m^3
    R = 8.31,
    T = 273.15 + 140 = 412.15 K
    n = 111.11

    So, we can get P.

    However, this gives an incredibly large value... 190275916.7 Pa = 190 MPa
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Sep 2008
    Posts
    607
    Thank you, On my steam table, it shows saturated vapour pressure, at 140C to be 0.3614 MPa. Also I know that  pressure_ {absolute} = Pressure_{gauge} + pressure_{barometric}

     1 MPa = 10^{6} Nm^{-2} = 10bar

    thanks, you have been a great help.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Nov 2007
    From
    Trumbull Ct
    Posts
    909
    Thanks
    27

    Heat Transfer Problem

    Hi Tweety,
    Part 1 is an enthalpy balance as shown by others. The pressure of saturated steam at 140 C is 53 psi absolute
    Part 2 requires the heating media ,hot oil . high pressure steam, or other fluids

    bjh
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: November 6th 2011, 05:50 PM
  2. Integral Calculus help for Heat Transfer Problem
    Posted in the Advanced Applied Math Forum
    Replies: 0
    Last Post: April 17th 2011, 12:40 AM
  3. Heat transfer question
    Posted in the Advanced Math Topics Forum
    Replies: 1
    Last Post: January 31st 2011, 01:46 PM
  4. Rearranging Equation (Heat Transfer)
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: November 28th 2009, 01:16 AM
  5. Heat Transfer through a lagged sphere
    Posted in the Advanced Applied Math Forum
    Replies: 2
    Last Post: June 12th 2008, 01:01 AM

Search Tags


/mathhelpforum @mathhelpforum