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Thermodynamics: Control Volume flow through radiatior

  1. Jan 20, 2009 #1
    1. The problem statement, all variables and given/known data

    Air enters a car radiator at 50 km/hr, 25 C. The radiator is 1 m2 facial area. 2 kg/s of water enters the radiator from the engine at 10 bar, 200 C and leaves at 10 bar, 180 C. What is the temperature of the air as it leaves the radiator?

    2. Relevant equations

    I did an energy balance for air and then for water, but I am not sure which terms can be ignored. After i obtain these eqns, i can set them equal to each other and solve for T (from
    Q =m(c_p)(Change in T))?

    3. The attempt at a solution

    For Air: W = Q + (m)(h_in + [(v_in)^2]/2 +g*z_in - h_out - [(v_out)^2]/2 -g*z_out)

    m = mass flow rate
    Q = heat flow rate
    W = work flow rate
    h = enthalpy
    v = velocity

    I assume W, z_in, z_out are zero and also that m_in = m_out

    So i get Q = m(h_in - h_out + [(v_in)^2]/2)

    For water:

    Q = m(h_out - h_in) = (0.2)(762.8 - 2827.9) = -2065.1 kW
     
  2. jcsd
  3. Jan 21, 2009 #2

    Q_Goest

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    Hi aznkid. You've almost got it. On the right track anyway.

    The problem gives you air velocity into the radiator but not out of the radiator. From that I'd simply assume the velocity out is equal to the velocity in, so your velocity terms in your equation:
    Q = m(h_in - h_out + [(v_in)^2]/2)
    also drop out and you're left with the change in enthalpy for the air.

    Now the only part of the problem you really haven't resolved is how to determine mass flow rate of air. How do you think you can determine mass flow rate of air? Note the variables you're given:
    - Velocity
    - cross sectional area
    - Inlet temp
     
  4. Jan 22, 2009 #3
    Can i assume its an ideal gas?

    Then i can use the ideal gas eqn to get specific volume v: pv = RT, where R = 8.314/MW

    Next, m = VA/v [(velocity*area)/specific volume]?
     
  5. Jan 22, 2009 #4
    You sure can.
     
  6. Jan 22, 2009 #5

    Q_Goest

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    You can assume ideal gas in order to calculate specific volume, or just look up density for air at standard conditions.
     
  7. Jan 23, 2009 #6
    So can i assume standard air pressure of 101.325 kPa?

    Then once i find Q, i can use Q = mc(change in T) to find T?

    What would my c be?
     
  8. Jan 23, 2009 #7

    Q_Goest

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    Sure... If they don't give you atmospheric pressure, just assume it's standard conditions.

    I think you're asking whether to use Cv or Cp.

    Consider it this way.

    mCp(dT) calculates a change in enthalpy.

    mCv(dT) calculates a change in internal energy.

    Do you know which one is applicable here?
     
  9. Jan 23, 2009 #8
    Since it's constant pressure, would it be C_p?

    v = RT/p = (8.314)(298)/(101.325) = 0.8458 m^3/g = 845.8 m^3/kg

    m = (13.89 m/s)(1 m^2) / (845.8 m^3/kg) = 0.0164 kg/s

    Then Q = m*c_p*(T_2 - T_1), where Q = 2065.1 kW (from h20 calculated before)

    2065.1 kW = (0.0164 kg/s)*(1.005 kJ/kg k)*(T_2 -298K)

    T_2 = 125592.26 K

    This seems wrong
     
    Last edited: Jan 23, 2009
  10. Jan 24, 2009 #9

    Q_Goest

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    For water at 10 barg, saturation temp is 184 C which means the water is going in superheated and coming out subcooled.

    For water at 10 bara, saturation temp is 179.4 C which means the water is going in and coming out superheated.

    Let me guess... this problem was made up by a grad student, right? lol

    Let's just assume the person that created this problem meant for the water to be liquid in and out. I'll use 170 C inlet and 150 C outlet at 10 barg.. You should get an energy removal of roughly 177 kW. (or 237 hp) so the power removed his HUGE. A very large truck might approach this much heat rejection at full tilt, but that’s way more than a car.

    Also, I’m getting an air mass flow rate of 16.5 kg/s and an outlet temperature of roughly 35.7 C.

    Give it another shot.
     
  11. Jan 24, 2009 #10
    Was there a particular reason you chose 170C and 150C?
     
  12. Jan 24, 2009 #11

    Q_Goest

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    um... yea. because it's less than the saturation pressure of water at 10 barg, and because specific heat is roughly constant over the range of pressure and temperature you're looking at so I would want to maintain the 20 degree C dT.
     
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