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Homework Help: Thermodynamics flow control?

  1. May 23, 2015 #1
    I have this problem and have been pulling whats left of my hair out:
    The Wivenhoe hydroelectric power station requires water to be pumped from Wivenhoe Dam to
    Splityard Creek Dam, using two pumps (each consuming 250 MW of power at 98% efficiency) to
    elevate the water approximately 70 m. At full power, Splityard Creek Dam can be filled from empty
    to its capacity of 28 700 ML in 14 hours. The water being supplied is stationary at 4 atm and 10°C; it
    is released at 1 atm through a pipe of 20 m diameter.
    Assume that there will be no net heat transfer to the water. What is the maximum
    temperature at which the water will exit?

    Values that I'm using
    State 1:
    P1 = 405.3kPa T1 = 10C; h1=42.022kj/kg
    KE1 = 0 ; PE1 = 0

    State 2:
    P2 = 101.325kPa

    Ive got a volume flow rate of 569.4x103 L/s hence a mass flow rate of 569.4x103 kg/s

    and doing some algebra from Mass flow rate = (density)*(Velocity)*(Cross-sectional Area) v = 1.8m/s
    and A = 314.16m2
    Win = 2x(250MW x 0.98) = 490,000kW

    From the 1st Law using

    my idea is to use the 1st law to find h2 and use that with the Pressure at state 2 get the temp.

    but it give me a value of -643.71kJ/kg

    There must be something I'm missing or completely off track..

    any asssistance would be greatly appreciated
  2. jcsd
  3. May 23, 2015 #2
    The term P/p is included in the definition of enthalpy in the first law, as it is defined by h = u + P/p. So yeah, your formula for the first law should not include the P/p terms.

    EDIT: I didn't realize you were using h for internal energy instead of enthalpy; In that case, your arithmetic is probably off.
    Last edited: May 23, 2015
  4. May 23, 2015 #3
    It looks like your approach is correct, so let's see your arithmetic.

  5. May 23, 2015 #4
    Net heat transfer to the water is 0; delta Q = 0

    -Wout = 0; Win = 2x(250MW x 0.98) = 490000kW

    m = 569444kg/s

    h1 = hf@10C = 42.022kJ/kg; h2=???

    ke2 = ½ v^2 = 1/2x(1.8m/s)^2 = 1.6m/s^2; ke1 = 0

    pe2 = gz2 = 9.81m/s^2 x 70m = 686.7; pe1 = 0

    Qin – Qout + Win – Wout = m(h2 – h1 + ke2 – ke1 + pe2 – pe1)

    Win = m(h2 – h1 + ke2 + pe2)

    Win/m = h2 – h1 + ke2 + pe2

    h2 = (Win/m) +h1 – ke2 – pe2

    h2 = (490000kJ/s / 569444kg/s) + 42.022kJ/kg - 1.6m/s^2 – 687.7m/s^2

    h2 = -646 kJ/kg
  6. May 23, 2015 #5
    2 things:
    - Again, the value of 42.022 kJ/kg is only the internal energy, not the enthalpy, which is what you need (although the error is quite small)
    - Your equation is not dimensionally consistent, you are subtracting J/kg from kJ/kg. This is the primary reason its failing.
  7. May 23, 2015 #6
    I'm getting the 42.022kJ/kg from the Sat. liquid hf@10C from table A-4

    I'm completely lost with your second point sorry - I think I've been looking this for so long that my pea brain had turned to pea mush

    Attached Files:

  8. May 23, 2015 #7
    or was i meant to have the ke2 & pe2 both be multiplied by (1kJ/kg / 1000m^2/s^2)
  9. May 23, 2015 #8
    The enthalpy h1 should be about 42.4 kJ/kg for water at p = 405.3 kPa and T = 10C. You can either look it up for water at 10C and 405.3 kPa, or calculate it from the internal energy as:

    h1 = u1 + P1/p1 = 42 [kJ/kg] + 405.3 [kPa] /1000 [kg/m^3] = 42.4 kJ/kg (this is just an approximation, but its consistent with an online calculator).

    For my second point, its very important that you make sure the units match:

    1 km + 1 m is not 2 km, but 1.001 km. Similarly:

    The first two terms are in kJ/kg, but the last 2 are in J/kg. Therefore, you should divide the last 2 terms by 1000 to convert them to kJ/kg:

    h2 = (490000kJ/s / 569444kg/s) + 42.4kJ/kg - (1.6/1000)[kJ/kg] – (687.7/1000)[kJ/kg]
  10. May 23, 2015 #9
    Thanks mate, I really appreciate that!!:smile:

    That definitely makes sense now. Its the little things that make a big difference!:cool:
  11. May 23, 2015 #10
    Also, I think you're using the wrong table for water. You are using the table for saturated water. In that condition, the thermodynamic properties of water are determined by either T and P, T and P are dependent in that state. However, the state the water is in here is not saturated water, so you have to use the correct table.
  12. May 23, 2015 #11
    Thanks,I'll go back and hit the books harder to get my basics up to scratch

    I'm studying Thermodynamics by correspondence - its a difficult subject as it is, but without the face to face interaction it makes it twice as hard.

    Thanks again for taking your time to help me out
  13. May 23, 2015 #12
    Thanks for pointing this out. I missed that in my post # 3.

  14. May 23, 2015 #13
    The steam tables I have only give values for saturation and superheating. However, the relationship you gave in post #8 can be accurately used, as long as you're not too close to the critical point.

  15. May 24, 2015 #14
    This problem could have been done so much more easily, without the use of the steam tables, simply by writing:

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