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  1. #1
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    Pressure Enthalpy



    This is a question on one of my L3 papers.
    A 3kw heat pump refrigeration system has a 2.5 cop in cooling. What is it's heating duty?

    im not looking for someone to give me the answer, believe it or not i do actually want to learn.
    just hoping someone can give me some direction on the formula and how to plot it on the charts.
    Thanks in advance



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    Re: Pressure Enthalpy

    Have you got any practice in doing the charts?

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    Re: Pressure Enthalpy

    a little, i can interprate the various values and get the relevant information from the diagrams

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    Re: Pressure Enthalpy

    Quote Originally Posted by james10 View Post
    This is a question on one of my L3 papers.
    A 3kw heat pump refrigeration system has a 2.5 cop in cooling. What is it's heating duty?

    im not looking for someone to give me the answer, believe it or not i do actually want to learn.
    just hoping someone can give me some direction on the formula and how to plot it on the charts.
    Thanks in advance
    COP is just a ratio of Work Done (WD) to Refrigeration Effect (RE) or Total Heat of Rejection (THR).

    So the question is dealing with a heat pump so THR / WD would be the formular.
    If RE of a heat pump is 3kw and COP is 2.5 what is the WD?

    (spoiler answer below)

    Rob
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    Where does the heat come from in the condenser?
    It comes from the heat absorbed in the evap and then is added to by the comp.
    So whatever the COP of refrigeration is COP of heating is +1

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    Last edited by Rob White; 05-07-2014 at 11:58 PM.
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  5. #5
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    Re: Pressure Enthalpy

    Quote Originally Posted by Rob White View Post
    COP is just a ratio of Work Done (WD) to Refrigeration Effect (RE) or Total Heat of Rejection (THR).

    So the question is dealing with a heat pump so THR / WD would be the formular.
    If RE of a heat pump is 3kw and COP is 2.5 what is the WD?

    (spoiler answer below)

    Rob
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    Where does the heat come from in the condenser?
    It comes from the heat absorbed in the evap and then is added to by the comp.
    So whatever the COP of refrigeration is COP of heating is +1

    .


    .
    Hi Rob, i already knew COP heat = COP cool+1
    I need to calculate the heating duty, in Kw I presume. i guess its too simple to assume that a "3kw " unit will be 3kw.

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    Re: Pressure Enthalpy

    Another question on the paper
    state the volume flow rate at compressor suction,
    i know volume flow rate is; mass flow x specific volume
    specific volume being the isochoric axis on the chart expreesed in m3/kg, but can i for the life of me figure out the mass flow, i thought it was specifc volume at evap in minus specific volume at evap out to give the mass flow in to hte compressor suction, but apparently not, MY HEAD HURTS

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    Re: Pressure Enthalpy

    you need to are we talking real or paper as real compressor displacement and rpm come to play.

    If on Paper it is saturated gass density per btu/pound

    another way to say it, how many pounds do we need to pump to achieve the rated thermal load

    So how many pounds (kg) do we need to move to get 3kw worth of cooling
    Now in Redvers Sask.

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    Re: Pressure Enthalpy

    To work out the Total heat rejected you need to find the hoc (heat of compression) then use the COPh which you know to be 3.5
    The difference between genius and stupidity is that genius has its limits.

    Marc

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    Re: Pressure Enthalpy

    Enthalpy of the vapor and enthalpy of the liquid at SST and SCT. Then minus one from the other.
    That gives us the refrigerating effect per kilo (KJ/kg)

    1kw (Kj/s) divide by the refrigerating effect.

    That gives us the mass flow rate required for a 1kw system (kg/s)
    The difference between genius and stupidity is that genius has its limits.

    Marc

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    Re: Pressure Enthalpy

    cheers fellas I will digest the info and report back

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    Re: Pressure Enthalpy

    The volume flow rate at compressor suction (of a particular compressor with particular operating speed) is swept volume of compressor and nothing else. The mass flow rate may vary with the supply liquid conditions i.e. how much sub-cooled is refrigerant before expansion devise and the suction gas super heat.

    For reciprocating compressor the actual suction volume will vary from swept volume according to volumetric efficiency of compressor. The volumetric efficiency of compressor varies with clearance volume.

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    Re: Pressure Enthalpy

    Quote Originally Posted by james10 View Post
    Hi Rob, i already knew COP heat = COP cool+1
    I need to calculate the heating duty, in Kw I presume. i guess its too simple to assume that a "3kw " unit will be 3kw.
    Some things do not need over complicating.

    If RE is 3kw and COP is 2.5

    3kw / 2.5cop = 1.2 (ish)
    WD = 1.2 (ish)
    THR = RE + WD

    It's all there but the answer.

    Rob

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    Re: Pressure Enthalpy

    Quote Originally Posted by Rob White View Post
    Some things do not need over complicating.

    If RE is 3kw and COP is 2.5

    3kw / 2.5cop = 1.2 (ish)
    WD = 1.2 (ish)
    THR = RE + WD

    It's all there but the answer.

    Rob

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    4.2kw then, top man as ever Rob

  14. #14
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    Re: Pressure Enthalpy

    Quote Originally Posted by Rob White View Post
    Some things do not need over complicating.

    If RE is 3kw and COP is 2.5

    3kw / 2.5cop = 1.2 (ish)
    WD = 1.2 (ish)
    THR = RE + WD

    It's all there but the answer.

    Rob

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    so is this the same equation the other way round ? e.g
    1. A refrigeration unit has a duty of 12kW in heating. What is its duty in cooling if the COP in heating is 3.8?
    If RE is 12kw and COP heat is 3.8
    12kw/3.8cop=3.15 WD
    THR=RE+WD 12+3.15 = 15.15KW Cooling duty

  15. #15
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    Re: Pressure Enthalpy

    Quote Originally Posted by james10 View Post
    so is this the same equation the other way round ? e.g
    1. A refrigeration unit has a duty of 12kW in heating. What is its duty in cooling if the COP in heating is 3.8?
    If RE is 12kw and COP heat is 3.8
    12kw/3.8cop=3.15 WD
    THR=RE+WD 12+3.15 = 15.15KW Cooling duty

    Just remember the RE is the power absorbed by the evaporator.
    This question asks about the Heating capacity which is heat rejected from the Condenser.

    This question is talking about heat pumps, so the 12kw is the THR.
    THR = RE + WD.
    So this time you devide 12 by 3.8 to find the WD...... 12 / 3.8 = 3.15
    and RE = THR - WD. THR = 12 - 3.15 WD, so RE = 12 - 3.15 (ish)

    Don't confuse RE and THR when talking about heatpumps.
    RE is alway heat in THR is always heat out WD is what makes up the difference
    and COP is the ratio between them all.


    Rob

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    Last edited by Rob White; 07-07-2014 at 04:33 PM.
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  16. #16
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    Re: Pressure Enthalpy

    thanks again Rob, i have to share the love though

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