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  1. #1
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    Basic questions from newbie



    Hello
    First of all,excuse my bad english
    Since some time i'm interested in refrigeration and air conditioning.Now i'm reading some books about that,but i'm an electrician by education and by occupation,so many aspects of AC/Refrig. are distant to me.That's why many questions arises in my head while i'm reading,but i can't find answers to all of them.That's why i need to ask someone which have knowledge/experience .To prove my understandings too.It's hard to me to formulate my questions without numbers,so i will take a real compressor data for example calculations.
    For example i'll take small split type air conditioner 18000BTU.Assume it is using this compressor - QXC-30K.
    My first question is about the needed airflow through the evaporator (assume we are heating the room,so the outdoor unit is the evaporator).The compressor capacity is 18000 BTU,it's displacement is 30.7cm3 and it is running at 2900rpm with R407C refrigerant.
    If i am correct the refrigerant flow through the evaporator for 1 minute is
    Code:
    Rfm=2900*30.7=89030cm3 or 0.089m3
    than the flow for 1 hour is
    Code:
    Rfh=0.089*60=5.34m3
    Assuming the outside temp is 2C and the temperature after the evaporator is 5C (arbitary numbers,i don't know if they are correct).From the properties of the R407 at 5C the density of vapor is Hello
    First of all,excuse my bad english
    Since some time i'm interested in refrigeration and air conditioning.Now i'm reading some books about that,but i'm an electrician by education and by occupation,so many aspects of AC/Refrig. are distant to me.That's why many questions arises in my head while i'm reading,but i can't find answers to all of them.That's why i need to ask someone which have knowledge/experience .To prove my understandings too.It's hard to me to formulate my questions without numbers,so i will take a real compressor data for example calculations.
    For example i'll take small split type air conditioner 18000BTU.Assume it is using this compressor - QXC-30K.
    My first question is about the needed airflow through the evaporator (assume we are heating the room,so the outdoor unit is the evaporator).The compressor capacity is 18000 BTU,it's displacement is 30.7cm3 and it is running at 2900rpm with R407C refrigerant.
    If i am correct the refrigerant flow through the evaporator for 1 minute is
    Code:
    Rfm=2900*30.7=89030cm3 or 0.089m3
    than the flow for 1 hour is
    Code:
    Rfh=0.089*60=5.34m3
    Assuming the outside temp is 2C and the temperature after the evaporator is 5C (arbitary numbers,i don't know if they are correct).From the properties of the R407 Hello
    First of all,excuse my bad english
    Since some time i'm interested in refrigeration and air conditioning.Now i'm reading some books about that,but i'm an electrician by education and by occupation,so many aspects of AC/Refrig. are distant to me.That's why many questions arises in my head while i'm reading,but i can't find answers to all of them.That's why i need to ask someone which have knowledge/experience .To prove my understandings too.It's hard to me to formulate my questions without numbers,so i will take a real compressor data for example calculations.
    For example i'll take small split type air conditioner 18000BTU.Assume it is using this compressor - QXC-30K.
    My first question is about the needed airflow through the evaporator (assume we are heating the room,so the outdoor unit is the evaporator).The compressor capacity is 18000 BTU,it's displacement is 30.7cm3 and it is running at 2900rpm with R407C refrigerant.
    If i am correct the refrigerant flow through the evaporator for 1 minute is
    Code:
    Rfm=2900*30.7=89030cm3 or 0.089m3
    than the flow for 1 hour is
    Code:
    Rfh=0.089*60=5.34m3
    Assuming the outside temp is 2C and the temperature after the evaporator is 5C (arbitary numbers,i don't know if they are correct).From the properties of the R407 at 5C the vapor density is 22.37kg/m3,so
    Code:
    Rmass=22.37*5.34=119.45kg
    of refrigerant are evaporated.
    Again at 5C the specific enthalpy of the refrigerant (its vapor state) is 416.6 kJ/kg.So to fully evaporate that amount (119.45kg) of refrigerant the energy required is
    Code:
    Hg=119.45*416.6=49 762kJ
    According this calculator
    at 2C outdoor temp and atmospheric pressure (1bar) the specific enthalpy is 275.416kJ/kg.
    So we need
    Code:
    Av=49 762/275.416=180m3
    of air.
    Does this sound right or not?
    Thanks in advance



  2. #2
    Join Date
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    Re: Basic questions from newbie

    You are going about it in the wrong way

    It might help you if you do an advanced search on the forum using the search function at the top of the page for Refrigeration 101 by member Gary
    I'm back on the Pale

  3. #3
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    Re: Basic questions from newbie

    Where is my mistake?

  4. #4
    Join Date
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    Re: Basic questions from newbie

    You need to determine your cooling load and work back from there...not pick a compressor and try to make it do the job

    Have you read the post I linked to?...it might help you
    I'm back on the Pale

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