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Cold weather performance of air sourced heat pumps

Any heat pump experts on here? I’m looking at air sourced heat pump performance for a friend’s installation. They have NIBE F2040’s for the house, and Waterco Ultra 31’s for pool. Both use same R410A refrigerant, but that seem to be perhaps the only thing in common. NIBE (made in Sweden) twin rotary variable speed compressors are inverter controlled as standard (I guess vector or torque type drives), whilst the Waterco units are direct on line, capacitor start induction motors (I thought that technology went out long ago), with no soft start options.
 I’m interested in COP performance in the colder weather, the Ultra 31’s are specified as generally for use in ambient greater than 10 degrees C, although they do rate this particular model down to zero, BUT do not provide COP performance below 5 ambient, but enquiries suggest a COP of around 1 (perhaps even lower, ie, less power out than in!). The NIBE manufactures claim very much better, performance down to at least -25 ambient, with COP of around 2 even at this low temperature (is that even possible?).
Anyway, any thoughts on the subject, much appreciated.

Parents
  • Good morning John,
    I'm not giving up on you. I think you know,
    The world is built on mathematical and physical calculations so you need to know and learn (I am also constantly learning new things).
    Please refer to my table and calculations (I also cited an example). I'm sure it will help you understand and know how to heat water in the pool and which water heater pump will regulate.
    I am at your service with incomprehensible or have any issues.
    Good luck....
    How to quickly calculate the amount of time needed to heat water at different power supplies
    The specific heat of water is 186.4 Joules per gram of Celsius
    The formula on which to base is:
    Q = c * m * (dT)
    (J / g K) Heat-Purple-c
    m - the mass to be heated (g (
    dT - The change in temperature in the Kelvin degrees
    Q - The energy required for the process (J (
    For example: to heat 200 liters of water from 45 ° C to 90 ° C, we use the formula: dT (cm = Q (
    Q = 4.186 J / g K * 200,000g * 45K = 37,674,000J
    Energy conversion (Q) power: 1 watt = 1 joule per second
    Therefore, if we want to calculate how long this process will take, consider this:
    For the calculation, let's say we have a power heater (P) 2000 watts.
    t (s) = Q / P
    so:
    t = 37,674,000J / 2000 (J / s) = 18,837s = 313.95min = 5.23h = 05: 14: 00
    This means that it takes about 5 hours to heat 200 liters of water from 45 ° C to 90 ° C at 2000 watts.
    Here are some examples of the amount of time needed to heat different volumes of water at different power levels:
    Assumptions: Optimal conditions, zero heat loss to the environment, full conversion of heater voltage to heat.

    Attached is a table,

    If you increase the power by two times - the heating time will be doubled (for the same volume), and so on.


Reply
  • Good morning John,
    I'm not giving up on you. I think you know,
    The world is built on mathematical and physical calculations so you need to know and learn (I am also constantly learning new things).
    Please refer to my table and calculations (I also cited an example). I'm sure it will help you understand and know how to heat water in the pool and which water heater pump will regulate.
    I am at your service with incomprehensible or have any issues.
    Good luck....
    How to quickly calculate the amount of time needed to heat water at different power supplies
    The specific heat of water is 186.4 Joules per gram of Celsius
    The formula on which to base is:
    Q = c * m * (dT)
    (J / g K) Heat-Purple-c
    m - the mass to be heated (g (
    dT - The change in temperature in the Kelvin degrees
    Q - The energy required for the process (J (
    For example: to heat 200 liters of water from 45 ° C to 90 ° C, we use the formula: dT (cm = Q (
    Q = 4.186 J / g K * 200,000g * 45K = 37,674,000J
    Energy conversion (Q) power: 1 watt = 1 joule per second
    Therefore, if we want to calculate how long this process will take, consider this:
    For the calculation, let's say we have a power heater (P) 2000 watts.
    t (s) = Q / P
    so:
    t = 37,674,000J / 2000 (J / s) = 18,837s = 313.95min = 5.23h = 05: 14: 00
    This means that it takes about 5 hours to heat 200 liters of water from 45 ° C to 90 ° C at 2000 watts.
    Here are some examples of the amount of time needed to heat different volumes of water at different power levels:
    Assumptions: Optimal conditions, zero heat loss to the environment, full conversion of heater voltage to heat.

    Attached is a table,

    If you increase the power by two times - the heating time will be doubled (for the same volume), and so on.


Children
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