A Warm WinterA Warm Winter

Q & A: How do you calculate the thermal efficiency of a heat transfer process?

Question by fkneg1: How do you calculate the thermal efficiency of a heat transfer process?
I need to find out how to calculate the thermal efficiency of a heat transfer process using the flow rate, temperature change and input power.
I know that the thermal efficiency is the power you put in divided by the power you get out but can’t find out how to wokr this from the values I’ve got.

Best answer:

Answer by Skywave
Quote: “I know that the thermal efficiency is the power you put in divided by the power you get out but can’t find out how to work this from the values I’ve got”.
In that case, it seems as if you have a simple problem of dividing two numbers. Perhaps we need a clearer description of your problem.

Powered by Yahoo! Answers

Know better? Leave your own answer in the comments!

2 Responses to Q & A: How do you calculate the thermal efficiency of a heat transfer process?

  1. Crude Oil Boiler

    In a cooling process, such as an air refrigeration unit in a home or office building:

    Heat is pulled out of the air at a certain rate
    Electricity is used to drive the compressor and fans in the evaporator and condenser

    Total up the power used to drive the process and put it into kW*hr
    Total up the total heat per time removed from the air, convert to kW*hr

    Divide the energy removed from the air by the energy used to run the process;
    This equals your true efficiency

    In an air heating process, simply change the sign; take the heat input into the air to heat it
    Divide that energy gained by the energy used to drive the process; = true efficiency

    In a heat exchanger: first, solve for the temperature driving force by calculating the LMTD
    This is the log mean temperature drive; look this up on the Googles

    Next calculate either your hot side heat transfer or your cold side heat transfer; if dealing with a phase change, it is usually easier to calculate the duty on the side that doesn’t change phase.

    I.E. using cooling water to condense naphtha vapor; naphtha vapor cools, then condenses, then sub cools; the cooling water (cold side) simply heats up with no phase change

    Non phase change duty = Q = mass * heat capacity * change in temperature

    Next either look up or calculate the total exchanger area in square feet

    Now you can calculate the overall heat transfer coefficient; this can be viewed as the ‘rate at which heat is transferred per area per degree driving force’ or simply, how well or efficiently are we transferring heat?

    OHTC = U = duty / area / LMTD = Q / (A*delta T) units are btu / hour / ft^2 / degree F

    You can look up heat transfer coefficient in the Googles also.

    In a steam boiler, there are two methods:

    Calculate the amount of excess air in the stack, then assume that that excess air (above the amount needed for total combustion of the fuel) must be heated from ambient to the temperature of the exiting gases out of the stack.

    Take this total heat rate, then subtract it from the total fired duty from the heating value of the fuel. Now: divide this discounted heat rate by the gross fired duty; the new fraction is your thermal efficiency. Gross fired duty is your fuel heating value in btu / cubic foot * fuel rate in cubic feet per hour; you’ll have btu / hour

    Another method is to calculate the total duty (mass rate x enthalpy) of the steam production and get it into btu per your. Now take this duty and divide it by your gross fired duty; this fraction is your fired efficiency.

    Where did the heat go that didn’t make steam? It heated the incoming excess air (and the 100% combustion air also) and also you lose heat by radiation and gas losses.

    In a coal boiler: same as steam; just get the heating value of the coal, then perform the previous method.

    In a freezer or refrigerator; this is exactly the same as the home refrigeration unit.

    Good luck with a microwave! I think you can estimate the efficiency by taking the voltage of your home (120 volts) and the microwave’s load in amperes and the product of that (over that period of time) to turn it into kW*hr. Then estimate how much heat was added to your food by using something close to water’s heat capacity and temperature change. Divide your food heat by the work done by the electricity to get an efficiency estimate.

    Good luck, I hope this helps some…

  2. kasab

    With heat transfer processes your idea of the efficiency is not applicable. You do not give a clue about the heat transfer process involved.
    With a refrigeration process the word efficiency is completely out.

Leave a Reply

Your email address will not be published. Required fields are marked *