1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Otto cycle exhaust heat rejection

  1. Jun 6, 2012 #1
    Hello All! I am new to the forum and need help. I have looked at multiple sites an looked at 328 threads titles and about 20 actual threads today to find the info I was looking for....no luck.

    Here is my question:

    I have a 12.5KW natural gas generator and want to know much heat and I can get from the exhaust, which of course leads to tube-in-shell heat exchanger sizing requirements. At 1800 RPM it uses 64 CFM of combustion air (cooling air is obviously a lot more). At 1800 RPM and full load it uses 230 CFM of natural gas p/hour. Each cubic foot of natural gas has 950-1150 BTUs and a net heating value of 850-1050 BTUs. I have no idea if this is right, but trying to be conservative I am assuming about 1000 BTUs per cubic foot of natural gas is being used for combustion. Simple math tells me that 23,000 BTUs of natural gas are being burned in the engine p/hour. How do I figure out how many BTUs are going out the exhaust? How much exhaust air flow should I have (I am assuming at least 64 CFM but does it expand when the natural gas and air burn)?

    Numbers:
    At 1800 RPM and full load
    64 CFM of combustion air charge
    1000 BTUs p/cubic foot of natural gas
    230 CF p/hour at full load


    Thank you in advance.

    Kevin
     
  2. jcsd
  3. Jun 7, 2012 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    We would need to know the efficiency of your generator. That depends upon the design of the device. With that, it is a simple matter to determine the heat rejected. It is just:

    Qc = Qh(1-η) where η is the efficiency (= Work output/energy input = W/Qh)

    AM
     
  4. Jun 7, 2012 #3
    Thank you for the reply. I will try to look for the efficiency of the engine; however, none of the literature I have gives my that number. I know it is rated at 42.5 BHP (31.71 KW) at 2700 RPM and when used for power generation it runs at 1800 RPM making 30.5 BHP (22.76 KW). Putting a little thought into I converted 22.76 KW to btu/hr and came up with 77 660.34 btu/hr. I also realized that 230 CF of natural gas contains 230,000ish btus and not 23,000 which I put int he original post.

    Would I be right in assuming that if the machine is producing 77,660 btu/hr of mechanical power and consuming 230,000 btu/hr of fuel that 77,660/230000 = 33.7% ? That would make sense for an Otto cycle engine efficiency number.

    Now how do I use your equation "Qc = Qh(1-η) where η is the efficiency (= Work output/energy input = W/Qh)" to isolate just the exhaust heat rejection?



    Thank you again.

    Kevin
     
  5. Jun 7, 2012 #4

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    You must be in an engineering course. These units drive physicists crazy.

    1 KJ = .948 BTU
    1 KW = 3412 BTU/hr
    .746 KW = 1 BHP
    1 cfm gas = 1034 BTU = 1090.7 KJ

    You don't really need to know the efficiency if you know how much work it is producing. The heat rejected is Qc = Qh-W. If you know the efficiency but not the work output you would use Qc = Qh(1-η)

    At 1800 rpm it produces 22.76 KW of work. It consumes 230 x 1090.7 KJ = 251,000 KJ/hour or 69.7 KW of energy. So its efficiency is 22.76/69.7 = 32.7%. But it is simpler here to simply use W = 22.76 KW and Qh = 69.7 KW and Qc = Qh-W.

    AM
     
  6. Jun 7, 2012 #5

    russ_watters

    User Avatar

    Staff: Mentor

    A good rule of thumb is that a third of the energy goes out via the output shaft, a third via the radiator and engine surface and a third goes out the tailpipe.
     
  7. Jun 7, 2012 #6
    Thank you Andrew and Russ. Not in an engineering course actually, but the numbers do drive me crazy. I am working on a co-gen project for an off grid home I am planning on building. The 12.5KW air cooled natural gas generator is pilot (cheap test subject) for a much nicer 10KW water cooled diesel generator which will go in my utilities shed.

    I appreciate the info. This helps me down the road on my project. Now I need to get a good exhaust gas flow rate and temperature at full load and start working on my delta Ts.

    I read somewhere that input of combustion air equals exhaust output. This mostly makes sense to me because the expansion of gas created when the combustion air charge is ignited goes into work on the crankshaft, then what is exhausted is what came in, just burned. Does this make sense?

    Thank you.

    Kevin
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Otto cycle exhaust heat rejection
  1. The Otto Cycle (Replies: 4)

Loading...