Calc 2 - Eq. 1 Eq. 2 Eq. 3 Eq. 4 Eq. 5 Eq. 6 Eq. 7 Eq. 8...

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Unformatted text preview: Eq. 1 Eq. 2 Eq. 3 Eq. 4 Eq. 5 Eq. 6 Eq. 7 Eq. 8 Eq. 9 Eq. 6 All equations used are based on the Schmidt Theory for Stirling Engines http://www.bekkoame.ne.jp/~khirata/academic/schmidt/schmidt.htm 1.3.1.1. Engine Speed = N Rt Where = / N engine frequency Hz RPS = ( ) R Number of revolutions RPM = ( ) t Time seconds From the Product Target Specifications Table, we require an RPM of 3000RPM = R 3000RPM As R is in revs per minute, the number of second is one minute is 60 = t 60 seconds Substituting these into eq. 1 we get: = N 300060 = N 50 Hz 1.3.1.2. Engines Energy Calculations = P Wi N Where = ( ) N Engine Frequency Hz = ( ) Wi Indicated energy Joules = ( ) P Power of engine Watts From eq. 1, we determined a value for N = N 50 Hz From the Product Target Specifications Table, we require an output power of 3000W = P 3000W Substituting into eq. 2 we get: = 3000 Wi 50 Rearranging we have: = Wi 300050 = Wi 60 J 1.3.1.3. Volume calculation = + -- Vse Wi1 1 c2Pmeanc1 tsina...
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This note was uploaded on 10/23/2009 for the course HES 2340 taught by Professor Tomedwards during the Three '09 term at Swinburne.

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Calc 2 - Eq. 1 Eq. 2 Eq. 3 Eq. 4 Eq. 5 Eq. 6 Eq. 7 Eq. 8...

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