john carruthers | 01/07/2015 18:43:54 |
![]() 617 forum posts 180 photos | Sorry if this has been asked before but what sort of power output would a Stuart Victoria (single) engine produce and at what pressue? |
roy entwistle | 02/07/2015 11:59:03 |
1716 forum posts | If I remember 2x pressure x length of stroke x area of piston x rpm / 33000 = horse power Thats from nearly 70 years ago Roy |
Ian S C | 02/07/2015 13:46:44 |
![]() 7468 forum posts 230 photos | John, you can work out the indicated HP using the formula P.L.A.N/33,000(see Roy's entry), and when you have built the engine you can measure the brake HP by measuring the torque in(for this size engine) inch oz, and RPM/ 1352 to give the power in Watts, or gram x cm x RPM x .00001026. I only had to remember back 53 yrs, it's about the only formula I remember from my mechanics class. Still got my copy of "Engineering Science" Vol 1, by Brown & Bryant, mines a 1957 edition, but it dates back to a first ed 1937, and I don't think it was updated, still I don't suppose physics changes. Ian S C |
JasonB | 02/07/2015 14:00:02 |
![]() 25215 forum posts 3105 photos 1 articles | Am I using the right figures as I get quite a high HP using Ians formula, My thoughts last night were about 0.15 to 0.2 actual HP using 100psi and 750rpm.
With the formula using 100psi, 2" travel, 0.78sq in, 750rpm 100x2x0.78x750 /33000 117000 / 33000 3.54HP Seems there is a decimal one place out Roy's with the extra 2x would give 7hp Edited By JasonB on 02/07/2015 14:34:12 |
roy entwistle | 02/07/2015 15:16:46 |
1716 forum posts | The 2x is because its double acting Roy |
JasonB | 02/07/2015 15:23:06 |
![]() 25215 forum posts 3105 photos 1 articles | Thanks Roy, I had a feeling that may be the reason though sitll seems to be a high result? |
JasonB | 02/07/2015 15:36:09 |
![]() 25215 forum posts 3105 photos 1 articles | ME Handbook to the rescue L should be in feet not inches So 2x100x0.167x0.78x750/33000 19539/33000 0.6HP Though whether the relatively long stroke of the victoria compared with its bore would suit such high revs is debatable, and pressure could be a bit high so lets say 2x 75psi x0.167x 325/33000 = 0.25HP Edited By JasonB on 02/07/2015 15:36:56 |
Nick_G | 02/07/2015 16:43:36 |
![]() 1808 forum posts 744 photos | . Slightly related and perhaps off topic a tad. (depending upon the OP's reason for asking) What sort of steam engine HP would be required to run this 30 amp alternator.? **LINK**
Nick |
john carruthers | 02/07/2015 16:54:46 |
![]() 617 forum posts 180 photos | Thank you gentlemen, I found PLAN but not the constant and was wondering why the HP value was so high. |
JasonB | 02/07/2015 17:27:10 |
![]() 25215 forum posts 3105 photos 1 articles | Nick basic calculation is W = V x A so 420w or .56HP you would need to allow for losses etc.
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