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Message-ID: <a460a0bc0706011335h3691b5edwa5c88d7290b42bb3@mail.gmail.com>
Date: Fri, 1 Jun 2007 15:35:48 -0500
From: "Blake Lewis" <blake.lewis@gmail.com>
To: "Rotary motors in aircraft" <flyrotary@lancaironline.net>
Subject: Re: [FlyRotary] PP Ve??? was Re: Intake CFM air flow
In-Reply-To: <list-2079013@logan.com>
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References: <list-2079013@logan.com>

12A  Pports can do 310 HP at 10,700 RPM at .667

If the .667 is BSFC then start with that.
density of air at sea level 15C = .0765 lbs/cf
air/fuel = 12

.667 bsfc x 310hp is 206.77 lbs/hr fuel

206.77lbs x 12 = 2481.24 lbs/hr air

2481.24 lbs/hr  /  60min/hr = 41.354 lbs/min

(41.354 lbs/min)  / (.0765 lbs/cf) * (1728 ci/cf) = 93411 ci /min

(93411ci/min) / (10000 rev/min) = 93.41 ci/rev

1300cc / (2.54cm * 2.54cm * 2.54cm)/ci = 79.33 ci volume

93.41ci / 79.33ci = 1.1774 = 117.74 % VE

Blake


On 6/1/07, Ed Anderson <eanderson@carolina.rr.com> wrote:
>
>
>
>
> 12A  Pports can do 310 HP at 10,700 RPM at .667 .
>
> Lynn E. Hanover
>
>
> Thanks for that Data point, Lynn.  I see if I can use it to get a gestimate
> on PP Ve.
>
>  So if a 12A is 73 cubic inches then at 10700 rpm and 100%Ve it would flow
>
> 73*10700/(1728) =   452 CFM air flow.
>
>  So if at this 100% Ve we get less power than 310HP then we can assume the
> Ve of the 12A must be greater than 100%Ve to give us more power.
>
> At sea level standard day 1 cubic foot of air = 0.076 lbms.  So for that
> flow we would have 0.076 * 452 =  34.35 lbs/min.   Now I don't have any idea
> what Air Fuel ratio a rotary racer uses but I best power is reportedly to
> close to 12:1
>
>  Assuming a race air fuel ratio of around 12:1 then the fuel needed for that
> ratio at that airflow.  Then  the fuel needed would be 34.35 /12 :=  2.8624
> lb of gasoline per minute.
>
>   A lb of gasoline has 19000 BTU depending on octane.  Higher octane has
> less  so assuming 19000 BTU/Lbm gasoline, we can next calculate the power
> being produced in the engine.
>
> Converting 2.8624lb/min of gasoline into lb/sec we have 2.8624/60 = 0.047708
> lb/sec.
>
> To find the BTU we have .048*19000 = 912 BTU/sec. IF ALL this energy were
> converted to torque it would give  912 *778 = 709536 ft-lbs of torque.  or
> divide by 550 = 1290 HP!!!
>
> Unfortunately, we know approx 50% goes out the tail pipe as heat and another
> 25% (more or less) is Waste heat rejected by our coolers leaving us
> somewhere around 25-30% depending on whose estimate you use for efficiency
> of a rotary engine of  1290 *.25 = 322 HP or using 30% 1290 *.30 = 387 HP
>
>
> Hummm, since the 12A PP is producing 310, but the calculations shows it
> should be getting closer to 322 HP that would suggest a PP port 12A flows
> less than 100% Ve.  If fact, it would suggest that the Ve of the 12A at
> 10700 rpm is closer to 310/322 *100 = 96.27%Ve.
>
> But, this is instantaneous BHP,  I have not subtracted for mechanical or
> other inefficiencies so taking a guess that amounts to around 5% of the
> total.  Then to get a dyno of 310 HP the engine would need to produce
> 310*1.05 = 325 HP.  So here we would get 325/322 =  101% Ve for the 12A at
> 10700 rpm.
>
> Given we know that some racers are restricted by the size of the intakes
> permitted (is this true for the PP, Lynn?) perhaps that is why the VE seems
> a bit on the low side.  But, that's just a guess.  Well, that was my best
> crack at trying to determine the efficiency of a PP.
>
> So anybody else having an idea or source of information or opinion - jump
> in.
>
> Ed