Subject: [FlyRotary] Re: Cooling oil
----- Original Message -----
From: "Joseph Berki" <joseph.berki@grc.nasa.gov>
To: "Rotary motors in aircraft" <flyrotary@lancaironline.net>
Sent: Thursday, June 10, 2004 12:39
PM
Subject: [FlyRotary] Re: Cooling oil
> It would be interesting to measure flow in both
engines. I thought that
> both Lycoming and Mazda engines rejected 2/3 heat
load through the
> oil that is why I started going down this
road. If the engines generated
> the same Hp than the heat load should be
similar.
>
> Joe Berki
Joe, both engines may generate the same heat load, but
the proportion
rejected through the coolant in case of the Mazda is
2/3 of its waste heat
while the oil rejects another 1/3 of the waste
heat. Neither engine rejects
anywhere near 2/3 of its waste heat through the
oil.
Most aircraft engines reject on the order of 300-600
BTU/Min through the
oil, the Mazda at 160HP rejects approx 2446 BTU/Min
through the oil.
Ed Anderson
Ed;
That number looked
a bit high to me, so I went in to my file to check. My data shows 28% of
the fuel burn energy in the rotary gets converted to HP, 18% goes to the
coolant, and about 7% to the oil. Most of the rest goes out the exhaust pipe.
For 160 HP output, I think that should be 1725 BTU/Min going to the oil
cooler. So about 3 times the comparable powered Lyc.
Double check me on
this.
Al
Thanks
Al,
I did double check and here are my calcuations - in case I've missed an error
in them somewhere and can't spot it.
I use a conservative 25% for HP, 25% for Coolant and 50% for
Exhaust. So I am interested in where you got the 28% for HP. That
would mean less waste heat to the coolers.
Here are my
calculations
1 Lb of gasoline
containts approx 19,000 BTU of energy. For 160HP I get 1.58 Lb/min of
fuel flow.
which is approx
1.58*19,000 = 30020 BTU/min. Using Hp = 25%, Waste = 25%, Exhaust =
50%
Waste = 0.25*30020
= 7505 BTU/Min. The oil dumps approx 1/3 of the waste heat of a
rotary,
so 0.3333*7505
= 2501 BTU/Min at 160HP. So unless my fuel flow is off I think my
initial
calculation is
approximately correct. I think I used 18,800 BTU in my orginal
calculations which
would give a
slightly lower value. I have seen the BTU content of a lbm of
gasoline vary between
about 18,800 to
20,000 BTU/Lbm.
Checking my fuel
flow for a 160HP engine.
I have air
flow = 80*6000/1728 = 277.77 CFM for a 2 rotor 13B
1 cubic foot of
air mass at sea level = approx 0.076 lbm. So that gives .076*277.77 = 21.11
lbm/minute of air flow for the 2 rotor at 6000 rpm at sea
level.
I used an air/fuel
ration for best power of 13.3 (just the value I happen to have in my
calculator). So my fuel needed for that air flow (21.11 lbm/minute) is
21.11/13.3= 1.587 lb/min of fue.
So 1.587*19,000 =
30157.14 BTU/Min of which 0.25*0.333*30157 = 2510 BTU/Min for the
cooler.
So unless I've
screwed up somewhere in the calculations or unless the my allocations of the
BTU for waste heat rejection are not valid, I think my calcuation is pretty
close.
Ed