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John,
Your pressure ratio formula is (Pa +
Pb)/Pa. First, I prefer using PSI, so taking that route
first.
At sea level we have at MAP 50 =
(50-30)/30*14.7 = 9.8 psi gauge of boost pressure
A check (14.7+9.8)/14.7 = 1.66 pressure ratio.
Pb is your differential gauge pressure or Boost pressure
Going to 10,000 MSL with the same boost reading we
have (10+9.8)/10 = 1.98 pressure ratio due to ambient pressure decrease. I
used 10 psi for atmospheric which is probably too high, but too lazy to look it
up. Stick in the correct figure..
You ambient pressure is an absolute
measurement like 29.92 "HG or 14.7PSI. You boost pressure is differential
or the amount above ambient (whatever ambient is).
I think you
have mixed MAP measured in inches of mercury and MAP measured in Lbs/in^2.
It helps to always state your units of measure.
Using your approach we have
MAP = 50 at sea level = (30+20 )/30 = 1.666,
so that checks Boost here in " Hg is 50" - 30" = 20" this is your
Pb. It can be stated in either inches Hg, PSI, Pascals, etc. The
important thing is that it is a differential pressure that is Pb = PT -
Pa. If PT is your Total absolute pressure (including boost) then your
differential pressure Pb = Pt-Pa. In your example at sea level your Pb =
50 - 30 = 20" Hg
At 10,000 ft altitude if you are going to use "hg
then - and if the air pressure is 10 psi (as we assumed) then the MAP
is 10/30*14.7 = 20" Hg for the ambient pressure in inches mercury not 10.
Each psi is approx 2 " Hg.
Our Boost pressure remains the same as at sea
level or 20 " Hg = 20"/30"*14.7 = 9.8 psi so that checks.
Using your MAP = 40 at 10,000 MSL,
we first find our boost Pb = 40"(Pt) - 20"(Pa)(not
10) = 20 "hg boost same as we had at sea level
Tanking these numbers and putting them in our
formula we have (20"Pa{not 10} + 20"Pb)/20"Pa{not 10) = (20+20)/20 = 2 or. Close
enough to my 1.98 and far away from your calculate pressure ratio of
4.0
Hope this helped. Just remember you
really need to keep track of your units of measurement - it helps clarify and
prevent confusion - I know I need all the help I can get, so hope this helped
you a bit.,
Summary:
You ambient pressure (Pa) is an absolute
measurement like 29.92 "HG or 14.7PSI. You boost pressure(Pb) is
differential or the amount above ambient (whatever ambient is). Your Total
pressure (PT) is the sum of the two.
Pressure ratio = (Pa+Pb)/Pa Pa = Ambient
atmospheric pressure (at whatever altitude). Pb is your differential boost
pressure
Pb = Pt - Pa Where Pt is your total absolute
pressure (Pt = Pa+Pb)
To convert from inches Hg to PSI, I use this
forumla Taking an absolute pressure X in " Hg, we have
X/30*14.7psi = " Hg.
For example to convert 30 " Hg we
have 30/30*14.7 = 14.7 psi. Conversely to convert
psi to " Hg we use X/14.7*30 = "
Hg
Taking 14.7 psi we have 14.7/14.7*30 = 30"
Hg.
In your case for altitude - if the ambient
pressure were 10 psi then to convert to " Hg, we have
10/14.7*30 = 20.48" Hg or 20" Hg rounded
off.
Ed
Ed Anderson
RV-6A N494BW Rotary
Powered
Matthews, NC
----- Original Message -----
Sent: Monday, August 23, 2004 9:45
AM
Subject: [FlyRotary] Re: Compressor
maps
Let's see if I understand this.
To
help me read the map, I made a tiny spreadsheet (attached) to show lbs/min at
various rpm
This
tells me I have 23lb/min at 6000. So...
At SL and MAP 50 I have a PR of (30+20)/30 = 50/30 =
1.66
So on the map I read up from
23 till I hit 1.66 on the y axis
and see that I'm on the bottom of the island.
Now
go to 10000ft and set the MAP at, say 40. My PS is now (10+30)/10 = 40/10
= 4.0 which is off the scale. :(
Looking at the (compressor) MAP again I see that my max PR at 23lb is
about 1.8, so working backwards my max MAP at 6000 ft is 18 if I want to stay
on the island. To get into higher PR's I need more rpm. I'd
need 9000+ rpm to reach the top of the island where the available PR
is about 2.5.
Doesn't sound right. Where did I go wrong?
Comments?
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