Return-Path: Received: from [24.25.9.102] (HELO ms-smtp-03-eri0.southeast.rr.com) by logan.com (CommuniGate Pro SMTP 4.1.8) with ESMTP id 2882155 for flyrotary@lancaironline.net; Mon, 08 Dec 2003 14:46:08 -0500 Received: from o7y6b5 (clt78-020.carolina.rr.com [24.93.78.20]) by ms-smtp-03-eri0.southeast.rr.com (8.12.10/8.12.7) with SMTP id hB8Jk2xt016719 for ; Mon, 8 Dec 2003 14:46:04 -0500 (EST) Message-ID: <000a01c3bdc3$7573c6a0$1702a8c0@WorkGroup> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: Air Density at altitude Date: Mon, 8 Dec 2003 14:42:47 -0500 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0007_01C3BD99.8C49D240" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2800.1106 X-MIMEOLE: Produced By Microsoft MimeOLE V6.00.2800.1106 X-Virus-Scanned: Symantec AntiVirus Scan Engine This is a multi-part message in MIME format. ------=_NextPart_000_0007_01C3BD99.8C49D240 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable =20 >=20 > > Found a table of air density vs Altitude > > > > Sea level Density =3D .00237 Slug/Ft^3 > > Density at 20,000 =3D 0.001267 Slug/Ft^3 or a 47% decrease > > > > So taking formula for air mass W =3D p*V*A with p 47% less than at = sea > level > > means you would get 47% less air mass flow (with the same cubic > feet/minute > > of air volume flow) at 20,000 ft compared to what you would get at = sea > level > > for the same volume flow. > > > > While cooler temps would help, it would not compensate for a 45% = less air > > mass flow. > > > > Ed >=20 > But Jim does have a point. > Indicated Air Speed should be an indication of mass air right? > So if it cools enough at X mph IAS, it will work at any altitude at X = mph > IAS? > And this should mean that the cooler air would give an advantage as = the mass > air is cooler. >=20 > P.S. > I, as many I am sure, deeply appreciate your work in our behalf. If = we may > ever assist you please let us know. >=20 Thanks, Eric! how much money can you spare? {:>) I don't really know the answer, doesn't seem unreasonable, but I think = we can check that theory fairly easily. Lets consider that point. Will IAS provide an indication of air mass = flow if the IAS is the same at two different altitudes? =20 The same indicated air speed at any altitude implies that the dynamic = pressure (as that is what the pitot tube measures) is the same. Again = looking at our two equations we have Pd =3D 1/2*p*V^2 (dynamic pressure) and W =3D p*V*A (for mass flow) Were p is density and V velocity and A area of our duct (1ft^2) The density at sea level is 0.00237 Slug/ft^3 whereas the density at = 20000 ft is 0.001267 slug/ft^3 according to the chart I have. So if we have the same indicated airspeed at both altitudes then that = means the dynamic pressure is the same at both altitudes. Dynamic = pressure at sea level for 120 mph TAS (has to be true airspeed for V as = that is the speed at which we are moving through the air mass) =3D = .00237*(176 ft/sec)^2 =3D.00237*30976 =3D 73.41312 lbf/ft^2 =3D 73.4/144 = =3D 0.51 psi dynamic pressure. So 0.51psi gives us an indicated airspeed of X IAS (would have to know, = temperature, pressure altitute, instrument and installation errors to = really get IAS from this) for 120MPH TAS at sea level. and at sea level the Mass flow =3D 0.00237 *(176 ft/sec)*(1 ft^3) =3D = 0.41712 Slug/Sec So lets now go to 20,000 altitude. Now while we don't know what X IAS was a sea level, but we know we want = it to be the same so that means the same dynamic pressure has to be = present in the pitot tube. So Pd =3D 0.51 psi =3D73.4 lbf/ft^2 has to be the same at 20,000 as it = was at sea level to give us the same IAS. So working backwards and = recalling that the density is now 0.001267 slug/ft^3 Pd =3D 1/2*p*V^2 and solving for V^2 =3D 2*Pd/p and V =3D = Squareroot(2*Pd/p) V (TAS) =3D Sqrt(2*Pd/p) =3D sqrt(2*73.4 lbf/ft^2/.001267 slug/ft^3) =3D = sqrt(115864) =3D340 ft/sec =3D 232 mph TAS. So to get the same indicated airspeed at sea level given by a true air = speed of 120mph we would have to be traveling at 232 MPH TAS at 20,000 = ft. So looking at out mass flow again at this true airspeed (232mph) and air = density(.001267)=20 W =3D .001267 slug/ft^3*(340 ft/sec)(1 ft^2) =3D 0.4307 slug/sec of mass = flow So comparing our sea level mass flow of 0.417 slug/sec with that at = 20,000 of 0.4307 slug/sec there is only an approx 3% difference (could be I lost it rounding = numbers, but in any case not signficantly different) So, unless I've screwed up the math badly, it does indeed appears that = indicated airspeed provides a fairly good indication of mass flow at any = reasonable altitude provided we are not so fast as to encounter = compressibility. Ed Anderson ------=_NextPart_000_0007_01C3BD99.8C49D240 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
 
 

>
> > Found a table of air = density vs=20 Altitude
> >
> > Sea level Density =3D .00237 = Slug/Ft^3
>=20 > Density at 20,000 =3D 0.001267 Slug/Ft^3 or a 47% decrease
>=20 >
> > So taking formula for air mass  W =3D p*V*A with = p 47% less=20 than at sea
> level
> > means you would get 47% less air = mass=20 flow (with the same cubic
> feet/minute
> > of air volume = flow)=20 at 20,000 ft compared to what you would get at sea
> level
> = >=20 for the same volume flow.
> >
> > While cooler temps = would=20 help, it would not compensate for a 45% less air
> > mass = flow.
>=20 >
> > Ed
>
> But Jim does have a point.
> = Indicated Air Speed should be an indication of mass air right?
> = So if it=20 cools enough at X mph IAS, it will work at any altitude at X mph
> = IAS?
> And this should mean that the cooler air would give an = advantage as=20 the mass
> air is cooler.
>
> P.S.
> I, as many = I am=20 sure, deeply appreciate your work in our behalf.  If we may
> = ever=20 assist you please let us know.
>
 
Thanks, Eric! how much money can you = spare?=20 {:>)
 
 
I don't really know the answer, doesn't = seem=20 unreasonable, but I think we can check that theory fairly = easily.

Lets consider that point. Will = IAS provide=20 an indication of air mass flow if the IAS is the same at two = different=20 altitudes? 
 
 The same indicated air speed at = any altitude=20 implies that the dynamic pressure (as that is what the pitot tube = measures) is=20 the same.  Again looking at our two equations we have
 
Pd =3D 1/2*p*V^2  =20 (dynamic pressure) and W =3D p*V*A (for mass=20 flow)
 
Were p is density and V velocity and A = area of our=20 duct (1ft^2)
 
The density at sea level is = 0.00237 Slug/ft^3=20 whereas the density at 20000 ft is 0.001267 slug/ft^3 according to the = chart I=20 have.
 
So if we have the same indicated = airspeed at both=20 altitudes then that means the dynamic pressure is the same at both=20 altitudes.  Dynamic pressure at sea level for 120 mph TAS (has to = be true=20 airspeed for V as that is the speed at which we are moving through the = air=20 mass) =3D .00237*(176 ft/sec)^2 =3D.00237*30976 =3D 73.41312 = lbf/ft^2 =3D 73.4/144=20 =3D 0.51 psi dynamic pressure.
 
So 0.51psi gives us an indicated = airspeed of =20 X IAS (would have to know, temperature, pressure = altitute, instrument and=20 installation errors to really get IAS from this) for 120MPH TAS at sea=20 level.
 
and at sea level the Mass flow =3D = 0.00237 *(176=20 ft/sec)*(1 ft^3) =3D 0.41712 Slug/Sec
 
So lets now go to 20,000 = altitude.
 
Now while we don't know what X IAS was = a sea level,=20 but we know we want it to be the same so that means the same dynamic = pressure=20 has to be present in the pitot tube.
 
So Pd =3D 0.51 psi =3D73.4 lbf/ft^2 has = to be the same=20 at 20,000 as it was  at sea level  to give us the same = IAS.  So=20 working backwards  and recalling that the density is now 0.001267=20 slug/ft^3
 
Pd =3D 1/2*p*V^2 and solving for V^2 = =3D 2*Pd/p and V =3D=20 Squareroot(2*Pd/p)
 
V (TAS) =3D Sqrt(2*Pd/p) =3D = sqrt(2*73.4=20 lbf/ft^2/.001267 slug/ft^3) =3D sqrt(115864) =3D340 ft/sec  =3D 232 = mph=20 TAS.
 
So to get the same indicated airspeed = at sea level=20 given by a true air speed of 120mph we would have to be traveling at 232 = MPH TAS=20 at 20,000 ft.
 
So looking at out mass flow again at = this true=20 airspeed (232mph) and air density(.001267)
 
W =3D .001267 slug/ft^3*(340 ft/sec)(1 = ft^2) =3D 0.4307=20 slug/sec of mass flow
 
So comparing our sea level mass flow of = 0.417=20 slug/sec with that at 20,000 of 0.4307 slug/sec
there is only an approx 3% difference = (could be I=20 lost it rounding numbers, but in any case not signficantly=20 different)
 
So, unless I've screwed up the math = badly,=20  it does indeed appears that indicated airspeed provides = a fairly=20 good indication of mass flow at any reasonable altitude provided we are = not so=20 fast as to encounter compressibility.
 
Ed Anderson
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