Return-Path: Received: from fed1mtao07.cox.net ([68.6.19.124] verified) by logan.com (CommuniGate Pro SMTP 4.1.5) with ESMTP id 2649764 for flyrotary@lancaironline.net; Fri, 24 Oct 2003 14:49:04 -0400 Received: from BigAl ([68.107.116.221]) by fed1mtao07.cox.net (InterMail vM.5.01.06.05 201-253-122-130-105-20030824) with ESMTP id <20031024184854.OAZM13061.fed1mtao07.cox.net@BigAl> for ; Fri, 24 Oct 2003 14:48:54 -0400 From: "Al Gietzen" To: "'Rotary motors in aircraft'" Subject: RE: [FlyRotary] Re: radiator tube sizing Date: Fri, 24 Oct 2003 11:49:05 -0700 Message-ID: <000201c39a5f$8004ba70$6400a8c0@BigAl> MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0003_01C39A24.D3A5E270" X-Priority: 3 (Normal) X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook, Build 10.0.4024 Importance: Normal In-Reply-To: X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2800.1106 This is a multi-part message in MIME format. ------=_NextPart_000_0003_01C39A24.D3A5E270 Content-Type: text/plain; charset="us-ascii" Content-Transfer-Encoding: 7bit > Maybe better check your flow rate units; gpm vs cfm. 20 cfm is > way too > high. For a 2-rotor you'd probably need a max of around 25 gpm, which > is about 3.5 cfm. In the immortal words of Homer Simpson: DOH!! Here is revised data using an absolute minimum flow rate of 20'g'pm. 25 should probably be a design minimum, but I from what I've read, 20 can handle the load except in the most extreme conditions. 14ft 6ft 2ft dia wt psi wt psi wt psi .5 4.76 46 2.04 19.7 .68 6.57 .625 7.44 15 3.19 6.46 1.06 2.15 .75 10.71 6 4.59 2.6 1.53 .87 .875 14.58 2.8 6.25 1.2 2.08 .4 1 19.04 1.44 8.16 .62 2.72 .21 1.125 24.1 .8 10.33 .34 3.44 .11 1.25 29.75 .47 12.75 .2 4.25 .07 This data makes my proposed radiator placement a little more feasible. Using a 7/8" tube I'd get less than 3psi backpressure and add 15lbs of water. Using the EWP will save me about 8 of those, so I'm looking at adding about 7lbs and eliminating seperate intakes for cooling. You might also want to check your units for the weight. I don't know what your including, but those seem too high. 1" dia of h20 6' long would weigh about 2 lbs. .785 x 72 = 56.5 cu. In. Water weighs about .037 lbs/cu. in. Al ------=_NextPart_000_0003_01C39A24.D3A5E270 Content-Type: text/html; charset="us-ascii" Content-Transfer-Encoding: quoted-printable

> Maybe better check your flow rate units; gpm vs cfm. 20 cfm is =

> way too

> high. For a 2-rotor you'd probably need a max of around 25 gpm, = which

> is about 3.5 cfm.

In the immortal words of Homer Simpson: DOH!!

Here is revised data using an absolute minimum flow rate of 20'g'pm. 25 = should probably be a design minimum, but I from what I've read, 20 can handle = the load except in the most extreme conditions.

&nbs= p; = 14ft &nb= sp; = 6ft &nbs= p; 2ft

dia = wt = psi = wt = psi = wt psi

.5 &n= bsp; 4.76 = 46 = 2.04 19.7 = .68 6.57

.625 = 7.44 = 15 = 3.19 6.46 = 1.06 2.15

.75 = 10.71 = 6 = 4.59 = 2.6 = 1.53 .87

.875 = 14.58 2.8 = 6.25 = 1.2 = 2.08 .4

1 &nb= sp; 19.04 1.44 = 8.16 = .62 = 2.72 .21

1.125 = 24.1 = .8 = 10.33 .34 = 3.44 .11

1.25 = 29.75 .47 = 12.75 = .2 = 4.25 .07

This data makes my proposed radiator placement a little more feasible. = Using a 7/8" tube I'd get less than 3psi backpressure and add 15lbs of water. Using the EWP will save me about 8 of those, so I'm looking = at adding about 7lbs and eliminating seperate intakes for = cooling.

You might also want to check your units for the weight. I don’t know what your including, but those seem too high. 1” dia of h20 6’ long would weigh about 2 = lbs.

<= /font>

.785 x 72 =3D = 56.5 cu. In.

<= /font>

Water weighs = about .037 lbs/cu. in.

<= /font>

Al

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