Return-Path: Received: from wb2-a.mail.utexas.edu ([128.83.126.136] verified) by logan.com (CommuniGate Pro SMTP 4.2) with ESMTP-TLS id 363476 for flyrotary@lancaironline.net; Fri, 13 Aug 2004 09:04:29 -0400 Received-SPF: none receiver=logan.com; client-ip=128.83.126.136; envelope-from=msteitle@mail.utexas.edu Received: (qmail 69327 invoked from network); 13 Aug 2004 13:03:57 -0000 Received: from dhcp-191-101.per.utexas.edu (HELO benefits3.mail.utexas.edu) (146.6.191.101) by wb2.mail.utexas.edu with RC4-SHA encrypted SMTP; 13 Aug 2004 13:03:57 -0000 Message-Id: <5.1.1.5.2.20040813080307.0260cdb0@localhost> X-Sender: msteitle@mail.utexas.edu@localhost X-Mailer: QUALCOMM Windows Eudora Version 5.1.1 Date: Fri, 13 Aug 2004 08:03:51 -0500 To: "Rotary motors in aircraft" From: Mark Steitle Subject: Re: [FlyRotary] Re: DeltaT Coolant was : [FlyRotary] Re: coolant temps In-Reply-To: Mime-Version: 1.0 Content-Type: multipart/alternative; boundary="=====================_4188234==.ALT" --=====================_4188234==.ALT Content-Type: text/plain; charset="us-ascii"; format=flowed OK, now I see! At 05:45 AM 8/13/2004 -0700, you wrote: >Mark, Ed took care of that right in the second line "So with Tracy's 30 >gpm flow of water = 240 lbs/min" > >Dave Leonard >Ed, >Please humor me (a non-engineer) while I ask a dumb question. If it takes >1BTU to raise 1lb of water 1 degree, and you factor in 30 gpm flow to come >up with a 2400 BTU requirement for a 10 degree rise for 1 lb of water, >where does the number of pounds of water figure into the equation, or do >we just ignore that issue? Water is 8.34 lbs/gal, and say you have 2 >gallons of coolant, that would be 16.68 lbs. Seems that we would need to >multiply the 2400 figure by 16.68 to arrive at a total system requirement >of 40,032 BTU/min. What am I missing here? > >Mark S. > > > > At 09:58 PM 8/12/2004 -0400, you wrote: >>Right you are, Dave >> >>Below is one semi-official definition of BTU in English units. 1 BTU is >>amount of heat to raise 1 lb of water 1 degree Fahrenheit. >> >>So with Tracy's 30 gpm flow of water = 240 lbs/min. Since its >>temperature is raised 10 degree F we have >> >>BTU = 240 * 10 * 1 = 2400 BTU/min >> >>I know I'm ancient and I should move into the new metric world, but at >>least I didn't do it in Stones and Furlongs {:>) --=====================_4188234==.ALT Content-Type: text/html; charset="us-ascii" OK, now I see!

At 05:45 AM 8/13/2004 -0700, you wrote:
Mark, Ed took care of that right in the second line "So with Tracy's 30 gpm flow of water = 240 lbs/min"
 
Dave Leonard
Ed,
Please humor me (a non-engineer) while I ask a dumb question.  If it takes 1BTU to raise 1lb of water 1 degree, and you factor in 30 gpm flow to come up with a 2400 BTU requirement for a 10 degree rise for 1 lb of water, where does the number of pounds of water figure into the equation, or do we just ignore that issue?  Water is 8.34 lbs/gal, and say you have 2 gallons of coolant, that would be 16.68 lbs.  Seems that we would need to multiply the 2400 figure by 16.68 to arrive at a total system requirement of 40,032 BTU/min.  What am I missing here?

Mark S.



     At 09:58 PM 8/12/2004 -0400, you wrote:
Right you are, Dave
 
Below  is one semi-official definition of BTU in English units.  1 BTU is amount of heat to raise 1 lb of water 1 degree Fahrenheit.  
 
So with Tracy's 30 gpm flow of water = 240 lbs/min.  Since its temperature is raised 10 degree F we have
 
BTU = 240 * 10 * 1 = 2400 BTU/min
 
I know I'm ancient and  I should move into the new metric world, but at least I didn't do it in Stones and Furlongs {:>)
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