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 363988 for flyrotary@lancaironline.net; Fri, 13 Aug 2004 15:22:01 -0400 Received-SPF: none receiver=logan.com; client-ip=128.83.126.136; envelope-from=msteitle@mail.utexas.edu Received: (qmail 2965 invoked from network); 13 Aug 2004 19:21:48 -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 19:21:48 -0000 Message-Id: <5.1.1.5.2.20040813133803.0260e1e8@localhost> X-Sender: msteitle@mail.utexas.edu@localhost X-Mailer: QUALCOMM Windows Eudora Version 5.1.1 Date: Fri, 13 Aug 2004 14:21:40 -0500 To: "Rotary motors in aircraft" From: Mark Steitle Subject: Re: [FlyRotary] Marks Question Answer to when is 2 gallons enough?Re: DeltaT Coolant was : [FlyRotary] Re: coolant temps In-Reply-To: Mime-Version: 1.0 Content-Type: multipart/alternative; boundary="=====================_26860015==.ALT" --=====================_26860015==.ALT Content-Type: text/plain; charset="us-ascii"; format=flowed Ed, If memory serves me correctly, the 124 mph figure was about what I got when plugging my numbers into your spreadsheet. With the airflow across the radiator being so important, I can see where cowl flaps could be a big help in solving the cooling problem when attempting to optimize for cruise and takeoff conditions. I have a cowl flap that I hope will help to accomplish that. I guess I'll have to fly it to find out for sure. Mark At 01:54 PM 8/13/2004 -0400, you wrote: >Wait!!! Mark, remember this was all for the example of Tracy Crook >burning 7.0 gallons/hour of fuel. Now, that is near the low end of what a >two rotor will see and as Tracy stated that was for a Cruise condition - >which is the regime of flight that normally is the least demanding so far >as cooling requirements. You really need to look at the most severe >regime which is generally take off and initial climb. There you are >putting out full power and the air speed is slowest (therefore you cooling >mass flow is lowest). > >But, yes what ever the two rotor BTU cooling needs are (all other things >being equal) for a specific power output then the 3 rotor's needs would >be approx 1.5 times that. > > > Remember this cooling stuff is far from as simple as you might have > surmised from my calculations. These equations will give you the coolant > and airflow capable of removing your waste heat - however, whether you > particular system provides that air mass flow through your radiators and > heat transfer is a different problem. Your ducting and type diffuser and > cooler core are significant factors as to whether > >1. you get your minimal airflow required and > >2. That your cores effectively transfer the heat to that cooling air mass. > >For instance you could hang your radiators out in the slip stream and >possible get the required minimum airflow through them (if you fly fast >enough) but the cooling drag would be extremely high - not desirable. To >get maximum cooling with minimum drag takes a bit of more. >Its generally accepted that you should ideally design your cooling system >to either > >1. Provide adequate cooling with an airflow velocity around 10% of CRUISE >velocity > >2 Or 30% of your climb velocity > >In some cases the two rules produce approx the same air velocity figure >for the cooler. > >So if your climb velocity was say 120 MPH IAS then your "Optimum" airflow >through your radiator would be approx 120 *.3 = 36 MPH > >If your cruise airspeed was 200 mph then your design optimum (assuming you >chose cruise as you design point) would be approx 20 MPH. > >So first you need to decide on which regime you want to design for and the >trade offs involved. That establishes your airspeed and power >parameters. I choose climb cruise which leaves me with a slight cooling >deficit until I hit approx 120 mph in climb. That means my temperature >has continue to climb until I hit my design airspeed. But, I accept I >will have increased temps for a short periods I spent very little time in >take off and initial climb (to pattern altitude or maybe 2500 AGL. > >Then you need to estimate your power produced during that phase of >operation, in my cause I use 170 HP - I may not make that but if its less >then my cooling problem is less severe {:>). Then you figure your waste >BTU you need to get rid of. The spreadsheet I sent will do that for you >for either a 2 rotor or 3 rotor - but you have to change the value from 2 >to 3 on the spread sheet. > >So you find the rpm you are operating at and the power/BTU rejection you >must achieve. Figure that the number of BTU you must reject through oil >and coolant to be the same amount of BTU as the HP produced (that's what >the spreadsheet does). It breaks out that into the coolant and oil BTU >rejection requirements. > >Then once you have the BTUs (for that set of operating parameters) you can >start getting a handle on your cooling challenge. > >Since your total surface area is appox 165 sq inches greater that 360 vs >195 (for 2 gm cores used on the two rotors), I would say you have the >basic radiator needed to cooling your 3 rotor provided you get the minimum >required airflow. > >Lets see if I use 170 HP for my climb power figure lets assume you will >use 1.5 times that or 255 HP for your 3 rotor. That would mean you need >to reject approx 7834 BTU (not the 2400 BTU of the two rotor at lean >cruise conditions) with your radiator (and another 3917 BTU from your oil). > >So we need an airmass flow through your 360 sq inch to carry away 7834 >BTU/min. > >So again turning to our heat transfer equation we have Q = W DeltaT >cp. Lets assume the deltaT (increased in air temp caused by dumping the >heat into it) is around 20 - 50F(it will be more when you are operating at >a higher temps of full power situations, so lets pick 30F). The air mass >flow required W = Q/(DeltaT *0.25) > >W = 7834 /(50*.25) = 626 lbms/min. 626 lbm/0.076 = 8236.8 CFM (already >factored in the 0.25 cp). So changing that to cubic ft/sec we divide by >60 , 8237/60 = 137 cubic ft/sec . Taking your radiator core surface are >of 360/144 = 2.5 sq ft and we have > >( 137 cubic ft/sec)/(2.5 sq ft) = 54.31 ft/sec air velocity required. = >37.4 mph air flow. Using the 30% climb rule this means you would need to >climb (while producing 255 hp) at an airspeed of around 37/0.3 = 124 mph >to maintain cooling at that power setting and your current radiator size. >Now I fudged a little bit by assuming a DeltaT of 50F. > >I don't know what delta T your system will produce having no data points >on your radiator core. I purposely picked 50F as I knew it would produce >an airflow close to what you had calculated. (See you can make the >figures provide any answer you want provided you get to make the >assumptions {:>)). > >However, lets say your core's delta T is closer to the delta Ts reported >by GM core users which is generally 20-30F. > >Generally the GM core users have been reporting a 20 -31F rise in air >temps at high power settings. Lets say your radiator produced the same >increase in temp of 30F and not the 50F I earlier assumed. > >Then W = 7834/(30*.25) = 1045 lbm/min of air flow needed. 1045/0.076 = >137434 Cubic ft/Min divide by 60 gives 229 cubic ft/Sec > taking your 2.5 sq ft V = 229/2.5 = 92 ft/sec of air velocity = 62.7 > MPH through your radiator. So this is approx double the required airflow > over the case where the air temp increase was assumed to be 50F. Still > all in all you radiator should have no problem getting rid of the heat - > although your cooling drag may be a bit higher than optimum. > >So just trying to point out that while these calculations can give you a >starting point, just some small differences in assumptions (as I did with >the deltaT ) can make the difference between success and not. But since >your radiator is sufficiently large, the worst penalty you are likely to >paid is a bit more drag. Assuming that your ducting and diffuser provides >airflow reasonably close to that calculated. > > >Hope this helps > >Ed >Ed Anderson >RV-6A N494BW Rotary Powered >Matthews, NC >----- Original Message ----- >From: Mark Steitle >To: Rotary motors in aircraft >Sent: Friday, August 13, 2004 11:02 AM >Subject: [FlyRotary] Re: Answer to when is 2 gallons enough?Re: DeltaT >Coolant was : [FlyRotary] Re: coolant temps > >Ed, >Wow, that was a great explanation. I think I even understood most of >it. ;-) I'm assuming that for a 3-rotor I would just multiply the 2400 >BTU figure by 1.5 to arrive at a BTU requirement of 3600 for the >coolant? Same for the air requirement (240 * 1.5 = 360 lbs, or 5524.75 * >1.5 = 8287 CFM). My single radiator core is approx. 18" x 20" or 360 sq. >in. or 2.5 cubic feet. Then 8287/2.5 = 3314.8 ft/min (55.25 ft/sec) air >velocity, which equals 37.67 mph. > >Does this look right? I can cool my engine at 37 mph airflow across the >core? How does one determine the airspeed across the radiator? > >Mark >(See what you started???) >>>Ed Anderson >>>RV-6A N494BW Rotary Powered >>>Matthews, NC --=====================_26860015==.ALT Content-Type: text/html; charset="us-ascii" Ed,
If memory serves me correctly, the 124 mph figure was about what I got when plugging my numbers into your spreadsheet.  With the airflow across the radiator being so important, I can see where cowl flaps could be a big help in solving the cooling problem when attempting to optimize for cruise and takeoff conditions.  I have a cowl flap that I hope will help to accomplish that.  I guess I'll have to fly it to find out for sure.

Mark  

At 01:54 PM 8/13/2004 -0400, you wrote:
Wait!!!  Mark, remember this was all for the example of Tracy Crook burning 7.0 gallons/hour of fuel.  Now, that is near the low end of what a two rotor will see and as Tracy stated that was for a Cruise condition - which is the regime of flight that normally is the least demanding so far as cooling requirements.  You really need to look at the most severe regime which is generally take off and initial climb.  There you are putting out full power and the air speed is slowest (therefore you cooling mass flow is lowest).
 
But, yes what ever the two rotor BTU cooling needs are (all other things being equal)  for a specific power output  then the 3 rotor's needs would be approx 1.5 times that.
 
 
 Remember this cooling stuff is far from as simple as you might have surmised from my calculations.  These equations will give you the coolant and airflow capable of removing your waste heat - however, whether you particular system provides that air mass flow through your radiators and heat transfer is a different problem.  Your ducting and type diffuser and cooler core are  significant factors as to whether
 
1.  you get your minimal airflow required and
 
2. That your cores effectively transfer the heat to that cooling air mass.
 
For instance you could hang your radiators out in the slip stream and possible get the required minimum airflow through them (if you fly fast enough) but the cooling drag would be extremely high - not desirable.  To get maximum cooling with minimum drag takes a bit of more.
Its generally accepted that you should ideally design your cooling system to either
 
1.  Provide adequate cooling with an airflow velocity around 10% of CRUISE velocity
 
2  Or 30% of your climb velocity
 
In some cases  the two  rules produce approx the same air velocity figure for the cooler. 
 
So if your climb velocity was say 120 MPH IAS then your "Optimum" airflow through your radiator would be approx 120 *.3 = 36 MPH
 
If your cruise airspeed was 200 mph then your design optimum (assuming you chose cruise as you design point) would be approx 20 MPH.
 
So first you need to decide on which regime you want to design for and the trade offs involved.  That establishes your airspeed and power parameters.  I choose climb cruise which leaves me with a slight cooling deficit until I hit approx 120 mph in climb.  That means my temperature has continue to climb until I hit my design airspeed.  But, I accept I will have increased temps for a short periods I spent very little time in take off and initial climb (to pattern altitude or maybe 2500 AGL.
 
Then you need to estimate your power produced during that phase of operation, in my cause I use 170 HP - I may not make that but if its less then my cooling problem is less severe {:>).  Then you figure your waste BTU you need to get rid of.  The spreadsheet I sent will do that for you for either a 2 rotor or 3 rotor - but you have to change the value from 2 to 3 on the spread sheet. 
 
So you find the rpm you are operating at and the power/BTU rejection you must achieve.  Figure that the number of BTU you must reject through oil and coolant to be the same amount of BTU as the HP produced (that's what the spreadsheet does).  It breaks out that into the coolant and oil BTU rejection requirements.
 
Then once you have the BTUs (for that set of operating parameters) you can start getting a handle on your cooling challenge.
 
Since your total surface area is appox 165 sq inches greater that 360 vs 195 (for 2 gm cores used on the two rotors), I would say you have the basic radiator needed to cooling your 3 rotor provided you get the minimum required airflow.
 
Lets see if I use 170 HP for my climb power figure lets assume you will use 1.5 times that or 255 HP for your 3 rotor.  That would mean you need to reject approx 7834 BTU (not the 2400 BTU of the two rotor at lean cruise conditions) with your radiator (and another 3917 BTU from your oil).
 
So we need an airmass flow through your 360 sq inch to carry away 7834 BTU/min. 
 
So again turning to our heat transfer equation we have  Q = W DeltaT cp.  Lets assume the deltaT (increased in air temp caused by dumping the heat into it) is around 20 - 50F(it will be more when you are operating at a higher temps of full power situations, so lets pick 30F).   The air mass flow required W = Q/(DeltaT *0.25)
 
W = 7834 /(50*.25) = 626 lbms/min.  626 lbm/0.076 = 8236.8  CFM (already factored in the 0.25 cp).  So changing that to cubic ft/sec we divide by 60 , 8237/60 = 137 cubic ft/sec .  Taking your radiator core surface are of 360/144 = 2.5 sq ft and we have
 
( 137 cubic ft/sec)/(2.5 sq ft) = 54.31 ft/sec air velocity required.  = 37.4 mph air flow.  Using the 30% climb rule this means you would need to climb (while producing 255 hp) at an airspeed of around 37/0.3 = 124  mph to maintain cooling at that power setting and your current radiator size. Now I fudged a little bit by assuming a DeltaT of 50F. 
 
I don't know what delta T your system will produce having no data points on your radiator core.  I purposely picked 50F as I knew it would produce an airflow close to what you had calculated.  (See you can make the figures provide any answer you want provided you get to make the assumptions {:>)).
 
However, lets say your core's delta T is closer to the delta Ts reported by GM core users which is generally 20-30F.
 
Generally the GM core users have been reporting a 20 -31F rise in air temps at high power settings.  Lets say your radiator produced the same increase in temp of 30F and not the 50F I earlier assumed.
 
Then W = 7834/(30*.25) = 1045 lbm/min  of air flow needed.  1045/0.076 = 137434 Cubic ft/Min divide by 60 gives 229 cubic ft/Sec
 taking your 2.5 sq ft V = 229/2.5 =  92 ft/sec of air velocity = 62.7 MPH through your radiator.  So this is approx double the required airflow over the case where the air temp increase was assumed to be 50F.   Still all in all you radiator should have no problem getting rid of the heat - although your cooling drag may be a bit higher than optimum.
 
So just trying to point out that while these calculations can give you a starting point, just some small differences in assumptions (as I did with the deltaT ) can make the difference between success and not. But since your radiator is sufficiently large, the worst penalty you are likely to paid is a bit more drag.  Assuming that your ducting and diffuser provides airflow reasonably close to that calculated.
 
 
Hope this helps
 
Ed
Ed Anderson
RV-6A N494BW Rotary Powered
Matthews, NC
----- Original Message -----
From: Mark Steitle
To: Rotary motors in aircraft
Sent: Friday, August 13, 2004 11:02 AM
Subject: [FlyRotary] Re: Answer to when is 2 gallons enough?Re: DeltaT Coolant was : [FlyRotary] Re: coolant temps

Ed,
Wow, that was a great explanation.  I think I even understood most of it.  ;-)   I'm assuming that for a 3-rotor I would just multiply the 2400 BTU figure by 1.5 to arrive at a BTU requirement of 3600 for the coolant?  Same for the air requirement (240 * 1.5 = 360 lbs, or 5524.75 * 1.5 = 8287 CFM).  My single radiator core is approx. 18" x 20" or 360 sq. in. or 2.5 cubic feet.  Then 8287/2.5 =  3314.8 ft/min (55.25 ft/sec) air velocity, which equals 37.67 mph. 

Does this look right?  I can cool my engine at 37 mph airflow across the core?  How does one determine the airspeed across the radiator?   

Mark
(See what you started???)
Ed Anderson
RV-6A N494BW Rotary Powered
Matthews, NC
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