Return-Path: Received: from [216.52.245.18] (HELO ispwestemail2.mdeinc.com) by logan.com (CommuniGate Pro SMTP 4.2.5) with ESMTP id 531311 for flyrotary@lancaironline.net; Sat, 13 Nov 2004 21:52:53 -0500 Received-SPF: none receiver=logan.com; client-ip=216.52.245.18; envelope-from=wschertz@ispwest.com Received: from 7n7z201 (unverified [67.136.145.196]) by ispwestemail2.mdeinc.com (Vircom SMTPRS 4.0.330.8) with SMTP id for ; Sat, 13 Nov 2004 18:52:35 -0800 Message-ID: <017001c4c9f4$f7c98980$fc918843@7n7z201> From: "William" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: EWP Testing Date: Sat, 13 Nov 2004 20:52:22 -0600 MIME-Version: 1.0 Content-Type: text/plain; format=flowed; charset="iso-8859-1"; reply-type=original Content-Transfer-Encoding: 7bit X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2900.2180 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.2180 Bob, Pressure drop through a core is usually linear with the square of the flow rate, both for pumps and radiators. The point marked "real rad test" has my two evap cores in parallel being pumped by the mazda pump. Pumps in series add pressure, pumps in parallel add flow rate. Bill Schertz KIS Cruiser # 4045 ----- Original Message ----- From: "Bob White" To: "Rotary motors in aircraft" Sent: Saturday, November 13, 2004 6:50 PM Subject: [FlyRotary] Re: EWP Testing > I'm going to add snips from a couple of other posts here since they are > all related. > > Snip from Rusty's other post: >>I presume you meant GPM rather than GPH in your >> chart, right? > > Yes, I did mean GPM. > > I'm an electronics guy, so I've been comparing the pump to a battery. I > will interchange the terminology between the two fields somewhat freely > here. In the battery, we have a voltage and an internal impedance. The > internal impedance limits the maximum current flow available from the > battery or the 'short circuit current'. In the pump we have pressure > (psi) and flow (GPM). I can deduce the internal impedance of the pump > by dividing pressure by flow. (I wonder if the fluid flow guys have a > unit for this?) For this pump pressure (6 PSI), and flow (16 GPH) gives > an internal impedance of 0.375. > > On Fri, 12 Nov 2004 20:56:09 -0600 > "Russell Duffy" <13brv3@bellsouth.net> wrote: > > >> Hi again. I was just thinking about this, and wondering how they get >> the flow ratings that they give. What would happen if you connected >> the inlet, and outlet together, with a hose, completely filled with >> water. Of course you could only measure the flow with a flow meter, >> but would this condition show a higher flow than pumping from one >> container to another? >> >> If it would be higher, then it could actually be a more relevant test, >> since that's about what we do with the engine. We certainly have more >> restriction than a simple pipe would have. >> > > snip from Ernest's post: >>Second, I think Rusty has a good point here. There is a significant >> amount of energy invested in accelerating the water in these static >> test. Energy that is lost when it drops in the bucket. In >>the closed circuit of the engine, SOME of it will be recovered, but the >>drag through the engine is unknown, varies with the amount of flow, and >>is very non-linear. > > NOTE: I don't know how non-linear the flow vs pressure is. For this > analysis, I'm making a grand assumption that it is linear. I'm sure > this is wrong so nothing here will give "the answer". The pressure vs > flow is linear in a non-turbulent system, but if I'm not mistaken, a > cooling system is designed to be turbulent. Also, the analysis should > be OK if the flow is restricted to a particular value. If all > measurements were made at 20 GPM for example, the math would be OK, but > not for comparing 20 GPM to 10 GPM. I'm also neglecting any restriction > caused by the 3 foot 3/4 inch ID hoses used on the input and output > > I Think Rusty does have a good point. The 'short circuit current' should > have the output connected directly to the input. This would likely > result in a higher flow rate. There is some indication that that may be > true in this data. The flow rate for two pumps in series is higher > than for the single pump, which indicate that the internal impedance is > less than twice the impedance for one pump. > > I would expect the current flow to decrease in the closed loop also as > the pump will be doing less work. That could give me the 20 GPM at 5 A. > I can set the system up while monitoring the current, then connect the > hoses together in the big container. If I see the current drop, that > will indicate we are on the right track. > > Looking at Bill Schertz' flow chart for the Mazda 13B pump (attached), > at 5594 rpm, the pump is flowing 29 GPM with an 11 PSI drop across the > pump. The zero flow pressure of the pump is 19 psi, therefore, total > impedance for the system would be 19/29 = 0.66, the internal impedance > of the pump would be (19-11)/29 = 0.28 and the external impedance of the > system would be 11/29 = 0.38. > > This is an important number. It will determine whether I want to put > the pumps in series or parallel for maximum flow. For the following, I > am going to make another assumption that Mezarie did their max flow test > as Rusty has suggested and got the 20 GPM flow rate they advertise. > This would give the Mezarie an internal impedance of 6/20 = 0.3. > > > Example 1: One WP136 > > Pressure (P) = 6 psi > internal impedance (Zi) = 0.3 > external impedance (Ze) = 0.38 > > expected flow (F) = 6/(0.3 + 0.38) = 8.8 GPM > > Example 2: Two WP136 in series > > P = 12 psi > Zi = 0.6 (0.3 + 0.3) > Ze = 0.38 > > F = 12/0.98 = 12.2 GPM > > Example 3: Two WP136 in parallel > > P = 6 psi > Zi = 0.15 > Ze = 0.38 > > F = 6/0.53 = 11.3 GPM > > Using these assumptions, the series pumps would provide slightly more > flow. There is enough uncertainty in all of this that it would be hard > to predict which one was actually better. If Ze were lower, it would > favor the parallel pumps, and higher would favor the series pumps. > > If there is a point to all of this, it's that choosing parallel vs > serial for max flow isn't a given. It all depends on the rest of the > system. > > In looking over some other information Bill has published, I don't > believe he had a radiator in the circuit when he did the flow tests. Is > that correct Bill? There is another chart showing the pressure drop > across two evaporator cores (calculated from one core). at 12 GPM, the > drop is about 1.6 psi. at 29 GPM, it's around 7.2 psi. Impedance of > radiator (Zr) at 12 GPM = 1.6/12 = .13, at 29 GPM Zr = 7.2/29 = .25. > > So much for linearity. Whatever the impedance, it will increase the > total impedance, and push the system toward series pumps. On the other > hand, the actual impedance at 12 psi is probably lower than the > impedance at 29 psi, because of the non-linearity in the pressure/flow > curves. > > OK, I haven't decided on series vs parallel yet, but I'm leaning toward > parallel for the first try. > > Now I will take this data and work from Bill's charts. All of Bill's > slides are available in a pdf file at: > http://www.bob-white.com/ACRE/Water_Cooling_an_Aircraft_Engine.pdf > > To see a similar analysis of Todd Bartram's EWP cooling system go to: > http://www.bob-white.com/ACRE/schertz/EWP_analysis.html > > I will assume I can get a 12 GPM flow rate. Looking at page 9 I see > that I will need a 50 deg temp differential across the radiator to cool > 150 HP with a 12 GPM flow. Bruce Turrentine has set an upper limit of > 200 deg with short periods at 210 deg. If I use the 200 deg limit, the > water needs to cool to 150 deg while passing through the radiator. > > The output air from the radiator has to be somewhat less than 150 deg. > For this analysis, let's say I can heat the incoming air to 140. On a > 90 deg day, that give me a 50 deg temp rise thru the radiator. (Bill's > chart only goes to 50 deg. anyway) According to the chart I need > slightly less than 5000 cubic feet per minute of airflow = 5000/60 > ft^3/m = 83 ft^3/s. > > Now I will work form the direction of how large the air intake will have > to be to accommodate that flow. If I'm climbing out at 120 mph and I > get 60% of that velocity through the input duct, that would be 72 mph or > 105 ft/s. Now I have 83 ft^3/s / 105 f/s = 0.8 ft^2. using two ducts > of 0.4ft^2 each, I need two opening .63 ft X 0.63 ft, or 7.6 in X 7.6 > in. Let's just say 8 in X 8 in. These are pretty large openings. > > I will have to hope that some of the assumptions have been too > conservative. Todd's success with his turbo installation leads me to > think that might be so. > > Finally, I would like to thank Bill for his slides and the analysis he > did on Todd's data. Without that, I couldn't have done any of the > above. I have used his analysis as a guide, but done some calculations > from a different direction. I've made so many assumptions, I'm up to my > ASSumption limit.:) I have no idea how many mistakes I've made, but I'm > sure many of you will find them, and I don't know how far I've strayed > from reality. One of these days I hope to fire up the engine and get > some real data. All of you who are flying your rotaries and shared your > experiences give me some hope. Thanks to all of you. > > Bob White > > > > -- > http://www.bob-white.com > N93BD - Rotary Powered BD-4 (soon) > > -------------------------------------------------------------------------------- >>> Homepage: http://www.flyrotary.com/ >>> Archive: http://lancaironline.net/lists/flyrotary/List.html >