I had to look at this topic again as a commenter posted on Unz this page of Aulis, the best Moon hoax site:
https://www.aulis.com/saturn_v_evaluation.htm
It is always a pleasure to find scientific arguments for any conspiracy topic, but they need checking. The article refers to an unpublished paper by a Russian Ph.D. Gennady Ivchenkov:
https://www.aulis.com/PDF/F-1_Evaluation.pdf
Though I have never studied design of rocket engines, or any engines for that matter (except for a flower watering robot I once tried to build, but gave finally up, the flower died before the robot was ready), I think the paper is not beyond the level of normal engineering education, which I still may have after all of those years. Thus, I will try to check this paper as finding today anybody who would read and check an unpublished paper, especially of a controversial issue, is nearly impossible. Aulis editors did not try to check it, neither did the commenters on Unz.
The question is if the F-1 engine in Saturn V first-stage rocket could give the performance as specified in documents. Ivchenkov claims that it could not and this would imply that Apollo 11 did not reach the Low Earth Orbit and the Moon travel was a hoax.
I have earlier looked at this topic and there is some indication that the Moon travels were indeed hoaxes. The strongest evidence of it is that some of the first Moon stones are not from the Moon and some photos seem to be forgeries. However, here I will only look at the argument of Ivchenkov.
Let us first notice that Ivchenkov is much more competent in rocket engine design than I could ever be. I copy a part of his biography in the beginning of the paper, directly relevant to engine design: Gennady Ivchenkov graduated from the faculty «Power Engineering» of the Bauman Moscow State Technical University(Bauman MSTU) in 1974, majoring in «Aircraft engines» (Rocket Engines Department) (3rdspecialization –solid propellant rocket motors; 1stspecialization –LRE (liquid rocket engines)). After graduation he enrolled in graduate school and worked at the Bauman Aircraft Engines Department. Scientific interests –study of heat transfer in rocket nozzles. In 1980 he defended his thesis for the degree of Candidate of Technical Sciences (Ph.D.). The topic of the thesis is a study of combustion in high-speed gas flow.”
That is kind of impressive, but as it is, unpublished papers on controversial topics are seldom reviewed at least in a positive way by anyone competent on the topic for obvious reasons (it may harm your career, you better be on pension if you want to do such things, and better not to have any friends you could lose), so I will try to do it as a fairly well educated (or maybe only a good willing) layman.
First Ivchenkov comments that the tubular combustion chamber (C/C) is fundamentally unable to provide the specified pressure and F-1 engine thrust and states that it is shown in detail in the A.Velyurov study. I will not look at this study now, but only at arguments in Ivchenkov’s paper.
The F-1 motor of Saturn V had thrust chamber cooling provided by fuel circulating in cooling tubes before the fuel was injected to the combustion chamber. Thus, the kerosene pump pressure and tube sizes determined the speed of the fluid and the pressure in the walls of the cooling tubes. The paper does not say if kerosene in the cooling tubes turns to gas or only heats, but that does not matter to us here.
The kerosene pump pressure of the older H-1 engine was 1020 psi and that of F-1 engine was 1856 psi, or 2000 psi. The pump pressure is the pressure on the walls of the cooling tubes, as the pressure must be the same everywhere in fluid. The pressure does not depend on the the diameters of the cooling tubes and how they bifurcate, nor on the degree the fluid expands in higher temperature. The pressure in the fluid must equal the pressure the pump gives, because if the pump pressure is lower, fluid does not flow, while if it is higher, fluid accelerates. As fluid flows with a constant speed, the pressure is the same on every wall of cooling tubes and on the pump.
The tubes in H-1 are stainless steel 347, while in F-1 they are of a nickel alloy, called Inconel X-750. The relevant strength measure for the tubes is here the yield strength, determining how much pressure the tube can take without deforming. The yield strength of the steel tubes of H-1 was σ= 1605 kg/cm2 at the temperature T= 740°C (the maximal temperature is 800-900°C, so 740°C is probably the highest operating temperature).
The yield strength of annealed Inconel X-750 is 2460 kg/cm2 at 704°C and 2250 kg/cm2 at 815°C. The author argues that Inconel X-750 in F-1 was only annealed, not heat hardened. I will accept his argument, which is based on recent photos of Saturn V F-1 found by Jeff Bezos in the ocean bottom. Thus, Inconel X-750 is about 2400/1605≈1.5 times stronger material than stainless steel 347 in the operating temperature of the engine, as the author states.
The author also mentions that Inconel X-750 may have problems with sulphur in kerosene causing cracks in nickel alleys and refers to a hypothesis by A.Velyurov that F-1 engine cooling tubes were made of steel instead of Inconel. I will ignore this comment as it may or may not have caused problems for Saturn V F-1s and the argument that the author presents is speculative. We cannot directly apply conclusions derived from other nickel alleys to a specific nickel alley Inconel X-750.
The strength of the tubes depends also on the thickness of the tube walls. The thickness of the walls of H-1 tubes is 0.25 mm. The author states that a source gives the wall thickness for the tubes of F-1 as 0.457 mm and adds his doubts. However, I will assume now that the source is correct.
Then the author makes an argument that I find very curious. He observes that if 30% of the fuel is directly fed to the injector feeding it to the combustion chamber, this injector must throttle the fuel so much that the pressure needed to pass to the combustion chamber is the same as needed for passing to the cooling tubes, where 70% of the fuel first goes. This conclusion is naturally correct. The pressure is equal everywhere in fluid. But then the author writes like this: “But in this case kerosene pump outlet pressure should increase 1.5 times (with the same tube hydraulic resistance), which, clearly, the tubes of the cooling jacket won’t withstand.” This I do not agree with. If the fuel pump gives the pressure 2000 psi, then this is the pressure in the cooling tubes and also in the injector.
We have to ask if the cooling tubes of F-1 can withstand the pressure of 2000 psi. If their wall thickness is 0.457/0.25=1.828 times the wall thickness of H-1 cooling tubes and the material is 1.49 times as strong, then they should stand 2.7 times the pressure that H-1 cooling tubes tolerate. The pressure on H-1 cooling tubes is the kerosene pump pressure 1020 psi, while the pressure of the F-1 fuel pump is 2000 psi. The cooling tubes of F-1 should stand this higher pump pressure. Notice also that the thrust chamber of F-1 has the pressure 1000 psi. This external pressure should in some considerations be subtracted from the inside pressure, which is the fuel pump pressure of 2000 psi, in order to get the pressure that may deform the cooling tubes. Subtracting the external pressure confirms that the cooling tubes can take the inside pressure. Thus, I discard the author’s first argument why the cooling system of F-1 could not work.
Apart of the pressure, we can also ask if the cooling system manages to cool the engine. It has to be so that as much heat as is coming to the cooling pipes must be transported away by the fluid. If this is not the case, the cooling pipes heat to the burning temperature of the oxygen-kerosene mixture, which will destroy the engine. The pump pressure and (possibly) the wall thickness (with thermal conductivity of the material) determine if the cooling system works. Usually you would imagine that the pipes are not too thick and cooling mainly depends on the cold fluid flowing fast enough.
Thermal conductivity of stainless steel 347 of the H-1 engine is given as 22.5 W/m°K and of Inconel X-750 as 22.4 W/m°K at 760°C. We can conclude that thermal conductivity of these two materials is about the same. The wall thickness of F-1 tubes is 1.828 times higher, but we also have to consider how fast the fluid moves. This determines how much heat is moved. It appears that the tubes were of similar diameter, so as F-1 pump pressure was twice as high as in H-1, fluid moved twice as fast.
We come to the author’s second argument of the cooling system insufficiency.
The combustion chamber pressure of F-1 is given as 1000 psi, which is about 70 kg/cm2, or 67.75 atm. The author tells that the Soviet engine RD-107/108 (developedin1954-56) had the combustion chamber pressure up to 60 atm, almost as high, but the author claims that the cooling system of F-1 could not manage with this thrust chamber pressure. He explicitly states that: “At the same time, practice has shown that ‘American technology’ is flawed, deadlocked, and doesn’t provide satisfactory engine characteristics, such as the chamber pressure (not more than 50 atm) and, accordingly, the specific impulse.”
I will ignore this comment claiming that the tube solution can only support 50 atm thrust chamber pressure as the only argument given so far by the author in the paper is that this American technology is not used any more and modern rockets use Soviet/Russian technology. There may be other reasons for it. I also skip his discussion of film cooling (spraying kerosene on combustion chamber walls to prevent the metal surface from overheating), as he has not explained why this would prevent the engine from reaching the thrust given in the specifications.
We continue to the author’s argument. He states that originally F-1 had the combustion chamber pressure of 46 atm. This is reasonable and I accept it. The materials of the engines H-1 and F-1 have the same allowed temperature range. The F-1 combustion chamber produces more heat, thus the cooling system must carry away more heat.
The author writes the equation for the Nusselt number Nu for the case when a wall (or combustion chamber) is cooled by liquid moving in a pipe (cooling tube).
Nu=N Rex Pry
where N is an empirical constant, here 0.023, and the exponents x and y depend on the case, the author sets them to x=0.85, y=0.4. Re is Reynolds number and Pr is Brandtl number, which the author sets to 1. The values the author uses are very close to the Dittus-Boelter equation for fluid being heated and the equation can be accepted.
The wall in this case is the combustion chamber with pressure and it causes a difference: the author derives the result that the higher pressure, 70 atm in the F-1 chamber versus 46 atm in the H-1 chamber, increases the heat transfer coefficient of F-1 by 1.22. This means that the same temperature difference results to 1.22 times heat produced by the combustion chamber.
Then the author writes the following paragraph: “The heat flow in the convective heat transfer is determined by the formula
Q=αΔT, where ΔT =Tch-Twall.
Thus, with the same as in H-1 temperature gradient ΔT, the heat flow Q in the F-1 increases about 1.22 times compared to H-1, or ΔT should decrease in the F-1 by 1.22 times while maintaining the same heat flux as that of H-1. In this case Twall increases up to 1220°K. But, according to the characteristics of both tube materials of the cooling jacket, they will not withstand such temperatures. That is, in any case, the heat flow increases in F-1 by 1.22 times compared with H-1 (at stated pressure 70 atm). This means that the tube wall must transfer this heat flow to the cooler (kerosene) by the process of heat conduction.”
This is his second argument, but there is a problem with it.
The Reynold number Re depends linearly on the diameter on the tubes and on the speed v of the liquid in the pipe. It also depends on the fluid, but in this case the fluid is the same. Assuming the cooling tube diameters are the same in F-1 and H-1 and the pump pressure in F-1 is twice as high as in H-1, we notice that fluid moves twice as fast (or let us say, increasing the pressure increases the fluid flow. It may not be a linear increase, but the author has given no data to determine it better, so us assume that the if the work to move fluid mass through the tube system doubles, then the amount of moved mass doubles. This is the case if the work of moving the fluid is mostly moving fluid down and the up the pipes and friction has minor effect. With double pressure there is double force, so in the same time you do double work and move twice as much fluid up and down), so v is twice as high for F-1 as for H-1 and consequently Re is twice as large for F-1 as for H-1. Consequently this increases the Nusselt number and the heat transfer coefficient for F-1 by the factor 1.8=20.85.
Thus, while the higher pressure in the combustion chamber produces 1.22 times more heat, the higher pump pressure causes a faster flow in the cooling tubes and 1.8 times as much heat can be removed from the combustion chamber.
The temperature of the combustion chamber close to the cooling tubes we can assume to be 720-740°C in both cases and that is the higher temperature Tch in the convection formula. There is no reason why the inside temperature of the cooling tubes (Twall) in the formula) should increase as there is heat produced by combustion and heat removed by the cooling system. The walls of the cooling tubes are 1.83 times thicker in F-1 than in H-1, but the walls are not the restricting factor in cooling. The thin metal tubes can transfer the heat to the liquid. The limiting factor in cooling is how fast the liquid in the tubes can transfer the heat away, and that is determined by the pump pressure.
The author, however, thinks that the only solution to the problem he thought he found is to make the cooling tubes thinner. He claims that with the same wall thickness the cooling tube temperature should go to 1160°K and the nickel alley would not stand it. Nothing like this happens: there is the cooling liquid inside the tubes carrying away the heat.
I did not read the paper further, but I very much doubt this is a definite proof that Apollo 11 did not fly to the Moon.
13 Comments
Where do you get the idea that twice the pressure equals twice the mass flow…?
I guess you are unfamiliar with the conservation of mass principle and the resulting continuity equation…
mdot = rho * V * A
Do you see a pressure term anywhere in there…?
The regenerative cooling system of an LRE nozzle is a ‘control volume’ in fluid flow terms…ie the flow velocity is FIXED by the AREA of inlet and outlet as well as the fluid density…which in the case of a liquid is INCOMPRESSIBLE…
The turbopump is a FIXED VOLUME machine…with fixed outlet area…you can increase pressure 100 times and your fluid velocity WILL NOT CHANGE…ONLY THE PRESSURE…
Also you seem to know so little about heat exchangers that you are unaware that considerable pressure loss occurs over the length of a pipe carrying fluid flow…that’s one of the reasons that combustion chamber pressure is much lower than pump pressure…
The pressure also drops in the hundreds of fuel spray nozzles as pressure energy is converted to kinetic [velocity energy] in the spray mist…
The kerosene mass flow of the F1 engine is stated in the literature…the number of tubes and their diameter lets us find exactly the flow per tube [as claimed]…from which a heat transfer analysis can be done…
You know absolutely nothing about fundamental thermodynamics and fluid mechanics…never mind heat transfer…
“Where do you get the idea that twice the pressure equals twice the mass flow…?”
I do not claim anywhere that twice the pressure equals twice the mass flow. Fluid is not compressible, which means that you get the same pressure everywhere in the tubes. Clearly, you know nothing of physics.
Here is what you claim…
‘Assuming the cooling tube diameters are the same in F-1 and H-1 and the pump pressure in F-1 is twice as high as in H-1, we notice that fluid moves twice as fast, so v is twice as high for F-1 as for H-1 and consequently Re is twice as large for F-1 as for H-1. Consequently this increases the Nusselt number and the heat transfer coefficient for F-1…’
So yes you are claiming twice the mass flow…since the velocity is doubled but the inlet and outlet areas of the control volume are the same [‘assuming the pipe diameters are the same’]…
By the continuity equation, twice the velocity equals twice the mass flow, if the inlet and outlet area is the same…
This is what you are EXACTLY saying…that the higher pressure means a higher velocity…if inlet and outlet areas are the same…which is pure nonsense…
Again I refer to the continuity equation…for incompressible fluid we can drop the density term…
mdot = V * A
PRESSURE HAS NOTHING TO DO WITH IT…
The entire system of turbopump and coolant passages is a control volume…it is designed to handle the mass flow that the combustion chamber consumes, so its flow area is sized to provide that mass flow, in kg/s…
This mass flow is determined entirely by the flow area of the inlet and outlet of the turbopump…the pressure could be 1000 times higher and you would still get the same mass flow coming out of the turbopump…
We can look at the cooling tubes as a separate control volume…if the total inlet and outlet diameter is bigger than that of the turbopump, the fluid velocity will slow down…if it is smaller the fluid flow must speed up, due to conservation of mass, not any change in pressure which us irrelevant…[the pressure must simply be sufficient to overcome the ‘hydraulic resistance’ which is the viscous losses of the fluid flowing in the tubes…]
Any increase in pressure beyond that WILL NOT CHANGE THE FLUID VELOCITY…if the inlet and outlet areas stay the same…
So your entire premise for hastily dismissing this very qualified rocket designer’s analysis is based on fantasy, not science…
Then in your rebuttal you say this…
‘Fluid is not compressible, which means that you get the same pressure everywhere in the tubes.’
First of all, only LIQUID fluid is not compressible…gaseous fluid being quite compressible…
Also compressibility has nothing to do with the PRESSURE LOSSES IN PIPE FLOW…start here…
https://www.nuclear-power.net/nuclear-engineering/fluid-dynamics/major-head-loss-friction-loss/pressure-loss-in-pipe-friction-loss-in-pipe/
Do you think even water from your town pumping station comes out of your tap at the same pressure at which it left the pump, after flowing through miles of pipe [and that flow is at low speed, so pressure drop is small]…?
Crack open a text on heat transfer…for instance Fundamentals of Heat Transfer by Bergman et al…Section 8.1.4 deals with basics of Pressure Gradient in internal flows…
Considering the huge mass flow of about 800 kg/s of kerosene…the flow velocity would have to have been very high…causing very high pressure losses…especially since the flow made two trips…one down, and another back up to the C/C…
Use your head for something other than a place to set your dunce cap…
Pressure is linear of force F is the area is fixed. Moving mass (fluid) through the tube system with this force F you do work W. Doing work W for time T you move mass m. Doing the same work W for time 2T you move mass 2m. Doubling force to 2F you double work to 2W and move 2m in the time T. (If you want to be exact, doubling the work does not necessarily double the moved mass, but the author of the paper has given no information that enables calculating it more exactly, so I decided to take it linear for a simple check if the paper makes any sense. It could be nearly linear and could is all that is needed to debunk the paper.) Thus, you move fluid with twice the speed. You complicate a simple thing. Try take a water hoe. Turn the tap and put your hand in front of the hoe. When the tap is more open, you feel more pressure and you have more water coming out of the hoe.
As for the pressure being equal in fluid, try this: take a metal bar and push it into a metal tube. The force you apply from one end to the bar is the same you get out in the other end, because the bar does not compress with your power. Change the bar to fluid, the result is the same. The force is equal as fluid does not compress, so pressure is equal.
You write in a very offensive way and you do not understand basic physics. Unless you change your style to a normal polite scientific style of discussing, I will not answer to your comments any more and I will delete them.
What you wrote is gibberish because it goes against conservation of mass…
And also your definitions of force and work are plain wrong…work is force times distance…NOT mass * distance…
Mass times distance is NOT a fundamental, nor a derived unit in physics…it is gibberish…
Work over time equals power…so force * distance / time = power
Mass flow, which applies only to fluid flow, is mass over time, kg/s…which when multiplied by velocity also gives FORCE…
kg/s * m/s = kg*m/s^2
We recall that the unit of force is kg*m/s^2 = 1 N
So your statement…
‘Doing work W for time T you move mass m. Doing the same work W for time 2T you move mass 2m.’
…is nonsensical…since work does NOT involve moving mass…again, work is a force applied over a distance…
But let’s talk about some fundamentals in fluid flow that do involve pressure…
Pressure energy in a flowing fluid is called ‘flow work’…this is covered in thermodynamic texts such as Cengel’s Thermodynamics…section 5-2…
In this section we learn that the flow work or ‘flow energy’ [pressure energy] of a flowing fluid is…
P * specific volume [inverse of density]
The flowing fluid also contains other types of energy…internal energy [heat]…kinetic energy [velocity energy]…and potential [height] energy…which we can ignore here…
But the combination of pressure energy plus internal energy is defined as enthalpy…so the total energy of a flowing fluid is…
enthalpy + V^2/2 [Eq 27]
Ie enthalpy plus kinetic energy…
Cengel writes…
‘By using the enthalpy instead of the internal energy to represent the energy of a flowing fluid, one does not need to be concerned about the flow work.
The energy associated with pushing the fluid into or out of the control volume is automatically taken care of by enthalpy. In fact, this is the main reason for defining the property enthalpy.
From now on, the energy of a fluid stream flowing into or out of a control volume is represented by Eq. 5–27, and no reference will be made to flow work or flow energy.
So your whole idea about flow pressure and work is an irrelevant non-sequitur…LOL
And what happens once that kerosene enters the control volume that is the series of tubes…?
It’s flow velocity DEPENDS ENTIRELY ON THE INLET AND OUTLET AREAS…AS PER CONSERVATION OF MASS…
The mass balance proves it…
Sum of mass 1 [entering control volume] = sum of mass 2 [exiting control volume]…
V * A 1 = V * A 2
Likewise an energy balance…
Sum of energy 1 = sum of energy 2…
Like I said the pressure entering the control volume makes no difference to the velocity in the control volume…
The only difference will be that the higher pressure will mean a higher energy balance…ie more energy going in and more energy coming out…BUT AT THE SAME SPEED…
The fluid flow Velocity cannot possibly be affected if the inlet and outlet areas are the same…
As for your denying that pressure loss DOES NOT OCCUR IN PIPE FLOW…
ARE YOU SERIOUS…?
Forget the friction. We have no information of the tube system in order to consider it. Most probably fluid flows just fine in those tubes and the main work is moving the fluid mass down and again up. Do it yourself: move the fluid in buckets going down the steps and then up the steps. It takes some work, independent on how fast you do it. It is not the friction that makes this work, it is mainly the steps, the fluid goes down and up. If you do it twice as fast you do in the same time twice as much work and you move trice as much fluid in your buckets.
In a closed tube system pressure of the fluid is the same everywhere. If the tube system is very large, like your city water system, then we have to take into account that pressure moves by waves, but not in an engine like here.
I must conclude that you have no idea of basic physics and you are totally confused. You claim that I have said something of mass times distance. I never said anything like that. As you are very clearly a total amateur in physics, I explained it to you with simple practical examples, but you do not even understand even with simple examples. I did study physics and theoretical physics and I can see that you did not, or you were very poor on both topics. You mix a simple thing with something that you totally do not understand. Are you, perhaps, the FB troll from the Unz review? You do sound like him.
This FB from the Unz forum, I think it is you. We once had a discussion of the Trade Center towers in unz. FB gave an analysis, where he mixed up force with torsion (force times length from the pivot point), showing that he (that is, you) know nothing of physics, just check that discussion if you think FB is correct in physics. FB (that is, you) additionally called Incinatus Inchworm, though Incinatus was quite correct that time. This FB wrote long comments which looked like there was some valid analysis, but when you went through them it turned out that FB made logical errors all the time. He did not understand basic physics. Here also you are quite incorrect: the pressure is equal in fluid if we ignore gravitation caused higher pressure on a lower level. You mix up things, just like FB or Unz. I do have a long education in physics, theoretical physics and mathematics.
You also have not changed your style of discussion. I think there is no sense to discuss with you any longer.
As for this statement of yours…
‘Turn the tap and put your hand in front of the ho[s]e. When the tap is more open, you feel more pressure and you have more water coming out of the ho[s]e.’
Yes there is more water coming out because you have increased the outlet area…
The pressure is less when the tap is opened only a little, due to THROTTLING…
This is also how the pressure is reduced in the F1 for the 30 percent of kerosene flow that doesn’t flow through the cooling tubes, but flows directly to the combustion chamber…
Ivchenko writes, starting on page 20…
‘…it’s necessary to equalize the pressure in the injector plate – that is to throttle the kerosene flow going directly to the injector manifold…’
This is necessary because the 70 percent flow going through the tubes has lost a lot of pressure due to friction…and Ivchenko points out that if the direct flow from the turbopump of 30 percent were not throttled to reduce its pressure…no flow would come from the tubes, due to its lower pressure…
Cengel says this about throttling valves…
‘Throttling valves are any kind of flow-restricting devices that cause a
significant pressure drop in the fluid. Some familiar examples are ordinary
adjustable valves, capillary tubes, and porous plugs (Fig. 5–32). Unlike turbines, they produce a pressure drop without involving any work.’
Which is why Ivchenko points out that the 30 percent flow through those throttling valves is a waste of turbopump work…
Besides that, your example is poorly constructed…
What exactly are the boundaries of the control volume here…?
To be realistic the control volume needs to be just the section of water pipe leading to that one single tap…nothing farther upstream…
When the tap is closed, there is water pressure in that pipe, but water is not flowing…when you open it a little, the mass flow at the pipe will be exactly equal to the mass flow coming out of the tap…
Therefore it is the FLOW AREA of the tap valve that determines velocity…as per conservation of mass…
mass flow 1 = mass flow 2
V * A 1 = V * A 2
where 1 is the pipe just upstream of the valve…
The velocity at the pipe will be a lot lower because its area is a lot bigger than the valve flow area…here is basic info on throttling valves…
https://www.controlglobal.com/assets/Media/MediaManager/RefBook_Cashco_Fluid.pdf
‘Forget the friction. We have no information of the tube system in order to consider it.’
Yes we do…
http://heroicrelics.org/info/f-1/f-1-thrust-chamber.html
This provides full details about the tubes…including pictures of the bifurcation…the diameters…EVERYTHING…
It is quite a simple thermal engineering exercise to get a first order approximation of the pressure drop, delta P…
First we know that the kerosene mass flow is ~800 kg/s…70 percent of that is 560 kg/s…
There are 178 tubes of 1 3/32 inch outside diameter…that then bifurcate into 356 tubes if 1 inch OD…
We can calculate just the pressure drop in that top section of 178 tubes to get a rough idea…
The thrust chamber is 11 ft long, so the circuit length of the passing fuel is 22 ft…actually more when accounting for the bends…
In SI units that is a total tube length [both up and down] of roughly 7 m…judging by the pictures the top part comprising the combustion chamber, nozzle throat, and the bell nozzle up to the 3:1 expansion plane would be about half the total length…so a total tube length of 3.5 meters…
The primary tube diameter is ~27 mm or 0.027 m…
We now have everything we need to get a rough calculation of the pressure drop in that top part of the thrust chamber…
First the mass flow per tube…only half of the tubes carry the output of the turbopump on the down trip…the other half carry it back up to the C/C…
So 560 kg/s divided by 89 tubes = 6.3 kg/s per tube…the density of kerosene is 820 kg/m^3…
Median fluid velocity is then…~10 m/s
[mass flow * 4 / density / pi / D^2]
Re number is then…~155,000
We next find the Darcy friction factor from the Moody Diagram for smooth tubes…or using the Petukhov empirical formula…which either way works out to about 0.02…
Now we have everything we need to find the pressure drop, using the Darcy-Weisbach formula…
https://en.wikipedia.org/wiki/Darcy%E2%80%93Weisbach_equation#Pressure-loss_form
We get a pressure drop of ~185 kPa…which is ~27 psi…
That’s just the tube flow in the top part…we then have the bottom part, plus manifolds, bends etc…total pressure-loss by the time of reaching the combustion chamber would be more than double this rough estimate…
That is about 3 percent of the pump pressure that is lost…the rest is from throttling the kerosene through the fuel spray nozzles…
Actually 3 percent pressure loss is quite good…most gas to gas or gas to liquid heat exchangers are lucky to get 5 percent per side…or 10 percent tta for both fluid streans…
Bottom line is Ivchenko has presented a solid analysis…your ‘critique’ and your comments here show that you have no business delving in thus subject…
“That is about 3 percent of the pump pressure that is lost…the rest is from throttling the kerosene through the fuel spray nozzles…”
As I wrote in the post, the article did not give enough data to know if the liquid turns to gas in the heat exchange process or if it stays as liquid, but should cause only a small difference, since gas must be in the same pressure as the liquid. However, the volume of the liquid changes when some of it turns to gas and this makes some difference, which I expected to be rather small. You searched for more data, calculated it and got 3% pressure loss. I will not check your calculation, let us assume you got it correct. Such small pressure changes you can have for many reasons. For instance, the tubes that are lower have a slightly higher pressure that tubes that are higher because on the Earth gravitation field liquid that is above some level causes pressure to the lower levels. But all such small effect are usually ignored in a simple analysis. So, forget 3% pressure loss.
3% is a small difference and does not much change the conclusions. If so little pressure is lost, it is almost no pressure lost and we can do a simple analysis with a model where the liquid stays as liquid and pressure is the same everywhere. It still is so that the liquid moves with alonst twice the speed and consequently cools almost twice as much. The work done by the force (pressure times volume) is to move the liquid against gravitation and friction (probably small component), and push it through the nozzles. The nozzles you can think for instance as hitting wooden sticks through a small hole with a hammer. It takes some work to hit a stick through, independent of the time it takes. Do it twice as fast, it takes twice as much work, or do twice as much work, you get twice as many through in a time unit. Consequently, doubling the pressure (=doubling the force=doubling the work) pushes fluid twice as fast through the hole, and then the cooling system cools (almost) twice as much. Ivchenko ignores the higher cooling power in his analysis, which invalidates his analysis. His analysis has other problems and the article is poorly written for a scientific article. It is not a solid piece of analysis.
‘…the article did not give enough data to know if the liquid turns to gas in the heat exchange process or if it stays as liquid…’
How can the kerosene boil when it is under more than 100 atm pressure…?
The boiling point of kerosene is very high…up to 290 C under simple atmospheric pressure…
Also you are sticking to your nonsense about twice the pressure equals twice the flow velocity…without showing any math…
I showed the math for finding the fluid velocity in each tube…
[mass flow * 4 / density / pi / D^2]
6.3 kg/s * 4 / 820 kg/m^3 / pi / .027^2 = 13.4 m/s
[Note…I had made a typo in my previous post, stating fluid velocity to be 10 m/s…however the calculation for pressure drop as stated previously is correct, because I used the correct velocity number in my spreadsheet…]
Note also that the above equation is simply the continuity equation but rearranged to use tube diameter instead of area…
We can alternatively find the area of the 0.027 m tube which is 0.000573 m^2…
And then use the standard continuity equation, but rearranged to solve for velocity…
Mass flow = density * V * A
V = mass flow / density / area
V= 6.3 / 820 / 0.00057 = 13.4
Now please go ahead and prove mathematically that doubling pressure is going to double the velocity…
You will notice there is no pressure term in the continuity equation…only density, which does not change with pressure in an incompressible fluid…
A compressible fluid like air is a different story…higher pressure means higher density…so in a jet engine compressor, the fluid flow passages get increasingly smaller with each compressor stage in order to maintain the design air velocity, which is about Mach 0.5…
I will not be continuing this discussion, unless you can show me some bona fide math to back up your nonsensical claim about velocity increasing with pressure…
I have already disproved your first claim that pressure will be the same everywhere in the tubes…doesn’t matter that it’s ‘only’ 3 percent…it shows your understanding was wrong…
“I have already disproved your first claim that pressure will be the same everywhere in the tubes…doesn’t matter that it’s ‘only’ 3 percent…it shows your understanding was wrong…”
The pressure is the same in liquid and it does not matter whether there is gas or not. The gas will also get the same pressure. Any material that can move must have zero total force affecting on it, else it accelerates to some direction. This is the simple reason why the pressure in liquid is the same everywhere. What you mean by the 3% pressure loss may be a pressure loss from the planned pressure as the pressure in the pump is the same as the pressure everywhere.
Pressure in a tube and flow through the tube are linearly related. You see it easily if you think of a water hoe. Open the tap more, there comes more water from the other end, put your hand in front of the water and you see there is more pressure. It is because the tap controls the aperture A. The ration of A to the diameter D of the tube determines both the pressure in the tube (a/d of the pressure in the other side of the tap, the city water pressure) and the flow through the pipe/ (A/D of what would flow if you open the tap fully). So, these are related linearly, just as I write. In a more fine-grained analysis you can find nonlinear forces, like friction, but in a simple analysis it is as I have it.
We stop this discussion now. Good luck.
I’m confused by all this tech jargon posted here. Can anyone tell me how many beans makle 5?
“Can anyone tell me how many beans makle 5?”
This is not from my text. You must be referring to the paper that claims to prove that Saturn rocket did not go to the Moon, the paper that I refute in my post. I found that paper very hard to read and after i had found sufficient errors to discard the arguments in the paper, I stopped reading. There were many unclear expressions. I wrote this post many years ago and cannot remember if I noticed the expressions “beans makle 5”. I probably would have understood makle as make and tried to understand it from the context. I reread what I wrote in the post and find my arguments quite easy to follow. They refute the paper and there is no need for you to read the paper more carefully: it is wrong.