Maths assistance required.

[ripcrow]

Racket in your original comment you quoted steam exiting a turbine at 40 degrees and 0.5 psia. That got me thinking. That’s 25 torr which is well below 40 degrees celcius boiling pressure. At 40 degrees celcius every kilogram of vapour still holds .677 kw of energy.

[ripcrow]

Jan 31, 2021 2:25:00 GMT -5 racket said:

You might need to redo your volume calculation again , its a "tad" optimistic

Missed a few decimal points. Does 0.0402 cubic metres per second look better using the pressure differential value as the difference between 7372 and 653 pascal I’m getting a different velocity also. My velocity calculated out to 1843 kph. Or 512 metres per second or 1679 ft per second. Could you check my maths.  Nothing looks right lol 

[ripcrow]

How do you do force calcs when there is no acceleration. The above problem has a velocity. Do we assume the velocity is accelerating an object from 0 to the speed of the fluid where no further acceleration can take place or do we just use velocity as the acceleration in the formula f = ma

[racket]

0.00714 cubic mts/sec

1/2 M X V2 kinetic energy , half mass times velocity squared ??

[ripcrow]

Jan 31, 2021 19:07:06 GMT -5 racket said:

0.00714 cubic mts/sec

1/2 M X V2 kinetic energy , half mass times velocity squared ??

Messed that up again. Lol. I’ll try another calculator. Thanks. 

[ripcrow]

Thanks for helping racket. We seem to have different values again but I reverse engineered your flow rate figure and found that you are using 1193 ft per second as the velocity. The online calculator I used calculated the pressure drop and pipe diameter to give me a velocity of 1679 ft per second. I then remembered that you stated the flow would have a choked sonic velocity of about 1000 ft per second so I checked and the velocity you are using is above the speed of sound so I’m thinking that’s not the problem either. I did find one problem in my calculations and did them again and got a mass flow rate of 0.0005152 kg per second and a volume flow rate of 0.01005056 cubic metres per second with a velocity of 1679 ft per second. Why are our velocity calculations so different.  

[racket]

My flow rate was calculated using your numbers of 1188 ft/sec and a 5mm dia , 1188/3.281=362 m/s , therefore the calc is PiX r squared which for your 5mm is 19.634953, this needs to be divided by 1,000,000 to get its area in sq mtrs , then that needs to be multiplied by 362 = 0.00710785 cubic mtrs/sec .

Another way is divide the 1,000,000 sq mms/sq mt by 19.634 = 1/50,932 of a sq mtr , then divide that 50,932 by your 362 for 1/140.69 cubic mtr or 0.007..... cubic mtrs /sec , just to be certain , lets multiply 0.007 by 140 and we get 0.98 close enough .................you need a new battery in your calculator :-)

All of this is very "theoretical" as the flow area will be reduced by boundary layer, and the actual velocity achieved will be less than the "theoretical" speed due to losses

[ripcrow]

Feb 1, 2021 16:07:42 GMT -5 racket said:

My flow rate was calculated using your numbers of 1188 ft/sec and a 5mm dia , 1188/3.281=362 m/s , therefore the calc is PiX r squared which for your 5mm is 19.634953, this needs to be divided by 1,000,000 to get its area in sq mtrs , then that needs to be multiplied by 362 = 0.00710785 cubic mtrs/sec .

Another way is divide the 1,000,000 sq mms/sq mt by 19.634 = 1/50,932 of a sq mtr , then divide that 50,932 by your 362 for 1/140.69 cubic mtr or 0.007..... cubic mtrs /sec , just to be certain , lets multiply 0.007 by 140 and we get 0.98 close enough .................you need a new battery in your calculator :-)

All of this is very "theoretical" as the flow area will be reduced by boundary layer, and the actual velocity achieved will be less than the "theoretical" speed due to losses

Because I messed up the area calc my velocity was wrong. It’s not 1188 ft per second. When you pointed out my area was a tad optimistic I calculated it again. The correct area gives me 1679 ft per second. 

[ripcrow]

My area calc lines up with yours now. 
I calculated a flow rate of 0.01005056 cubic metres per second which should give a mass flow of 0.0005152 kg per second. I used density at 0.051243 kg per cubic meter and volume of 19.515 cubic metres per kg. As per the vapour pressure table. 
Thus gave me a force of 67.52 Newton’s. With a 65 mm diameter turbine this gave me 2.1944 Newton metres torque. I calculated hp to be 10.79 hp at 35,000 rpm ,15.41 hp at 50,000 rpm and 3 hp at 10,000 rpm. I have assumed no losses and 100% efficiency. If my numbers are correct it’s not bad for such a small mass flow. Haven’t calculated energy use to keep the water at 40 degrees yet though. 

[racket]

Theres no way you'll get that sort of power output from such a miniscule mass flow ....................could you please provide your calculations and we'll see where you went wrong .


A rough guess would be closer to 1/40th HP

[ripcrow]

Calcs as follows. Force using 1/2 mass times velocity squared. Mass is 0.0005152 times 262144 ( 512 metres per second squared) equals 0.0002576 x 262144 = force of 67.52 Newton’s.

Torque is calculated on online calculator using force and distance. Force is 67.52 n acting on a 65 mm diameter so equation is 67.52 n and 32.5 mm radius equals 2.1944 nm or 1.618 ft lb

Hp calculated on this link. getcalc.com/mechanical-horsepower-calculator.htm My iriginal numbers calculate right if I use ft lbs in this calculator. If I use nm it actually says 6.27 hp at 15000 rpm instead of 4.62 which is the answer I got using ft lbs. At 25000 rpm using nm it gives 10.45 hp with nm and 7.7 hp using ft lbs. I went to rerun the numbers on the original calculator I used but found it said electrical horsepower calculator. So I found another calculator that said mechanical hp calculator. The electrical hp calculator gave the same results as using ft lbs in the mechanical calculator. I can’t work out why the difference using nm or ft lbs. I calculated nm to ft lb using online converter.

[ripcrow]

I just found the equations you supplied for calculating freepower hp when I built my jet with your assistance. I don’t think they can be used here. I researched pelton wheel power and found they calculate hp using torque and rpm. Which is what the online calculator used. There is no way a 5 mm diameter nozzle could power a 65 mm diameter turbine like the jet uses but an implies turbine only requires one inlet so a pelton wheel is a possible fit in this thought experiment. If a modified jet turbine were to be used then the equations involving mass flow and gas deflection would be relevant and the nozzle diameter would need to be 65 mm also.

[racket]

OK , your calculators can't do the right calculations

As an example , my 12/118 engine flows 1.5 kgs/sec with an exhaust velocity of say 550 m/s , this produces 226,875 Joules/Newtons but would produce say 250 HP from a freepower , so ~900 Newtons/HP , your 67N would produce 1/13.5 - 0.074 HP , but very small turbines have poor efficiency and along with all the other losses from scale effects the power would be even less.

[ripcrow]

Absolutely agree. It couldn’t even turn a turbine. I have found a formula for shaft horsepower from an impulse turbine. If the torque calculations are correct the horsepower output lines up with my earlier calcs. Thanks for your assistance racket. Much appreciated. The physics gurus had no idea where to start. I’ll continue thinking about this. Thanks again.

[ripcrow]

Feb 6, 2021 23:44:35 GMT -5 racket said:

OK , your calculators can't do the right calculations

As an example , my 12/118 engine flows 1.5 kgs/sec with an exhaust velocity of say 550 m/s , this produces 226,875 Joules/Newtons but would produce say 250 HP from a freepower , so ~900 Newtons/HP , your 67N would produce 1/13.5 - 0.074 HP , but very small turbines have poor efficiency and along with all the other losses from scale effects the power would be even less.

Done the calcs again. The mistake is the area. I missed a step and didn’t divide by 1,000,000 to get square metres. So that threw everything out by a long way. I also didn’t calculate the kinetic energy (1/2 m times velocity squared ) as joules. I calculated it as Newton’s then calculated torque using Newton’s on a radius then multiplied torque by rpm divided by 5252 to get hp. Lots of mistakes by me.