newrobert » Tue Nov 18, 2014 9:53 am wrote:...
In fact I have one amusement park project where I required to lift heavy water at 250+ feet height, in other words simply 100+ meter. Water is required for some amusement games/rides and water will drop in ground pool again from where it requires to be lifted up again. So traditional pumps cost is very very heavy and impossible to bear as compare to water volume.
...
complete 1 round in 1015 minutes that's enough for us. I mean buckets will be filled from ground pool and drop at 100m+ height pool by installing a small mechanisim. Water like 400k500k gallon per round.
CanadysPeak » November 19th, 2014, 10:57 pm wrote:You should listen to Marshall. Since you have put friction (inefficiency) aside for the moment, you have a conservative problem, that is, it is path independent. Thus, it doesn't matter what you use to do the lifting at 100 % efficiency, and it doesn't matter whether it is straight or circular. You may have some trouble with those ideas, so I urge you to do some experiments. I have built several large linear and circular machines. I have seen the equations work out. I have used a circular pump utilizing water cups probably several thousand times in my life.
There is no magic in engineering. It's all careful attention to detail.
newrobert » Wed Nov 19, 2014 11:11 pm wrote:CanadysPeak » November 19th, 2014, 10:57 pm wrote:You should listen to Marshall. Since you have put friction (inefficiency) aside for the moment, you have a conservative problem, that is, it is path independent. Thus, it doesn't matter what you use to do the lifting at 100 % efficiency, and it doesn't matter whether it is straight or circular. You may have some trouble with those ideas, so I urge you to do some experiments. I have built several large linear and circular machines. I have seen the equations work out. I have used a circular pump utilizing water cups probably several thousand times in my life.
There is no magic in engineering. It's all careful attention to detail.
Thank you for posting and I also listened to Marshall.
1. I know there will be friction and inefficiency.
2. For a moment please ignore that water is loaded on train, you just assume that there is mass on train that is moving in circular path like roller coaster. Mass is equally distributed on train. Then what power will be required to calculate power.
Are my calculations above are correct?
Robert
newrobert » Sat Nov 22, 2014 2:09 pm wrote:I understand that difference is that filled buckets are going up and empty buckets are coming down. So if roller coaster is moving clockwise then it means from right side empty buckets are coming down and from left side filled buckets are going up.
Left Side = Kinetic Energy (filled buckets moving up, around 300 ton weight)
Right Side = Potential Energy (empty buckets coming down, might be 10 ton weight of empty buckets only)
So I think a solution of this so how to balance this.
What if we can manage to put 100200 ton weight on topright side of roller coaster (from where empty buckets are coming down). For example one 20 ton concrete block is fixed in a steel cage (hanging with strong steel wires,just not let him fall) and there are heavy tyres (like plane) those can bear load of concrete and then these tyres are transferring potential energy (by of 20 ton concrete block) to roller coaster chain/belt by tyre friction? Just like many aumusement park rides are tyre driven (frictional drive).
So at left side there is weight of water and at topright side heavy concrete blocks are trying to balance weight.
Will this balance something?
Robert
CanadysPeak » Thu Nov 20, 2014 4:28 am wrote:For a circular motion, the mass is not the same. You have the wheel PLUS water going up and the wheel only coming down. The power will be calculated based on the rate at which you raise the water. At 100 % efficiency, the actual mechanism is insignificant. The amount of power is the same in all instances.

newrobert » Sun Nov 23, 2014 4:50 am wrote:Dear Marshall / CanadysPeak,
Thank you for replying to this post for my better understanding.
My calculation in this thread
In this thread as I calculated and you people did not find any mistake on these calculations i.e. this method of calculation is correct.
Mass moved up in 1 round: 863,228 kg
Mass moved up of one side train: 849,428 kg (water weight + buckets weight)
Mass equal to gallon: 441,499 gallon
50% of total water: 220,750 gallon (we calculated one side here @ 50% time i.e. 7.5 minutes)
Water lifted up: 220,750 gallon (equals to 849,428 kg mass)
Water lifting Time: 7.5 minutes (50% of 1 round time, from bottomtop)
Water lifting height: 107 meter (net height of water lifted)
Per Minute Water Lifted: 220,750 gallon / 7.5 minutes = 29,433 GPM
Motor Power: 26.50 KW (excluding frictions) or we can consider this 50 KW
So as per calculations here 26.50 KW motor is rotating this roller coaster and in 7.5 minutes 220,750 gallon water will be lifted at 107 meters in 7.5 minutes @ 29,433 GPM.
Simple Calculations
If same mass is lifted at same height then here are simple calculations.
Mass: 849,428 kg
Height: 107 meters
Time: 450 seconds (7.5 minutes x 60)
Speed: 107/450 = .24 m/s
Force: Mass x Gravity = 849,428 x 9.81 = 8,332,889 N
Work: Force x Height = 8,332,889 x 107 = 891,619,089 Joule
Power: Work / Time = 891,619,089 / 450 = 1,981,376 Watts (1,981 KW)
Water Pumping Calculations
Here are water pumping formula that I found on Internet.
Water Flow: 29,433 GPM
Height: 107 meter
Height: 351 Feet
Power Required: (Feet x gpm)/3960 = (351 x 29,433)/3960 = 2,609 HP = (1,945 KW)
1. What's wrong in our calculation?
2. I did not know anything about "perpetual motion machines" but I will now I will know more about these.
3. You understand correctly that I think that "due to circular motion less energy can be consumed to lift same amount of water as compare to traditional water". There is no "centripetal force" involved in simple water pump and this force can help in saving some energy.
4. I want to understand this scientifically. To make a water wheel or roller coaster how much energy will be required.
Marshall / CanadysPeak, can you please give your comments on Point # 1 + 3 + 4.
I am making a diagram of what's in my mind regarding recent idea of adding some potential energy at right side of roller coaster from where empty water buckets are coming down.
Robert

Marshall » Wed Nov 19, 2014 8:43 pm wrote:This part is wrong. The work done in moving something is not equal to the K.E. multiplied by the distance.
...
==mistake==
Work (J) = K.E. x Height
Total circumference is 336 meter and distance of bottomtop (1 side) is 168 meter
Work (J) = (K.E.) 59,249 x (distance) 168 meter
Work (J) = 59,249 x 168 = 9,958,211 Joule
It means 9,958,211 Joule is required to move this mass at 168 meter.
==endquote==
...
newrobert » Wed Nov 19, 2014 11:24 pm wrote:...I mean I also calculated correct Joule.
Work (J) = 59,249 x 168 = 9,958,211 Joule
...:)
K.E (J) = 59,249 Joule
Work (J) = 59,249 x 168 = 9,958,211 Joule
Power = Work (J) / Time
Power = 9,958,211 / 450
Power: 22,129 watts
Motor inefficiency: 10% (enough or add more)
All frictions: 10% (enough or add more)
Total Power Added: 20%
Net Power: 22,129 x 4425.80 = 26,554 Watts
So I think a solution of this so how to balance this.
What if we can manage to put 100200 ton weight on topright side of roller coaster (from where empty buckets are coming down). For example one 20 ton concrete block is fixed in a steel cage (hanging with strong steel wires,just not let him fall) and there are heavy tyres (like plane) those can bear load of concrete and then these tyres are transferring potential energy (by of 20 ton concrete block) to roller coaster chain/belt by tyre friction? Just like many aumusement park rides are tyre driven (frictional drive).
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