help needed in my project "compressed air driven vehicle" full information given?

Question

the engine of my proposed vehicle is a pneumatic impact wrench ,the one we see in automobile garages for turning nuts.. these

are always run by a compressors operating parallely for the supply of compressed air ( fuel ).the specifications of the

pneumatic wrench that is thought to be as the prime mover (engine) are....
1.air consumption --- 6cfm(cubic feet per minute)
2.operating pressure ---6.2 bar (90 psi,pounds per square inch)
3.max torque --- 500 lb-ft (678N-m)
4.weight --- 4.6 Kg

the project should not include a compressor ,instead vehicle should be incorporated with a air storage tank .so it is

thought appropriate to use a well capacitated tank that can store compressed air at atleast 30 bar so that the vehicle could

run some distance..say abt 400 to 500 meters.
the storage tank specifications are
diameter-40 cm
length - 1 m (even though the tank has hemispherical ends the additional volume is not included)

volume of cylinder= 3.14(pi) x rad^2 x length =3.14*.04*1=0.125cu m (cubic meter)
1cu m = 100 litre (lt)==>0.125 cu m = 125 lt (this 125 lt is the capacity of cylinder)
the compressed air is thought to be stored in the tank at the pressure of 30 bar.
so the total amount of air available is equal to 125 * 30 = 3750 lt ( this is called as total air capacity)
but... the wrench does not run if pressure falls below 6.2 bar right!so the air capacity at this pressure ie 6.2 bar is
125 * 6.2 = 875 lt ( rounded to 875 ). so the total air available for the wrench to function is ( 3750-875 )lt = 2875 lt take

it as 2800 lt (for minimum time for which the wrench can be operated) .the air consumption of wrench is 6 cfm = 171.4

lt/min.(round it of to 175 lt/min for min time)so the time for which the wrench can be operated is (2875 lt)/ (175lt/min)=16

minutes.
the total work(W.A) that can be done with the compressed air in the tank is equal to (V * dp);

V = capacity of cylinder & dp
is the pressure difference i.e (30 bar - 6.2 bar)=23.8 bar.so W.A = 0.125 cu m * 23.8 bar = 297500N-m (bcoz 1 bar = 10 ^5
N/sq m).so work available W.A = 297500 N-m .
now consider the total loads on vehicle
1.weight of structure of vehicle = 200 kg
2.wt of person = 100 kg
3.wt of cylinder = 125 kg ( i guess)
4.wt of wrench = 5 kg
5 weight of compressed air at 30 bar= 4.5 kg
6.extra fittings wt = 25 kg (say)
total weight = 459.5 kg (say it as) 475 kg
so this mass is acting vertically downwards in "y - axis" so the normal reaction (N) acting on vehicle is (m*g)==> 475 * 10 =
4750 N .THE VEHICLE HAS TO MOVE IN " X - AXIS " right! so it has to have a mininum force so as to overcome frictional force which is equal to ( u * N ); where u = coeff of friction between wheel and tyre .since the frictional force is rolling
friction it is very less; also once it starts moving the frictional force ( F ) to be overcomed for the movement decreases

the F = 0.15 * 4750 = 712.5 N say it 720 N.

AM I right ??? does my pneumatic wrench helps in moving the vehicle ???

what i think is......
1.the value of F is much less than 720 N when in motion.work = force * distance ;the above calculated work is equated to F *
d (d = distance moved )force is 720 N so.... d = 297500/720 = approx 400 m . but in actual i thing it is still more cos of

gear arrangement bcos of which speed increases and hence distance ,also it is rolling friction and not static so once the

body is set to motion ,there is no need to apply 720 N even lesser force would be enough!! am i right????????????

2.even though 3 gear shifts are proposed for speed increments of vehicle ,for convinience and for finding out whether the
initial torque provided by wrench is sufficient in moving the vehicle , i assume that the wrench is directly coupled to axle
by a gear arrangement such that the direction of rotation of the wrench's drive is turned by 90 degrees

so that the axle

rotates in the direction of rotation of wheels . ok!!!!! now due to such a SUN - PLANET gear arrangement let the torque fall

to a value like.... say around 550 N-m (i.e from 678 N-m to 550 N-m )so now the axle is driving both the rear wheels with 550

N-m.this torque is constrained to the curcumference of axle only!!??? got ittt!!!! i mean ... torque = force *

distance ok!!this force acts tangential to the axle's circumference and distance is equal to radius of axle ( say it .02 m)

hence what i meant was the above torque is torque acting on axle only ;

3.the force required to move the vehicle overcomming the friction is 720 N which acts tangential to the rear wheels!!! am i
right??????so the torque required to move vehicle is = ( 720 * radius of wheel ) ==> 720 N * 0.2 m = 144 N-m okkkk but the

torque available at axle is 550 N-m i.e at distance of 0.02 m from center of axle ..so the torque with which the axle twists

the wheel is

equal to ????? it can be found from equation torque (T) = F * distance ==> T is directly proportional to

distance ( radius) so Ta/ Ra = Tw/ Rw (Ta = torque at axle, Rw = rad of wheel ,Tw = torque at wheel ,Ra = radius of axle );

so 550/.02 = Tw/.2 ==>
Tw=5500N-m;that is huge amount of torque is available for the movement; actual required torque is 144 N-m which is veryyyy

small to what is available ,even if the torque required is more say around 300N-m ,it can also be overcomed .

?????? am i right ???????if not plsssss suggest meee and correct mee

?????? am i right ???????if not plsssss suggest meee and correct mee

Answer 1

I would think the impact wrench would spend a lot of time and use a lot of air impacting and not turn the drive mechanism continuously.

http://www.bluetools.com/Merchant2/merchant.mv?Screen=CTGY&Store_Code=bluetools&Category_Code=air_motors2
http://www.tonson-motor.com/?OVRAW=air%20motors&OVKEY=air%20motor&OVMTC=standard

Answer 2

I tired the same thing........and after years of trying to prefect it, I gave up. I suggest the same. Give up. Your wasting your time.