Angular torgue and this can be turned into a normal torque

[herpamad]

I have been searching the net for well over an hour, and i can't seem to find an equation that gives me a result.

What it is, i know that a torque is 19.5Nm (may be rounded), and i know other variables are input rpm = 9700rpm and power = 2.3kw.

What i am trying to do is calculate the torque that an electric engine produces when the engine is producing 2.3kw @ 9700rpm.

There seems to be loads of torque equations, and some in imperial and some metric.

Can be torque be calculated as an Angular Value (Omega)?

I keep getting a value around 2.2n/m, so i was wondering if i am calculation torque wrong, or if i am getting an angular torgue and this can be turned into a normal torque as such?

Thanks

 

[Dr.D]

Your problem statemnt indicates that you have all of your values, so what you want is a bit unclear.

The several quantities are related as follows:
N = shaft speed, rpm = 9700
P = power = 2.3 kw = 2300 w
omega = shaft speed, rad/sec = 2*pi*N/60
T = shaft torque, Nm
omega*T = P is the basic relation

You can rearrange it, or substitute in the numbers and do the arithmetic to get your numbers to suit yourself.

Torque is an angular quantity, and there is no such thing as a "normal torque."

 

[herpamad]

Think i am getting confused with Shaft Torque and Input Torque?

Here is the one equation i have been using, but i am not sure what this torque output is?

T = Power Input (kw) * 9950 / R.P.M

What i need to do is establish input torque and the shaft torque? Not 100% sure on how to do this.

Variables I have that will be of use

1) Input RPM = 9700
2) Output RPM = 1950
3) RATIO = 4.97
3) Power Transmitted = 2.3kw

The notation of input torque was T(Omega), so this is torque in radians?

Torque increases with power right?

Power = Omega * Torque, but why is this higher that the transmitted power? and is this in units of K/W?

The input torque and shaft torque i am looking for are denoted as been in units of N/M

I have figured out why the figure was 19.5, this was because it was in FT/ib rather than been N/M.

If you get more torque from the gear step down, what is the loss? as you can't have something for nothing right? Is it a loss of velocity?

 

[Dr.D]

Shaft torque is the torque in a shaft. Input torque is the torque input to a system, which will necessarily come through a shaft in most cases. Output torque is the torque out of a system, also usuually a shaft torque.

Torque has units of Force * Length, not Force / Length as you have written several places.

I have no idea where you 9950 came from, but I gave you the basic relations in the previous post. That is all that there are.

 

[stewartcs]

Think i am getting confused with Shaft Torque and Input Torque?

Here is the one equation i have been using, but i am not sure what this torque output is?

T = Power Input (kw) * 9950 / R.P.M

What i need to do is establish input torque and the shaft torque? Not 100% sure on how to do this.

Variables I have that will be of use

1) Input RPM = 9700
2) Output RPM = 1950
3) RATIO = 4.97
3) Power Transmitted = 2.3kw

The notation of input torque was T(Omega), so this is torque in radians?

Torque increases with power right?

Power = Omega * Torque, but why is this higher that the transmitted power? and is this in units of K/W?

The input torque and shaft torque i am looking for are denoted as been in units of N/M

I have figured out why the figure was 19.5, this was because it was in FT/ib rather than been N/M.

If you get more torque from the gear step down, what is the loss? as you can't have something for nothing right? Is it a loss of velocity?

The torque would be:

$$T = \frac{P_{shaft} \cdot 7.04}{n_r}$$

where,

T is the torque in lb-ft
P_shaft is the power in watts
n_r is the rotor speed in rpm

Which yields about 1.66 lb-ft or 2.25 N-m.

Now, the real question is if the power you have stated is the input power to the motor or the shaft power (remember that all machines have efficiencies due to irreversibilities).

CS