Which speed results in maximum engine efficiency

descriptionFormula symbolunitUnit designation
Power (at the shaft)P2[W]watt
TorqueM.[Nm]Newton meter
rotation speedn[min-1], [rpm],
[Rpm], [rpm]
revolution
per minute
jobW.[Nm]Newton meter
forceF.1630Newton
Weight forceF.G1630Newton
Angular
speed
w
(small omega)
[1 / s], [s-1]revolution
per second
Dimensionsm[kg]kilogram
accelerationa[m / s²]Meters per second squared
Acceleration due to gravityG[m / s²]is factor -> 9.81 m / s²

POWER

P2 = (M ·  2 · · n) / 60
P2 = (M * n) / 9550
P2 = W / tW. = F * s
P2 = (F * s) / t
F = m * a
F.G = m * g

The speed v is replaced by the speed n times the circumference d .
The speed is measured per minute, the speed per second. This gives the division by 60.

If the power is to be output in kW, the factor 1000 is added under the fraction line.
1000 · 60 / 2 · then results in the conversion factor 9550, if one as 3.14 and rounds down accordingly.

 

Example 1:
A crane lifts a load of 90 kN to a height of 4 m in 25 seconds
Lifting motor without friction losses?

Solution: P = (M * n) / 9550 = (90000 N * 4 m) / 25 s = 14 kW

 

Example 2:
A shaft has to transmit a power of 15 kW at a speed of 180 1 / min. How large is
the torque?

Solution: M = (9550 xP) / n = (9550 * 15 kW) / 180 = 795.833 Nm

 

Example 3:
Brake dynamometer (Prony's bridle brake), To determine the effective power of an engine, we use a brake dynamometer of l = 2.2 m lever length with FG= 81 N at speed n = 30 1 / s loaded, P =? kW.

Solution:

P = M * w = FG* l * 2 * p * n
P = 81 N * 2.25 m * 2p

P = 34353 Nm / s = 34.35 kW

TORQUE

9550 * P [kW]
Md [Nm] = ------------
n [min-1]

Example 4: electric motor
A three-phase standard motor has 1420 min-1 Nominal speed according to the nameplate and a power of
15 kW.

  1. What is the torque of the motor at the nominal speed?

  2. Solution:

9550 * 15 kW
Md [Nm] = ------------ = 102 Nm
1420 min-1


  1. This motor is to be controlled electronically. What torque does it have at 100 min-1
    and at 1 min-1 ?
     

    Solution:
    The torque with electronic control is the max.torque at nominal speed! It remains roughly the same until the speed is zero.

    -> 102 Nm at 100 min-1
    -> 102 Nm at 1 min-1

    Danger:
    The motor has no or insufficient cooling below approx. 900 rpm! External ventilation must be installed.

Example 5:
Electric motor A high-speed motor should have a maximum of 15,000 min-1. Are needed
at 9000 min-1 a torque of 1000 Nm. What service is required?

Solution:

9550 * P n * Md 9000 min-1 * 1000 Nm
Md = -------- => P = ------ => P = ------------------- = 942 kW
n 9550 9550

The order must read: three-phase motor 942 kW at 9000 rpm-1; max.speed 15000 min-1.

Example 6: gear motor
A worm gear motor is supposed to pull a cart. The tensile force was determined using the lever arm and weight. At the point of application of the gear shaft, a lever arm 1 m in length requires 1 kg of weight in order to turn the wheel properly and thus to pull the car properly. The necessary torque is therefore 1kg * 1m = 1kpm => 10 Nm!
The speed of the wheel should be 10 min-1!

What gear motor power is required?

Solution:

9550 * P n * Md 10 min-1 * 10 Nm
Md = -------- => P = ------ => P = ---------------- = 0.01 kW
n 9550 9550

0.06 kW is chosen from our lists!

Reason:

  1. this is the smallest, cheapest standard motor with gearbox
  2. Worm gears have poor efficiency. So you always have to choose a higher output.

Sample:
The torques are given in the technical lists. E.g. with type VFR44 with 9.6 min-1 and 0.09 kW; i = 140 the torque of 4.9 daNm = 49 Nm is specified! 0.06 / 0.09 kW = factor 0.66
0.66 * 49 Nm = 32 Nm at 0.06 kW is fully sufficient

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