# How to determine the compressor power in a refrigeration system?

The picture below represents a refrigeration unit In order to calculate the power produced by the motor, two quantities must be measured

1. Angular speed of the motor. This parameter can be observed from the tachometer, which indicates the rotational speed of the the compressor. The belt pulley ratio is used to find the rotational speed of the motor.
2. The torque produced by the motor. In my lab's manual, it is written that the spring balance reading times the troque arm is equal to the torque produced by the motor. Can you help me understand how does this spring balance reading indicates the torque produced by the motor? (I know that the the torque is equal to Fxd, but I can't understand how does the spring balance reading represent the force produced by the motor)

Also, it is written in the manual "Assume typical compressor friction force =5N"

Does this value represent the friction between the belt and the motor? If not, then What does this value represent?

And I think that since there is friction, then power dissipation will occur due to friction and thus Power delivered to the compressor = Power produced by the motor minus power dissipated by friction"

Please, if there is something wrong, correct me

• Eman, I want to help- is your problem that you do not understand how the spring gauge measures torque? – niels nielsen Nov 2 '17 at 5:31
• @user40292 yes this is my problem – Eman.suradi Nov 2 '17 at 7:58

The motor frame isn't bolted to anything. The pillow block bearings locate the entire motor. The torque arm is the only thing stopping the motor frame from spinning instead of the drive pulley.

I don't know what the 5N refers to. If I had to guess, It would say it's the piston rings on the cylinder wall.

Assumptions:

• length(arm) is known (dynomometer reaction arm)
• Current to motor (I) is known
• Voltage to motor (V) is known

Measurements:

• Angular momentum of motor (rpm)
• Force applied to load cell of dynomometer

Computations:

• Torque = sum(F) X length(arm) --> T = (F(load cell) - F(spring)) x length(arm) --> T = (F(load cell) - 5N) x length(arm)

• P(mech) = T x ω(rad/s) where, ω(rad/s) = ω(rpm) x (2*pi)/60

• Compressor Power = P(mech) = (F(load cell) - 5N) x length(arm) x ω(rpm) x (2*pi) / 60

• P(elec) = I * V

• P(elec) = P(mech) + P(loss)
• P(loss) = P(elec) - P(mech)

Therefore: P(loss) = (I * V) - ((F(load cell) - 5N) x length(arm) x ω(rpm) x (2*pi)/60)

Balance the work/power of the system, while accounting for losses due to inefficiencies (frictional, electrical and thermodynamic).

• the load cell of the dynamometer? - all the ones I used had a system of weights (masses...) that had to be hung to find the value then multiplied by the length of the arm... – Solar Mike Nov 1 '17 at 6:58
• The photo provided shows a lever arm, and what appears to be a mass with a spring attached. The photo is a bit non-specific, but labels a dynomometer, and load cell. So, by extension a weight is implied. Your thoughts? – takala Nov 1 '17 at 13:06
• The Op said that the manual gives "spring balance reading times the torque arm" - but, it does show a star with load cell... To believe the manual or not... – Solar Mike Nov 1 '17 at 13:17
• There are many stars in the photo provide not asked about in the users question not inquired about. The sum of the forces applied at the dyno lever arm would be load cell reading - 5N. i will edit the equations. – takala Nov 1 '17 at 13:23

OP has messaged me that his problem is understanding how the spring force setup measures torque. Let's see if this furnishes the answer he wants:

Imagine an electric motor which is connected to a load. Assume that it is performing work on the load, which means that

1. the motor is exerting a force on the load

Assuming a constant force, the net work done by the motor on the load is then w = (force) x (distance) and the power is p = (force x distance)/time.

Since velocity = distance/time, we can rearrange the power equation to read p = (force) x (velocity) and since we are talking about rotary motion, the force involved is a twisting action and the velocity is a rotational speed. Thus the power equation is translated to

shaft horsepower = (torque) x RPM, taking care to keep our units consistent.

Now we are ready to look at the motor and the apparatus connected to it. First we note that the electric motor is applying a twisting force (torque) to the load which means the load is applying an equal and opposite twisting force to the motor. this means the motor casing is applying that same twisting force to the foundation it is bolted to, which means those bolts are transferring the torque reaction. In turn, this means that if we can determine the stresses in those mounting bolts, we can solve for the torque being produced by the motor, multiply that torque by the RPM at which the motor is running, and then know the power being produced by the motor and absorbed by the compressor.

The diagram above shows the motor is mounted on a pivot so that anything that applies a torque to the motor will cause the entire motor to rotate about that pivot. To prevent this from happening, a spring force gauge is also connected to the motor mount through a lever. the spring force gauge is simply a spring which deflects a known distance under the influence of a force and hence lets you measure that force.

And here is the solution: the spring force gauge can be thought of as a mounting bolt which lets us measure the stress in it which is exerted upon it by the motor. the force that the spring gauge senses is being applied to the motor via a lever arm as specified in the drawing, and since torque = force x lever arm, the spring force gauge reading can be converted to the torque being applied to the load by the motor.

Let me know if this answers your question. Please note that this arrangement of a force-gauge-and-lever-arm plus an RPM gauge is known as a dynamometer and is the standard tool for measuring the horsepower being transferred by rotating shafts. You can find more examples of dynamometers on the web.