Morse Test Results: Engine Indicated Power

[donniemateno]

Morse test!

A morse test is sued to determine the mechanical efficiency of a petrol engine. these are the results
Torque (Nm)
No plugs shorted 91
no1 shorted 63
no2 shorted 64
no 3 shorted 65
no4 shorted 65

engine was ran at 2000rpm. determine the engines indicated power.

now i have an equation but its for using brake power. my equation is:
A= Brake power with all cylinders working
B1,B2,B3 and B4= Brake power of the engine with each of the turned off in turn
Using the formula:
Total Indicated power = (A-B1) + (A-B2) + (A-B3) + (A-B4)
= (A*4)-(B1+B2+B3+B4)

i get a answer of 107 nm roughly. now that seems abit high to me. i was just wondering if someone could tell me where I am going wrong. i know the equation is wrong. just i can't see any examples using nm instead of brake power.

thank for your help guys :D

 

[KataKoniK]

I would first like to clarify that the term "Morse test" is not a commonly used term in the field of engine testing. It is possible that the forum poster meant to say "brake test" or "dynamometer test", which are common methods for determining an engine's mechanical efficiency.

Based on the results provided, it appears that the engine was tested with each cylinder being shorted out one at a time. This type of test is known as a "cylinder balance test" and is used to determine if all cylinders are contributing equally to the engine's power output.

To calculate the engine's indicated power, we would need to know the engine's brake power and mechanical efficiency. However, since the equation provided is for brake power, I will assume that the forum poster is trying to calculate the brake power of the engine.

Using the formula provided, the calculated brake power would be approximately 107 Nm, which seems high for a petrol engine. This could be due to several factors such as measurement error, incorrect calculation, or a highly efficient engine. Without more information, it is difficult to determine the exact cause.

If the forum poster is looking for a more accurate way to calculate brake power, the following formula can be used:

Brake power = (Torque x RPM) / 9549.27

Using this formula with the given torque values and an assumed RPM of 2000, the calculated brake power would be approximately 79 kW. This seems like a more reasonable value for a petrol engine.

In conclusion, it is important to use the correct terminology and equations when conducting engine tests. It is also important to consider all factors that may affect the results and to double check calculations for accuracy.

 

[Mene]

I would like to point out that the equation you have provided is for calculating brake power, not indicated power. Brake power is the power output of an engine measured at its output shaft, while indicated power is the theoretical power output calculated from the pressure inside the cylinders. Therefore, the equation you have used is not applicable for determining indicated power.

To calculate indicated power, we need to use the following formula:

Indicated power = (2πNT)/60

Where:
N = engine speed (in RPM)
T = torque (in Nm)

Based on the results of the morse test, the torque values for each cylinder are:
No plugs shorted = 91 Nm
No1 shorted = 63 Nm
No2 shorted = 64 Nm
No3 shorted = 65 Nm
No4 shorted = 65 Nm

To determine the indicated power, we need to calculate the average torque value by adding all the torque values and dividing it by the number of cylinders tested (5 in this case).

Average torque = (91+63+64+65+65)/5 = 69.6 Nm

Now, we can plug this value into the formula for indicated power:

Indicated power = (2π * 2000 * 69.6)/60 = 459.7 kW

Therefore, the engine's indicated power at 2000 RPM is approximately 459.7 kW. It is important to note that this is a theoretical value and may not be the same as the actual power output of the engine, as there are many factors that can affect engine performance. The morse test is just one method of determining engine efficiency and should be used in conjunction with other tests for a more accurate assessment.