Troubleshooting Thermodynamics: Calculating Heat Input & BSFC for an Engine

In summary, the conversation revolves around calculating heat input and brake specific fuel consumption for an engine. The formula for heat input involves reaction enthalpy, fuel flow rate per injector, and engine specifications. The issue of engine speed is addressed, with the conclusion that it is not necessary to consider it. For BSFC, the goal is to get units of kg/MJ, and it is suggested to multiply the fuel flow rate by time and divide by the reaction enthalpy. The conversation ends with a request for further ideas.
  • #1
morry
136
0
Hi guys,

Im trying to calculate the heat input and brake specific fuel consumption for an engine. However, I am having troubles with units lining up.

Heat Input= reaction enthalpy (J/kg) x fuel flow rate per injector (kg/s) x 4 (4 cyl engine) x 0.5 (4 stroke)

Doing it like that gives me J/s = power, which is what I want, but shouldn't the engine speed have something to do with this?

Next problem is the BSFC. I am after units of kg/MJ. It looks simple to just multiply the flow rate by time and multiply the reaction enthalpy by mass. Then divide one by the other. This would give me the correct units, however I don't really know what to do.

Can someone point me in the right direction?

Thanks.
 
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  • #2
As the flowrate of fuel is already mentioned, you need not worry about the engine speed(alternatively, it is mass of fuel per one power stroke times the rpm incase of a two stroke engine and mass of fuel times the rpm/2 incase of a 4 stroke engine).

The reciprocal of the reaction enthalpy (I take it as calorific value) gives you the specific consumption in terms of Indicative Power. When you apply efficiency, you will get specific consumption in terms of brake power.
 
  • #3
Ok, so the fuel flowrate is good.

The BSFC still confuses me a bit. It says in my book that it is fuel used per joule of output. So units of kg/MJ. I know the fuel flow rate, but this is in kg/pulse. Not kg. This is what confuses me. To work out the other half of the equation is simply the reciprocal of the calorific value like you mentioned.

Thanks quark.
 
  • #4
Anyone else got any ideas?
 

1. What is thermodynamics?

Thermodynamics is a branch of physics that deals with the study of heat and its relationship with energy, work, and temperature. It also explains how energy is transferred between different forms, such as heat, work, and internal energy.

2. What are the laws of thermodynamics?

The laws of thermodynamics are fundamental principles that govern the behavior of energy in a system. They are:
1. The first law - energy cannot be created or destroyed, only transferred or converted from one form to another.
2. The second law - the total entropy of a closed system will always increase over time.
3. The third law - it is impossible to reach absolute zero temperature through a finite number of steps.

3. What are some common applications of thermodynamics?

Thermodynamics has many practical applications in everyday life. It is used in the design and operation of engines, refrigerators, air conditioners, and power plants. It is also important in metallurgy, chemical reactions, and material science.

4. What is the difference between heat and temperature?

Heat is a transfer of energy from one object to another due to a difference in temperature. Temperature, on the other hand, is a measure of the average kinetic energy of the particles in a substance. In simpler terms, heat is the energy being transferred, while temperature is a measure of how hot or cold something is.

5. What are some common misconceptions about thermodynamics?

One common misconception is that thermodynamics only deals with heat and temperature. In reality, it also encompasses other forms of energy, such as work and internal energy. Another misconception is that the laws of thermodynamics only apply to closed systems, when in fact they can also be applied to open systems. Lastly, many people believe that thermodynamics only applies to large-scale systems, but it is also relevant to microscopic systems at the atomic and molecular level.

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