Troubleshooting Thermodynamics: Calculating Heat Input & BSFC for an Engine

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Calculating heat input for an engine involves using the formula that incorporates reaction enthalpy, fuel flow rate, cylinder count, and engine cycle type, resulting in power output in J/s. Engine speed is not directly necessary for this calculation, as the fuel flow rate already accounts for it, especially in four-stroke engines. For brake specific fuel consumption (BSFC), the desired units are kg/MJ, which can be derived from the fuel flow rate and calorific value. The confusion arises from the fuel flow rate being expressed in kg per pulse rather than kg per second. Clarifying these unit conversions and understanding the relationship between fuel consumption and power output is essential for accurate calculations.
morry
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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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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.
 
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.
 
Anyone else got any ideas?
 
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