We were
discussing Diesel cycle, the ideal cycle for the
operation of internal combustion compression ignition reciprocating engines, in our recent post. We have also
discussed basic operations and arrangements of various components of ideal cycle for the operation of internal combustion compression
ignition reciprocating engines. We have also seen there the PV
diagram for a Diesel cycle.

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Today we will see here the calculation of efficiency of Diesel cycle with the help of
this post and we will also see here the TS diagram for a Diesel cycle.

###
**Diesel
cycle: Efficiency**

Diesel
cycle is one type of air standard cycle which is designated as the ideal cycle for the operation of internal combustion
compression ignition reciprocating engines. Before understanding the method for determination of the
efficiency of the Diesel cycle, we will have to remind here various processes
involved.

Therefore
first let us see an overview of a Diesel cycle with the help of PV diagram and
TS diagram as displayed here in following figure. As we can see in below
figure, there will be two isentropic or adiabatic processes, one constant
volume process and one constant pressure process. We will determine the various
properties for unit mass of working fluid.

###
**Process
1-2: Compression stroke**

Let us use
the concept of first law of thermodynamicsÂ

Q= Î”U + W

As we have
already seen that process 1-2 will follow constant entropy process and
therefore Q= 0

Hence we
will have from first law of thermodynamics Î”U = - W

Î”U = C

_{V}Â (T_{2}-T_{1})
W = C

_{V}Â (T_{1}-T_{2})###
**Process
2-3: Combustion stroke**

As we have
already seen that process 2-3 will follow constant pressure process and
therefore

W = P

_{2}(V_{3}-V_{2})
Î”U = C

_{V}Â (T_{3}-T_{2})
Hence,
heat energy addition to the system Q = Q1 = C

_{P}Â (T_{3}-T_{2})###
**Process
3-4: Expansion or power stroke**

As we have
already seen that process 3-4 will follow constant entropy process and
therefore Q= 0

Hence we
will have from first law of thermodynamics Î”U = - W

Î”U = C

_{V}Â (T_{4}-T_{3})
W = C

_{V}Â (T_{3}-T_{4})###
**Process
4-1: Blow down**

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As we have already seen that process 4-1 will follow constant volume process and therefore
W= 0

Q= Î”U = C

_{V}Â (T_{1}-T_{4})
Hence,
heat energy rejection from the system Q = Q2 = C

_{V}Â (T_{4}-T_{1})###
**Efficiency
of the Otto cycle**

Efficiency
of the Otto cycle will be determined with the help of following formula

Do you
have any suggestions? Please write in comment box.

We will
see another topic in our next post in the category of thermal engineering.

###
**Reference:**

Engineering
thermodynamics by P. K. Nag

Engineering
thermodynamics by Prof S. K. Som

Image
courtesy: Google

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