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

Today we will see here the calculation
of efficiency of Otto cycle with the help of this post.

###
**Otto
cycle: Efficiency **

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

Therefore first let us see an overview
of an Otto 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 and two constant volume processes. 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
thermodynamics

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}(T_{2}-T_{1})
W = C

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

As we have already seen that process 2-3
will follow constant volume process and therefore W= 0

Q= Î”U = C

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

_{V}(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}(T_{4}-T_{3})
W = C

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

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

Q= Î”U = C

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

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

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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