If Tmax and Tmin be the maximum and minimum temperatures in an Otto cycle, then for the ideal conditions, the temperature after compression should be

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BPSC AE Paper 5 (Mechanical) 2012 Official Paper
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  1. \(\dfrac{T_{max} + T_{min}}{2}\)
  2. \(\sqrt{\dfrac{T_{max}}{T_{min}}}\)
  3. \(\sqrt{T_{max}\times T_{min}}\)
  4. \(T_{min} + \dfrac{T_{max}-T_{min}}{2}\)

Answer (Detailed Solution Below)

Option 3 : \(\sqrt{T_{max}\times T_{min}}\)
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Detailed Solution

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Explanation:

Otto cycle:

RRB JE ME 49 15Q TE CH 4 HIndi - Final Diag(Shashi) images Q4

(1-2) - Reversible adiabatic compression

(2-3) - Constant volume of heat addition.

(3-4) - Reversible adiabatic Expansion.

(4-1) - Constant volume of heat rejection.

Calculation:

Given:

During  Ideal conditions, we get maximum work output (T2=T4)

T= TMin and T= TMax

Calculation:

(1-2) Reversible adiabatic compression:

\(\left( {\frac{{{T_2}}}{{{T_1}}}} \right) = {\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^{\gamma - 1}}\)

(3-4) Reversible adiabatic expansion:

\(\left( {\frac{{{T_3}}}{{{T_4}}}} \right) = {\left( {\frac{{{V_4}}}{{{V_3}}}} \right)^{\gamma - 1}}\)

In the otto cycle: V= V1, V= V2

\((\frac{{T_2}}{{T_1}} )=(\frac{{T_3}}{{T_4}})\)

\(T_2×T_4=T_3×T_1\)

\({\left( {{T_2}} \right)^2} = \left( {{T_3} \times {T_1}} \right)\)

\(\left( {{T_2}} \right) = \left( {\sqrt {{T_3} \times {T_1}} } \right)=\sqrt{T_{max}\times T_{min}}\)

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