MPPSC SES Mechanical study materials
MPPSC SES Mechanical study resources

MPPSC SES Mechanical: Thermodynamics, Heat Transfer and Energy Systems: Cycles and Thermal Resistance

MPPSC SES Mechanical · Work through your exam, one topic at a time.

All guide sections

Thermodynamics, Heat Transfer and Energy Systems: Cycles and Thermal Resistance

Laws and cycles

First law for a closed system: Q − W = ΔU; for steady flow: q − w = Δh + ΔKE + ΔPE. The second law limits performance: no engine exceeds Carnot efficiency 1 − T_L/T_H, with temperatures in kelvin. Entropy of an isolated system never decreases; reversible adiabatic processes are isentropic.

Air-standard efficiencies:

  • Otto: 1 − 1/r^(γ−1)
  • Diesel: 1 − [1/(r^(γ−1))] × [(ρ^γ − 1)/(γ(ρ − 1))], with cut-off ratio ρ
  • Brayton: 1 − 1/r_p^((γ−1)/γ)

Heat transfer

  • Conduction: Q = kAΔT/L; slab resistance L/kA.
  • Convection: Q = hAΔT; radiation Q = εσA(T⁴ − T_sur⁴), temperatures in K.
  • Heat exchangers: LMTD = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂); for condensers and evaporators the LMTD equals the constant difference.
  • Refrigerator COP = Q_L/W; heat pump COP = Q_H/W = COP_R + 1.

Worked example

An Otto cycle with compression ratio 8 and γ = 1.4. η = 1 − 1/8^0.4 = 1 − 1/2.297 = 1 − 0.435 = 0.565, i.e. 56.5%. Raising r improves efficiency but is limited by knock; the Diesel cycle at the same r is less efficient because of the cut-off ratio.

Traps

  • Using °C in Carnot or radiation formulae.
  • Applying 1 − 1/r^(γ−1) to a Diesel cycle - it overestimates efficiency.
  • Assuming COP must be below 1; it is usually above 1.
  • Using arithmetic mean temperature difference where LMTD is required.