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ENGR5003 / Week 4
ENGR217 2019 Q A2Past paperOld spec ENGR2172:125 marks30 min

ENGR217 Summer 2019 Q A2[VALID]

Answer ALL parts (a) - (f).

In an air standard Otto cycle the compression ratio is 8 and the pressure and temperature of the air at the start of compression are 1 bar and 300 K.

Formulas you may need
  • Adiabatic ideal gas: pVγ=constpV^\gamma = \text{const}, TVγ−1=constTV^{\gamma-1} = \text{const} (on the formula sheet)
  • Constant-volume heat transfer: dQ=mcv dTdQ = m c_v\, dT; isochoric p/T=constp/T = \text{const} (Gay-Lussac) (on the formula sheet)
  • Isothermal ideal gas: dU=0dU = 0 so dQ=dWdQ = dW, ∣W∣=mRgTln⁡(V1/V2)|W| = m R_g T \ln(V_1/V_2) (on the formula sheet)
  • Thermal efficiency: ηth=1−QoutQin\eta_{th} = 1 - \dfrac{Q_{out}}{Q_{in}}; Otto η=1−1rγ−1\eta = 1 - \dfrac{1}{r^{\gamma-1}} (on the formula sheet)
  • Rg=0.287 kJ kg−1 K−1R_g = 0.287\ \mathrm{kJ\,kg^{-1}\,K^{-1}} for air (on the formula sheet)
  1. (a)
    Sketch and label the processes of an Otto cycle on a PV diagram.
    [4]Third
  2. (b)
    Taking γ=1.4\gamma = 1.4 during the isentropic compression compute the pressure and temperature at the end of the compression stroke.
    [3]
  3. (c)
    The heat added at constant volume in the cycle is 1080 kJ kg−11080\ \mathrm{kJ\,kg^{-1}}. Taking the heat capacity as 0.880 kJ kg−1 K−10.880\ \mathrm{kJ\,kg^{-1}\,K^{-1}} while this heat is added, determine the maximum pressure and temperature of the cycle.
    [4]
  4. (d)
    If γ\gamma and the heat capacity are unchanged during the isentropic expansion and heat rejection respectively, determine the efficiency.
    [3]
  5. (e)
    Taking γ=1.33\gamma = 1.33 during the isentropic expansion and taking the heat capacity as 0.74 kJ kg−1 K−10.74\ \mathrm{kJ\,kg^{-1}\,K^{-1}} while heat is rejected determine the heat rejected and hence the efficiency.
    [6]
  6. (f)
    If it were possible to make the compression process isothermal rather than isentropic, and assuming the same heat addition as before, γ=1.33\gamma = 1.33 for expansion, cv=0.880 kJ kg−1 K−1c_v = 0.880\ \mathrm{kJ\,kg^{-1}\,K^{-1}} for heat addition and cv=0.74 kJ kg−1 K−1c_v = 0.74\ \mathrm{kJ\,kg^{-1}\,K^{-1}} for heat rejection, compute the new thermal efficiency.
    [5]First