written 7.8 years ago by | • modified 7.8 years ago |
Mumbai University > Mechanical Engineering > Sem 5 > IC Engines
Marks: 10M
Year: Dec 2014
written 7.8 years ago by | • modified 7.8 years ago |
Mumbai University > Mechanical Engineering > Sem 5 > IC Engines
Marks: 10M
Year: Dec 2014
written 7.8 years ago by |
Fig 1.3 (a) and (b) shows the theoretical p-V diagram and T-s diagrams of this cycle respectively.
• The point 1 represents that cylinder is full of air with volume V1 pressure P1 and absolute temperature T1.
• Line 1-2 represents the adiabatic compression of air due to which p1, V1 and T1 change to p2, V2 and T2, respectively.
• Line 2-3 shows the supply of heat to the air at constant volume so that p2 and T2 change to p3 and T3. (V3 being the same as V2).
• Line 3-4 represents the adiabatic expansion of the air. During expansion p3, V3 and T3 change to a final value of p4. V4 or V1 and T4 respectively.
• Line 4-1 shows the rejection of heat by air at constant volume till original state (point 1) reaches. Consider 1 kg of air (working substance):
Heat supplied at constant volume = Cv(T3-T2).
Heat rejected at constant volume = Cv(T4-T1).
But, Work done = Heat supplied – Heat rejected
$$= Cv(T3-T2) - Cv(T4-T1)$$
$$Efficiency = \frac{{Work{\text{ done}}}}{{Heat{\text{ supplied}}}} = \frac{{{C_v}({T_3} - {T_2}) - {C_v}({T_4} - {T_1})}}{{{C_v}({T_3} - {T_2})}}$$
$$ = 1 - \frac{{{T_4} - {T_1}}}{{{T_3} - {T_2}}}$$