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INTRODUCTION TO FIRST LAW OF THERMODYNAMICS. EXAMPLES FOR FIRST LAW OF THERMODYNAMICS. EXAMPLES FOR FIRST LAW OF THERMODYNAMICS. EXAMPLES FOR FIRST LAW OF THERMODYNAMICS. PROBLEMS. dU = ∫ δ Q - ∫ δ W U 2 –U 1 =∫ δ Q - ∫ δ W U 2 = U 1 ( Closed Cycle ) 0 = ∫ δ Q - ∫ δ W
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PROBLEMS dU = ∫δQ - ∫δW U2 –U1 =∫δQ - ∫δW U2 = U1 ( Closed Cycle ) 0 = ∫δQ - ∫δW ∫δQ = ∫δW So if ∫δQ = ∫δW , FLT is fulfilled. ∫δQ =7000-3500+17500+0 =21000 J/s ∫δW =5300+9100+8700-2100 =21000 J/s Thus FLT is satisfied. Net rate of work = Power = 21000 J/s =21 kW Thermal Efficiency = Net Work / Heat In put Heat Input = Sum of all positive quintiles = 7000 + 17500 =24500 J/s Thermal Efficiency = Net Work / Heat In put = 21000 / 24500 =85.71 %