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11) T h e C a r n o t efficiency a s given b y Eq. 12) is t h e simplest expression of t h e C a r n o t efficiency. By equating Eq. 2) to Eq. 14) since Q2 has a negative value, indicating that heat is being rejected by the system. F r o m Eq. 14) it can b e d e d u c e d that the sum of the ratios of the heat transfer t o the t e m p e r a t u r e at which the heat is transferred is equal t o zero. I n C h a p t e r 5 , it will b e d e m o n s t r a t e d that Eq. 14) is directly c o n n e c t e d to the second law of t h e r m o d y n a m i c s .
5 BTU of heat. Determine the change in the internal energy in the cylinder. What is the specific internal energy change? 1-12. An air compressor, considered to be a closed system, requires 1 HP to drive it. During the compression process 30 BTU/min are carried away by the water cooled jacket. Determine the time rate of change of the internal energy of the air during the compression process. 1-13. A one pound mass of a perfect gas is expanded by a constant energy 3 3 process from Vx = 1 ft to V2 = 3 ft .
5) into E q . 6) Vdp) = -pdV U p o n simplification and replacing CJCV get, by y, the adiabatic constant, we . 7) By dividing Eq. 8) is the perfect gas law for an adiabatic process. N o w , substituting for ρ in Eq. 10) &U = -Ç—(V2- -Vl - ) γ— 1 y T h e constant C may be evaluated from either pxVx or p2V2 , depending upon w h e t h e r the initial or final conditions are k n o w n . Let us n o w discuss the validity of the a b o v e results. T h e original assumption w a s that w e w e r e dealing with a reversible p r o c e s s , that is, the path of the process was reversible.