q The work output or input during an adiabatic process can be determined from the following The change in internal energy is M=(h2-P2V2)-(h-Pv1) The technical work available h Fig. 6-8 Re (Isentropic Process K Example 6-11 Steam expands isothermally at t=300'Cfrom P,=1 MPa to p2=0.1MPa Determine the amount of heat added to the steam, the change in internal energy and the expansion (Solution] From the steam table, we can get to know the water is superheated vapor at state"I 1=300 C and P,=1 MPa. The other properties of th state V1=0.25793m3/kg,h=30504kJkg,S1=71216kJkg Draw an isothermal line through the initial point"Iand it meets the isobar of p,=0. 1 MPa at point We can find h,=3 073. 8 kJ/kg,s,=8.214 8 kJ/kg. K)when P2=0. 1 MPa and t=300 C The amount of the heat addition can be determined by q=7(s2-s1)=573×(82148-7.1216)=6264kJkg e change in internal energy of the steam is equal to △a=(h2-P2V2)-(h-P11) (3073.8 0.1×10°×26388 )-(305041×10×0.25793 1000 1000 The expansion work output is =q-A=6264-17.5=6089kJ/ks K Example 6-21 Superheated vapor at 8 MPa and 500 Experiences an isentropic expansion process to P2=0.1 MPa. Determine the final state of the steam, the change in internal energy and the work Solution: We can find on the h-s diagram that at the end of expansion process, the steam becomes saturated liquid-vapor mixture with the quality of x=0.84 The initial properties of the superheated vapor at 8 MPa and 500C are h=3400kJ/kg,n1=0045m3/kg106 Fig. 6-8 Reversible Adiabatic Process Fig. 6-9 Irreversible Adiabatic Process (Isentropic Process) The work output or input during an adiabatic process can be determined from the following equation ( ) ( ) 1 2 1 1 1 2 2 2 w = u −u = h − p v − h − p v The change in internal energy is ( ) ( ) 2 2 2 1 1 1 u = h − p v − h − p v The technical work available is wt = −h 【Example 6-1】 Steam expands isothermally at t = 300 ℃from p1 =1 MPa to p2 = 0.1 MPa. Determine the amount of heat added to the steam, the change in internal energy and the expansion work output. 【Solution】From the steam table, we can get to know the water is superheated vapor at state “1”, t = 300 ℃ and p1 =1 MPa. The other properties of this state are 3 1 v = 0.257 93 m /kg , 1 h = 3 050.4 kJ/kg , 1 s = 7.121 6 kJ/kg . Draw an isothermal line through the initial point “1”and it meets the isobar of 2 p = 0.1 MPa at point “2”.We can find 2 h = 3 073.8 kJ/kg ， 2 s =  8.214 8 kJ/(kg K) when 2 p = 0.1 MPa and t = 300 ℃， The amount of the heat addition can be determined by 2 1 q T s s = − =  − = ( ) 573 (8.214 8 7.121 6) 626.4 kJ/kg The change in internal energy of the steam is equal to 2 2 2 1 1 1 6 6 ( ) ( ) 0.1 10 2.638 8 1 10 0.257 93 (3 073.8 ) (3 050.4 ) 1 000 1 000 17.5 kJ/kg  = − − − u h p v h p v     = − − − = The expansion work output is w= q −u = 626.4−17.5= 608.9 kJ/kg 【Example 6-2】 Superheated vapor at 8 MPa and 500℃experiences an isentropic expansion process to p2 = 0.1 MPa. Determine the final state of the steam, the change in internal energy and the work output by using the h s − diagram Solution: We can find on the h s − diagram that at the end of expansion process, the steam becomes saturated liquid-vapor mixture with the quality of x = 0.84 The initial properties of the superheated vapor at 8 MPa and 500℃ are 1 h = 3 400 kJ/kg; 1 v = 0.045 m3 /kg q = 0