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Homework I Solution Problem 1 1.1 Natural Uranium : X (g/yr) Enriched Uranium :Y (g/yr) Discarded Uranium: Z (g/yr) Mass Conservation: X=Y+Z Mass Conservation of U-235: 0.71X=4.4Y+0.25Z Using Eqs. (1) and (2), we get Y=0.1108X Eq.(1) Eq.(2) Eq.(3) Hence, we lose almost 90% of the natural uranium in the enrichment process. Since U-235 in the enriched Uranium is 4.4/100*Y, we can represent the used Uranium for fission from the natural Uranium as 4.4/100*0.111X=0.004877X. Eq.(4) Hence, only 0.4877% of the natural Uranium is actually used for fission. Then, power can be represented as 0.004877X*η rankine *η nuclear *1MW-day/g=1000MW*365 day*η capacity where η rankine =0.35, η nuclear =0.95 and η capacity =0.9. Then, we get X=2.026*108 g/yr=202.6 ton/yr=0.56ton/day. 1.2 Average daily amount of coal used: X (kg/day) X*η steam *27800BTU/kg*1J/9.48*10-4BTU*1day/(24*3600sec) =1000*106 W*η capacity where η steam =0.471. Then, we get X=6.04*106kg/day=5640 ton/day. 1 M.M. El-Wakil, Power Plant Technology, McGraw Hill, 2984, page 72 1.3 Area for the flat panel : X (cm2) 500cal/(cm2 day)/(0.239cal/J)*η conv *X*1day/(24*3600sec)=1000*106W where η conv =0.12. Then, X=3.441*1011cm2. Also, total area required is 2X=68.8km2. Problem 2 2.1 P1=1atm, T1=300K → Isentropic compression I → P2,T2 P2,T2→ Intercooling → P2’=P2,T2’=T1=300 P2’,T2’ → Isentropic compression II → P3=100atm,T3 In the compressions I and II, T and P have the following relationships P2 T2 = P1 T1 k /( k −1) P T and 3 = 3 P2 T1 k /( k −1) Eq. (5) Also, work can be represented as w = c p (T3 − T1 ) + c p (T2 − T1 ) Eq.(6) Using Eqs. (5) and (6), we get P w = c p T1 2 P1 ( k −1) / k P + 3 P2 ( k −1) / k − 2 Eq.(7) By differentiating w by P2 , we get P2 = P1 P3 =10atm. 2.2 From Eq. (5), we get T2=T3=579K. Using Eq. (6) and cp~1kJ/kg K at T~450K, we get 558kJ/kg. 2.3 P1=1atm, T1=300K → Isentropic compression → P2=100atm,T2 Using Eq. (5) again, we get T2=1118K. Using w = c p (T3 − T1 ) and cp~1.1kJ/kg K at T~700K, we get w = 900kJ/kg. It is clear that intercooling reduces the work required for compression significantly. 2.4 w=vdP=0.001m3/kg*(10130kPa-101.3kPa)=10kJ/kg. The compression work of liquids is typically 1-2% of that of gases. 2.5 Assuming η c ~0.7-0.9, work in 2.2 and 2.3 should increase by 10-40%. Problem 3 3.1 η thermal =0.5 : thermal efficiency η voltage =0.7/1.23=0.569 : Voltage drop η H 2 =237/286=0.829 : H2 to electric power conversion efficiency η tot =η thermal *η voltage *η H 2 =0.236 3.2 η tot =0.96*η ICE η ICE =0.246 3.3 0.7*η H 2 *η voltage =0.330 0.330=0.96η ICE η ICE =0.344 Without considering the refinery efficiency of fuels, it becomes η ICE =0.330