hw10 - Prob 8.3 PROBLEM 8.3 PROBLEM STATEMENT An electric...

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Prob 8.3 PROBLEM 8.3 PROBLEM STATEMENT: An electric utility company has a plant with a capacity of 1000 MW of electricity for sale to its customers. The overall thermal efficiency of the plant is 25 percent and toil (the fuel) costs 25 cents a gallon. The heating value of the oil is 4.65 x 10 6 kJ/kg and there are 3.4 kg of oil per gallon. What is the cost of running at maximum capacity for one week? GIVEN: NET ΗΕ W =1000 MW, = 25 % η & Fuel cost = 25¢/gallon, 1 gallon = 3.4 kg, × 6 HV = 4.65 10 kJ/kg FIND: Total operational cost for one week ($). ASSUMPTIONS: GOVERNING RELATIONS: Thermal efficiency, HE NET IN =W Q & & η QUANTITATIVE SOLUTION: From the thermal efficiency relation, the total power input into the plant is NET IN HE 6 W 1000 Q = = = 4000 MW 0.25 =4 10 kJ/s η × & & One week is equivalent to 6.048 x 10 5 s. Thus, the total energy consumed is () ( ) ( )( ) 65 IN IN 12 Q = Q time = 4 10 kJ/s 6.048 10 s =2.4192 10 ±kJ ×× × × × & The required mass and volume to produce this energy is ( ) × × × ×× × 12 5 IN 6 5 5 2.4192 10 kJ Q m= = =5.203 10 kg HV 4.65 1 g a l V V = m = 5.203 10 kg m3 . 4 k g o =1.530 10 gal i l The total cost is computed as follows: ( ) ( ) × 5 cost = volume oil price =1.530 10 gal 0.25 $/gal =$38,250 DISCUSSION OF RESULTS: Thermal efficiency is not just of thermodynamic interest; it is directly related to the economics of the power plant. The power company produces kilowatt-hours (W NET ) for which it charges its customers, and one of the biggest factors in the cost of production is fuel (Q IN ). Thus the efficiency, which is the ratio of W NET to Q IN , can be thought of as the power plant’s return on its energy investment.
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Prob 8.3 PROBLEM 8.4 PROBLEM STATEMENT: Plot the P-v, T- s , P-h, and h-s cycle diagrams for the air- standard Carnot cycle operating with 1 lbm of working fluid between 60°F and 400 °F, if the maximum volume reached in the cycle is 1 ft 3 , and the maximum pressure in the cycle is 2000 psia. Assume that the specific heat of air can be taken as temperature-independent. DIAGRAM DEFINING SYSTEM AND PROCESS: W IN Q H Q C 3 1 2 4 Adiabatic Reversible Compression W OUT W IN Isothermal Cooling & Compression Isothermal Heating & Expansion Adiabatic Reversible Expansion W OUT GIVEN: State 1: T 1 = 60 o F State 2: T 2 = 400 o F, P 2 =2000 psia State 3: T 3 = 400 o F State 4: T 4 = 60 o F, V 4 = 1 ft 3 m = 1 lbm FIND: Specific enthalpies (h), pressures (P), and specific entropies (s) for each state and plot P-v, T-s, P-h, and h-s diagrams ASSUMPTIONS: 1. Air Standard Carnot cycle 2. Constant mass 3. Constant specific heats (evaluate at 60 ° F for air standard) 4. Non-flow processes GOVERNING RELATIONS: 1. PV = mRT
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Prob 8.3 2. For adiabatic and reversible (isentropic) process, constant specific heat:  =   k1 k bb aa TP PROPERTY DATA: From Table 5e at 60 ° F k = 1.4 c P = 0.24 Btu/lbm o R h 1 (60 o F) = h 4 = 124.5 Btu/lbm h 2 (400 o F) = h 3 = 206.74 Btu/lbm QUANTITATIVE SOLUTION: −− =⇒ = = =   k 22 2 11 . 4 1 .
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hw10 - Prob 8.3 PROBLEM 8.3 PROBLEM STATEMENT An electric...

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