Heating Then Continues Until the Water is Saturated Vapor Show the Processes of the W

6E-1 : Heat, Work and Efficiency of a Water Vapor Power Cycle 8 pts One-half kilogram of water executes a Carnot power cycle. During the isothermal expansion, the water is heated until it is saturated vapor from an initial state where the pressure is 15 bar and the quality is 25%. The vapor then expands adiabatically to a pressure of 1 bar while doing 403.8 kJ/kg of work. a.) Sketch the cycle on PV coordinates. b.) Evaluate the heat and work for each process, in kJ c.) Evaluate the thermal efficiency. Read : Apply the 1st Law (for a closed system) to get Q and W. Use the first law applied to step 2-3 to determine U ₃ and x ₃ . The trick is to get Q ₃₄ . Use T _C , T _H , Q ₁₂ and the Carnot efficiency of this reversible cycle to determine Q ₃₄ . Given : P ₁ 15 bar P ₃ 1 bar x ₁ 0.25 Q ₂₃ = Q ₄₁ 0 kJ/kg P ₂ 15 bar W ₂₃ 403.8 kJ/kg x ₂ 1 m 0.5 kg Find : Part (a) PV Diagram Part (b) Q ₁₂ ,Q ₂₃ ,Q ₃₄ ,Q ₄₁          ? kJ W ₁₂ ,W ₂₃ ,W ₃₄ ,W ₄₁      ? kJ Part (c) h ? Diagram : Part a.) Assumptions : - The system undergoes a Carnot cycle. - Steps 1-2 and 3-4 are isothermal - Steps 2-3 and 4-1 are adiabatic - All steps are reversible - The water inside the cylinder is the system and it is a closed system. - Changes in kinetic and potential energies are negligible. - Boundary work is the only form of work interaction wuring the cycle. Equations / Data / Solve : Part b.) Begin by applying the 1st law for closed systems to each step in the Carnot Cycle.  Assume that changes in kinetic and potential energies are negligible. Eqn 1 Step 1 - 2 Apply the 1st Law, Eqn 1 , to step 1-2 : Eqn 2 Boundary work at for a constant pressure process, like step 1-2, can be determined from : Eqn 3 Now, we can substitute Eqn 3 into Eqn 1 to get : Eqn 4 The definition of enthalpy is: Eqn 5 For isobaric processes, Eqn 5 becomes : Eqn 6 Now, combine Eqns 4 and 6 to get : Eqn 7 We know the pressure and the quality of states 1 and 2, so we can use the Saturation Table in the Steam Tables to evaluate V and H for states 1 and 2 so we can use Eqns 3 and 7 to evaluate Q ₁₂ and W ₁₂ . Properties are determined from NIST WebBook : At P ₁ and x ₁ : V _{sat liq, 1} 0.001154 m ³ /kg V _{sat vap,} 0.13171 m ³ /kg V ₁ 0.033793 m ³ /kg U _{sat liq, 1} 842.83 kJ/kg U _{sat vap,} 2593.4 kJ/kg U ₁ 1280.5 kJ/kg H _{sat liq, 1} 844.56 kJ/kg H _{sat vap,} 2791 kJ/kg H ₁ 1331.2 kJ/kg Saturated vapor at P ₂ : V ₂ 0.13171 m ³ /kg U ₂ 2593.4 kJ/kg W ₁₂ 73.438 kJ H ₂ 2791 kJ/kg Q ₁₂ 729.92 kJ Step 2 - 3 Apply the 1st Law, Eqn 1 , to step 2-3 : Eqn 8 The specific heat transferred and specific work for process 2-3 are given in the problem statement. W ₂₃ 201.9 kJ Q ₂₃ 0 kJ We can plug these values into Eqn 8 to determine D U ₂₃ : DU ₂₃ -201.9 kJ We already determined U ₂ , so we can now determine U ₃ : Eqn 9 U ₃ 2189.6 kJ/kg We can use this value of U ₃ to determine the unknown quality, x ₃ , using : Eqn 10 Properties are determined from NIST WebBook : At P ₃ : U _{sat liq, 3} 417.4 kJ/kg U _{sat vap, 3} 2505.6 kJ/kg x ₃ 0.849 At P ₃ and x ₃ : V _{sat liq, 3} 0.001043 m ³ /kg V _{sat vap, 3} 1.6939 m ³ /kg V ₃ 1.4377 m ³ /kg H _{sat liq, 3} 417.5 kJ/kg H _{sat vap, 3} 2674.9 kJ/kg H ₃ 2333.3 kJ/kg Step 3 - 4 Apply the 1st Law, Eqn 1 , to step 2-3 : Eqn 11 Because step 3-4 is isobaric, just like step 1-2, Eqn 7 is the simplified form of the 1st Law : : Eqn 12 To determine the properties at state 4, we make us of the relationship between the absolute Kelvin temperature scale and heat transferred in a Carnot Cycle. Eqn 13 Solve Eqn 13 for Q ₃₄ : Eqn 14 T _H = T _sat (P ₁ ) : T _H 471.44 K T _C = T _sat (P ₃ ) : T _C 372.76 K Q ₃₄ -577.13 kJ Q ₃₄ -1154.26 kJ/kg Now, we can use Q ₃₄ and Eqn 12 to determine H ₄ as follows: Eqn 15 or : Eqn 16 H ₄ 1179.03 kJ/kg Now that we know the values of two intensive properties at state 4, T ₄ and H ₄ , we can evaluate all the other properties using the Saturation Tables in the Steam Tables. Properties are determined from NIST WebBook : At P ₄ : H _{sat liq, 4} 417.5 kJ/kg H _{sat vap, 4} 2674.9 kJ/kg Eqn 17 x ₄ 0.337 At P ₄ and x ₄ : V _{sat liq, 4} 0.0010432 m ³ /kg V _{sat vap, 4} 1.6939 m ³ /kg V ₄ 0.5721 m ³ /kg U _{sat liq, 4} 417.4 kJ/kg U _{sat vap, 4} 2505.6 kJ/kg U ₄ 1121.9 kJ/kg At last we have U ₄ and we can plug it into Eqn 11 to evaluate W ₃₄ : Eqn 18 W ₃₄ -43.26 kJ/kg Step 4 - 1 The heat transferred for Process 4-1 is given in the problem statement. Apply the 1st Law, Eqn 1 , to step 4-1 : Eqn 19 Solve Eqn 19 for W ₄₁ : W ₄₁ -79.31 kJ Part c.) The efficiency of a Carnot cycle is defined by: Eqn 20 Where : Eqn 21 Eqn 22 Q _in 729.9 kJ W _cycle 152.8 kJ h 0.209 Or the efficiency can be determined in terms of reservoir temperatures: Eqn 23 h 0.209 Verify : The assumptions made in the solution of this problem cannot be verified with the given information. Answers : Process Q W 1-2 729.9 73.4 2-3 0 201.9 3-4 -577.1 -43.3 4-1 0 -79.3 Cycle 152.8 152.8 The thermal efficiency of the process is : 21%

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