This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Problem 5.20 Known: Data are provided for four cases involving reversible power cycles operating between hot and cold reservoirs. Schematic & Given Data: Fig. P.5.20 Symbols: η : efficiency
Wcycle : net amount of energy transfer by work QH : amount of energy transfer by heat at hot reservoir QL : amount of energy transfer by heat at cold reservoir
Analysis: Thermal efficiency of a system undergoing a reversible power cycle is ηmax = 1 − TC TH (Eq.5.9) (a) TH = 1200K, TC =300K. Find η With Eq.5.9, η = 1 − (b) TH = 500°C, TC =20°C, Wcycle =1000kJ. Find QC and QH With Eq.5.9 and η = Wcycle QH , which applies for any power cycle, TC 300K = 1− = 0.75 (75%) TH 1200K Wcycle QH = 1− W TC 1000kJ ⇒ QH = cycle = = 1610kJ 293K TC TH 1− 1− 773K TH For any power cycle, Wcycle = QH − QC ; thus, QC = QH − Wcycle = 1610kJ − 1000kJ=610kJ (c) η = 60%, TC =40°F. Find TH With Eq.5.9, η = 1− TC 500°R ⇒ 0.60 = 1 − ⇒ TH = 1250°R (790°F) TH TH (d) η = 40%, TH =727°C. Find TC With Eq.5.9, η = 1− TC TC ⇒ 0.40 = 1 − ⇒ TC = 600K (327°C) TH 1000K Problem 5.38 Known: Steady‐state operating data are provided for a power plant discharging energy by heat transfer to a river. Find: Determine the increase in the temperature of the river traceable to such heat transfer for each of two cases. Schematic & Given Data: Model and Assumptions: 1. The power plant is modeled as a power cycle operating at steady state. 2. The combustion products provide energy by heat transfer to the power cycle. Energy is discharged by heat transfer to the river. These are the only heat transfers. 3. The river water is modeled as incompressible with constant specific heat C. Symbols: η : efficiency
Wcycle : net rate of work Qin : recieved heat transfer rate Qout : discharged heat transfer rate
Analysis: For a control volume enclosing a section of the river, as shown in the sketch an energy rate balance reduces to give Qout = m [ h2 − h1 ] where Qout is a positive number denoting the energy discharged from the power cycle to the river. With Eq.3.20b, this becomes Qout = mc [T2 − T1 ] (1) For the power cycle, η = Wcycle Qin , where Qin = Wcycle + Qout . That is, η= Wcycle Wcycle + Qout ⎡1 ⎤ ⇒ Qout = Wcycle ⎢ − 1⎥ ⎣η ⎦ (2) Combining Eqs. (1) and (2), ⎡1 ⎤ Wcycle ⎢ − 1⎥ ⎣η ⎦ (T2 − T1 ) = mC
where C = 4.2 kJ/kg ⋅ K from Table A‐19. (a) η = η max = ⎢1 − (3) ⎡ ⎣ TC ⎤ ⎡ 290 ⎤ = 0.508 . Inserting values in Eq.(3) ⎥ = 1− TH ⎦ ⎢ 590 ⎥ ⎣ ⎦ 103 kJ/s ⎡ 1 ⎤ ( 750MW ) ⎢ 0.508 − 1⎥ 1MW ⎣ ⎦ = 1.05K ( = 1.05°C ) (T2 − T1 ) = 5 (1.65 ×10 kg/s ) ( 4.2kJ/kg ⋅ K )
(b) η = ηmax = 0.339 2 3 103 kJ/s ⎡ 1 ⎤ ( 750MW ) ⎢ 0.339 − 1⎥ 1MW ⎣ ⎦ = 2.11K ( = 2.11°C ) (T2 − T1 ) = 5 (1.65 ×10 kg/s ) ( 4.2kJ/kg ⋅ K ) Comment: An effect of irreversibilities within the power plant is to increase the amount of energy discharged by heat transfer to the river and thus increase the temperature rise of the river traceable to such heat transfer. Problem 5.44 Known: A reversible refrigeration cycle operates between cold and hot reservoirs at TC and TH. Find: For each of 5 sets of data, obtain a specific result. Symbols: β : coefficient of performance for a refrigerator Analysis: The coefficient of performance of any system undergoing a reversible refrigeration cycle is β max = TC TH − TC (Eq.5.10) (a) β = 3.5, TH = 80°F , find TC . Since the cycle is reversible, Eq.5.10 applies: β max = TC TC = 3.5 ⇒ = 3.5 ⇒ TC = 420°R ( −40°F ) TH − TC 540°R − TC
TC 243K = = 4.05 TH − TC ( 303 − 243) K (b) TC = −30°C, TH = 30°C , find β . β = β max = (c) QC = 500Btu, QH = 800Btu, TC = 20°F , find TH . with Eqs.5.5 and 5.10, β = β max ⇒ QC TC = QH − QC TH − TC 500Btu 480°R = ⇒ TH = 768°R ( 308°F ) 800Btu − 500Btu TH − 480°R (d) TC = 30°F, TH = 100°F , find β . β = β max = TC 490°R = =7 TH − TC 560°R − 490°R (e) β = 8.9, TC = −5°C , find TH . β max ⇒ TC = 8.9 TH − TC 268K ⇒ = 8.9 ⇒ TH = 298K ( 25°C ) TH − 268K Problem 5.77 Known: One kg of air undergoes a Carnot cycle for which η = 50%. Find: Determine the minimum and maximum temperatures, the pressure and volume at the beginning of the isothermal expansion, the work and heat transfer for each process, and sketch the cycle on P‐V coordinates. Schematic & Given Data: Summary: Process 1‐2 2‐3 3‐4 4‐1 Cycle Q [kJ] 50 0 ‐25 0 25 W [kJ] 50 220.7 ‐25 ‐220.7 25 η = 50%
Q12 = 50kJ P 2 = 574kPa Model and Assumptions: V2 = 0.3m3 1. The system shown in the schematic consists of air modeled as an ideal gas. 2. Volume change is the only work mode. Analysis: (a) Using the ideal gas model equation of state, T2 = PV2 mR . Then 2 ⎛ 3N⎞ 3 ⎜ 574 × 10 2 ⎟ ( 0.3m ) m⎠ T2 = ⎝ = 600K . ⎛ 8314 N ⋅ m ⎞ ⎜ 28.97 kg ⋅ K ⎟ (1kg ) ⎝ ⎠
Then, since η = 1 − TC ⇒ TC = TH (1 − η ) . With TH = T2 , TH TC = 600 (1 − 0.5) = 300K , where TC = T3 = T4 . (b) For process 1‐2, Q12 = 50kJ (given). An energy balance reads m ( u2 − u1 ) = Q12 − W12 , but since internal energy of an ideal gas depends on temperature and T1 = T2 , W12 = Q12 . Further, W12 = ∫ PdV = ∫
1 2 2 1 mRTH dV = mRTH ln V2 V1 . V Solving and inserting values ln V2 W12 V = ⇒ ln 2 = V1 mRTH V1 50kJ = 0.2904 ⇒ V1 = 0.224m3 ⎛ 8.314 kJ ⎞ (1kg ) ⎜ ⎟ ( 600K ) ⎝ 28.97 kg ⋅ K ⎠ Since T1 = T2 , PV1 = mRT , PV2 = mRT 1 2 ⇒ PV1 = PV2 , P = P2 (V2 V1 ) = 574kPa 0.3 1 2 1 ( 0.224 ) = 769kPa. (c) For process 2‐3: Q23 = 0. An energy balance reduces to give W23 = m ( u2 − u3 ) . With data from Table A‐22, W23 = (1kg ) ( 434.78 − 214.07 ) = 220.7kJ For process 3‐4, W34 = Q34 (as for process 1‐2). Also, Eq.5.7 is applicable: Q34 Q12 ⎛ 300 ⎞ = ⇒ Q34 = ⎜ ⎟ ( 50kJ ) = 25kJ ⇒ Q34 = −25kJ, W34 = −25kJ. TC TH ⎝ 600 ⎠ (d) For process 4‐1: Q41 = 0. An energy balance reduces to give W41 = m ( u4 − u1 ) . Since u1 = u2 , u4 = u3 , W41 = −220.7kJ . The work and heat transfers are summarized in the table. Problem 5.84 Known: A simple vapor power plant operates at steady state as shown below. Process 1‐2 and 3‐4 are adiabatic. Find: In each of the following three cases, determine whether the cycle is possible, internally reversible, or impossible using Eq. 5.13 (Clausius inequality). (a) Process 4‐1: constant pressure at 1 MPa from saturated liquid to saturated vapor. Process 2‐3: constant‐pressure at 20 kPa from x2 = 88% to x3 = 18% . (b) Process 4‐1: constant pressure at 8 MPa from saturated liquid to saturated vapor. Process 2‐3: constant‐pressure at 8 kPa from x2 = 67.5% to x3 = 34.2% . (c) Process 4‐1: constant pressure at 0.15 MPa from saturated liquid to saturated vapor. Process 2‐ 3: constant‐pressure at 20 kPa from x2 = 90% to x3 = 10% . Schematic & Given Data: Model and Assumptions: 1. 2. 3. 4. A control volume at steady state encloses each of the four components Power and heat transfer rates are in the directions of the arrows. Stray heat transfer can be ignored. Kinetic and potential energy changes can be ignored. Analysis: The Clausius inequality can be expressed equivalently as ∫⎜ ⎝ ⎛ δQ ⎞ ⎟ = −σ cycle T ⎠b (5.13) where σ cycle can be interpreted as representing the “strength” of the inequality. σ cycle = 0 : no irreversibility present within the system σ cycle > 0 : irreversibility present within the system σ cycle < 0 : impossible Property data from Table A‐3, a b c Expressed on a rate basis, Eq.5.13 takes the form, P41 (MPa) 1 8 0.15 T41 (°C) 174.9 195.1 111.4 (hg‐hf) (kJ/kg) 2015.3 1441.3 2226.5 P23 (kPa) 20 8 20 T23 (°C) 60.06 41.51 60.06 (hg‐hf) (kJ/kg) 2358.3 2403.1 2358.3 Q41 Q23 − = −σ cycle T41 T23
where Q41 = m ( h1 − h4 ) = m h fg () 41 and Q23 = m ( h2 − h3 ) = m ⎡( h f + xh fg ) − ( h f + xh fg ) ⎤ 2 3⎦ ⎣ = m ( x2 − x3 ) ( h fg )
23 Accordingly σ cycle
m = ( x2 − x3 ) ( h fg )23 ( h fg )41
T23 − T41 (a) x2 = 0.88, x3 = 0.18 σ cycle
m = ( 0.88 − 0.18 )( 2358.3) − 2015.3 = 0.5 kJ
333.21 453.05 kg K possible (b) x2 = 0.675, x3 = 0.342 σ cycle
m = ( 0.675 − 0.342 )( 2403.1) − 1441.3 ≈ 0 kJ
314.66 568.25 kg K internally reversible (c) x2 = 0.9, x3 = 0.10 σ cycle
m = ( 0.9 − 0.10 )( 2358.3) − 2226.5 = −0.13 kJ
333.21 384.55 kg K impossible ...
View Full
Document
 '08
 TORRANCE

Click to edit the document details