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Assignment__8

# Assignment__8 - Assignment#8 PROBLEM 8.1 PROBLEM STATEMENT...

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Assignment #8 PROBLEM 8.1 PROBLEM STATEMENT: A thermodynamic cycle using air as the working fluid in a closed system is outlined below. The specific heats of air can be assumed constant throughout the cycle with values taken at the initial conditions. The four processes that make up the cycle are: Process 1-2: Air at standard conditions (State 1: 1 atm, 298 K) is heated at constant volume to state 2, where 2 1 P =10P. Process 2-3: The air at State 2 is expanded isobarically to State 3, where 3 2 v = 3v . Process 3-4: The air is expanded isentropically until it returns to pressure P 1 at State 4. Process 4-1: The air is compressed isobarically to return it to State 1. (a) Sketch the P- v and T - s diagrams of this cycle. Pay special attention to the relative slopes of the process lines. (b) Find the numerical value of the power cycle efficiency of this heat engine. GIVEN: Air, P 1 = 1 atm, T 1 = 298 K Process 1-2: Constant volume to P 2 =10P 1 Process 2-3: Constant pressure (isobaric) to v 3 =3v 2 Process 3-4: Constant entropy (isentropic) to P 4 =P 1 Process 4-1: Constant pressure (isobaric) to v 1 FIND: . HE P- v and T - s diagram, η ASSUMPTIONS: Ideal gas, constant specific heats. GOVERNING RELATIONS: 1. For constant specific heats, ( 29 ( 29 V P V P T s = T c and T s = T c 2. For an ideal gas, R = c P c V and k = c P / c V 3. Ideal gas equation, Pv =RT 4. P-T relation for isentropic process, constant specific heats: ( 29 ( 29 (k-1) k b a b a T T = P P 5. Thermal efficiency, HE NET IN = w q η QUANTITATIVE SOLUTION: a). The P - v diagram can be derived from the known processes. For the T - s diagram, note that for the constant volume process, V V T T = s c and for the isobaric process, P P V T T T = = s c R + c Thus, the slope for the constant volume process (Process 1-2) is steeper than the isobaric processes. The diagrams for this cycle are shown below.

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Assignment #8 b). The temperature of each state can be found in terms of T 1 . Also, the thermal efficiency can be expressed in terms of T 1 and k, where k is 1.4001 from Table 5s for T = 25°C (298 K). For State 2: (from the ideal gas equation) P v R 1 P v = T R 2 2 1 1 1 2 P T = T =10T P T For State 3: P v R 3 P = T v R 3 3 2 1 1 2 2 v T = T = (10T ) 3 = 30T v T × For State 4: k-1 1.4001-1 k 1.4001 4 4 3 1 1 3 P 1 T = T = 30T =15.54T P 10 Now apply the first law to determine the work and heat transfer for each process: Process 1-2: , , ( ) IN 12 IN 12 2 1 V 2 1 w 0(constant volume) q u u c T T = = - = - Process 2-3: Constant pressure , , , ( ) ( ) ( ) ( ) ( ) ( ) 3 OUT 23 2 3 2 3 3 2 2 3 2 2 IN 23 OUT 23 3 2 3 2 V 3 2 P 3
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