ENGRD 2210 - Homework 8 Solution - ENGRD 2210...

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ENGRD 2210 – THERMODYNAMICS HOMEWORK 8 SOLUTION Problem 1: Using thermodynamics tables: (a) At T=48 o C, and p = 520 kPa, find the specific entropy of R-22. (b) At T=200 o C, and p = 500 kPa, find the specific entropy of water (c) Refrigerant 134a expands isothermally from an initial state at p=300kPa, T=40 o C to a final state at p = 150 kPa. Find the change in specific entropy of Refrigerant 134a that occurs in this process. (d) Calculate the change in entropy for air between the states p 1 =1 atm, T 1 =50 ° C and p 2 =10 atm, T 2 =1000 o C. (e) Calculate the change in specific entropy of carbon dioxide as an ideal gas between the state T1 = 820 o F, p1 = 1 amt, T2=77 o F, p2=3 atm. Solution: (a) For R-22 at 48 °C and 520 kPa = 5.2 bar, we utilize Table A-9 and interpolation to calculate the specific entropy. For T = 48 °C, p = 5 bar, we can interpolate the specific entropy to be: For T = 48 °C, p = 5.5 bar, we can interpolate the specific entropy to be: Now we can interpolate again to get the specific entropy at 5.2 bar: (b) For water at 200 °C and 500 kPa = 5 bar, we can determine that we are dealing with water vapor by obtaining the saturation temperature at P = 5 bar. T sat = 151.86 °C at 5 bar for water (Table A-
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4). This confirms that we are dealing with superheated water vapor. From Table A-4 we can directly obtain the specific entropy for the given temperature and pressure: (c) For R-134a refrigerant at 40 °C and 300 kPa = 3 bar, we can see from Table A-12 that T sat is much less than 40 °C. Thus, we are dealing with superheated R-134a. For the specific entropy of the initial state, we interpolate between P = 2.8 bar and P = 3.2 bar for 40 °C, and get: For the specific entropy of the final state, we interpolate between P = 1.4 bar and P = 1.8 bar at 40 °C, and get: It follows that: (d) We are given: T 1 = 50 °C = 323.15 K p 1 = 1 atm T 2 = 1000 °C = 1273.15 K p 2 = 10 atm And furthermore, we know: Using Table A-22 and interpolation for air at the given temperatures , s°(T):
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(e) We are given: T 1 = 820 °F = 1280 R p 1 = 1 atm T 2 = 77 °C = 537 R p 2
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This note was uploaded on 03/15/2010 for the course ENGRD 2210 at Cornell University (Engineering School).

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ENGRD 2210 - Homework 8 Solution - ENGRD 2210...

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