0.5 kpa (1000 Pa/kPa) A. A = 709.81 N/500 N/m 2 = 1.37 m 2

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1 AREN 2110 Spring 2010 Homework #2: Due Friday, Jan 29, 6 PM Solutions 1. Consider a 70-kg woman who has a total footprint area of 400 cm2. She wishes to walk on the snow, but the snow cannot withstand pressures greater than 0.5 kpa. Determine the minimum size of the snowshoes (imprint area per snowshoe) to enable her to walk on the snow without sinking. Assumption: walking in snowshoes requires that one snowshoe at a time carries the entire load. Solution: FBD of snowshoe. Resistance of snow before loading by woman = atmospheric pressure force, where P = net pressure of snow loaded by women on snowshoes = 0.5 kpa*a, where A = area of one snowshoe 70 kg * 9.81 m/s kpa (1000 Pa/kPa) *A A = 70*9.81 N/500 N/m 2 = 1.37 m 2 (these would be enormous snowshoes typically snowshoes have much smaller area and snowshoer sinks somewhat into the snow, which in turn packs the snow and increases resistance) 2. Determine the pressure exerted on a diver at 30 m below the free surface of the ocean. Assume the barometric pressure is 101 kpa and the specific gravity of seawater is Solution: P atm = 101 kpa P w (z = 30m) P w = gz = 1.03*(1000 kg/m 3 )*9.81 m/s 2 *30m * 10-3 kpa/pa = 303 kpa P diver = P atm + P w = kpa = 404 kpa

2 3. Is it possible to have water vapor at -10 o C (263K)? Explain. Yes. Saturated ice and saturated water vapor are in equilibrium when T is lower than the triple point temperature for the triple point pressure of the system. From appendix table A-8, at T = -10 o C, corresponding pressure for saturated ice and vapor is kpa. 4. Does the amount of heat absorbed as 1 kg of saturated liquid water boils to saturated vapor have to the same as the heat released when 1 kg saturated water vapor condenses at 100 o C? Explain. Yes. Absorbed energy during vaporization has to equal the energy released during condensation at the same T and P. If you could release/absorb more energy in either direction, then you would be creating or destroying energy a violation of the first law of thermodynamics. 5. Complete the following table for H 2 O P (kpa) T ( o C) v (m 3 /kg) phase Saturated mixture Saturated vapor Superheated vapor Compressed liquid 1: v f < v < v 50 o C means saturated mixture and P = P 50 o C (Table A-4) 2: T = T 200 kpa and v = v g,@ 200 kpa (Table A-5) 3: T > T 400 kpa (143.6 o C). Superheated vapor; get v from Table A T < T 400 kpa (158.8 o C), v ~ v 110 o C (Table A-4) 6. Complete the following table for H 2 O: P (kpa) T ( o C) h (kj/kg) x phase Saturated mixture Saturated mixture Saturated liquid NA Compressed liquid NA Superheated vapor

3 1: 0 < x < 1 means saturated mixture. h = x*h fg + h 200 kpa (Table A-5) 2: h f < h < h 140 o C, means saturated mixture and P = P 140 o C and calculate x = (h h f )/h 140 o C (Table A-4) 3: x = 0 is saturated liquid, T = T sat and h = h 950 kpa (Table A-5) 4: T < T 500 kpa (151.8 o C) is compressed liquid. h ~ h o C (Table A-4) 5: h > h 800 kpa is superheated vapor. Get h from Table A-6, P = 0.8 MPa, T=350 o C. 7. Complete the following table for the various substances substance P (kpa) T ( o C) v (m 3 /kg) x* phase 1 H 2 O Saturated mixture 2 H 2 O NA Superheated vapor 3 R-134a Saturated mixture 4 R-134a NA Compressed liquid * use "na" for "not applicable" where quality does not apply 1: P = P sat, v = x(v fg ) + v 150 o C (Table A-4) 2: v > v 150 o C ( m 3 /kg) is superheated vapor. Find P corresponding to v = m o C (Table A-6) 3. v f < v < v 0 o C C, is saturated mixture. P = P sat and x = v v f /v 0 o C (Table A-11) 4. T < T 400 kpa (143.6 o C) is compressed liquid (Table A-5) and v ~ 0 o C (Table A-4) 8. A piston-cylinder device contains 0.85 kg of refrigerant 134a at -10 o C (263 K). The piston has a mass of 12 kg and a diameter of 25 cm. The local atmospheric pressure is 88 kpa. Now, heat is transferred to the refrigerant until the temperature is 15 o C. Determine: a. The final pressure b. The change in volume of the cylinder space c. The change in enthalpy of the refrigerant Solution: piston cylinder keeps constant pressure during expansion process. Force balance (FBD) on piston at state 1 to determine pressure of refrigerant at states 1 and 2. 88kPa*A p (kn) 12 kg*9.81m/s 2 *10-3 kn/n P 1 *A p (kn) A p = *(0.25 m) 2 /4 = 0.05 m 2 a) P 1 = (88*0.05 kn + 12*9.81* 10-3 kn)/0.05 m 2 = 90.4 kpa = P 2 (final pressure)

4 b) 90.4 kpa < -10 o C, so superheated refrigerant. Use Table A-13. Find v 1 (@ -10 o C) by interpolation between 60 kpa and 100 kpa. v 1 = m 3 /kg to find v 2, must first interpolate to find 15 o C at P = 60 kpa and P = 100 kpa. v 15, 60 = ( )/2 = m 3 /kg v 15, 100 = ( )/2 = m 3 /kg v 2 = m 3 /kg V = m*(v 2 v 1 ) = 0.85 kg*( ) m 3 /kg = m 3 c) Similar procedure as b) for H h 1 = kj/kg to find h 2, must first interpolate to find 15 o C at P = 60 kpa and P = 100 kpa. h 15, 60 = ( )/2 = kj/kg h 15, 100 = ( )/2 = kj/kg h 2 = kj/kg H = m*(h 2 h 1 ) = 0.85 kg*( ) kj/kg = 18.4 kj/kg 9. A rigid tank with a volume of 2.5 m 3 initially contains 15 kg of saturated liquidvapor mixture of water at 75 o C. The water is slowly heated until all the water is saturated vapor. a. What is the quality of the mixture at the initial state (before heating)? b. Determine the temperature at with the liquid in the tank is completely vaporized to saturated vapor.

5 c. What is the pressure in the tank? d. Show the process on the T-v diagram on the next page. Solution: this is a constant volume (isochoric) phase change process. Both temperature and pressure will increase as heat is added to closed system rigid tank. a) v = V/m = 2.5 m 3 /15 kg = m 3 /kg. Since v f < v < v 75 o C, system is saturated mixture. x = 0.04 b) at state 2, water is saturated vapor, and v 2 = v 1 = m 3 /kg = v T 2 where 185 < T 2 < 190 o C (Table A-4). Interpolate c) interpolate to find P 2 : T 2 = 187 o C P 2 = 1176 kpa

6 T ( o C) Water Liquid-Vapor PhaseT-v Diagram v (m 3 /kg) 10. One kilogram (1 kg) of water vapor at 200 kpa fills the left chamber of a partitioned system shown below. The volume of this chamber is m 3. The right chamber has twice the volume of the left chamber and is evacuated at the initial state. 1 kg water 200 kpa m 3 Now the partition is removed and heat is transferred so that the temperature of the water is 3 o C. a. What is the initial temperature of the water (before the partition is removed)? b. What is the pressure of the water after the partition is removed and heat transferred? c. What is the quality of water at the final equilibrium state? Solution: State 1 Calculate v 1 from given: V and m: v 1 = m 3 /1 kg = m 3 /kg. At P 1 = 200 kpa, v 1 > vg ( m 3 /kg) so system is superheated vapor. a) From Table A-6, v 1 corresponds to T = 250 o C

7 b) At T = 3 o C, v 2 = V 2 /m where V 2 = 3V 1 = 3* m 3 and v 2 = 3* m 3 /1kg = m 3 /kg. Interpolate to find v 3 o C: v 3 o C = m 3 /kg OR notice that m 3 /kg is less than vg at either 0.01 or 5 o C, so system is saturated mixture at state 2. Interpolate to find P sat at 3 o C: P 2 = kpa c) Find x at state 2 using v f and v g at 3 o C, and m 3 /kg, respectively. x = 0.02

Chapter 3 PROPERTIES OF PURE SUBSTANCES. Thermodynamics: An Engineering Approach, 6 th Edition Yunus A. Cengel, Michael A. Boles McGraw-Hill, 2008

Chapter 3 PROPERTIES OF PURE SUBSTANCES Thermodynamics: An Engineering Approach, 6 th Edition Yunus A. Cengel, Michael A. Boles McGraw-Hill, 2008 Objectives Introduce the concept of a pure substance. Discuss