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Solution p2

problem sheet thermo 2 solution

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457 views25 pages

Solution p2

problem sheet thermo 2 solution

Uploaded by

Sumra Maqbool
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© © All Rights Reserved
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1. Determine the phase for each of the following water states using the tables in Appendix and indicate ‘the relative position in the P-v, T-v, and P-T diagrams. 2.120°C, S00 kPa b. 120°C, 0.5 m*/kg, [Subcooled liquid, (b) two phase mixture] Eicon epee crore Ole ween hia indicate the relative position in the P-v, T-v, and P-T diagrams. a) 120°C, 500 kPa b) 120°C, 0.5 m/kg, Enter Table 252 with 220°C. The saturation, pressure is 198.54 kPa, so we have acompressed — Hither pressures (P> Py ata given 1) fepuid, polit aia Figure 500 KPa LP | Lower tempreatures (T'< Ty ata given P) Lower specific volumes (v < vata given P or 7 “That is above the saturation line for 20°C. We m wsumas } could also have entered Table 2.5, with 500 kPa and found the saturation temperature as 151.84°C, so we would say itis subcooled liquid. Lower internal enemies (u v, = 0.08401 m?/leg Fw so from the P-v diagram the | state is superheated vapor. We” can find the state in Table 2.7.3, 3. Determine the temperature and quality (if defined) for water at a pressure of 300 kPa and at each of, ‘these specific volumes: 2.0.5 m'/kg b.1.0m*/kg {(a) t=133.6 °C, x=0.825, (b) t=379.8°C, superheated ] Determine the temperature and quality (if defined) for water at a pressure of 400 KPa and at each of these specific volumes: 2.0.4 m/kg 7 peste b.o.7 mi/kg a. By comparison with the values in Figure, the ““""* state at which vis 0.4 m’/kg is seen to be in the j liquid-vapor two-phase region, at which T= came v, = 0.1272 mY/kg so we have superheated vapor. Proceed to Table 2.5.3 at any subsection with 200°C; say we start at 200 kPa. There the v = 1.0804, which is too large so the pressure must be higher. For 400 kPa, v = 0. 5343, and for 600 kPa, v = 0.3520 At 200°C and 200 kPa v, = 1.0804 m%kg At 200°C and 400 kPa v, = 0.5343 m%/kg, i v=o4m/kg. At 200°C and 600 kPa v, = 0.3520 m°kg A linear interpolation, Figure, between the two pressures is done to get P at the desired v. g P = 400 + (600 — 400) » (0.4-0.5343) {oas2-o5e4a) 54734 KPa yes iar as asa amo 8. Are the pressures in the tables absolute or gauge pressures? Solution: The behavior of a pure substance depends on the absolute pressure, so P in the tables is absolute 9. At higher elevations, as in mountains, air pressure is lower; how does that affect the cooking of food? Athigher altitudes, the air pressure is decreased, which forces water's boiling point to lower. This lowered boiling point of water requires an increase in cooking times or temperature and possibly even alterations of ingredients. This effect becomes relevant above altitudes of 2,000 feet (600 meters). At that altitude, water boils at approximately 208°F (98°C) and adjustments sometimes need to be made to compensate for the reduced air pressure and water boiling point. Beginning around 2,500 feet above sea level, altitude starts to affect cooking in three different ways: + The higher the elevation, the lower the boiling point of water. When water boils at lower temperatures, it takes longer for foods to cook in or over water. + The higher the elevation, the faster moisture evaporates, + The higher the elevation, the faster leavening gases (air, carbon dioxide, and water vapor) expand. 10. If have 1 liter of ammonia at room pressure and temperature (100 kPa, 20°C), what is the mass? [0.706 g] 37 IfThave 1 L ammonia at room pressure and temperature (100 kPa, 20°C) how much ‘mass is that? Ammonia Tables B.2: B.2.1 Poy = 857.5 kPa at 20°C so supetheated vapor. B22 v=14153 m¥skg under subheading 100 kPa 0.001 m3 OI — 9,000 706 kg = 0.706 V7 [4153 mike & & 11. Why are most compressed liquid or solid regions not included in the printed tables? 3.12 Why are most of the compressed liquid or solid regions not included in the printed tables? For the compressed liquid and the solid phases the specific volume and thus density is nearly constant. These surfaces are very steep nearly constant v and there is then no reason to fill up a table with the same value of v for different P and T. 12. Give the phase for the following states: =0.5 MPa [Superheated vapor] a. COzat ‘and P b. Air at t= 20 °C and P = 200 kPa [Superheated vapor] NH, at t= 170 °C and P = 600 kPa [Superheated vapor} os Give the phase for the following states. ‘Solution: a. CO, T=267C P=0SMPa TableA2 superheated vapor assume ideal gas Table A.5 b. Air T=20°C — P=200KPa Table A2 superheated vapor assume ideal gas Table A.5 c. NH; T= 170°C P=600kPa TableB.2.2 or A2 T>T,—> superheated vapor 13, Determine the phase of the substance at the given state using Appendix tables. a. Water: 100 =C, 500 kPa b. Ammonia: -10 «C, 150 kPa cc. R-410a: 0 =C, 350 kPa Determine the phase of the substance at the given state using Appendix B tables a) Water 100°C, $00 kPa b) Ammonia =10°C. 150 kPa ©) RAD 0°C. 350 kPa Solution: 8) From Table B11 Pyy(100°C) = 101.3 kPa S500 KPa> Pia themit is compressed liquid OR from Table B.1.2 Tyap(500 kPa) = 152°C 100°C Pgge(0 °C) = 309 kPa Compressed liquid. The $-L fusion line goes slightly to the left for water. It tilts slightly to the right for most other substances. 14, Determine whether water at each of the following states is a compressed liquid, @ superheated vapor, or a mixture of saturated liquid and vapor: 2. 10 MPa, 0.003m*/kg, b, 1 MPa, 190 =C c. 200°C, 0.1 m*/kg d. 10 kPa, 10 «C 3.30 Determine whether water at each of the following states is a compressed liquid, a superheated vapor. or a mixture of saturated liquid and vapor. a. P=10MPa, v= 0.003 m'/kg b. 1 MPa, 190°C ce 200°C, 0.1 mk d. 10 kPa, 10°C Solution: For all states start search in table B.1.1 (if T given) or B.1.2 (if given) a, P=10MPa,v=0.003 mg so look in B.1.2 at 10 MPa 0.001452 0.01803 m/kg, oye sv sve so mixture of liquid and vapor. b. 1MPa,190°C : Only one of the two look-ups is needed BALL: PTegp= 179.91°C so it is superheated vapor ©. 200°C. 0.1 mikg: look in B11 vp = 0.001156 mk : vy = 0.12736 mike. > wy cv P= 1.2276 kPa so compressed liquid From B.L.2: Tvg => superheated vapor Look in B22 b) B21 P=Pyay= 2033 kPa w= vp +x vfg = 0.001777 + 0.5 x 0.06159 = 0.0326 miikg. ee 17. Give the missing property of P, T, v, and x for R-134a at a. T=-20-C, P= 150 kPa b.P= 300kPa, y= 0.072 mi/kg 3.40 Give the phase and the missing property of P. T. v and x. a. Rel3da T=-20°C, P= 150 kPa b. Rel3da P= 300 kPa, v= 0.072 m/kg e. CH, T=155K, v=0.04 m/kg do CHy =: T=350K. v 25 mk. Solution: a)BS.1 P> Py =133.7kPa > compressed liquid 0.000738 mikg b)BS2 — v>vgat300kPa => superheated vapor = (cio) (2072007111 10* (20-10) (9 o7441 - 0.07111 x= undefined = 127°C ©) B71 04892 m3/kg 2-phase ME 2877 Ve 0.04605 295.6 kPa = 0.806 P sat ABT T>T, and v>>v, = superheated vapor B.7.2 located between 600 & 800 kPa 0.25-0.30067 P= 600+ 200 =734kPa 0.2251-0.30067 18. Fill out the following table for substance water PikPal TPC] vim'/kg] 500 20 500 0.20 3.38 Fill out the following table for substance water: Solution: P [kPa] TLC] vik] x a) 500 20 0.001002 Undefined b) 500 181.86 0.20 0.532 ©) 1400 200 0.14302 Undefined d) 8581 300 0.01762 «08 a) Table B11 P> Psat so itis compressed liquid => Table B14 b) TableB2 vgcvTygy= 195°C Table B.1.3 subtable for 1400 kPa gives the state properties 4) Table B.1.1 since quality is given itis two-phase Va vet Nx Vp = 0.001404 + 0.8 x 0.02027 = 0.01762 m¥/kz_ 19, Place the four states a-d listed in Problem 18 as labeled dots ina sketch of the P-v and T-v diagrams. 3.34 Place the four states aed listed in Problem 3.33 as labeled dots in a sketch of the Pew and T-v diagrams. Solution: 20. Determine the specific volume for R-410a at these states: a. -15 «C, 500 kPa b. 20°C, 1000 kPa €. 20°C, quality 25% Determine the phase and the specific volume for ammonia at these states using ‘the Appendix B table. a, -10°C. 150 kPa b. 20°C, 100 kPa ©. 60°C, quality 25% Solution: Ammonia, NH3, properties from Table B.2 a) Table B.2.1: P< Pyge(-10 °C) = 291 kPa Supesheated vapor B22 0.8336 mg b) Toble B.2.1 at given Puy = S47 SkPa 50 PEP Superheated vapor B.2.2 v= 14153 m‘/kg s) Table B.2.1 enter with T (this is two-phase L + V) Vg 7X Vfg = 0.001834 + x x 0.04697 = 0.01388 in"/kg v 21. You want a pot of water to boil at 105 -C, How heavy a lid should you put on the 15-cm-diameter pot when Patm = 101 kPa? [35.7ke) 3.44 You want a pot of water to boil at 105°C. How heavy a lid should you put on the 15 em diameter pot when Patm = 101 kPa? Solution: Table BLL at 108°C: Pyyy= 120.8 kPa A 0.01767 m* sat Patin) A = (120.8 = 101) kPa x 0.01767 mi? = 0.3498 EN = 350N Poet = Mpa 2 350 9 g07 735-7 ke m= 22. Water at 120 °C with a quality of 25% has its temperature raised 20 =C in a constant-volume process. What is the new quality and pressure? 3.13 Water at 120°C with a quality of 25% has its temperature raised 20°C in a constant volume process. What is the new quality and pressure? Solution: State 1 from Table B.1.1 at 120°C vy vg +X vag = 0.001060 + 0.25 x 0.8908 = 0.22376 m?, State 2 has same v at 140°C also from Table B.1.1 N= %f _ 0.22376 - 0.00108 x = 0.4385. Vig 0.50777 P= Psat = 361.3 kPa Pop T op 361.3 140 120 198.5 + 23. A sealed rigid vessel has volume of 1 m3 and contains 2 kg of water at 100°C. The vessel is now heated. if @ safety pressure valve is installed, at what pressure should the valve be set to have 3 maximum temperature of 200:C? 3 A scaled rigid vessel has volume of 1 m} and contains 2 kg of water at 100°C. The vessel is now heated, Ifa safety pressure valve is installed, at what pressure should the valve be set to have a maximum temperature of 200°C? Solution: Process: State 1: v,=1/2=0.5 m3/kg T cr . * 4 500 kPa from Table B.1.1 we. itis 2-phase 400 kPa State 2: 200°C, 0.5 m? ke roe Table B.1.3 between 400 \ and 500 kPa so interpolate = _ 0.5-0,53422 _ P2400+ Tayqrg ssiay * (500-400) = 431.3 kPa 24. Saturated water vapor at 200 kPa is in a constant pressure piston/cylinder assembly. At this state the ‘on is 0.1 m from the cylinder bottom. How much is this distance and what is the temperature if the water is cooled to accupy half of the original volume? [0.05m| 3.48 Saturated water vapor at 200 kPa is in a constant pressure piston cylinder. At this state the piston is 0.1 m from the eylinder bottom. How much is this distance and the temperature if the water is cooled to occupy half the original volume? Solution: State 1: B12 vj=v,(200KPa)= 0.8857 mckg, T= 120.2°C Process: P= constant = 200 kPa State2: PLvy =vy/2=0.44285 mike Table B..2 vy = he 3 Vtot = Va + Vp = 3.1606 mr = 3, ‘tot Brot = 05746 m3 kg 27. Briefly answer following questions; a. Is iced water a pure substance? Why? 2.1 Is Iced water a pure substance? Why? ‘Yes, because it has a uniform chemical composition, H20. b. What is the difference between saturated liquid and compressed liquid? 2.2. what te aterence between sautratd ald and compressed qui? ‘A saturated quis one thats abou to evaporate, Any Increase In tmpertue (@ fixed P) or decrease In pressure (@ fie) wil ing about c. What is the difference between saturated vapor and superheated vapor? above the aturon 3th presse) wl nk bing bok d, Isit true that water boils at higher temperatures at higher pressures? Explain. Closely packed tapetherrequring Higher energy to separate them from eachother to cause a phase change. . If the pressure of a substance is increased during a boiling process, will the temperature also increase ‘or will it remain constant? Why? 2.7 Woe pressure of 9 substance i icrased dig a biling process, wl the temperature als increase a lt remain canta? Why? f, Why are the temperature and pressure dependent properties in the saturated mixture region? The pressure and tapers are dependent nis reson becaure the att heat of vapotton (or candeneation) depends on bth pressure g. What is the difference between the critical point and the triple point? 2.9 What isthe difference betwen the ctical point and th ‘The cial point isthe polnt at which the phase change pro cfc volume increases continuously. h. Is it possible to have water vapor at -10°C? |. What is the physical significance of hig? Can it be obtained from a knowledge of he and hy? How? mm, Is ittrue that it takes more energy to vaporize 2 kg of saturated liquid water at 100°C than it would at 120°C? When the temperature is greater than 100 deg C the water boils and tums into vapor. To get ‘the highest temperature In that phase the pan would need to be covered with a heavy Ud. This will give the shortest cooking time. ji How does the boiling process at supercritical pressures differ from the boiling process at subcritical pressures? 342C At supercritical pressures, there is no distinct phase change process. The liquid uniformly and gradually expands into a vapor. At subcritical pressures, there is always a distinct surface between the phases. k, Does the amount of heat absorbed as 1 kg of saturated liquid water boils at 100°C have to be equal to the amount of heat released as 1 kg of saturated water vapor condenses at 100°C? 3-17€ Yes, Otherwise we can create energy by altemately vaporizing and condensing a substance. |. What isthe physical significance of hig? Can it be obtained from a knowledge of he and hy? How? mm. [sit true that it takes more energy to vaporize 1 kg of saturated liquid water at 100°C than it would at 120°C? 3:19C The term he, represents the amount of energy needed to vaporize a unit mass of saturated liquid at a specified temperature or pressure, It can be determined from i= ig li mm. Is it true that it takes more energy to vaporize 1 kg of saturated liquid water at 100°C than it would at 120°C? 3-20 Yes; the higher the temperature the lower the hg value. 1n, What is quality? Does it have any meaning in the superheated vapor region? 3:21C Quality is the finction of vapor in a saturated liguid-vapor mixture. It has no meaning in the superheated vapor region. 0. Which process requires more energy: completely vaporizing 1 kg of saturated liquid water at 1 atm. pressure or completely vaporizing 1 kg of saturated liquid water at 8 atm pressure? 3.22C Completely vaporizing 1 kg of saturated liquid at 1 atm pressure since the higher the pressure, the lower the hi. p. Does hig change with pressure? How? 3.23C Yes. It decreases with increasing pressure and becomes zero at the exitical pressure. 28, A 1.8-m? rigid tank contains steam at 220°C, One third of the volume is in the liquid phase and the rest is in the vapor form. Determine (a) the pressure of the steam, (b) the quality of the saturated mixture, and (c) the density of the mixture, (a) Two phases coexist in equilibrium, thus we have a saturated liquid-vapor mixture. The pressure of the steam is the saturation pressure at the given temperature. Thus, P= Teaozmc= 2320 KPa (b) The total mass and the quality are determined as 4% W3x08m') oy myo = 504.2 7 vy 0.001190 ake is 23% n my on 2213) os o4tg U, 0.08609 mike 504.2+13.94 = S18 1ke my +m, x= 712 3% 0.0280 5181” (c) The density is determined from wavy +3(v, ~vy) = 0.001190 + (0.0269}(0,08609) = 0.003474 mike 1 3 a > = pomsary = 287-8 kom Pp 29. Water is to be boiled at sea level in a 30-cm-diameter stainless steel pan placed on top of a 3-kW. ‘electric burner. If 60 percent of the heat generated by the burner is transferred to the water during boiling, determine the rate of evaporation of water. Power supplied = 3 kW Power transferred to the water during boiling = 60% of 3 KW = 1.8 KW. To convert water into steam, heat supplied = Q= mL, where L = Latent heat of vaporization. “The specific latent heat of vaporization is the amount of heat required to convert unit ‘mass of a liquid into the vapour without a change in temperature.” For water at its normal boiling point of 100 °C, the latent specific latent heat of ‘vaporization is 2260 kJ.ke", This means that to convert I kg of water at 100°C to I kg of steam at 100 °C, 2260 KJ of heat must be absorbed by the jwater. Assume a time petiod of | s Then, 1.8*1=m*2260 ‘Therefore, m= 1.8/2260 = 7.97°10* kgs = 0.797 grams/s, 30. Water is boiled at 1 atm pressure in 2 25-cm-internaldiameter stainless steel pan on an electric range, IF tis observed that the water level in the pan drops by 10 cm in 45 min, determine the rate of heat transfer to the pan, 3-43 Water is boiled at | atm pressure in a pan placed on an electric burner. The water level drops by 10 tem in 45 min during boiling, The rate of heat transfer to the water is to be determined, Properties ‘The properties of water at | atm and thus at a saturation temperature of Tay = 100°C are ig 2256.5 kl/kg and y ~ 0.001043 m'ky (Table A-4), Analysis The ate of evaporation of water is a [(0.25 m) m) H, (DL _[wl0.25m°/ 410.10) _ 5 yg py no yy 0.001083 Ftggg = Most = A TONKS _ 991742 e's “Ar 45x60s. ‘Then the rate of heat transfer to water becomes ry bg, = (0.001742 kas (2256.5 kW) = 3.93 KW 431, Saturated steam coming off the turbine of a steam power plant at 30°C condenses on the outside of 12 3.cm outer-diameter, 35-m-long tube at a rate of 45 kg/h. Determine the rate of heat transfer from the steam to the cooling water flowing through the pipe, 3-48 Saturated steam at Tyr = 30°C condenses on the outer surface of a cooling tube at a rate of 45 kg/h. The rate of heat transfer from the steam to the cooling water isto be determined, Assumptions 1 Steady operating conditions exist. 2 The condensate leaves the condenser as a saturated liquid at 30°C. Properties The properties of water at the saturation temperature of 30°C are fig = 2429.8 klikg (Table A- 4). Analysis Noting that 2429.8 kI of heat is released as 1 ka of saturated ‘vapor at 30°C condenses, the rate of heat transfer from the steam to the 30°C ‘cooling water in the tube is determined directly from Cn [= g = Hilerapl ge = (45 kglh)(2429.8 ki/kg) = 109,341 kivh = 30.4 KW 35m ~ 32. Water in a S-cm-deep pan is observed to boil at 98°C. At what temperature will the water in a 40- em-deep pan boil? Assume both pans are full of water. Scepnimtobedceme Properties The density of liquid water is approximately p = 1000 kg/m? Analysis The pressure at the bottom of the 5-cm pan is the saturation 40 om ‘ae apd btng wpe of SEP Pore Se tate The rar difrcs bree te bam of pn A A 1000 kyim-s? Brg totais Som P= pgh=(Hhon 9307 WH 038 mf a3. Tooitng = Tanasr 2 era 133. Acoaking pan whase inner diameter is 20 cm is filed with water and cavered with a 4-kg lid. Ifthe local atmospheric pressure is 101 kPa, determine the temperature at which the water starts bolling when itis heated. Pea = 101 KPA yg = 4 kg a 3-48 A cooking pan is filled with water and covered with a 4-kg lid, The boiling temperature of water is 10 be determined, Analysis The pressure in the pan is determined from a foree balance on the lid, PA = Pal + We Py 1 y= ENA a 102.25 kPa wotm i ‘The boiling temperature is the saturation temperature corresponding to this pressure, T =Tynaroz2s ra =100.2°C (Table A-S) 34. Water is boiled in a pan covered with 2 poorly fitting lid at a specified location. Heat is suppli the pan by a 2-kW resistance heater. The amount af water in the pan is observed ta decrease by 1.19 kg in 30 minutes. If it is estimated that 75 percent of electricity consumed by the heater is transferred to the water as heat, determine the local atmospheric pressure in that location, ‘3-49 Water is boiled in a pan by supplying eletical heat. The local atmospheric pressure isto be estimated. _Assiomptions 75 pesceat of electricity consumed bythe heater i rasferred to the water Analysis The amnovat of uettraasfer to the water during ths period is (0.752 11830608) = 270013 ‘Using the data by a wial-etor approach in saturation table of water (Table A5) of using BES as we did, the saturation pressure that corresponds to an enthalpy of vaporization value of 2269 Kirke xs

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