Chater 3 3.19 Aarent molecular weight of mixture 1kmolofCO 2 (12 + 32) 44 kg 1kmolofN 2 28kg otal mass of mixture Number of kilometers in mixture 100 cm 3 of mixture will contain 95 cm 3 of CO 2 and 5 cm 3 of N 2 If V is volume occuied by 1 kmol of any gas: and, Number of kmols of CO 2 in 100 cm 3 of mixture 95 V Number of kmol of N 2 in 100 cm 3 of mixture 5 V µ µ 95 5 V 44 + V 28 Aarent molecular wt of mixture (M a ) 95 V + 5 V M a 43.2 Gas constant for 1 kg of mixture R 8.3143 103 M a 43.2 192.46 J deg 1 kg 1 3.20 If water vaor comrises 1% of the volume of the air (i.e., if it accounts for 1% of the molecules in air), what is the virtual temerature correction? Answer v ' 1 K(or1 C) Solution: he aarent molecular weight of air that contains 1% by volume (i.e., by number of molecules) of water vaor is, M moist (Aarent Molecular Weight of dry air) (Fraction of Dry Air) +(Molecular Weight of Water Vaor) (Fraction of Water Vaor) For moist air: Also, for moist air we could write: (28.97 0.99) + (18 0.01) 28.86 R d ρ moist v R moist ρ moist v R moist M d R d M moist 12
(Since R s are inversely roortional to molecular weights.) If we take 288 K v 28.97 28.86 1.0038 v 0.0038 v (0.0038) 288 1.09 C ' 1 C 3.21 Archimedes buoyancy rincile asserts that an object laced in a fluid (liquid or gas) will be lighter by an amount equal to the weight of the fluid it dislaces. Provide a roof of this rincile. [Hint: Consider the vertical forces that act on a stationary element of fluid rior to the element being dislaced by an object.] Consider an element of the fluid with mass m. he downward force on this element is mg. Since the element of fluid is stationary (before it is dislaced) the net uward force acting on it due to the surrounding fluid must be mg. If this element is dislaced by an object, the surrounding fluidwillexertthesamenetuwardforceontheobjectasitdidonthe element (namely, mg). herefore, the object will be lighter a mass, that is, by the mass of the fluid dislaced. 13
3.22 ρ(volume of balloon)g ρ 0 (volume of balloon)g (ρ ρ 0 )(volume of balloon)g. For hydrostatic equilibrium of balloon Also, R d ρand,since 0, or From (1) and (2) m gross (ρ ρ 0 )(volume of balloon) (1) m gross ρ ρρ 0 0 ρ 0 ρ 0 (2) µ1 0 (volume of balloon) (3) Now, Density of air at 273 K and 1000 hpa 1.275 kg m -3 From (3) and (4) and density varies at Density of air at 293 K and 900 hpa is 900 273 ρ (1.275) kg m-3 1000 293 1.069 kg m -3 (4) m gross 1.069 µ 1 293 0 (volume of balloon) 14
or, or, µ 600 1.069 1 293 0 3000 0 293 293 600 1 0.1871 1 3000 1.069 360.43 K 0 87.43 C 3.23 For no vertical acceleration of balloon m gross (ρ ρ 0 )(volume of balloon) Since m gross is same for the two balloons, (ρ ρ 0 )(volume of balloon) constant herefore, ρ (volume of helium balloon) (density of helium)(volume of helium balloon) ρ (volume of hot air balloon) (density of hot air)(volume of hot air balloon) (1) Also, R unit ρ mass 15
and, therefore, Also, From (1) and (2) ρ 0 1 R unit (MW) helium mass helium ambient (2) or, or, (MW) air (volume of helium balloon) (MW) helium (MW) air (volume of hot air balloon) (MW) air hot air (volume of balloon) (MW) air (MW) helium Ã! 1 (volume of hot air balloon)(mw) air 1 hot air (volume of helium balloon) (volume of hot air balloon) herefore, 3.24 From (3.29) (volume of balloon) (MW) air (MW) helium (MW) air µ hot air (volume of hot air balloon) (volume of helium balloon) 28.97 2 28.97 hot air volume of hot air balloon (volume of helium balloon) µ 26.97 363 1000 28.99 100 3377 m 3 1 2 ex (Z 2 Z 1 ) µ H ' 2 1+ (Z 2 Z 1 ) H 16 363 273 (volume of hot air balloon) µ 363 26.97 363 28.99 100
1 2 ' 2 (Z 2 Z 1 ) H (1) H 29.3 v v ' 253K at 500 hpa From (1) and (2) H 7.413 km (2) Z (in km) near 500 hpa ' 500 7.413 If 1hPa, Z ' 14.8 ' 15 m. 3.25 he error is due to the virtual temerature ( v ) measurement recorded by the radiosonde being assigned to the wrong ressure level. his will lead to an error in the magnitude of Z 1 Z 2 (and therefore Z 2, the height of the 500 hpa level) calculated from (3.24). On average, between µ sea level and 500 hpa each measurement of v will 1 be assigned 2 5 hpa (11 m er hpa) 27.5 m higher in altitude than it really is. With an average lase rate of 7 Ckm 1,thetemerature between sea level and 500 hpa will be, on average, 7 Ckm -1 27.5 10 3 km ' 0.2 C too high. Hence from (3.29), the thickness between sea level and 500 hpa will be too high by about 29.3 mdeg 1 (0.2 deg)ln 1000 500 4m. (Note: his uncertainty is relatively small.) 3.26 From (3.29) Z 2 Z 1 R d v g o ln 1 2 For layer outside the hurricane: For center of hurricane: Z 2 Z 1 (287) (270) 9.81 ln (287) (270) 9.81 ln µ 1010 200 µ 1010 287 µ v 940 200 9.81 ln 200 17
v µ 101 ln 20 270 µ 94 ln 20 282.38 K 3.27 Average temerature difference between center of the hurricane and its surroundings (282.38 270) K ' 12 K 12 C ¾ z? 500 hpa v 0 C 1000 hpa From the hysometric equation: For 1000 500 hpa: z (500 hpa) z (1000 hpa) R d g v µ 1000 500 29.26 v (ln 2) z (500 hpa) z (1000 hpa) 29.26 273 0.693 5536.8 m For 1020 1000 hpa layer: z (1000 hpa) z (1020 hpa) R d g v ln µ 1020 1000 29.26 288 ln 1020 1000 29.26 288 0.02 167 m (Note: We are not given v between 1000 and 1020 hpa, however, because the height difference between these two ressure levels is small, v will not differ greatly from the temerature at 1020 hpa, namely, 15 C or 288 K.) Now, herefore, z (1000 hpa) z (sea level) (167 50) m 117 m Height of 500 hpa level above sea level 5537 + 117 5654 m 5.654 m (Note: Answer given in 1st Ed. of book is based on taking R d /g 29.3 instead of the more accurate value of 29.26 used above.) 18
3.28 500 hpa mass m 5 10 6 J 1000 hpa Area 1 m 2 Difference in ressure between the two layers herefore, 500 hpa 5 10 4 Pa mg 5 10 4 m 5 104 0.51 10 4 kg 9.81 If is change in temerature of layer due to heating: 0.51 10 4 c 5 10 6 5 10 6 (0.51 10 4 ) (1004) 0.976 C 0.98 C Also, 3.29 From eqn. (3.29) in text: Z 500 Z 100 R d v ln ( 1 / 2 ) g o (Z 500 Z 100 ) R µ d 1000 ln v g o 500 287 (ln 2)(0.976) 9.81 (Z 500 Z 100 ) 19.8 m20m or, Z 2 Z 1 Z R d v ln 1 g o 2 ( Z) R d ln 1 v g o 2 19
v ( Z) R d ln 1 g o 2 180 287 1000 ln 9.81 500 180 20.279 8.9 C (Note: Since lase rate does not change, change in surface temerature is equal to change in mean value of v.) 3.30 Derive a relationshi for the height of a given ressure surface () in terms of the ressure 0 and temerature 0 at sea level assuming that the temerature decreases uniformly with height at a rate Γ Kkm 1. Answer z o 1 µ o RΓ g Solution: Let the height of the ressure surface be z; thenits temerature is given by o Γ z (1) combining the hydrostatic equation (3.17) with the ideal gas equation (3.2) yields d g dz (2) R From (1) and (2) d g R ( o Γz) dz Integrating this equation beween ressure levels o and and corresonding heights 0 and z and neglecting the variation of g with z, weobtain or Z o d g R Z z o dz ( o Γz) ln g µ o RΓ ln o Γz o 20
herefore, " z µ # RΓ/g o 1 (3) Γ o his equation forms the basis for the calibration of aircraft altimeters. An altimeter is simly an aneroid barometer that measure ambient air ressure. However, the scale of the altimeter is exressed at the height z of the aircraft, where z is related to by (3) with values of o, o and Γ aroriate to the U.S. Standard Atmoshere, namely, o 288 K, o 1013.25 hpa, and Γ 6.50 Kkm 1. 3.31 Consider a hiker who gets to 1 km without any change in ressure due to synotic conditions. hen, altimeter reads correct height of 1 km. Now, if ressure falls by 8 hpa, the altimeter (which is calibrated assuming that 1 hpa decrease in ressure corresonds to 8 m increase in height) will read 1000 m + (8 8) m1064 m. 3.32 Work done by a system in changing its volume from v 1 to v 2 is given by: 3.33 (a) Z v2 work dv (1) v 1 he gas equation for 2 kg of dry air can be written as From (1) and (2): Work done on a system For an isothermal transformation: Since v 2 v 1 10 v (2R d ) (2) Z v2 v 1 2R d dv v Work done on a system 2R d ln v 2 v 1 and 288 K, Work done on a system 2(287)288 ln 1 10 2(287)(288)(2.3026) 3.806 10 5 J ' 3.8 10 5 J dq du + dw (1st Law for unit mass) du + dα µ µ u u d + dα + dα α α 21
dq c v d + dα µ u for an ideal gas, since 0. α Also, α R dα + αd R d From (i) and (ii) for an adiabatic transformation (dq 0) (i) (ii) Also for an ideal gas: c v (dα + αd)+rdα 0 or Integrating, Hence, or, for volume v, R c c v c v dα + c v αd (c c v ) dα 0 d + γ dα α 0 where c /c v γ ln + γ ln α constant α γ constant v γ constant (b) For the isothermal transformation: 1 v 1 2 v 2 (1000)(7.5) 2 (2.5) 2 3000 hpa For the adiabatic transformation ( 2 v γ 2 3 v γ 3 2 v 2 3v 3 2 3 3 2 µ v2 v 3 γ 1 µ γ 1 µ 2.5 2.5 290 290 7.5 7.5 3 186.8 Kor 86.2 C 0.4 22
From 2 v γ 2 3 v γ 3 3 2 µ v2 3000 v 3 γ µ 2.5 7.5 γ 3000 (0.333) 1.4 3 643 hpa 3.34 If the balloon in Exercise 3.22 is filled with air at the ambient temerature of 20 C at ground level where the ressure is 1013 hpa, estimate how much fuel will need to be burned to lift the balloon to its cruising altitude of 900 hpa. Assume that the balloon is erfectly insulated, and that the fuel releases energy at a rate of 5 10 7 Jkg 1. Answer 5.23 kg Solution: 23
For the air in balloon at ground level: i 1013 hpa i 20 C 293 K For the air in balloon at 900 hpa f 900 hpa f 87.43 C (from solution to Exercise 3.22) 360.43 K Suose the air in the balloon goes from its initial to its final state in two stes: (1) Heat is added at constant ressure at ground level until the temerature reaches f 0. (2) Air in the balloon exands adiabatically as dros from 1013 to 900 hpa, while dros from f 0 to f. For ste (2): v γ constant and v constant 1 γ γ constant 1 γ i 0 γ f 1 γ f ( f ) γ 0 f µ f i µ i f µ f i 1 1.4 1.4 1 γ γ f 0.286 (360.43) f µ 0.286 1013 (360.43) 900 (1.034) (360.43) 372.83 K Heat required for heating of air (mass of air)(c )( ) Joules µ mass of fuel 5 7 Jkg -1 (mass of air)(c needed in kg ) 24
Mass of air in balloon (in kg) volume of air in balloon at 900 hpa 3000 m 3 µ R d 3000 m 3 µ 900 10 2 287 360.43 2610.12 kg density of air at 900 hpa and 360.43 K Mass of fuel needed (in kg) (2610.12 kg) 1004 JK -1 kg 1 (372.83 273 K) 5 10 7 Jkg 1 (2610.12) (1004) (99.83) 5 10 7 52322110.5 10 7 kg 5.23 kg 3.35 H 2 H 1 3(L f )+3 Z 373 273 c w d Z 373 3(3.34) 10 5 +3 (4183.9 + 0.125 d) 273 10.02 10 5 +3 4183.9 + 0.125 2 373 2 273 " Ã! 10.02 10 5 0.125 (313) 2 +3 (4183.9 313) + 2 Ã!# 0.125 (273) 2 (4183.9 273) 2 10.02 10 5 + 3 (168821) 10.02 10 5 + 5.06 10 5 15.1 10 6 J 3.36 We have to rove that for an adiabatic change hat is, to rove µ R/c µ o o 1 2 1 2 1 R/c 1 25 2 R/c 2 R/c (1)
We have for an adiabatic transformation ( 1 γ 1 2v γ 2 1 v 1 2v 2 1 2 Hence, or v 1 v 2 2 2 1 1 2 µ 1 1 γ µ 2 1 1 γ 2 1 µ γ 2 or, But, Hence, or From (3.54) 1 γ γ Hence, for 1 and 1, and for 2 and 2, From (1), (3) and (4), 2 µ 1 2 1 γ γ 1 c /c v c /c v 1 R/c 1 µ 1 2 1 2 1 c v c c R/c 2 1 2 R/c 2 θ,whichis R c µ R/c o (2) µ R/c o θ 1 1 (3) 1 µ R/c o θ 2 2 (4) 2 θ 1 θ 2 26
3.37 Accurate answer (by calculation): For an adiabatic transformation v γ constant But herefore, v constant µ γ constant 1 γ γ constant (1) 1 200 hpa 1 60 C 273 K 2 1000 hpa 2 θ From (1), θ µ 1 γ 200 γ 1 1000 µ 1 γ 1 γ 213 5 µ 1 c /c v µ Rd /c 1 c /c v 1 213 213 (2) 5 5 From c and c v values given in 1st Ed. of Wallace & Hobbs we get R d /c 0.286, therefore θ (0.2) 0.286 (213) 337.46 K 64.5 C he versus ln chart uses c /c v 1.40449 or R d /c 0.288. If this valueisusedin(2),weget θ 65.59 ' 66 C (i.e., same as from versus ln chart on the book web site) 27
3.38 (a) Macroscoic kinetic energy of an air arcel with mass m is K m 1 2 mc2 s 1 h 2 m (γr d ) 1/2i 2 1 2 m c (c c v ) c v 1 µ 2 mc c 1 c v 1 H (γ 1) 2 K m 1 H (0.40) 0.20 H 2 (b) K m 0.20 H dk m 0.20 dh 0.20(m c d ) (1) From (1) and (2), Also, K m 1 2 mc2 s dk m mc s dc s (2) mc s dc s 0.20 mc d dc s 0.20 c d c s c 2 s dc s 0.20 c c s d c 2 0.20 c s γr d dc s c s d 0.20 c v R d 3.39 c erson M erson m eva L eva m eva c erson M erson L eva 4.2 10 3 5 (2.5 10 6 ) 0.0084 0.84% 28
3.40 wenty liters of air at 20 C and a relative humidity of 60% are comressed isothermally to a volume of 4 liters. Calculate the mass of water condensed. he saturation vaor ressure of water at 20 Cis23hPa. (Density of air at 0 C and 1000 hpa is 1.28 kg m 3.) Answer 0.14 g Solution: comression. We must find mass vaor in the air before and after the Mass of water vaor in air initially Initially, the artial ressure of the vaor 60 100 (SVP of water at 20 C 60 (23.371 hpa) 100 14.02 hpa Since Rρ ρ / herefore, for air: ρ 20 C293 K and 14.02 hpa ρ 0 C273 K and 1000 hpa herefore, or, 14.02/293 1000/273 14.02 273 1000 293 ρ 20 Cand14.02 hpa 14.02 1000 ρ 20 Cand14.02 hpa 0.01666 kg m 3 273 (1.2754) kg m 3 293 Provided the water vaor and air are at the same temerature and ressure (i.e., 14.02 hpa and 20 Cinthiscase). Density of water vaor 18.016 (density of air) 28.97 0.62 (density of air) herefore, the mass (in kg) of water vaor that occuies 20 liters under these conditions is (0.62) (density of air at 20 C and 14.02 hpa) volume of 20 liters in m 3 5 8 (0.01666) 2,0000 10 6 2.0825 10 4 kg (1) 29
Mass of water vaor in air after comression o saturate air at 20 C it would have to be comressed until the vaor ressure is raised from 60% to 100% of the saturation value. From Boyle s Law, this means that it becomes saturated when its volume is reduced to 60% of initial value. his comression is exceeded when volume is reduced from 20 to 4 liters. herefore, after comression the air is saturated and its artial ressure is equal to the saturation vaor ressure of water at 20 C 23.37 hpa. Proceeding as before, For air: herefore, ρ 293 K and 23.371 hpa 23.371/293 ρ 273 K and 1000 hpa 1000/273 ρ 293 K and 23.371 hpa 23.371 1000 273 293 (ρ 273 K and 1000 hpa) Hence, density of air at 293 K and 23.371 hpa is: ρ 293 K and 23.371 hpa 23.371 1000 273 (1.2754) kg m 3 293 herefore, mass of water vaor that occuies 4 liters when air is saturated is µ 5 8 ρ 293 K and 23.371 hpa 4,000 10 6 kg µ 5 8 23.371 1000 273 293 1.2754 4,000 10 6 6.943 10 5 kg herefore, mass of water condensed (mass of water vaor in air initially) (mass of water vaor in air after comression) 2.0825 10 4 6.943 10 5 kg 0.139 grams 0.14 grams An alternative solution: Relative humidity e e s 100 herefore, for initial state: 60 e 23.37 100 30
or, e 14.022 hpa 1402.2 Pa Let us now assume system is comressed isothermally and consider (fictionally) a fixed mass of water vaor (i.e., no condensation), then for vaor: (1402.2)20 2 4 herefore, 2 7011.0 Pa 70.11 hpa his exceeds the SVP at 20 C, which is 23.37 hpa, by 46.74 hpa. herefore, water must condense to bring vaor ressure to saturation at 20 C ( 23.37 hpa). Mass fraction of water that condenses is 70.11 23.37 0.67 70.11 But, mass of water vaor in air before condensation is (from first solution given above) 2.0825 10 4 kg herefore, mass of water condensed 0.67 (2.0828 10 4 ) kg 1.39 10 4 kg 0.14 grams 3.41 Secific humidity (q) where r mixing ratio m v /m d m v m v /m d r m v + m d 1+m v /m d 1+r If q 0.0196 0.0196(1 + r) r r 0.02 kg/kg 20 g/kg Also, v (1 + 0.61r) 303(1 + 0.61 0.02) 306.697 K v 33.7 C R d ρ moist v for moist air 1014 hpa 1014 10 2 Pa 1.014 10 5 (298)(ρ moist )(306.7) 1.014 105 ρ moist 287 306.7 ρ moist 1.15kgm 3 kg m-3 31
3.42 where, e (kilomol fraction of water vaor) m v /M w m a + m w v ε + w M a M w ε M w M a R a R w 287 461 0.623 w 1.80 10 3 kg kg 1 975 hpa e e 1.80 10 3 975 0.623 + 1.80 10 3 2.81 hpa v ' (1 + 0.61 w) 288 1+0.61 1.80 10 3 288 (1.001) 288.3 K v 15.3 C 3.43 Referring to above diagram, the heat released due to moist air being cooled from to w is: ( w )(c + wc v ) Heat required to evaorate water is: L (w 0 w) ( w )(c + wc v )L (w 0 w) 32
w Lw0 ( w ) c c v ( w )+L 2.25 10 6 8.7 10 3 6 1004 1952 6+2.25 10 6 5.99 10 3 kg/kg 5.99 g/kg 6.0 g/kg 3.44 µ du du c v d + dv du c v d. herefore, du c v µ du d µ dq d d + v v µ du d µ du dv dv. But for an ideal gas du mc v d v dv µ du dv 0, therefore, (du) water (du) air [mc v ( 2 1 )] water [mc v ( 2 1 )] air 4 4218 ( 2 1 ) 1000 717 ( 2 1 ) 0.0235 2.4% 3.45 See skew ln chart on the book web site. Note that the moist adiabats on the skew ln chart diverge with increasing height (or decreasing ressure). Hence, if the lase rate follows a moist adiabat, a 1 K rise in temerature at the Earth s surface will be accomanied by a temerature rise of greater than 1 K in the uer trooshere. o estimate how much larger the temerature rise in the uer trooshere would be, one need only estimate the rate of sreading of the moist adiabats. On the chart rovided on the book web site the sacing between successive moist adiabats (in terms of temerature at the 1000 hpa level) is 5 K. o estimate the rate of sreading, identify the moist adiabat that asses through 25 C at the 1000 hpa level and the moist adiabats on either side of it. Follow these three contours u the the 250 hpa level and note the sacing between them (again in terms of temerature on a secific ressure level). he sacing is about 12.5 K. Hence, the multilication factor is around 12.5 K er 5 K, or 2.5 K er degree K at the 1000-hPa-level. 33
3.46 See on the book web site skew ln chart. (a) Mixing ratio w s at d 5.1gkg 1 RH w s at d w s at 5.1 10.8 47% Wet-bulb temerature: Found, using Norman s Rule, to be: 9.3 C. Potential temerature: Since air arcel is at 1000 hpa: θ 15 C 288 K Wet bulb otential temerature (θ w ): Since air arcel is at 1000 hpa θ w w 9.3 C (b) If the arcel rises to 900 hpa: Mixing ratio remains the same as 5.1 g kg 1 Since arcel is below LCL: Relative humidity is w s at d w s at 5.1 6.8 100 75% Wet-bulb temerature is obtained by alying Normand s rule at 900 hpa to give 4.5 C Potential temerature remains unchanged at 288 K Wet-bulb otential temerature is also conserved at 9.3 C (c) If the arcel rises to 800 hpa, it is now lifted above its LCL, therefore air becomes saturated at LCL. Mixing ratio is changed to 4.3 C Since air is saturated RH is 100% Since air is saturated w 1.1 C Potential temerature (θ) from chart is 290 K (not conserved) Wet-bulb otential temerature (θ w ) is conserved at 9.3 C 34
(d) LCL 847 hpa 3.47 Solutions are found as follows: (a) Locate the oint (A) at 1000 hpa and 25 C, and the oint (b) at 1000 hpa and w 20 C. Find the LCL of this air by locating the oint (C) at which the θ line through A intersects the θ e (or θ w ) line through B. C is found to be at 900 hpa. Now take the w s line through C and extraolate it to 1000 hpa (D). hen, from Normand s Rule, the temerature of D, namely 18 C, is the dewoint of the air. (b) and (c) By definition of θ e this is the temerature that the air would obtain if it were exanded until all of the moisture in it condensed and fell out and the air were then comressed to 1000 hpa. Run u the saturated adiabat through C until this saturated adiabat runs arallel to a dry adiabat. hen, at this oint, all of moisture has condensed and fallen out. he θ value of this dry adiabat is, by definition of θ, the temerature the dry air arcel would have if comressed adiabatically to 1000 hpa. his θ value is about 62 C. (Note: By definition, this is the θ e value of the air at A.) (d) Wet-bulb otential temerature (θ w ) corresonding to the saturated adiabats are given by the temerature (in C) along the200 hpa line in the skew ln chart. he saturated adiabat that asses through C, has a θ w value of about 20 C. 3.48 See skew ln chart on the book web site. d emerature at 900 hpa 14 C ' 19.8 C 3.49 (a) Show that when a arcel of dry air at temerature 0 moves adiabatically in ambient air with temerature, the temerature lase rate following the arcel is given by d 0 dz 0 (b) Exlain why the lase rate of the air arcel in this case differs from thedryadiabaticlaserate(g/c ). [Hint: Start with eqn. (3.54) with 0. ake the natural logarithm of both sides of this equation and then differentiate with resect to height z.] g c 35
Solution: From (3.54) with 0 we have for the air arcel µ R /c θ 0 o ln θ ln 0 + R c (ln o ln ) Differentiating, 1 dθ θ dz 1 d 0 0 dz R 1 d c dz But for the ambient air we have, from the hydrostatic equation, (1) d gρ (2) dz From (1) and (2): 1 dθ θ dz 1 d 0 0 dz R 1 c ( gρ) For an adiabatic rocess θ is conserved (i.e., dθ 0). herefore, dz or, But, for the ambient air, 0 1 0 d 0 dz + R gρ c d 0 dz R ρ 0 g c (3) R ρ (4) From (3) and (4), d 0 dz g 0 c he dry adiabatic lase rate µγ d gc is determined under the assumtion that the air arcel develo no macroscoic kinetic energy (i.e., that 0 ), so Γ d d 0 dz g c. 3.50 Derive an exression for the rate of change in temerature with height (Γ s ) of a µ arcel of air undergoing a saturated adiabatic rocess. Assume ws that ρl v is small comared to 1. 36
Answer Γ s l 1+ L v c Γ d µ ws Solution: Substituting (3.20) into (3.51) yields dq c d + gdz (1) If the saturation ratio of the air with resect to water is w s,thequantity of heat dq released into (or absorbed from) a unit mass of dry air due to condensation (or evaoration) of liquid water is L v dw s,whenl v is the latent heat of condensation. herefore, L v dw s c d + gdz (2) If we neglect the small amounts of water vaor associated with a unit mass of dry air, which are also warmed or cooled) by the release (or absortion) of the latent heat, then c in (2) is the secific heatatconstantressure of dry air. Dividing both sides of (2) by c dz and rearranging terms, we obtain " d 1+ L v dz c d dz µ # ws L v dw s c dz g c " L µdws v c dz d g 1+ L v c g µ ws µ # ws d + d g c d (3) dz Or, using the hydrostatic equation on the last term on the right side of (3) µ g ws Γ s d 1 ρl v dz c 1+ L µ ws c or µ ws Γ s d 1 ρl v dz Γ d " 1+ L µ # (4) v ws c In Exercise (3.51) we show that ρl v µ ws ' 0.12 37
herefore, from (4) Γ s dz ' 1+ L v c Γ d µ ws 3.51 In deriving the exression for the saturated adiabatic lase rate in the revious exercise, it is assumed that ρl v ( w s / ) is small comared to 1. Estimate the magnitude of ρl v ( w s / ). Show that this last exression is dimensionless. [Hint: Use the skew ln chart on the book web site to estimate the magnitude of ( w s / ) for a ressure change of, say, 1000 to 950 hpa at 0 C.] Answer About 0.12 Solution: Estimation of magnitude of ρl v µ ws ake ρ ' 1.275 kg m 3 L v 2.5 10 6 Jkg 1 Suose ressure changes from 1000 to 950 hpa, so that d 50 hpa 5000 Pa. hen, from skew ln chart we find that: Hence, ρl v µ ws dw s ' (4 3.75) 0.25 g/kg ' 0.25 10 3 kg/kg ' 1.275 kg m 3 2.5 10 6 Jkg 1 µ 0.25 10 3 kg kg 1 ' 0.12 µ ws he units of ρl v are kg m 3 Jkg 1 kg kg 1 µ 1 Pa µ /kg /m / /3 µ Ã! /kg /m /2 /s / /2 /k /g / /1 1 /k /g /m / /1 /s / /2 which is dimensionless. 3.52 In deriving the exression (3.71) for equivalent otential temerature it was assumed that µ L v c dw Lv w s s w d c 38 5000 Pa
Justify this assumtion. [Hint: Differentiate the right-hand side of the above exression and, assuming L v /c is indeendent of temerature, show that the above aroximation holds rovided d dw s w s Verify this inequality by noting the relative changes in and w s for small incremental dislacements along saturated adiabats on a skew ln chart.] Solution: Show that: µ L v c dw Lv w s s ' d c (1) Differentiate RHS assuming L v /c is a constant: L v c If d then, 1 dw d s w s 2 L v d dw s w s L vd dws c c d w s dw s w s From (2) and (3): (which can be verified from skew ln chart) dw s d À w s (2) (3) R.H.s of (1) L v c dw s LHSof(1) QED 3.53 Solution: See skew ln chart on the book web site. Answer (a) AB, unstable; BC, neutral; CD, neutral; DE, stable; EF, stable, FG, stable. (b) All layers are convectively unstable excet CD, which is convectively neutral. 3.54 Potential density D is defined as the density that dry air would attain if it were transformed reversibly and adiabatically from its existing conditions to a standard ressure 0 (usually 1000 hpa). (a) If the density and ressure of a arcel of the air are ρ and, resectively, show that µ cv /c 0 D ρ where c and c v are the secific heats of air at constant ressure and constant volume, resectively. 39
(b) Calculate the otential density of a quantity of air at a ressure of 600 hpa and a temerature of 15 C. Answer 1.17 kg m 3 (c) Show that 1 dd D dz 1 (Γ d Γ) where Γ d is the dry adiabatic lase rate, Γ the actual lase rate of the atmoshere, and the temerature at height z. [Hint: akethe natural logarithms of both sides of the exression given in (a) and then differentiate with resect to height z.] (d) Show that the criteria for stable, neutral, and unstable conditions in the atmoshere are that the otential density decreases with increasing height, is constant with height, and increases with increasing height, resectively. [Hint: Use the exression given in (c).] (e) Comare the criteria given in (d) with those for stable, neutral, and unstable conditions for a liquid. Solution: (a) For a reversible, adiabatic transformation of an ideal gas V γ constant where, γ c /c v. For a unit mass of a gas, V 1/ρ, whereρ is the density of the gas. herefore, µ ρ γ constant Hence, if the initial ressure and density of a gas are and ρ, and the final ressure and density are o and D, and the gas undergoes an adiabatic transformation, ρ γ o D γ or, µ 1/γ µ cv/c o o D ρ ρ (3.122) (b) From the ideal gas equation for a unit mass of air From (3.122) and (3.123) D R d ρ (3.123) R d 40 µ cv /c o (3.124)
For 600 hpa 6 10 4 Pa, (273 15) K 258 K, o 1000 hpa 10 5 Pa, R d 287 Jdeg 1 kg 1, c v 717 Jdeg 1 kg 1 and c 10004 Jdeg 1 kg 1, we have from (3.124) 6 10 4 µ 717 10 5 1004 D (287) (258) 6 10 4 D 1.17 kg m 3 (c) aking logarithms of (3.122), we have ln D lnρ + c µ v o ln c Differentiating with resect to height z, 1 dd D dz 1 ρ dρ dz c v 1 d c dz (3.125) Also, for a unit mass of dry air considered as an ideal gas R d ρ herefore, ln lnr d +lnρ +ln and, differentiating this last exression with resect to height z, From (3.125) and (3.126) 1 dd D dz 1 d dz 1 dρ ρ dz + 1 d dz 1 d dz + 1 µ d 1 c v dz c 1 d dz + R d d c dz (3.126) (3.127) wherewehaveusedtherelationc c v R d (see eqn. (3.45) in text). From the hydrostatic equation and the ideal gas equation, d dz From (3.127) and (3.128), gρ g R d 1 dd D dz 1 µ g + d c dz (3.128) (3.129) 41
But, the dry adiabatic lase rate Γ d is given by (see eqn. (3.53) in the text) Γ d g c (3.130) From (3.129) and (3.130) 1 dd D dz 1 (Γ d Γ) (3.131) where, Γ is the lase rate in the atmoshere. z (d) From Section 3.6.1 in the text, we have for a stable atmoshere: for an unstable atmoshere: for a neutral atmoshere: Γ < Γ d Γ > Γ d Γ Γ d Hence, from (3.131), for a stable atmoshere: dd is negative dz (i.e., D decreases with increasing height) dd for an unstable atmoshere: is ositive dz (i.e., D increases with increasing height) dd for a neutral atmoshere: dz 0 (i.e., D is constant with height) 3.55 A necessary condition for the formation of a mirage is that the density of the air increase with increasing height. Show that this condition is realized if the decrease of atmosheric temerature with height exceeds 3.5 Γ d,whereγ d is the dry adiabatic lase rate. [Hint: ake the natural logarithm of both sides of the exression for D given in Exercise 3.54a, then differentiate with resect to height z. Follow the same two stes for the gas equation in the form ρr d. Combine the two exressions so derived with the hydrostatic equation to show that 1 dρ ρ dt 1 (d/dz + g/r d). Hence, roceed to the solution.] Solution: From the solution to Exercise 3.54(a) D ρ µ cv /c 0 herefore, ln D lnρ + c µ v 0 ln c 42
and, differentiating with resect to height z, 1 dd D dz 1 dρ ρ dz c v 1 d c dz From the ideal gas equation for a unit mass of dry air, (3.132) R d ρ therefore, ln lnr d +lnρ +ln and, differentiating with resect to height z, 1 d dz 1 dρ ρ dz + 1 d dz From (3.132) and (3.133), 1 dd D dz 1 d dz + R d d c dz From the hydrostatic equation and the ideal gas equation, d g gρ dz R d From (3.133) and (3.135), 1 dρ ρ dz 1 µ d dz + g R d (3.133) (3.134) (3.135) (3.136) For a mirage to occur, dρ must be ositive. herefore, from (3.136), dz for a mirage to occur d dz + g < 0 R d or, d dz > g c g c Γ d (3.137) R d R d c R d where, in the last ste, we have used eqn. (3.53) in the text, that is, g Γ d. Substituting c 1004 Jdeg 1 kg 1 and R d 287 Jdeg 1 c kg 1 into (3.137), we find that a necessary condition for a mirage to occur is that d dz > 3.5 Γ d hat is, the temerature of the air must decrease with height at more than 3.5 Γ d 3.5 9.8 Ckm 1 34 Ckm 1. Such a stee lase rate generally occurs only over strongly heated surfaces, such as deserts and roads. 43
3.56 Assuming the truth of the Second Law of hermodynamics, rove the following two statements (known as Carnot s heorems): (a) No engine can be more efficient than a reversible engine working between the same limits of temerature. [Hint: he efficiency of any engine is given by eqn. (3.87); the distinction between a reversible (R) and an irreversible (I) engine is that R can be driven backward but I cannot. Consider a reversible and an irreversible engine working between the same limits of temerature. Suose initially that I is more efficient than R and use I to drive R backwards. Show that this leads to a violation of the Second Law of hermodynamics, and hence rove that I cannotbemoreefficient than R.] (b) All reversible engines working between the same limits of temerature have the same efficiency. [Hint: Proof is similar to that for art (a).] Solution: (a) o rove that no engine can be more efficient than a reversible engine working between the same limits of temerature consider a reversible (R) and irreversible (I) engine working between θ 1 and θ 2.Assume I is more efficient than R. hen, if R takes heat Q 1 from source and yields heat Q 2 to sink. herefore, if I takes Q 1 from source it must yield heat Q 2 q (q ositive) to sink. Now let us use I to drive R backward. his will require I to do work Q 1 Q 2 on R. But, in one cycle, I develos work Q 1 (Q 2 q) (Q 1 Q 2 )+q. Hence, even when I is drawing R backwards, mechanical work q is still available. But, in one cycle of the combined system, heat Q 2 (Q 2 q) q is taken from colder body. his violates 2nd Law. (b) ake two reversible engines oerating between θ 1 and θ 2,andassumeoneengineismoreefficient than the other. hen follow same rocedure as in (a) above to show this would violate 2nd Law. 44
3.57 Lord Kelvin introduced the concet of available energy, which he defined as the maximum amount of heat that can be converted into work by using the coldest available body in a system as the sink for an ideal heat engine. By considering an ideal heat engine, that uses the coldest available body as a sink, show that the available energy of the universe is tending to zero and that loss of available energy 0 (increase in entroy) where 0 isthetemeratureofthecoldestavailablebody. Solution: For an ideal reversible engine Q 1 Q 2 Q 1 1 2 1 Work done in 1 cycle Q 1 Q 2 1 2 1 Q 1. If engine oerates with sink at o ( 2 ): Available energy 1 0 1 Q 1 Let Q ass from 1 to 2 ( 1 > 2 ) by, say, conduction or radiation. hen, µ µ 1 0 2 0 Loss of available energy Q Q 1 2 µ 1 2 Q o 1 2 Since 1 > 2, there is a loss of available energy for natural rocesses Loss of available energy o µ Q 2 Q 1 o (increase in entroy) 3.58 η Q 1 Q 2 1 2 Q 1 1 1 373 K, 2 273 K η 100 373 Work done in 1 cycle Q 1 Q 2 Q 1 η 20 100 373 5.36 J Work done in 10 cycles 53.6 J 45
Heat rejected to sink in 1 cycle Q 2 Q 1 (1 η) Heat reflected to sink in 10 cycles 10Q 1 (1 η) µ (10) (20) 1 100 373 146.4 J 3.59 Q 1 Q 2 Q 2 1 2 2 17 273 (Note this is not the same as efficiency η of engine Q 1 Q 2 )Forevery Q 1 Q 2 joules taken from water, Q 1 Q 2 joules ( Q 2 17/273) of work has to be done by the motor that drive the refrigerator. Heat required to freeze 20 kg of water 20 L m 20 3.34 10 5 J 6.68 10 6 J Hence, work done by motor to freeze 20 kg of water 6.68 10 6 17 273 J 4.16 10 5 J. Since 1 watt 1 J s 1 a 1 kw motor does 1000 J of work er second. ime needed to do 4.16 10 5 Jofworkis 4.16 105 1000 secs 416 secs 6.93 mins 3.60 A Carnot engine oerating in reverse (i.e., as an air conditioner) is used to cool a house. he indoor temerature of the house is maintained at i and the outdoor temerature is o ( o > i ). Because the walls of the house are not erfectly insulating, heat is transferred into the house at a constant rate given by µ dq K( o i ) dt leakage where K (> 0) isaconstant. (a) Derive an exression for the ower (i.e., energy used er second) required to drive the Carnot engine in reverse in terms of o, i and K. (b) During the afternoon, the outdoor temerature increases from 27 to 30 C. What ercentage increase in ower is required to drive the Carnot engine in reverse to maintain the interior temerature of the house at 21 C? 46
Answer (a) K( o i ) 2 / i ; (b) 125% Solution: Figure 3.27 (a) See Figure 3.27. Let Q 1 and Q 2 be the heats that a Carnot engine running in reverse (i.e., serving as an air conditioner) takes in and rejects, resectively, in one cycle to kee a house at a temerature i that is below the outside temerature of o. If i is to remain constant, µ dq dt umed out of house by air conditioner µ dq dt heat leakage into house Hence, Q 1 t K ( o i ) (3.138) where, t is the time eriod for one cycle of the air conditioner. he ower (i.e., work er unit time) needed to drive the air conditioner is Work done to drive air conditioner er cycle P t Q 2 Q 1 t Since the air conditioner is ideal (3.139) or herefore, Q 2 Q 1 Q 1 Q 2 Q 1 o i o i i µ o i Q 2 Q 1 Q 1 i (3.140) 47
From (3.139) and (3.140) P Q 1 t µ o i i (3.141) From (3.138) and (3.141) P K ( o i ) 2 Á i (3.142) (b) For outside temeratures of o1 and o2, the owers P 2 and P 1, resectively, needed to drive the air conditioner to maintain the house at temerature i are, from (3.142) in (a) above, P 2 P 1 K ( o2 i ) 2 Á i K ( o1 i ) 2 Á i ( o2 i ) 2 ( o1 i ) 2 For i 294K, 01 300 Kand o2 303 K, P 2 92 P 1 6 2 2.25 herefore, the increase in ower needed to kee the house at 21 C or 294 Kis125%. 3.61 Increase in entroy in warming ice from 263 to 273 K: Z 273 d m ice c ice 263 m ice c ice ln 273 263 (0.002) (2106) ln 273 263 4.212 (0.0373) 0.1572 Jdeg 1 Increase in entroy on melting ice at 273 K (0.002) 3.34 10 5 273 2.447 Jdeg 1 Jdeg 1 Increase in entroy on heating water from 273 K to 373 K (0.002) (4218) ln 373 273 2.633 Jdeg 1 48
Increase in entroy on changing water to steam at 373 K: (0.002) 2.25 10 6 373 12.06 Jdeg 1 otal increase in entroy (0.1572 + 2.447 + 2.633 + 12.06) 17.297 Jdeg 1 17.3 J deg 1 3.62 In general, For an ideal gas: Since, S 2 S 1 S 2 S 1 Z 2 1 Z 2 1 c v d dq c v ln 2 1 + Z 2 + 1 Z 2 1 dv R dv v c v ln 2 1 + R ln v 2 v 1 1 v 1 1 2v 2 2 S 2 S 1 c v ln 2 1 + R ln µ 1 2 2 1 c v ln 2 1 + R ln 1 2 + R ln 2 1 Since, R c c v S 2 S 1 c ln 2 1 + R ln 1 2 For a diatomic gas c /c v γ 1.4. Also, c c v + R c γ + R 49
or c γr γ 1 (1.4 8.3145) /0.4 Jdeg 1 mol 1 29.10 Jdeg 1 mol 1 µ 373 286 S 2 S 1 29.10 ln +8.3145 ln µ 1 2 (29.10) (0.2656) + 8.3145 ( 0.6931) 7.729 5.763 1.966 Jdeg 1 2.0 J deg 1 3.63 Show that the exression numbered (3.118) in the solution to Exercise 3.50 above can be written as (1 + w s L v /R d ) Γ s Γ d (1 + w s L 2 v/c R v 2 ) Solution: Comaring exression (3.118) in Exercise 3.50 with the exression given inµ the statement of this exercise, we see that we have to dws rove (a) ρl v w µ sl v d R d,and(b)l v dws w sl 2 v c d c R v 2 (a) o show: µ dws ρl v w sl v d R d From eqn. (3.63) w s ' 0.622 e s (3.143) Also, R d ρ (3.144) From (3.143) and (3.144), µ dws dw s d d (R d ρ) or, since is constant, But, µ dws 1 dw s d R d d 50 w s ρ sv ρ (3.145)
herefore, dw s dρ ρ d sv dρ From (3.145) and (3.146), µ dws d From (3.146) and (3.147), µ dws d herefore, ρ 1 ρ sv ρ 2 (3.146) 1 ρ sv R d ρ 2 (3.147) w s R d ρ µ dws w s ρl v L v d R d (3.148) QED (b) o show: L v c From (3.143) in (a) above: µ dws d µ dws w sl 2 v d c R v 2 d µ 0.622 e s d 0.622 de s d (3.149) Using the Clausius-Claeyron eqn. (3.93) and the ideal gas equation for saturated water vaor, de s d L v α 2 From (3.149) and (3.150) µ dws d L v R v e s 0.622 L v e s 2 R v L v e s 2 R v (3.150) and using (3.143) herefore, L v c µ dws d w sl v 2 R v w sl 2 v c R v 2 51
with 3.64 From (3.99) B B ( 2 1 ) atmos L v d atmos (1013 600) hpa 413 hpa 4:13 10 4 Pa B 373 K at 1013 hpa L v 2:25 10 6 J K 1 ) d B (373 K) 1:66 0:001 m 3 kg 1 4:13 10 4 Pa 11:30 deg Hence, new boiling oint 88.7 C 2:25 10 6 J K 1 3.65 he change M in the melting oint of ice due to a change in ressure is, by analogy with (3.99), M M ( w i ) L M where w and i are the seci c volumes of water and ice, resectively, and L M is the latent heat of melting of ice. Substituting M 273 K, ( w i ) (1:0010 1:0908) 10 3 m 3 kg 1 0:0898 10 3 m 3 kg 1, L M 3:34 10 5 J kg 3, and 1 atm 1:013 10 5 Pa we obtain M 273 0:0898 10 3 3:34 10 5 1:013 10 5 C ) M 0.0074 C herefore, an increase in ressure of 1 atm decreases the melting oint of ice by 0.0074 C. (Ice is unusual in this resect; the seci c volumes of the liquid forms of most materials is greater than the seci c volumes of the solid forms. Consequently, the melting oint of most materials increases with increasing ressure.) 3.66* By di erentiating the enthaly function, de ned by eqn. (3.47), show that @ @s @ @ s where s is entroy. Show that this relation is equivalent to the Clausius- Claeyron equation. @h @h [Hint: dh ds + d @s @ s 1
and, since h is a function of state, µ h s s Solution: From (3.47) herefore µ h ] s h u + α dh du + dα + αd (dq dα)+dα + αd dh ds + αd (1) h h (s, ) µ h dh s ds + µ h d (2) s From (1) and (2), µ h and s µ h α (3) s But, From (3) and (4), or, µ h s s µ s µ s µ h s µ α ds µ s dα (4) (5) Since ds dq L v for a hase change from liquid to vaor at, and if the vaor is saturated, so that e s,anddα α 2 α 1 (5) becomes µ es L v (α 2 α 1 ) which is the Clausius-Claeyron equation. s 53