1 / 23

Mass Transfer 2

Mass Transfer 2. Section # 9. Sheet#2 , Problem#1. Givens: Air-water vapor mixture T=65 o C, T w =35 o C, P T = 1 atm The radiation coefficient can be considered negligible Antoine constants for water The steam table Required: The humidity of the air without using the chart.

basil
Download Presentation

Mass Transfer 2

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Mass Transfer 2 Section # 9

  2. Sheet#2 , Problem#1 Givens: Air-water vapor mixture T=65oC, Tw=35oC, PT = 1 atm The radiation coefficient can be considered negligible Antoine constants for water The steam table Required: The humidity of the air without using the chart

  3. Sheet#2 , Problem#1 Solution • w - ww = - hg [ 1 /Ky Mgλv] (1) t - tw • For air-water system, the term (hg /Ky Mg) = 0.26 BTU / lb oF = 1.09 j / g oC • From the steam table, table 1, you will find that at Tw = 350C λ = 2418.8 j/g , saturation pressure = 0.056 atm • Note: you can also calculate the saturation pressure using Antoine equation at Tw=35oC. • Ww = Ps * MV = 0.056 * 18 = 0.0368 kg water P- Ps Mg 1 - 0.056 29 kg dry air • Sub. in (1) get w = 0.0229

  4. Sheet#2 , Problem#2 Givens: Air-Toluene vapor mixture T=60oC, w = 0.05 kg vapor / kg air D= 9.2 * 10-6 m2/sec, μ = 1.95 * 10-5 kg/ m sec, k = 0.0277 W / m K A table for toluene properties hc = Sc 0.567 = Le 0.567 Ky Cs pr Required: • The wet bulb temperature of the mixture • The adiabatic saturation temperature of the mixture

  5. Sheet#2 , Problem#2 Solution • Sc = μ , Pr = μ * Cs ρ * D k • ρ = PM = 1 * 29.1 = 1.065 g / L Sc = 1.99 RT 0.082 * (60+273) • Cs = Cg + w Cv = 1.005 + (0.05* 1.82) = 1.096 j / g K Pr = 0.772 • But, hc = Sc 0.567 = 1.71 (hc/ Ky ) = 1.71 * 1.096 = 1.875 Ky Cs Pr • w-ww = - hg [ 1 / Ky Mgλv ] 0.05-ww = - 1.875 (1) t-tw 60 - twλv • Now, a trial and error procedure is needed. We will perform the following: a) Assume tw b) Get λv , and the toluene’s saturation pressure from the toluene’s table c) Calculate ww using the got saturation pressure d) Check that the L.H.S and R.H.S of eqn. (1) has the same value.

  6. Sheet#2 , Problem#2 • 1st assumption: • Tw = 30oC λv = 409.56 kj /kg , and Ps = 0.04972 bar • Ww = Ps * MV = 0.04972 * 92 = 0.166 kg toluene P- Ps Mg 1 -0.04972 29 kg dry air • L.H.S of eqn. (1) = -0.003866, R.H.S of eqn. (1) = -0.004577 • 2nd assumption: • Tw = 35oC λv = 406.66 kj /kg , and Ps = 0.06292 bar • Ww = Ps * MV = 0.06292 * 92 = 0.213 kg toluene P- Ps Mg 1 -0.06292 29 kg dry air • L.H.S of eqn. (1) = -0.00652, R.H.S of eqn. (1) = -0.00461 • We can get the true wet bulb temp. using interpolation: Tw – 30 = 0 – (0.004577-0.003866) 35 -30 (0.00461 – 0.00652) – ((0.004577-0.003866) Get Tw = 31.356 oC

  7. Sheet#2 , Problem#2 • w-ws = - Cs / λs 0.05-ws = - 1.096 (2) t-ts 60 - tsλs • You can use the same technique performed in estimating the wet-bulb temperature here to estimate the adiabatic saturation temperature. After trial and error, you can get that: Ts = 27oC

  8. Sheet#2 , Problem#7 Givens: Dehumidification tower, counter-current Liquid rate = 1000 lb/hr ft2 , TL2=60oF , TL1=90oF Tv1=100oF, % Humidity1=90% Gas rate = 27000 std ft3/hr ft2 hLa / kya = 200 Required: The outlet air conditions

  9. Sheet#2 , Problem#7 Solution • Hv2 -Hv1 = (L avg CL ) / S Hv2 -61= 1000*1 (1) TL2 –TL1 V’ / S 60 – 90 G’ Where, Hv1 was calculated by knowing 2 information about the entering gas. • Since gas rate = 27000 std ft3/ hr ft2, we want to convert it into lb/ hr ft2 to be able to sub. In the above operating line equation. • For ideal gases : PV = nRT 14.7 * 27000 = n*10.72*492 get n1 = 75.25 lbmole / hr ft2 Note:(std means that T = 0oC, and pressure = 1 atm) • n = 75.25 lbmole air / hr ft2 G1 = 75.25 * 29 = 2182.25 lb air / hr ft2 • G1= G’* (1+w1) G’ = G1 / (1+0.041) = 2096.3 lb dry air / hr ft2 • Sub. in (1) Get Hv2 = 46.689 BTU / lb dry air • As TL2=45oC, and TL1=29oC, then the temperature range that we will take to draw the equilibrium curve will be (50, 25 oC) • Steps: a) Draw the equilibrium curve on a graph paper b) From Tw1 andT1, get Hv1 from the chart c) As we have TL1 , and Hv1 , then we can locate point (1) on the operating line d) Draw a locus for TL2 e) Match point (1) with the (intersection between the equilibrium curve and the locus of TL2 ) to get the pinch point.

  10. Sheet#2 , Problem#7 • The required was the outlet air conditions. The humid enthalpy is not enough. • We will apply Mickley’s method so as to get Tv2 • To apply Mickley’s method, we should have the slope of the tie-lines. • It is given that hLa / kya = 200. This is a very large value for the slope. In this case, the slope is vertical. • After performing Mickley’s method Get Tv2 = 87oF • You will find that the point describing the outlet air conditions here lies on the eqm. curve % RH2 = 100 % , and Tv2=Tw2

  11. Sheet#2 , Problem#7

  12. Sheet#2 , Problem#7

  13. Sheet#2 , Problem#7

  14. Sheet#2 , Problem#8 Givens: Dehumidification tower, counter-current Liquid rate = 900 lb/hr ft2 , TL2 = 82oF, TL1 = 100oF Tv1=125oF, Tw1= 111oF Tv2=96oF , Tw2=95oF Z = 8 ft Required: • hca, hLa, and kya • In the same tower, if the water inlet temperature, air and water flow rates would not change, but the inlet gas conditions become 140oF dry bulb, and 111oF wet bulb. Determine the outlet air conditions. You can use the coefficients calculated from part (a)

  15. Sheet#2 , Problem#8 Solution • When Z, TL1, TL2, Tv1, Tv2, additional information on the inlet air, and an additional information about the exiting air are given, “Reverse Mickley” technique can be used to be able to calculate hca, hLa, and kya. • How to perform Reverse Mickley’s method? a) By knowing Tv1 and Hv1,you can locate the point representing the entering gas. You can name it (1). b) In this case, you can also represent the exiting gas by knowing Tv2 and Hv2 where the point representing it is point (2). c) Assume a tie-line slope, then from point (1), apply Mickley’s technique as we have done before d) Repeat the steps till you reach the locus of Hv2. If the point got on this locus is (point 2), then the assumed tie line slope is true. If not, reassume the slope.

  16. Sheet#2 , Problem#8

  17. Sheet#2 , Problem#8 • 1st trial: a) Assume (- hLa / kya) = -1.5 b) From point (1), start Mickley c) Continue Mickley’s method till you reach the locus of Hv2. If you cut the locus in Tv2, then the assumed slope is ok. d) You will find that this slope will cut the Hv2 locus in 95oF (not 96oF) • 2nd trial: a) Assume (- hLa / kya) = -11 b) From point (1), start Mickley c) Continue Mickley’s method till you reach the locus of Hv2. If you cut the locus in Tv2, then the assumed slope is ok. d) You will find that this slope will cut the Hv2 locus in 96.5oF (not 96oF) • You will find that the true tie-line slope is -4.66 BTU / lb oF

  18. Sheet#2 , Problem#8 Solution • Hv2 -Hv1 = (L avg CL ) / S 57-89= 900*1 (1) TL2 –TL1 V’ / S 82 – 100 G’ Where, Hv1 and HV2 were calculated by knowing 2 information about the entering and exiting gas respectively. • Get G’ = 506.25 lb / hr ft2 • Z = HTG * NTG 8 = HTG*NTG (2) Where: Hv2 • NTG = ∫ d Hv ( It can be calculated as we know everything) Hv1 Hi – Hv • HTG = (G’ / kya) kya is unknown

  19. Sheet#2 , Problem#8 Hv2 • NTG = ∫ d Hv We need the tie-line equation Hv1 Hi – Hv • Hv – Hi = - hL a = - 4.66 BTU / lb o F TL – T I kya • Get NTG = 2.1145 • Sub. in (2) get HTG = 3.7833 ft 3.7833 = 506.25 / kya kya = 133.81 lb / hr ft3 • But - hL a = - 4.66 BTU / lb o F hLa = 623.56 lb / hr ft2 kya

  20. Sheet#2 , Problem#8 • For air-water system, r = 1 where: r = psychrometric ratio = hca / kya Cs (3) • Cs = Cg + wavg Cv = 0.24 + wavg (0.45) • wavg = (w1 +w2) / 2 = (0.074 + 0.036) / 2 = 0.055 Cs = 0.2652 BTU lb oF • Sub. in (3) by r = 1 hca = 35.48 BTU / hr ft3oF

  21. Sheet#2 , Problem#8 Solution b) • TL2, Lavg, G’, Z, kya, hLa are the same (1) • Tw1 is also the same Hv1 is also the same = 89 BTU / lb dry air • HTG = (G’ / kya) From (1), HTG is the same = 3.7833 ft • Since Z and HTG aren’t changed, NTG will not also be changed Hv2 Therefore, NTG = ∫ d Hv = 2.1145 Hv1 Hi – Hv • Since the tie-line slope and Hv1 are the same, then Hv2 will not be changed. Hv2 = 57 BTU / lb dry air • Hv2 -Hv1 = (L avg CL ) / S 57-89= 900*1 TL1 = 100oF TL2 –TL1 V’ / S 82 – TL1 506.25

  22. Sheet#2 , Problem#8 • Therefore, the operating line will not change. Of course, the equilibrium curve also will not change. • The required was the outlet air conditions. The humid enthalpy (Hv2) is not enough. • We will apply Mickley’s method so as to get Tv2 • In this case, point (1) is (Tv1=140oF, Hv1= 89 BTU / lb dry air) • The tie-line slope is the same as got in the 1st requirement (= - 4.66) • After performing Mickley, get Tv2 = 97oF • From the chart, at Tv2 = 97oF, and Hv2 = 57 BTU/lb dry air w2 = 0.0385

  23. Sheet#2 , Problem#8

More Related