ch11 - CHAPTER ELEVEN 11.1 a The peroxide mass fraction in...

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Unformatted text preview: CHAPTER ELEVEN 11.1 a. The peroxide mass fraction in the effluent liquid equals that in the tank contents, which is: x M M p p = Therefore, the leakage rate of hydrogen peroxide is ¡ / m M M p 1 b. Balance on mass: Accumulation = input – output E = − = = dM dt m m t M M ¡ ¡ , 1 0 (mass in tank when leakage begins) Balance on H O 2 2 : Accumulation = input – output – consumption E = − F H G I K J − = = dM dt m x m M M kM t M M p p p p p p ¡ ¡ , 1 0 11.2 a. Balance on H 3 PO 4 : Accumulation = input Density of H 3 PO 4 : ρ = 1834 . g / ml . Molecular weight of H 3 PO 4 : M = 98 00 . g / mol . Accumulation = dn dt (kmol / min) Input = 20.0 L 1000 ml 1.834 g mol 1 kmol min L ml 98.00 g 1000 mol kmol / min dn dt n kmol p p p0 = E = = = × = 03743 03743 150 0 05 75 . . , . . t b. dn dt n t p n t p p 7 5 03743 75 03743 . . . . ) z z = ⇒ = + (kmol H PO in tank 3 4 x n n n n n n t t p p p p p = = + − = + + 75 03743 150 03743 . . . kmol H PO kmol 3 4 c. 015 75 03743 150 03743 471 . . . . . = + + ⇒ = t t t min 11- 1 11.3 a. b g & ¡ m a bt w = + t m w = = 750 , ¡ t m m t w w = = ⇒ = + 5 1000 750 50 , ¡ ¡ b g b g kg h h b g Balance on methanol: Accumulation = Input – Output M dM dt m m t dM dt t t M f w = = − = − + E = − = = kg CH OH in tank kg h kg h kg h kg 3 ¡ ¡ , 1200 750 50 450 50 750 b g b g b. dM t dt M t 750 450 50 z z = − b g E − = − E = + − M t M t 750 450 25 750 450 25 2 2 t t Check the solution in two ways : ( ) , 1 750 450 50 t M t = = ⇒ = − ⇒ kg satisfies the initial condition; (2) dM dt reproduces the mass balance. c. dM dt t M = ⇒ = = ⇒ = + − = 450 50 9 750 450 9 25 9 2775 2 h kg (maximum) ( ) ( ) M t t = = + − 750 450 25 2 t = − ± + − ⇒ 450 450 4 25 750 2 25 2 b g b gb g b g t = –1.54 h, 19.54 h d. 3.40 m 10 liter kg 1 m 1 liter kg 3 3 3 0792 2693 . = (capacity of tank) M t t = = + − 2693 750 450 25 2 t = − ± + − − ⇒ 450 450 4 25 750 2693 2 25 2 b g b gb g b g t = 719 1081 . , . h h Expressions for M(t) are: M(t) = 750+ 450t - 25t and (tank is filling or draining) (tank is overflowing) (tank is empty, draining as fast as methanol is fed to it) 2 719 1081 1954 2693 719 1081 1954 2054 ≤ ≤ ≤ ≤ ≤ ≤ ≤ ≤ R S | T | | t t t t . . . ( . . ) ( . . ) b g 11- 2 11.3 (cont’d) 500 1000 1500 2000 2500 3000 5 10 15 20 t(h) M(kg) 11.4 a. Air initially in tank: N 492 00258 = ° ° = 10.0 ft R 1 lb - mole 532 R 359 ft STP lb - mole 3 3 b g . Air in tank after 15 s: P V PV N RT N RT N N P P f f f f 0 0258 0 2013 = ⇒ = = = . . lb- mole 114.7 psia 14.7 psia lb- mole Rate of addition: ¡ . . n = − = 0 2013 00258 b g lb- mole air 15 s .0117 lb- mole air s b. Balance on air in tank: Accumulation = input dN dt = 00117 . lb- moles s b g ; t N = = 00258 , . lb- mole c. Integrate balance: dN n dt N t N t 0 0258 0 0258 0 0117 ....
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This note was uploaded on 06/07/2008 for the course CHE 2171 taught by Professor Wetzel during the Spring '08 term at LSU.

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ch11 - CHAPTER ELEVEN 11.1 a The peroxide mass fraction in...

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