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Sample Problems on Crystallization
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SAMPLE PROBLEM Calculate the yield OF Mg 504 7H120 solution Data: Given: MW last by evaporation during cooling. OF MgSO4 at 350 K = MgSO4 Ot 303X F F= 1000 kg TF = 353 K 64.12 kg MgSO4 100 kg water MW Reqd: C Soin. OMB: CMB Solubility Solubility OF XF = % at 35BK ug soy = 120.3 Masoy 7H₂0= 246.3 and yield F= L + V + C Determine V: 1000 kg = 10000 Xc = 44.2 100 + 64.12 Lt C = IS V= (0.10) (1000 kg) (. -MG 504 Yield = V Lt vtc cooled 10% OF Lt 60.9013 TC 598.4350 1000 ama.0987 kg 100 100 + (04.12) : YFF = XLL + XC C at water From 0.3910 x 100 = T= 303 K = MW Mg604 NW MgSO4.7H2 (0.3910) (1000 kg) = (0.2898) (L) + (0.4884) (C) L = 340-66 37 kg C = 598. 4350 kg 1120. 303 K с MgSO4.7H20 TC = 303 K Eaul 246.3 154.8495 64.12 kg /100kg when crystals assuming 10% Feed XL = 60. 9013 kg 40.8 kg/100 kg water 0.4884 1 40.8 100 + 40-8 water TL- 80BK L 40.8 kg Ng 504 100 кО Н20 1000 kg Ean. (2 OF 0.2898 the OF saturated WCHer 15 SAMPLE PROBLEM 2: OF are solution 12.5% by weig ht precipitated. At 100 kg OF water. original water in Given: F = 5000 kg Nar Cog XF = A not MW MW and H₂2₂0 0.125 Nan (09 Nanco-10 H2O 286 = OMB: f= L + V + C = 100 4812.5 = CMB NO₂ COB: XL= Reqa: yield OF Narco crystals soin: LTC containing 5000 g OF Na₂CO3 is cooled at газ к solubility Calculate V= (0.05) (0.75) (5000) = 187.5 kg 5000 = Lt 187.5 TC XFF 21.5 100 + 21.5 (0.125) (5000) = 1 % yield (NaR (03) = = 0.1770 the me yield OF Na₂CO3 the system evaporates on cooling. V 5% OF Water L= 2755-7464 kg C= 2056.7536 kg Egn. = XLL + XcC Xc (c) XF (F) (0.1770) (1) + cowe For me yield...
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Alternative transcript:
Of Nancos: % yield (NaLL (03) amt of Nanco in C amt. Of Nan Cog in F Xc = *100 = 60. 9784 % is 286 (0.3100) (C) Nanco ૧૦૭ from F 106 T= 293 K 21.5 kg Na₂CO3- 10H226 X 100 = 0.3706 and Ean (2) and (0.3706) (2056.7536 (0.15) (5000) crystals obtained if water L with x 100 crystals OF annydrous Na2003 per concentration Na₂ (08.10H₂0 21.5 kg Na₂CO3 100 kg H₂0 5% OF SAMPLE PROBLEM A not solution containing 2000 kg OF Mg 604 and and with a concentration OF MgSO4 THRO crystals are 30% by weight MgSOy is The solubility at 293 K 330 K Cooled 10 293K is 85.5 xg and for med. Mg 604 / 100 kg water. The average heat capacity OF the Feed solution is 293 K is -13.31 X103 kJ/kmol Ng SON. TH₂0. 2.93 KJ/xg.K. The песн balance. Assume Calculate no water Given: F = 2000 kg MgSO4 TF = 330 K XF= 0.80 CPF = 2.93 KJ Reqd: Soiri. XL = MW MgSO4 = 120.3 MW Mg SOY. TH2O = 246.3 the is vaporized. kg.K ONB: Yield q (loss) solution at yield of crystals and make the 35.5 OF crystals 100 35.5 F = L + C 2000 = CMB- Solute : L +C Ltc = 2000 heat XFF = Yield of crystals = 0.26 201 - XLL Ean. Ⓒ + XcC * C. T= 243 K (0.30) (2000) = (0.2620) L + (0.2620) + (0.4884) C 335, 6890 C MgSO4. THRO TC= 298 K 2000 hs= Xc = F LE HOUÝ, B10 ng 335. 6890 kg 120. 246.3 L -1B. 31 x10³ kJ Knol NgSON. TH₂0 x 100 = 16.7845 TL= 293 K sol. = (0.4884) (C) (0.30) (2000) = 0.4884 J water at 35.5 kg Mg 504 100 kg H₂₂0 Ean. Ⓡ q+ VAL = PCPF (TF - TL) + Cnc hc -ns q= (2000 kg) (2.93 •) (380-293)X + (335.6890 kg)(- KJ kg.K 9 = 204₁ 960 5627 kg SAMPLE PROBLEM cooled water 333K From OF to 2831 in an 150 kg. At 283 K, the solubility OF water Nansou crystallizes as by evaporation and yield is lost must be A solution containing 500 kg OF Nan Soy and agitated mild He at C Latent Given: - (13.31 x10³ k5/kmol MgSO4. 7H₂O) 13. 31 x 10³ kJ/mol Mg 604. TH20 Aaditional data: capacity of Heat capacity of неон OF TF= CPF E 333K removed F= 500kg Na2SO4 + 12500 kg H₂0 4: Ht cat B.6 kJ/kg.K mild soin = -78.5 NJ/kmol Read: Heat Msteel vessel = 750 kg Cp wild steel= Xv: 2395 KJ/xg soin = 3. KJ/Kg.K V OF vaporization OF OF 0.5 kJ/kg.K NaRSO4 = 1412 MW MW NarsOy - 10H120= 13212 lost yield of hydrated kg during cooling, steel = 0.5 KJ/kg.K (2% OF original water) hydrated crystals T= 283 K 1 kmol MgSO4. THRO 246.3 kg Ng SOY. TH₂U salt is anny drous salt. IF Glauber's calculate crystals water с Tc = 28 3 к Glauber's salt Na₂604 10 H₂O 2500 kg OF steel L 2395 kJ/kg TL = 283 K Sol. = hs= -78.5 NIS/ KALOJ In 2% water is vessel weighing in kg. 13.31 X108 KJ Ikmul MgSO4. TH ₂0/ 8.9 kg per 100 kg The original ne at that OF the 8.9 kg anny drous 100 kg H₂0 soin: • Determine V= AL= hc= Solve XF= 500 ONB: CMB- q + VAL (0.02) (2500 kg) = 50 kg Av = 2395 KJ (Kg 500 f 2500 = 3000 kg - ns = -(-78.5 MJ/kmol) = 78.5 For C: ૧ : 2500 +500 solute : FCPF (TF- TL) + Cnc + M vessel Cpvessel (Tp-TL) = 0.1647 F = L + V +C 3000 = L + 50 + C Lt C = 2950 kg XFF XL = MJ/kmol 8.9 100+ 8.9 = XLL + XCC (0.1667)(3000) = (0.0817) (L) + (0.4410) (C) (0.0817) + (0.4410) C = (0.1647) (3000) L = 2229. 1957 kg C = 720 8043 kg q + VAL = FCPF (TF-TL) + Chc + q + (50 kg) ( 23 95 kJ/xg) = (3000 = 0.0817 9 = 614 724. 0297 KJ Xc= 142 312122 nivessel Cpvessel (TF-TL) kg) (3.6 kJ/xg.x) (3313 - 283) K + (720.8043 xg) (78500 KJ/kmol) (¹ km01/31212kg) + (750 kg) (0.5 kJ/xg.k) (333- 283) K 0.4410
These notes focus on crystallization and energy conversions. It includes converting joules (J) and kilojoules (kJ), joules per mole (J/mol) to kilojoules per mole (kJ/mol), and understanding conversions like kJ to J and kJ/kg to BTU/lb, and conversions between British Thermal Units (BTU) and pound-force feet (lbf ft).
These notes focus on crystallization and energy conversions. It includes converting joules (J) and kilojoules (kJ), joules per mole (J/mol) to kilojoules per mole (kJ/mol), and understanding conversions like kJ to J and kJ/kg to BTU/lb, and conversions between British Thermal Units (BTU) and pound-force feet (lbf ft).
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SAMPLE PROBLEM Calculate the yield OF Mg 504 7H120 solution Data: Given: MW last by evaporation during cooling. OF MgSO4 at 350 K = MgSO4 Ot 303X F F= 1000 kg TF = 353 K 64.12 kg MgSO4 100 kg water MW Reqd: C Soin. OMB: CMB Solubility Solubility OF XF = % at 35BK ug soy = 120.3 Masoy 7H₂0= 246.3 and yield F= L + V + C Determine V: 1000 kg = 10000 Xc = 44.2 100 + 64.12 Lt C = IS V= (0.10) (1000 kg) (. -MG 504 Yield = V Lt vtc cooled 10% OF Lt 60.9013 TC 598.4350 1000 ama.0987 kg 100 100 + (04.12) : YFF = XLL + XC C at water From 0.3910 x 100 = T= 303 K = MW Mg604 NW MgSO4.7H2 (0.3910) (1000 kg) = (0.2898) (L) + (0.4884) (C) L = 340-66 37 kg C = 598. 4350 kg 1120. 303 K с MgSO4.7H20 TC = 303 K Eaul 246.3 154.8495 64.12 kg /100kg when crystals assuming 10% Feed XL = 60. 9013 kg 40.8 kg/100 kg water 0.4884 1 40.8 100 + 40-8 water TL- 80BK L 40.8 kg Ng 504 100 кО Н20 1000 kg Ean. (2 OF 0.2898 the OF saturated WCHer 15 SAMPLE PROBLEM 2: OF are solution 12.5% by weig ht precipitated. At 100 kg OF water. original water in Given: F = 5000 kg Nar Cog XF = A not MW MW and H₂2₂0 0.125 Nan (09 Nanco-10 H2O 286 = OMB: f= L + V + C = 100 4812.5 = CMB NO₂ COB: XL= Reqa: yield OF Narco crystals soin: LTC containing 5000 g OF Na₂CO3 is cooled at газ к solubility Calculate V= (0.05) (0.75) (5000) = 187.5 kg 5000 = Lt 187.5 TC XFF 21.5 100 + 21.5 (0.125) (5000) = 1 % yield (NaR (03) = = 0.1770 the me yield OF Na₂CO3 the system evaporates on cooling. V 5% OF Water L= 2755-7464 kg C= 2056.7536 kg Egn. = XLL + XcC Xc (c) XF (F) (0.1770) (1) + cowe For me yield...
SAMPLE PROBLEM Calculate the yield OF Mg 504 7H120 solution Data: Given: MW last by evaporation during cooling. OF MgSO4 at 350 K = MgSO4 Ot 303X F F= 1000 kg TF = 353 K 64.12 kg MgSO4 100 kg water MW Reqd: C Soin. OMB: CMB Solubility Solubility OF XF = % at 35BK ug soy = 120.3 Masoy 7H₂0= 246.3 and yield F= L + V + C Determine V: 1000 kg = 10000 Xc = 44.2 100 + 64.12 Lt C = IS V= (0.10) (1000 kg) (. -MG 504 Yield = V Lt vtc cooled 10% OF Lt 60.9013 TC 598.4350 1000 ama.0987 kg 100 100 + (04.12) : YFF = XLL + XC C at water From 0.3910 x 100 = T= 303 K = MW Mg604 NW MgSO4.7H2 (0.3910) (1000 kg) = (0.2898) (L) + (0.4884) (C) L = 340-66 37 kg C = 598. 4350 kg 1120. 303 K с MgSO4.7H20 TC = 303 K Eaul 246.3 154.8495 64.12 kg /100kg when crystals assuming 10% Feed XL = 60. 9013 kg 40.8 kg/100 kg water 0.4884 1 40.8 100 + 40-8 water TL- 80BK L 40.8 kg Ng 504 100 кО Н20 1000 kg Ean. (2 OF 0.2898 the OF saturated WCHer 15 SAMPLE PROBLEM 2: OF are solution 12.5% by weig ht precipitated. At 100 kg OF water. original water in Given: F = 5000 kg Nar Cog XF = A not MW MW and H₂2₂0 0.125 Nan (09 Nanco-10 H2O 286 = OMB: f= L + V + C = 100 4812.5 = CMB NO₂ COB: XL= Reqa: yield OF Narco crystals soin: LTC containing 5000 g OF Na₂CO3 is cooled at газ к solubility Calculate V= (0.05) (0.75) (5000) = 187.5 kg 5000 = Lt 187.5 TC XFF 21.5 100 + 21.5 (0.125) (5000) = 1 % yield (NaR (03) = = 0.1770 the me yield OF Na₂CO3 the system evaporates on cooling. V 5% OF Water L= 2755-7464 kg C= 2056.7536 kg Egn. = XLL + XcC Xc (c) XF (F) (0.1770) (1) + cowe For me yield...
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SuSSan, iOS User
Love this App ❤️, I use it basically all the time whenever I'm studying
Alternative transcript:
Of Nancos: % yield (NaLL (03) amt of Nanco in C amt. Of Nan Cog in F Xc = *100 = 60. 9784 % is 286 (0.3100) (C) Nanco ૧૦૭ from F 106 T= 293 K 21.5 kg Na₂CO3- 10H226 X 100 = 0.3706 and Ean (2) and (0.3706) (2056.7536 (0.15) (5000) crystals obtained if water L with x 100 crystals OF annydrous Na2003 per concentration Na₂ (08.10H₂0 21.5 kg Na₂CO3 100 kg H₂0 5% OF SAMPLE PROBLEM A not solution containing 2000 kg OF Mg 604 and and with a concentration OF MgSO4 THRO crystals are 30% by weight MgSOy is The solubility at 293 K 330 K Cooled 10 293K is 85.5 xg and for med. Mg 604 / 100 kg water. The average heat capacity OF the Feed solution is 293 K is -13.31 X103 kJ/kmol Ng SON. TH₂0. 2.93 KJ/xg.K. The песн balance. Assume Calculate no water Given: F = 2000 kg MgSO4 TF = 330 K XF= 0.80 CPF = 2.93 KJ Reqd: Soiri. XL = MW MgSO4 = 120.3 MW Mg SOY. TH2O = 246.3 the is vaporized. kg.K ONB: Yield q (loss) solution at yield of crystals and make the 35.5 OF crystals 100 35.5 F = L + C 2000 = CMB- Solute : L +C Ltc = 2000 heat XFF = Yield of crystals = 0.26 201 - XLL Ean. Ⓒ + XcC * C. T= 243 K (0.30) (2000) = (0.2620) L + (0.2620) + (0.4884) C 335, 6890 C MgSO4. THRO TC= 298 K 2000 hs= Xc = F LE HOUÝ, B10 ng 335. 6890 kg 120. 246.3 L -1B. 31 x10³ kJ Knol NgSON. TH₂0 x 100 = 16.7845 TL= 293 K sol. = (0.4884) (C) (0.30) (2000) = 0.4884 J water at 35.5 kg Mg 504 100 kg H₂₂0 Ean. Ⓡ q+ VAL = PCPF (TF - TL) + Cnc hc -ns q= (2000 kg) (2.93 •) (380-293)X + (335.6890 kg)(- KJ kg.K 9 = 204₁ 960 5627 kg SAMPLE PROBLEM cooled water 333K From OF to 2831 in an 150 kg. At 283 K, the solubility OF water Nansou crystallizes as by evaporation and yield is lost must be A solution containing 500 kg OF Nan Soy and agitated mild He at C Latent Given: - (13.31 x10³ k5/kmol MgSO4. 7H₂O) 13. 31 x 10³ kJ/mol Mg 604. TH20 Aaditional data: capacity of Heat capacity of неон OF TF= CPF E 333K removed F= 500kg Na2SO4 + 12500 kg H₂0 4: Ht cat B.6 kJ/kg.K mild soin = -78.5 NJ/kmol Read: Heat Msteel vessel = 750 kg Cp wild steel= Xv: 2395 KJ/xg soin = 3. KJ/Kg.K V OF vaporization OF OF 0.5 kJ/kg.K NaRSO4 = 1412 MW MW NarsOy - 10H120= 13212 lost yield of hydrated kg during cooling, steel = 0.5 KJ/kg.K (2% OF original water) hydrated crystals T= 283 K 1 kmol MgSO4. THRO 246.3 kg Ng SOY. TH₂U salt is anny drous salt. IF Glauber's calculate crystals water с Tc = 28 3 к Glauber's salt Na₂604 10 H₂O 2500 kg OF steel L 2395 kJ/kg TL = 283 K Sol. = hs= -78.5 NIS/ KALOJ In 2% water is vessel weighing in kg. 13.31 X108 KJ Ikmul MgSO4. TH ₂0/ 8.9 kg per 100 kg The original ne at that OF the 8.9 kg anny drous 100 kg H₂0 soin: • Determine V= AL= hc= Solve XF= 500 ONB: CMB- q + VAL (0.02) (2500 kg) = 50 kg Av = 2395 KJ (Kg 500 f 2500 = 3000 kg - ns = -(-78.5 MJ/kmol) = 78.5 For C: ૧ : 2500 +500 solute : FCPF (TF- TL) + Cnc + M vessel Cpvessel (Tp-TL) = 0.1647 F = L + V +C 3000 = L + 50 + C Lt C = 2950 kg XFF XL = MJ/kmol 8.9 100+ 8.9 = XLL + XCC (0.1667)(3000) = (0.0817) (L) + (0.4410) (C) (0.0817) + (0.4410) C = (0.1647) (3000) L = 2229. 1957 kg C = 720 8043 kg q + VAL = FCPF (TF-TL) + Chc + q + (50 kg) ( 23 95 kJ/xg) = (3000 = 0.0817 9 = 614 724. 0297 KJ Xc= 142 312122 nivessel Cpvessel (TF-TL) kg) (3.6 kJ/xg.x) (3313 - 283) K + (720.8043 xg) (78500 KJ/kmol) (¹ km01/31212kg) + (750 kg) (0.5 kJ/xg.k) (333- 283) K 0.4410