Chemical Engineering Calculations | Chemical Engineering Exam

A soap plant produces raw soap containing 50% moisture. This is to be dried to 20% moisture before it is pressed into cakes for sale. How many 100-g soap pieces can be obtained from 1000 kg of original raw soap?

Given:


          Mass Fraction of Moisture from Raw Soap (MF_{H₂O, RS}) = 0.50 

          Mass Fraction of Moisture from Dried Soap (MF_{H₂O, DS}) = 0.20 

          Raw Soap (RS) = 1000 kg

Find:

Number of 100-g soap pieces = ?

Solution:


          Mass Fraction of Soap from Raw Soap (MF_{Soap, RS}) = ? 

          MF_{Soap, RS} = 1 - MF_{H₂O, RS} 

          MF_{Soap, RS} = 1 - 0.50 

          MF_{Soap, RS} = 0.50 



          Soap = ? 

          Soap = (MF_{Soap, RS})(RS) 

          Soap = (0.50)(1000 kg) 

          Soap = 500 kg 



          Dried Soap (DS) = ? 

          Mass Fraction of Soap from Dried Soap (MF_{Soap, DS}) = ? 

          MF_{Soap, DS} = 1 - MF_{H₂O, DS} 

          MF_{Soap, DS} = 1 - 0.20 

          MF_{Soap, DS} = 0.80 



          Soap = (MF_{Soap, DS})(DS) 

          500 kg = 0.80(DS) 

          625 kg = DS 



          Number of 100-g soap pieces = (DS)(100 g)^-1 

          Number of 100-g soap pieces = (625 kg)(100 g)^-1(1000 g)(1 kg)^-1 

          Number of 100-g soap pieces = 6250

Answer:

Number of 100-g soap pieces = 6250

Formulate the independent material balance equations for determining the flow rate of benzene in the feed under the following circumstances. An aqueous acetic acid solution containing 80% acetic acid and the rest water is charged into a still along with pure benzene. 400 kg/h pure acetic acid is withdrawn as product from the still. The top product leaving the still consists of 11.0% acetic acid, 21.5% water and the rest benzene.

Given:


          Mass Fraction of Acetic Acid from Acetic Acid Solution (MF_{CH₃COOH, Solution}) = 0.80 

          Pure Acetic Acid Product (CH₃COOH) = 400 kg/h 

          Mass Fraction of Acetic Acid from Top Product (MF_{CH₃COOH, TP}) = 0.110 

          Mass Fraction of Water from Top Product (MF_{H₂O, TP}) = 0.215

Find:


          Acetic Acid Solution (CH₃COOH_Solution) = ? 

          Pure Benzene (C₆H₆) = ? 

          Top Product (TP) = ?

Solution:


          Component Material Balance: CH₃COOH 

          (MF_{CH₃COOH, Solution})(CH₃COOH_Solution) = CH₃COOH + (MF_{CH₃COOH, TP})(TP) 

          0.80(CH₃COOH_Solution) = 400 kg/h + 0.110(TP) 

          CH₃COOH_Solution = 500 kg/h + 0.1375(TP) 



          Mass Fraction of Benzene (MF_{C₆H₆, TP}) = ? 

          MF_{C₆H₆, TP} = 1 - MF_{CH₃COOH, TP} - MF_{H₂O, TP} 

          MF_{C₆H₆, TP} = 1 - 0.110 - 0.215 

          MF_{C₆H₆, TP} = 0.675 



          Component Material Balance: C₆H₆ 

          C₆H₆ = (MF_{C₆H₆, TP})(TP) 

          C₆H₆ = 0.675(TP) 



          Mass Fraction of Water from Acetic Acid Solution (MF_{H₂O, Solution}) = ? 

          MF_{H₂O, Solution} = 1 - MF_{CH₃COOH, Solution} 

          MF_{H₂O, Solution} = 1 - 0.80 

          MF_{H₂O, Solution} = 0.20 



          Component Material Balance: H₂O 

          (MF_{H₂O, Solution})(CH₃COOH_Solution) = (MF_{H₂O, TP})(TP) 

          0.20(CH₃COOH_Solution) = 0.215(TP) 

          0.9302325581(CH₃COOH_Solution) = TP 



          CH₃COOH_Solution = 500 kg/h + 0.1375(0.9302325581(CH₃COOH_Solution)) 

          CH₃COOH_Solution = 500 kg/h + 0.1279069767(CH₃COOH_Solution) 

          0.8720930233(CH₃COOH_Solution) = 500 kg/h 

          CH₃COOH_Solution = 573.3333333 kg/h 



          (0.9302325581)(573.3333333 kg/h) = TP 

          533.3333333 kg/h = TP 



          C₆H₆ = (0.675)(533.3333333 kg/h) 

          C₆H₆ = 360 kg/h

Answer:


          Acetic Acid Solution = 500 kg/h + 0.1375(Top Product) 

          Pure Benzene = 0.675(Top Product) 

          Top Product = 0.9302325581(Acetic Acid Solution) 



          Acetic Acid Solution (CH₃COOH_Solution) = 573.3333333 kg/h 

          Pure Benzene (C₆H₆) = 360 kg/h 

          Top Product (TP) = 533.3333333 kg/h

Urea, phosphoric acid and potassium chloride are mixed together to obtain a mixed fertilizer having NPK content 10 : 26 : 26 as %N, %P₂O₅ and %K₂O by weight, balance being the weight of filler materials. Calculate the quantities to be mixed to get 1000 kg of mixed fertilizer.

Given:


          x_N = 0.10 

          x_P₂O₅ = 0.26 

          x_K₂O = 0.26 

          Mixed Fertilizer (MF) = 1000 kg

Find:


          Urea ((NH₂)₂CO) = ? 

          Phosphoric Acid (H₃PO₄) = ? 

          Potassium Chloride (KCl) = ?

Solution:


          N = (x_N)(MF) 

          N = (0.10)(1000 kg) 

          N = 100 kg 



          Molar Mass of (NH₂)₂CO (MM_(NH₂)₂CO) = 60.05837 kg kmol^-1 

          Molar Mass of N in (NH₂)₂CO (MM_N) = 28.01456 kg kmol^-1 

          Mass Fraction of N in (NH₂)₂CO (MF_N) = ? 

          MF_N = (MM_N)(MM_(NH₂)₂CO)^-1 

          MF_N = (28.01456 kg kmol^-1)(60.05837 kg kmol^-1)^-1 

          MF_N = 0.4664555498 



          N = (MF_N)((NH₂)₂CO) 

          100 kg = 0.4664555498((NH₂)₂CO) 

          214.3826996 kg = (NH₂)₂CO 



          P₂O₅ = (x_P₂O₅)(MF) 

          P₂O₅ = (0.26)(1000 kg) 

          P₂O₅ = 260 kg 



          Molar Mass of P₂O₅ (MM_P₂O₅) = 141.94609 kg kmol^-1 

          Molar Mass of P in P₂O₅ (MM_{P, P₂O₅}) = 61.94724 kg kmol^-1 

          Mass Fraction of P in P₂O₅ (MF_{P, P₂O₅}) = ? 

          MF_{P, P₂O₅} = (MM_{P, P₂O₅})(MM_P₂O₅)^-1 

          MF_{P, P₂O₅} = (61.94724 kg kmol^-1)(141.94609 kg kmol^-1)^-1 

          MF_{P, P₂O₅} = 0.4364138526 



          P = (MF_{P, P₂O₅})(P₂O₅) 

          P = (0.4364138526)(260 kg) 

          P = 113.4676017 kg 



          Molar Mass of H₃PO₄ (MM_H₃PO₄) = 97.99703 kg kmol^-1 

          Molar Mass of P in H₃PO₄ (MM_{P, H₃PO₄}) = 30.97362 kg kmol^-1 

          Mass Fraction of P in H₃PO₄ (MF_{P, H₃PO₄}) = ? 

          MF_{P, H₃PO₄} = (MM_{P, H₃PO₄})(MM_H₃PO₄)^-1 

          MF_{P, H₃PO₄} = (30.97362 kg kmol^-1)(97.99703 kg kmol^-1)^-1 

          MF_{P, H₃PO₄} = 0.3160669257 



          P = (MF_{P, H₃PO₄})(H₃PO₄) 

          113.4676017 kg = 0.3160669257(H₃PO₄) 

          358.9986565 kg = H₃PO₄ 



          K₂O = (x_K₂O)(MF) 

          K₂O = (0.26)(1000 kg) 

          K₂O = 260 kg 



          Molar Mass of K₂O (MM_K₂O) = 94.19637 kg kmol^-1 

          Molar Mass of K in K₂O (MM_{K, K₂O}) = 78.1966 kg kmol^-1 

          Mass Fraction of K in K₂O (MF_{K, K₂O}) = ? 

          MF_{K, K₂O} = (MM_{K, K₂O})(MM_K₂O)^-1 

          MF_{K, K₂O} = (78.1966 kg kmol^-1)(94.19637 kg kmol^-1)^-1 

          MF_{K, K₂O} = 0.8301445162 



          K = (MF_{K, K₂O})(K₂O) 

          K = (0.8301445162)(260 kg) 

          K = 215.8375742 kg 



          Molar Mass of KCl (MM_KCl) = 74.5553 kg kmol^-1 

          Molar Mass of K in KCl (MM_{K, KCl}) = 39.0983 kg kmol^-1 

          Mass Fraction of K in KCl (MF_{K, KCl}) = ? 

          MF_{K, KCl} = (MM_{K, KCl})(MM_KCl)^-1 

          MF_{K, KCl} = (39.0983 kg kmol^-1)(74.5553 kg kmol^-1)^-1 

          MF_{K, KCl} = 0.5244201284 



          K = (MF_{K, KCl})(KCl) 

          215.8375742 kg = 0.5244201284(KCl) 

          411.5737759 kg = KCl

Answer:


          Urea ((NH₂)₂CO) = 214.3826996 kg 

          Phosphoric Acid (H₃PO₄) = 358.9986565 kg 

          Potassium Chloride (KCl) = 411.5737759 kg

Air is dehumidified at a constant pressure of 101.3 kPa. The partial pressure of water in the air admitted to the dehumidifier is 7 kPa and the temperature is 350 K. The partial pressure of water in the air leaving is 1.5 kPa. How much water (in kilograms) is removed from 100 cubic meters of air admitted?

Given:


          Constant Pressure (P) = 101.3 kPa 

          Partial Pressure of Admitted Water (PPAW) = 7 kPa 

          Temperature of Admitted Water (TAW) = 350 K 

          Partial Pressure of Leaving Water (PPLW) = 1.5 kPa 

          Volume of Admitted Air (VAA) = 100 cubic meter

Find:

Water Removed (WR) = ? in kilograms

Solution:


          Moles of Admitted Water (nAW) = ? 

          (PPAW)(VAA) = (nAW)(R)(TAW) 

          (7 kPa)(100 m³) = (nAW)(8.314 kPa m³ kmol^-1 K^-1)(350 K) 

          700 = (nAW)(2909.9 kmol^-1) 

          0.2405580948 kmol = nAW 



          Partial Pressure of Admitted Gas (PPAG) = ? 

          P = PPAW + PPAG 

          101.3 kPa = 7 kPa + PPAG 

          94.3 kPa = PPAG 



          Partial Pressure of Leaving Gas (PPLG) = ? 

          P = PPLW + PPLG 

          101.3 kPa = 1.5 kPa + PPLG 

          99.8 kPa = PPLG 



          Moles of Admitted Gas = Moles of Leaving Gas 

          Temperature of Admitted Air (TAA) = TAW = 350 K 

          Temperature of Leaving Air (TLA) = ? 

          Volume of Leaving Air (VLA) = ? 

          (PPAG)(VAA)(R)^-1(TAA)^-1 = (PPLG)(VLA)(R)^-1(TLA)^-1 

          (94.3 kPa)(100 m³)(350 K)^-1 = (99.8 kPa)(VLA)(TLA)^-1 

          26.94285714 m³ K^-1 = 99.8(VLA)(TLA)^-1 

          0.2699685084 m³ K^-1 = (VLA)(TLA)^-1 



          Moles of Leaving Water (nLW) = ? 

          (PPLW)(VLA) = (nLW)(R)(TLA) 

          (PPLW)(R)^-1(VLA)(TLA)^-1 = nLW 

          (1.5 kPa)(8.314 kPa m³ kmol^-1 K^-1)^-1(0.2699685084 m³ K^-1) = nLW 

          0.04870733252 kmol = nLW 



          Moles of Water Removed (nWR) = ? 

          nAW = nLW + nWR 

          0.2405580948 kmol = 0.04870733252 kmol + nWR 

          0.1918507623 kmol = nWR 



          Molar Mass of Water (MMW) = 18.01599 kg kmol^-1

          WR = (nWR)(MMW) 

          WR = (0.1918507623 kmol)(18.01599 kg kmol^-1) 

          WR = 3.456381415 kg

Answer:

Water Removed (WR) = 3.456381415 in kilograms

Wet sewage sludge enters a continuous thickener at a rate of 100 kg per hour and dehydrated sludge leaves the thickener at a rate of 75 kg per hour. Determine the amount of water removed in the thickener in one hour, assuming steady-state operation.

Given:


          Wet Sludge (WS) = 100 kg per hour 

          Dehydrated Sludge (DS) = 75 kg per hour

Find:

Water Removed (WR) = ? in one hour

Solution:


          Basis: One hour 


          Input = Output 

          WS = DS + WR 

          100 kg = 75 kg + WR 

          25 kg = WR

Answer:

Water Removed (WR) = 25 kg in one hour

A soap plant produces raw soap containing 50% moisture. This is to be dried to 20% moisture before it is pressed into cakes for sale. How many 100-g soap pieces can be obtained from 1000 kg of original raw soap?

Urea, phosphoric acid and potassium chloride are mixed together to obtain a mixed fertilizer having NPK content 10 : 26 : 26 as %N, %P2O5 and %K2O by weight, balance being the weight of filler materials. Calculate the quantities to be mixed to get 1000 kg of mixed fertilizer.

Wet sewage sludge enters a continuous thickener at a rate of 100 kg per hour and dehydrated sludge leaves the thickener at a rate of 75 kg per hour. Determine the amount of water removed in the thickener in one hour, assuming steady-state operation.

Urea, phosphoric acid and potassium chloride are mixed together to obtain a mixed fertilizer having NPK content 10 : 26 : 26 as %N, %P₂O₅ and %K₂O by weight, balance being the weight of filler materials. Calculate the quantities to be mixed to get 1000 kg of mixed fertilizer.