Question

A saturated solution of potassium chloride at 90 °C containing 700 kg potassium chloride is cooled to 20 °C in an open tank by circulation of water. The cooling water enters the tank at 15 °C through a jacket and leaves at 25 °C. The solubility of potassium chloride is 53.9 and 34.2 gm/100ml of water at 90 and 20 °C respectively, The average specific heat and density of the solution are 0.48 cal/g.°C and 1.2 g/cm', respectively. Assuming the overall heat transfer coefficient kcal/(hr)(m)(°C), determine (i) the capacity of the tank, (ii) the weight of crystals obtained, and (iii) the cooling surface of the tank. to be 120

Answer 1

i)

Saturated solution of KCl is present at 90°C

Mass of KCl present = 700 Kg

Solubility of KCl at 90°C = 53.9 g/100 ml water

1 ml = 1 g water

Weight fraction of KCl in solution at 90°C = (53.9/(53.9+100) ) = 0.35022

Mass of solution present at 90°C in tank = (700/0.35022) = 1998.7436 Kg

Density of solution = 1.2 g/ml = 1200 Kg/m³

Volume of solution = volume of tank = (1998.7436/1200) = 1.6656 m³

ii)

Solubility of KCl at 20°C = 34.2 g/100 ml water

Weight fraction of KCl in solution at 20°C = (34.2/(34.2+100) )= 0.25484

F - feed solution at 90°C

C- crystals formed  

L- solution at 20°C

Doing material balance we get

1998.857 = C + L

Doing KCl balance we get

1998.857(0.35022) = C(1) + L(0.25484)

Solving both equations we get

C = 255.852 Kg

L = 1743.004 Kg

Weight of crystals (C) = 255.852 Kg

iii)

Q = FCp(90-20) + C(latent heat of fusion)

Cp - 0.48 Kcal/Kg°C

From handbook latent heat of fusion of KCl = 31.307 KJ/Kg = 7.489 Kcal/Kg

Q = 1998.857(0 . 48) (90-20) +(255.852) (7.489)

Q =69077.871 Kcal

U = 120 Kcal/h m²°C

∆T(water) = 25-15 = 10°C

Q = UA(∆T)_lmtd

For 1 h, Q = 69077.871 Kcal/h

∆T lmtd =( ∆T1 - ∆T2) /ln(∆T1/∆T2)

∆T1 = 90-15 = 75

∆T2 = 25-20= 5

∆T lmtd = (75-5) /ln(75/5) = 25.84°C

69077.871 = 120 (A) (25.84)

A = 22.27 m²

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