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Heat Transfer - Engineering

Q1:

It is desired to concentrate a 20% salt solution (20 kg of salt in 100 kg of solution) to a 30% salt solution in an evaporator. Consider a feed of 300 kg/min at 30°C. The boiling point of the solution is 110°C, the latent heat of vaporisation is 2100 kJ/kg and the specific heat of the solution is 4 kJ/kg.K. The rate at which the heat has to be supplied in (kJ/min) to the evaporator is

A 3.06 x l05

B 6.12 x 105

C 7.24 x 105

D 9.08 x 105

ANS:A - 3.06 x l05

To determine the rate at which heat must be supplied to the evaporator, we can use the principle of energy balance. The heat required for evaporation can be calculated using the latent heat of vaporization, and the heat required to heat up the solution to its boiling point can be calculated using the specific heat capacity. The total heat supplied is the sum of these two components. Let's break down the calculation:

  1. Heat required to heat up the solution to its boiling point: 1=×ΔQ1​=m×c×ΔT Where:
    • m is the mass of the solution (300 kg/min),
    • c is the specific heat of the solution (4 kJ/kg.K),
    • ΔT is the temperature difference from the feed temperature to the boiling point (110°C - 30°C = 80°C).
  2. Heat required for evaporation: 2=×ℎlatentQ2​=m×hlatent​ Where:
    • ℎlatenthlatent​ is the latent heat of vaporization (2100 kJ/kg).
  3. Total heat required: total=1+2Qtotal​=Q1​+Q2​
Let's calculate:
  1. 1=300 kg/min×4 kJ/kg.K×80 K=96000 kJ/minQ1​=300kg/min×4kJ/kg.K×80K=96000kJ/min
  2. 2=300 kg/min×2100 kJ/kg=630000 kJ/minQ2​=300kg/min×2100kJ/kg=630000kJ/min
  3. total=96000 kJ/min+630000 kJ/min=726000 kJ/minQtotal​=96000kJ/min+630000kJ/min=726000kJ/min
So, the rate at which the heat has to be supplied in kJ/min to the evaporator is 726000 kJ/min726000 kJ/min, which can be approximated as 7.26×105 kJ/min7.26×105kJ/min. Therefore, the closest option is: c) 7.24×105 kJ/min7.24×105kJ/min