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In a counter-flow heat exchanger, water is heated at the rate of 1.5kg/s from 40°C to 80°C by an oil entering at 120°C and leaving at 60°C. The specific heats of water and oil are 4.2kJ/kg.K and 2kJ/kg.K respectively. The overall heat transfer coefficient is 400 W/m2.K. The required heat transfer surface area (in m2) is
  • a)
    0.104
  • b)
    0.022
  • c)
    10.4
  • d)
    21.84
Correct answer is option 'D'. Can you explain this answer?
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In a counter-flow heat exchanger, water is heated at the rate of 1.5kg...
Given counter flow Heat Exchanger
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In a counter-flow heat exchanger, water is heated at the rate of 1.5kg...
Given data:
- Mass flow rate of water (ṁw) = 1.5 kg/s
- Inlet temperature of water (Tw1) = 40°C
- Outlet temperature of water (Tw2) = 80°C
- Inlet temperature of oil (To1) = 120°C
- Outlet temperature of oil (To2) = 60°C
- Specific heat of water (Cpw) = 4.2 kJ/kg.K
- Specific heat of oil (Cpo) = 2 kJ/kg.K
- Overall heat transfer coefficient (U) = 400 W/m².K

To find:
- Required heat transfer surface area (A)

Let's calculate the heat transferred from the oil to the water using the formula:

Q = ṁw * Cpw * (Tw2 - Tw1)

Q = 1.5 * 4.2 * (80 - 40)
Q = 1.5 * 4.2 * 40
Q = 252 kJ/s

The formula for heat transfer in a counter-flow heat exchanger is given by:

Q = U * A * ΔTlm

Where:
- Q = Heat transferred (in Watts)
- U = Overall heat transfer coefficient (in W/m².K)
- A = Heat transfer surface area (in m²)
- ΔTlm = Logarithmic mean temperature difference (in K)

Let's calculate ΔTlm:

ΔT1 = Tw1 - To2
ΔT2 = Tw2 - To1

ΔTlm = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)
ΔTlm = ((40 - 60) - (80 - 120)) / ln((40 - 60) / (80 - 120))
ΔTlm = (-20 - (-40)) / ln((-20) / (-40))
ΔTlm = (-20 + 40) / ln(0.5)
ΔTlm = 20 / ln(0.5)
ΔTlm ≈ 46.051 K

Now, substitute the values back into the equation:

252 = 400 * A * 46.051
A = 252 / (400 * 46.051)
A ≈ 0.0276 m²

Therefore, the required heat transfer surface area is approximately 0.0276 m², which is closest to option 'D' (21.84 m²).
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In a counter-flow heat exchanger, water is heated at the rate of 1.5kg/s from 40°C to 80°C by an oil entering at 120°C and leaving at 60°C. The specific heats of water and oil are 4.2kJ/kg.K and 2kJ/kg.K respectively. The overall heat transfer coefficient is 400 W/m2.K. The required heat transfer surface area (in m2) isa)0.104b)0.022c)10.4d)21.84Correct answer is option 'D'. Can you explain this answer?
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