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A balanced counter flow heat exchanger has a surface area of 20 m2 and overall heat transfer coefficient of 20 W/m2K. Air (cp = 1000 J/kgK) entering at 0.4 kg/s and 280 K is to be preheated by the air leaving the system at 0.4 kg/s and 300 K. The temperature (in K) of the preheated air is
[2015]
  • a)
    290
  • b)
    300
  • c)
    320
  • d)
    350
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
A balanced counter flow heat exchanger has a surface area of 20 m2 and...
Counter flow heat exchanged Surface Area A = 20 m2, mass flow rate = 0.4 k g/s
 Temperature Tci = 280K

Since m is same for both flow = 0.4 kg/s
Assume Cp is same = 100 J/kg.K
Hence

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Most Upvoted Answer
A balanced counter flow heat exchanger has a surface area of 20 m2 and...
To determine the temperature of the preheated air in the balanced counter flow heat exchanger, we can use the energy balance equation.

Energy balance equation:
Q = m * cp * (T2 - T1)

Where:
Q is the heat transfer rate
m is the mass flow rate
cp is the specific heat capacity
T1 is the inlet temperature
T2 is the outlet temperature

In this case, the heat transfer rate is given by the surface area of the heat exchanger and the overall heat transfer coefficient:
Q = U * A * (T2 - T1)

Where:
U is the overall heat transfer coefficient
A is the surface area of the heat exchanger

Given that U = 20 W/m2K and A = 20 m2, we can substitute these values into the equation:
Q = 20 * 20 * (T2 - T1)

Since the heat exchanger is balanced, the heat transfer rate is the same for both sides. Therefore, we can equate the two energy balance equations:

m1 * cp * (T2 - T1) = m2 * cp * (T2 - T1)

Given that m1 = m2 = 0.4 kg/s, cp = 1000 J/kgK, and T1 = 280 K, we can substitute these values into the equation:

0.4 * 1000 * (T2 - 280) = 0.4 * 1000 * (300 - T2)

Simplifying the equation:

0.4 * 1000 * T2 - 0.4 * 1000 * 280 = 0.4 * 1000 * 300 - 0.4 * 1000 * T2

0.4 * 1000 * T2 + 0.4 * 1000 * T2 = 0.4 * 1000 * 300 + 0.4 * 1000 * 280

0.8 * 1000 * T2 = 0.4 * 1000 * (300 + 280)

0.8 * 1000 * T2 = 0.4 * 1000 * 580

T2 = (0.4 * 1000 * 580) / (0.8 * 1000)

T2 = 290 K

Therefore, the temperature of the preheated air is 290 K.
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A balanced counter flow heat exchanger has a surface area of 20 m2 and overall heat transfer coefficient of 20 W/m2K. Air (cp = 1000 J/kgK) entering at 0.4 kg/s and 280 K is to be preheated by the air leaving the system at 0.4 kg/s and 300 K. The temperature (in K) of the preheated air is[2015]a)290b)300c)320d)350Correct answer is option 'A'. Can you explain this answer?
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