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The logarithmic mean temperature difference (LMTD) of a counter flow heat exchanger is 20°C. The cold fluid enters at 20°C and the hot fluidenters at 100°C. Mass flow rate of the cold fluid is twice that of the hotfluid. Specific heat at constant pressure of the hot fluid is twice that ofthe cold fluid. The exit temperature of the cold fluid 
  • a)
    is 40°C
  • b)
    is 60°C
  • c)
    is 80°C
  • d)
    Cannot be determined
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
The logarithmic mean temperature difference (LMTD) of a counter flow h...
Ans (c) As mhch = mccc. Therefore exit temp. = 100 – LMTD = 100 – 20 =80°C.
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The logarithmic mean temperature difference (LMTD) of a counter flow h...
Logarithmic Mean Temperature Difference (LMTD) in a counter flow heat exchanger is given by the formula:

LMTD = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)

where ΔT1 is the temperature difference between the hot fluid inlet and the cold fluid outlet, and ΔT2 is the temperature difference between the hot fluid outlet and the cold fluid inlet.

The given LMTD is 20°C.

Hot Fluid: Inlet temperature = 100°C
Cold Fluid: Inlet temperature = 20°C

Since the mass flow rate of the cold fluid is twice that of the hot fluid, we can assume the mass flow rate of the hot fluid as 'm' and the mass flow rate of the cold fluid as '2m'.

Given that the specific heat at constant pressure of the hot fluid is twice that of the cold fluid, we can assume the specific heat capacity of the cold fluid as 'C' and the specific heat capacity of the hot fluid as '2C'.

The exit temperature of the cold fluid can be determined using the LMTD formula and the known values.

Now, let's calculate ΔT1 and ΔT2:

ΔT1 = (100 - T2) - (20 - T1)
ΔT1 = 80 - T2 + T1

ΔT2 = (100 - T1) - (20 - T2)
ΔT2 = 80 + T2 - T1

Substituting these values into the LMTD formula:

20 = (80 - T2 + T1) / ln[(80 - T2 + T1) / (80 + T2 - T1)]

Simplifying the equation:

ln[(80 - T2 + T1) / (80 + T2 - T1)] = (80 - T2 + T1) / 20

Since the answer is option 'C' (exit temperature of the cold fluid is 80°C), we can substitute T2 = 80 into the equation and check if it satisfies the equation.

ln[(80 - 80 + T1) / (80 + 80 - T1)] = (80 - 80 + T1) / 20

ln[(T1) / (160 - T1)] = T1 / 20

Taking exponential on both sides:

(T1) / (160 - T1) = e^(T1 / 20)

Simplifying the equation, we get:

T1 = 60

Therefore, the exit temperature of the cold fluid is 60°C, which contradicts the given correct answer option 'C'. Thus, the correct answer cannot be determined.
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The logarithmic mean temperature difference (LMTD) of a counter flow heat exchanger is 20°C. The cold fluid enters at 20°C and the hot fluidenters at 100°C. Mass flow rate of the cold fluid is twice that of the hotfluid. Specific heat at constant pressure of the hot fluid is twice that ofthe cold fluid. The exit temperature of the cold fluida)is 40°Cb)is 60°Cc)is 80°Cd)Cannot be determinedCorrect answer is option 'C'. Can you explain this answer?
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