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Question

A cold liquid enters a counter flow heat exchanger at 15 deg at a rate of 8 kg/s. A hot stream of the same liquid enters the heat exchanger at 75 deg at 2 kg/s. Assuming the specific heat of the fluid as 4 kJ/kg.°C, determine the maximum heat transfer rate.

The correct answer is

480 kW

Understanding Heat Exchanger Maximum Heat Transfer

This problem involves calculating the maximum possible heat transfer rate in a heat exchanger. The maximum heat transfer rate is limited by the stream that has the smaller heat capacity rate.

Identifying Given Parameters

  • Cold stream inlet temperature: $T_{c,in} = 15$ °C
  • Cold stream mass flow rate: $\dot{m}_c = 8$ kg/s
  • Hot stream inlet temperature: $T_{h,in} = 75$ °C
  • Hot stream mass flow rate: $\dot{m}_h = 2$ kg/s
  • Specific heat of the liquid: $C_p = 4$ kJ/kg.°C

Calculating Heat Capacity Rates

The heat capacity rate ($C$) for a fluid stream is calculated by multiplying its mass flow rate ($\dot{m}$) by its specific heat ($C_p$).

For the cold stream:

$$ C_c = \dot{m}_c \times C_p $$

$$ C_c = 8 \text{ kg/s} \times 4 \text{ kJ/kg.°C} $$

$$ C_c = 32 \text{ kJ/s.°C} $$

For the hot stream:

$$ C_h = \dot{m}_h \times C_p $$

$$ C_h = 2 \text{ kg/s} \times 4 \text{ kJ/kg.°C} $$

$$ C_h = 8 \text{ kJ/s.°C} $$

Determining Minimum Heat Capacity Rate

The maximum heat transfer is governed by the stream with the minimum heat capacity rate ($C_{min}$). Comparing the two streams:

  • $C_c = 32$ kJ/s.°C
  • $C_h = 8$ kJ/s.°C

Therefore, the minimum heat capacity rate is:

$$ C_{min} = \min(C_c, C_h) = 8 \text{ kJ/s.°C} $$

Calculating Maximum Heat Transfer Rate

The maximum possible heat transfer rate ($Q_{max}$) in a heat exchanger occurs when the minimum heat capacity rate stream undergoes the maximum possible temperature change, which is the difference between the hot stream inlet temperature and the cold stream inlet temperature.

The formula is:

$$ Q_{max} = C_{min} \times (T_{h,in} - T_{c,in}) $$

Substituting the values:

$$ Q_{max} = 8 \text{ kJ/s.°C} \times (75 \text{ °C} - 15 \text{ °C}) $$

$$ Q_{max} = 8 \text{ kJ/s.°C} \times 60 \text{ °C} $$

$$ Q_{max} = 480 \text{ kJ/s} $$

Since 1 kJ/s is equal to 1 kW, the maximum heat transfer rate is:

$$ Q_{max} = 480 \text{ kW} $$

Conclusion

The maximum heat transfer rate possible in this counter flow heat exchanger setup is 480 kW.

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Important Questions from Heat Exchanger Analysis

  1. The fin effectiveness can be enhanced by selecting _____ value of heat transfer co-efficient.

  2. NTU, which is a measure of effectiveness of heat exchanger, stands for _________.

  3. LMTD stands for _______.

  4. Water (Cp = 4.18 kJ/kg.K) at 80°C enters a counter flow heat exchanger with a mass flow rate of 0.5 kg/s. Air (Cp = 1 kJ/kg.K) enters at 30°C with a mass flow rate of 2.09 kg/s. If the effectiveness of the heat exchanger is 0.8, the LMTD (in °C) is

  5. For a heat exchanger, ΔTmax is the maximum temperature difference and ΔTmin is the minimum temperature difference between the two fluids. LMTD is the log mean temperature difference. Cmin and Cmax are the minimum and the maximum heat capacity rates. The maximum possible heat transfer (Qmax) between the two fluids is

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