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.
480 kW
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.
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:
$$ C_h = \dot{m}_h \times C_p $$
$$ C_h = 2 \text{ kg/s} \times 4 \text{ kJ/kg.°C} $$
$$ C_h = 8 \text{ kJ/s.°C} $$
The maximum heat transfer is governed by the stream with the minimum heat capacity rate ($C_{min}$). Comparing the two streams:
Therefore, the minimum heat capacity rate is:
$$ C_{min} = \min(C_c, C_h) = 8 \text{ kJ/s.°C} $$
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:
$$ Q_{max} = 480 \text{ kW} $$
The maximum heat transfer rate possible in this counter flow heat exchanger setup is 480 kW.
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