During open-heart surgery, a patient’s blood is cooled down to 25 °C from 37 °C using a concentric tube counter-flow heat exchanger. Water enters the heat exchanger at 4 °C and leaves at 18 °C. Blood flow rate during the surgery is 5 L/minute.

Use the following fluid properties:

Fluid Density (kg/m3) Specific heat (J/kg-K) Blood 1050 3740 Water 1000 4200

Fluid	Density (kg/m3)	Specific heat (J/kg-K)
Blood	1050	3740
Water	1000	4200

Effectiveness of the heat exchanger is _________ (round off to 2 decimal places).

The correct answer is 0.42.

1. Identification of Given Parameters:
From the problem statement and the schematic diagram provided in the image, we identify the parameters for the hot fluid (blood) and the cold fluid (water):
- Hot fluid (blood) inlet temperature:
$T_{\mathrm{hi}}=37\text{°C}$
- Hot fluid (blood) outlet temperature:
$T_{\mathrm{he}}=25\text{°C}$
- Cold fluid (water) inlet temperature:
$T_{\mathrm{ci}}=4\text{°C}$
- Cold fluid (water) outlet temperature:
$T_{\mathrm{ce}}=18\text{°C}$
- Volumetric flow rate of blood:
${\dot{V}}_h=5\text{L/min}=\frac{5\times {10}^{-3}}{60}{\text{m}}^3\text{/s}$
- Density of blood:
${\rho }_h=1050{\text{kg/m}}^3$
- Specific heat of blood:
$c_{p,h}=3740\text{J/kg-K}$
- Specific heat of water:
$c_{p,c}=4200\text{J/kg-K}$

2. Mass Flow Rate and Heat Capacity Rate of Blood (Hot Fluid):
The mass flow rate of blood (${\dot{m}}_h$) is calculated using its density and volumetric flow rate as shown in the image equations:
${\dot{m}}_h={\rho }_h\times {\dot{V}}_h=1050\times \frac{5\times {10}^{-3}}{60}=0.0875\text{kg/s}$
Now, the heat capacity rate of the hot fluid ($C_h$) is:
$C_h={\dot{m}}_h{c}_{p,h}=0.0875\times 3740=327.25\text{W/K}$

3. Heat Capacity Rate of Water (Cold Fluid) via Energy Balance:
The schematic diagram in the image visually represents the counter-flow configuration, showing the hot fluid temperature profile decreasing from 37 °C to 25 °C, and the cold fluid temperature profile flowing in the opposite direction and increasing from 4 °C to 18 °C. Applying the conservation of energy, the heat lost by the blood must equal the heat gained by the water:
$Q_{\text{lost}}=Q_{\text{gained}}$
$C_h\left(T_{\mathrm{hi}}-T_{\mathrm{he}}\right)=C_c\left(T_{\mathrm{ce}}-T_{\mathrm{ci}}\right)$
Substituting the temperatures and $C_h$:
$327.25\times \left(37-25\right)=C_c\times \left(18-4\right)$
$327.25\times 12=C_c\times 14$
$3927=14C_c$
$C_c=\frac{3927}{14}=280.5\text{W/K}$

4. Determining the Minimum Heat Capacity Rate ($C_{\min }$):
Comparing the heat capacity rates of both fluids:
- Hot fluid: $C_h=327.25\text{W/K}$
- Cold fluid: $C_c=280.5\text{W/K}$
Since $C_c<C_h$, the cold fluid capacity rate is the minimum capacity rate:
$C_{\min }=C_c=280.5\text{W/K}$

5. Calculation of Effectiveness ($\epsilon$):
The effectiveness of a heat exchanger is defined as:
$\epsilon =\frac{Q}{Q_{\max }}=\frac{C_c\left(T_{\mathrm{ce}}-T_{\mathrm{ci}}\right)}{C_{\min }\left(T_{\mathrm{hi}}-T_{\mathrm{ci}}\right)}$
Because $C_{\min }=C_c$, the capacity rate terms cancel out, simplifying the equation to:
$\epsilon =\frac{T_{\mathrm{ce}}-T_{\mathrm{ci}}}{T_{\mathrm{hi}}-T_{\mathrm{ci}}}$
Substituting the temperature values into the simplified expression:
$\epsilon =\frac{18-4}{37-4}=\frac{14}{33}\approx 0.4242$
Rounding to two decimal places yields:
$\epsilon \approx 0.42$