The cardiovascular counter-current heat exchanger has an overall heat transfer coefficient of \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Arterial blood enters at \(37^{\circ} \mathrm{C}\) and exits at \(27^{\circ} \mathrm{C}\). Venous blood enters at \(25^{\circ} \mathrm{C}\) and exits at \(34^{\circ} \mathrm{C}\). Determine the mass flow rates of the arterial blood and venous blood in \(\mathrm{g} / \mathrm{s}\) if the specific heat of both arterial and venous blood is constant and equal to \(3475 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and the surface area of the heat transfer to occur is \(0.15 \mathrm{~cm}^{2}\).

We first need to calculate the logarithmic mean temperature difference between the arterial and venous blood, which we will use in the heat transfer equation. The formula is: \(\Delta T_{LM} = \frac{(\Delta T_1 - \Delta T_2)}{\ln(\frac{\Delta T_1}{\Delta T_2})}\) where \(\Delta T_1 = T_{Ain} - T_{Vin}\) and \(\Delta T_2 = T_{Aout} - T_{Vout}\). The given temperature values are: \(T_{Ain} = 37^\circ C\) \(T_{Aout} = 27^\circ C\) \(T_{Vin} = 25^\circ C\) \(T_{Vout} = 34^\circ C\) Plugging in these values, we have: \(\Delta T_1 = 37^\circ C - 25^\circ C = 12^\circ C\) \(\Delta T_2 = 27^\circ C - 34^\circ C = -7^\circ C\) Now, we can find the \(\Delta T_{LM}\): \(\Delta T_{LM} = \frac{(12 - (-7))}{\ln(\frac{12}{-7})} = 9.378^\circ C\)

Now that we have the \(\Delta T_{LM}\), we can plug it into the heat transfer equation along with the given overall heat transfer coefficient \(U = 100 \frac{W}{m^2 K}\) and the surface area \(A = 0.15 cm^2 = 1.5 \times 10^{-5} m^2\) to find the rate of heat transfer \(Q\): \(Q = U \cdot A \cdot \Delta T_{LM} = 100 \frac{W}{m^2 K} \cdot 1.5 \times 10^{-5} m^2 \cdot 9.378 K = 0.014 W\)

The heat transfer between arterial and venous blood can be related to the mass flow rates of each blood type and the specific heat using the equation, \(Q = m_a \cdot c_p \cdot (T_{Ain} - T_{Aout}) = m_v \cdot c_p \cdot (T_{Vin} - T_{Vout})\) Where \(m_a\) is the mass flow rate of arterial blood, and \(m_v\) is the mass flow rate of venous blood. The specific heat for both arterial and venous blood is \(c_p = 3475\frac{J}{kgK}\). Since we know the specific heat and temperature differences for both blood types, we can divide the equation by \(c_p\) and plug in the temperature values: \(\frac{Q}{c_p} = m_a (37-27) = m_v (25-34)\)

Now we can substitute the value of \(Q\) calculated in step 2 and solve for the mass flow rates: \(\frac{0.014W}{3475\frac{J}{kgK}} = m_a (10K) = m_v (-9K)\) Solving for \(m_a\) and \(m_v\), we get: \(m_a = \frac{0.014W}{3475\frac{J}{kgK} \cdot 10K} = 4.03 \times 10^{-6} kg/s\) \(m_v = -\frac{0.014W}{3475\frac{J}{kgK} \cdot 9K} = 4.47 \times 10^{-6} kg/s\) Finally, we convert these values to grams per second: \(m_a = 4.03 \times 10^{-6} kg/s \cdot \frac{1000g}{1kg} = 4.03 \frac{g}{s}\) \(m_v = 4.47 \times 10^{-6} kg/s \cdot \frac{1000g}{1kg} = 4.47 \frac{g}{s}\) #Conclusion# The mass flow rate of the arterial blood is approximately \(4.03 \frac{g}{s}\), and the mass flow rate of the venous blood is approximately \(4.47 \frac{g}{s}\).