Question

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 \(0.15 \mathrm{~cm}^{2}\).

Short answer

Answer: 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 \(4.47 \frac{g}{s}\).

Step-by-step solution

(Step 1: Calculate the logarithmic mean temperature difference)

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 \(\Delta T_{LM}\): \(\Delta T_{LM} = \frac{(12 - (-7))}{\ln(\frac{12}{-7})} = 9.378^\circ C\)

(Step 2: Calculate the rate of heat transfer)

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 \(Q\): \(Q = U \cdot A \cdot \Delta T_{LM} = 100 \frac{W}{m^2 K} \cdot 1.5 \times 10^{-5} m^2 \cdot 9.378 K = 0.014 W\)

(Step 3: Relate mass flow rates to the rate of heat transfer and specific heat)

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: \(\frac{Q}{c_p} = m_a (37-27) = m_v (25-34)\)

(Step 4: Calculate mass flow rates)

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 \(4.47 \frac{g}{s}\).

Key concepts

Logarithmic Mean Temperature Difference

The Logarithmic Mean Temperature Difference (LMTD) is crucial in understanding heat exchangers, especially when dealing with non-linear temperature profiles. When heat is transferred between two fluids, such as blood in a heat exchanger, the temperature difference across the exchanger isn't constant. This makes direct calculation of heat transfer rates complex without a suitable approximation.
To solve this, we use the LMTD. It acts as an average temperature difference that enables calculations for the rate of heat exchange. You find it by using:

• Initial temperature difference across the ends: \( \Delta T_1 \)
• Final temperature difference across the ends: \( \Delta T_2 \)

The formula for LMTD is: \( \Delta T_{LM} = \frac{(\Delta T_1 - \Delta T_2)}{\ln(\frac{\Delta T_1}{\Delta T_2})} \). This formula helps in approximating the effectiveness of the heat exchanger by averaging out temperature differences. Understanding the LMTD boosts the efficiency and design of heat exchanger systems.

Specific Heat Capacity

Specific Heat Capacity is a measure of how much heat energy a substance can absorb for a given temperature rise. This property is intrinsic to materials, including biological substances like blood.
In the context of heat exchangers, specific heat is denoted as \( c_p \), and it expresses the amount of energy required to raise the temperature of 1 kilogram of the substance by 1 Kelvin.
For our exercise, both arterial and venous blood have a specific heat of \( 3475 \frac{J}{kg \cdot K} \). This means that is the amount of energy each kilogram of blood requires to undergo a 1-degree temperature change. It is crucial to understand specific heat capacity when calculating energy transfer, as it directly affects how much heat is required for a temperature change during the heat exchange process.

Counter-Current Heat Exchanger

Counter-Current Heat Exchangers are highly effective configurations where two fluids move in opposite directions. This setup allows for more efficient heat transfer compared to co-current flow, where fluids move in the same direction.
In a counter-current heat exchanger, the hottest part of the fluid is in contact with the hottest part of the routed structure. As the fluids traverse in opposite directions, it allows for more extensive temperature change and energy transfer.

• Maximizes heat transfer: due to continuous contact between hot and cold fluids across the heat exchanger length.
• Efficiency: typically higher thermal performance, as it exploits entire temperature gradients along the system.

In the case of our problem, the blood's counter-current flow allows effective heat exchange between arterial and venous blood, optimizing temperature regulation and overall energy exchange.

Overall Heat Transfer Coefficient

The Overall Heat Transfer Coefficient, denoted by \( U \), quantifies the heat transfer capability of a heat exchanger. It encompasses all modes of heat transfer, including conduction, convection, and sometimes radiation, through the material of the exchanger.
The coefficient reflects how well heat transfers across the differing phases or structures involved in the exchanger. In our problem, the \( U \) value is \( 100 \frac{W}{m^2 \cdot K} \), signifying efficient exchange under the given conditions.

• Higher \( U \) values: Indicate better heat transfer efficiency between the fluids across the exchange surface.
• Depends on material properties: Material conductivity, flow velocities, and general heat exchanger design impact \( U \).

Understanding \( U \) is essential because it helps predict the performance of a heat exchanger system, optimizing temperatures and energy costs.