Question

The cardiovascular countercurrent heat exchanger mechanism is to warm venous blood from \(28^{\circ} \mathrm{C}\) to \(35^{\circ} \mathrm{C}\) at a mass flow rate of \(2 \mathrm{~g} / \mathrm{s}\). The artery inflow temperature is \(37^{\circ} \mathrm{C}\) at a mass flow rate of \(5 \mathrm{~g} / \mathrm{s}\). The average diameter of the vein is \(5 \mathrm{~cm}\) and the overall heat transfer coefficient is \(125 \mathrm{~W} / \mathrm{m}^{2}\). K. Determine the overall blood vessel length needed to warm the venous blood to $35^{\circ} \mathrm{C}$ if the specific heat of both arterial and venous blood is constant and equal to \(3475 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

Short answer

Answer: The length of the blood vessel needed to warm the venous blood to 35 degrees Celsius is approximately 0.525 meters (52.5 cm).

Step-by-step solution

Write down the known values

Here are the given values: Venous blood initial temperature: \(T_{v_i} = 28^{\circ} \mathrm{C}\) Venous blood final temperature: \(T_{v_f} = 35^{\circ} \mathrm{C}\) Arterial blood inflow temperature: \(T_{a} = 37^{\circ} \mathrm{C}\) Mass flow rate of venous blood: \(m_v = 2 \mathrm{~g} / \mathrm{s}\) Mass flow rate of arterial blood: \(m_a = 5 \mathrm{~g} / \mathrm{s}\) Blood vessel diameter: \(D = 5 \mathrm{~cm}\) Heat transfer coefficient: \(U = 125 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) Specific heat of blood: \(c_p = 3475 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\)

Convert units to SI units

Change mass flow rate from grams per second to kg per second: \(m_v = 0.002 \mathrm{~kg} / \mathrm{s}\) \(m_a = 0.005 \mathrm{~kg} / \mathrm{s}\) Change blood vessel diameter from cm to meters: \(D = 0.05 \mathrm{~m}\)

Write down the energy balance equation and calculate the heat transfer

The energy balance equation for heat transfer between venous and arterial blood is: \(Q_v = Q_a\) We can write the heat transfer equations for venous and arterial blood as: \(Q_v = m_v \cdot c_p \cdot (T_{v_f} - T_{v_i})\) \(Q_a = m_a \cdot c_p \cdot (T_{a} - T_{v_f})\) Equalizing them, we get: \(m_v \cdot c_p \cdot (T_{v_f} - T_{v_i}) = m_a \cdot c_p \cdot (T_{a} - T_{v_f})\)

Solve for unknown temperature T_v_f

Next, we will solve this equation for \(T_{v_f}\): \(T_{v_f} = \frac{m_v \cdot (T_{v_i} - T_{v_f})}{m_a - m_v} + T_a\) Now plug in the known values: \(T_{v_f} = \frac{0.002 \cdot (28 - 35)}{0.005 - 0.002} + 37\) \(T_{v_f} = 35^{\circ} \mathrm{C}\)

Calculate heat transfer and the length of the blood vessel

Now we have the heat transfer for both venous and arterial blood. We can use the heat transfer formula for convection: \(Q = U \cdot A \cdot \Delta T\) Where \(A\) is the surface area of the blood vessel, which can be determined by: \(A = \pi \cdot D \cdot L\) Where \(L\) is the length of the blood vessel. Plugging in the values, we get: \(Q_v = 125 \cdot (\pi \cdot 0.05 \cdot L) \cdot (35 - 28)\) Now, we will solve for the length of the blood vessel: \(L = \frac{Q_v}{125 \cdot \pi \cdot 0.05 \cdot 7}\) Finally, we will substitute \(Q_v\) from the previously equalized heat transfer formula and find \(L\): \(L = \frac{0.002 \cdot 3475 \cdot (35 - 28)}{125 \cdot \pi \cdot 0.05 \cdot 7}\) \(L \approx 0.525 \mathrm{~m}\) The overall blood vessel length needed to warm the venous blood to 35 degrees Celsius is approximately 0.525 meters (52.5 cm).