The cardiovascular counter-current 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 \(125 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). 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

Expert verified
Answer: The required length of the blood vessels to warm up the venous blood to 35°C is 0.071 meters (or 7.1 cm).

Step by step solution

01

Determine the mass flow rates in kg/s and temperature difference

First, we need to convert the mass flow rates from g/s to kg/s: Mass flow rate of venous blood: \(2 \mathrm{~g/s} = 0.002 \mathrm{~kg/s}\) Mass flow rate of arterial blood: \(5 \mathrm{~g/s} = 0.005 \mathrm{~kg/s}\) Next, we find the temperature difference between the venous and arterial blood: Temperature difference: \(37^{\circ} \mathrm{C} - 35^{\circ} \mathrm{C} = 2 \mathrm{K}\)
02

Calculate the heat transfer rate

Using the specific heat capacity and mass flow rates, we can find the heat transfer rate (Q) with the formula: \(Q = m \cdot c \cdot \Delta T\) where m is the mass flow rate, c is the specific heat capacity, and \(\Delta T\) is the temperature difference. \(Q = 0.002 \mathrm{~kg/s} \cdot 3475 \mathrm{~J/kg \cdot K} \cdot 2 \mathrm{K} = 13.9 \mathrm{~W}\)
03

Calculate the area of the blood vessels' wall

To find the area of the blood vessel wall, we need to use the diameter of the vein (D) given as 5 cm, which we first convert to meters: Diameter of the vein: \(5 \mathrm{~cm} = 0.05 \mathrm{~m}\) The area (A) can be calculated using the formula for the surface area of a cylinder: \(A = 2 \cdot \pi \cdot r \cdot L\) where r is the radius (half of the diameter) and L is the length of the blood vessel we want to find. We will rearrange the formula to solve for L later.
04

Calculate the required length of the blood vessels using the overall heat transfer coefficient

To determine the required length of the blood vessels, we can use the overall heat transfer equation: \(Q = U \cdot A \cdot \Delta T\) where U is the overall heat transfer coefficient. We can rearrange this formula to solve for L as follows: \(L = \frac{Q}{U \cdot (\pi \cdot D \cdot \Delta T)}\) Now, we can plug in the values and calculate the required length: \(L = \frac{13.9 \mathrm{~W}}{125 \mathrm{~W/m^{2} \cdot K} \cdot (\pi \cdot 0.05 \mathrm{~m} \cdot 2 \mathrm{K})} = 0.071 \mathrm{~m}\) Therefore, the required length of the blood vessels to warm up the venous blood to \(35^{\circ} \mathrm{C}\) is 0.071 meters (or 7.1 cm).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Transfer Rate
Understanding the heat transfer rate is crucial when examining thermal systems, such as the counter-current heat exchanger found in the cardiovascular system. Essentially, the heat transfer rate, denoted as Q, represents the amount of heat energy transferred per unit of time. It is measured in Watts, where one Watt is equivalent to one Joule per second.

For our application, the formula to calculate the heat transfer rate is:
\[Q = m \cdot c \cdot \Delta T\]
where:
  • \(m\) stands for the mass flow rate of the fluid (in this case, blood), measured in kilograms per second (kg/s).
  • \(c\) is the specific heat capacity, an intrinsic property of the substance that indicates how much heat energy is required to raise the temperature of one kilogram of the substance by one degree Kelvin (J/kg·K).
  • \(\Delta T\) represents the temperature change the fluid undergoes in degrees Kelvin (K).
The result, in our cardiovascular example, involves the transfer of heat from the arterial blood to the venous blood, in order to warm the latter to a desired temperature. This transfer rate gives us insight into the effectiveness of the heat exchanger and also determines the required dimensions of the system, like the length of the blood vessels needed to achieve the temperature change.
Overall Heat Transfer Coefficient
The overall heat transfer coefficient, denoted by U, is a measure of a heat exchanger's ability to conduct heat from one fluid to the other. In particular, it quantifies the energy transferred per unit area per unit temperature difference and is expressed in Watts per square meter-Kelvin (W/m²·K).

For a given system, U is influenced by several factors including the materials of the heat exchanger, the nature of the fluids involved, and the flow dynamics. The higher the U value, the more efficient the heat exchanger at transferring heat.
To connect this to our counter-current heat exchanger exercise, the formula:\[Q = U \cdot A \cdot \Delta T\]highlights that the heat transfer rate (Q) is directly proportional to both the overall heat transfer coefficient (U) and the temperature difference (\(\Delta T\)). By rearranging the formula to solve for the blood vessels' length (L), we tie in the physical dimensions of our system, allowing us to design a heat exchanger that meets the required thermal performance.
Specific Heat Capacity
The specific heat capacity (c) plays a pivotal role in thermodynamics and heat transfer exercises. It is defined as the amount of heat energy required to increase the temperature of a unit mass of a substance by one degree in temperature. Its unit of measurement is Joules per kilogram-Kelvin (J/kg·K).

In the human body, blood has a specific heat capacity, meaning it requires a certain amount of heat to change its temperature. The constancy and equality of specific heat capacities of arterial and venous blood, as assumed in our example, simplifies the calculations significantly. It assures that the same amount of heat energy will cause the same temperature change in a given mass of either type of blood.
When calculating the length of the blood vessels required to achieve temperature regulation in the counter-current heat exchanger, the constant value of the specific heat of blood ensures uniformity in the heat absorption or release process as temperatures adjust, leading to predictable and reliable outcomes for the biological system.

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Most popular questions from this chapter

Consider a shell and tube heat exchanger in a milk be heated from \(20^{\circ} \mathrm{C}\) by hot water initially at \(140^{\circ} \mathrm{C}\) and flowing at a rate of \(5 \mathrm{~kg} / \mathrm{s}\). The milk flows through 30 thin-walled tubes with an inside diameter of \(20 \mathrm{~mm}\) with each tube making 10 passes through the shell. The average convective heat transfer coefficients on the milk and water side are \(450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(1100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. In order to complete the pasteurizing process and hence restrict the microbial growth in the milk, it is required to have the exit temperature of milk attain at least \(70^{\circ} \mathrm{C}\). As a design engineer, your job is to decide upon the shell width (tube length in each pass) so that the milk exit temperature of \(70^{\circ} \mathrm{C}\) can be achieved. One of the design requirements is that the exit temperature of hot water should be at least \(10^{\circ} \mathrm{C}\) higher than the exit temperature of milk.

Oil is being cooled from \(180^{\circ} \mathrm{F}\) to \(120^{\circ} \mathrm{F}\) in a 1 -shell and 2-tube heat exchanger with an overall heat transfer coefficient of \(40 \mathrm{Btu} / \mathrm{h} \mathrm{ft} 2{ }^{\circ} \mathrm{F}\). Water \(\left(c_{p c}=1.0 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) enters at \(80^{\circ} \mathrm{F}\) and exits at \(100^{\circ} \mathrm{F}\) with a mass flow rate of \(20,000 \mathrm{lbm} / \mathrm{h}\), determine (a) the log mean temperature difference and \((b)\) the surface area of the heat exchanger.

Hot water coming from the engine is to be cooled by ambient air in a car radiator. The aluminum tubes in which the water flows have a diameter of \(4 \mathrm{~cm}\) and negligible thickness. Fins are attached on the outer surface of the tubes in order to increase the heat transfer surface area on the air side. The heat transfer coefficients on the inner and outer surfaces are 2000 and \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. If the effective surface area on the finned side is 10 times the inner surface area, the overall heat transfer coefficient of this heat exchanger based on the inner surface area is (a) \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(857 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(1075 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(2150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Explain how you can evaluate the outlet temperatures of the cold and hot fluids in a heat exchanger after its effectiveness is determined.

Consider a water-to-water double-pipe heat exchanger whose flow arrangement is not known. The temperature measurements indicate that the cold water enters at \(20^{\circ} \mathrm{C}\) and leaves at \(50^{\circ} \mathrm{C}\), while the hot water enters at \(80^{\circ} \mathrm{C}\) and leaves at \(45^{\circ} \mathrm{C}\). Do you think this is a parallel-flow or counterflow heat exchanger? Explain.

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