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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

In a chemical plant, a certain chemical is heated by hot water supplied by a natural gas furnace. The hot water \(\left(c_{p}=\right.\) \(4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K})\) is then discharged at \(60^{\circ} \mathrm{C}\) at a rate of \(8 \mathrm{~kg} / \mathrm{min}\). The plant operates \(8 \mathrm{~h}\) a day, 5 days a week, 52 weeks a year. The furnace has an efficiency of 78 percent, and the cost of the natural gas is \(\$ 1.00\) per therm ( 1 therm \(=105,500 \mathrm{~kJ})\). The average temperature of the cold water entering the furnace throughout the year is \(14^{\circ} \mathrm{C}\). In order to save energy, it is proposed to install a water-to-water heat exchanger to preheat the incoming cold water by the drained hot water. Assuming that the heat exchanger will recover 72 percent of the available heat in the hot water, determine the heat transfer rating of the heat exchanger that needs to be purchased and suggest a suitable type. Also, determine the amount of money this heat exchanger will save the company per year from natural gas savings.

Can the temperature of the hot fluid drop below the inlet temperature of the cold fluid at any location in a heat exchanger? Explain.

Oil in an engine is being cooled by air in a crossflow heat exchanger, where both fluids are unmixed. Oil \(\left(c_{p h}=\right.\) \(2047 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K})\) flowing with a flow rate of \(0.026 \mathrm{~kg} / \mathrm{s}\) enters the tube side at \(75^{\circ} \mathrm{C}\), while air \(\left(c_{p c}=1007 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the shell side at \(30^{\circ} \mathrm{C}\) with a flow rate of \(0.21 \mathrm{~kg} / \mathrm{s}\). The overall heat transfer coefficient of the heat exchanger is \(53 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the total surface area is \(1 \mathrm{~m}^{2}\). If the correction factor is \(F=\) \(0.96\), determine the outlet temperatures of the oil and air.

Saturated water vapor at \(100^{\circ} \mathrm{C}\) condenses in the shell side of a 1 -shell and 2-tube heat exchanger with a surface area of \(0.5 \mathrm{~m}^{2}\) and an overall heat transfer coefficient of \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Cold water \(\left(c_{p c}=4179 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) flowing at \(0.5 \mathrm{~kg} / \mathrm{s}\) enters the tube side at \(15^{\circ} \mathrm{C}\), determine the outlet temperature of the cold water and the heat transfer rate for the heat exchanger.

Consider a recuperative cross flow heat exchanger (both fluids unmixed) used in a gas turbine system that carries the exhaust gases at a flow rate of \(7.5 \mathrm{~kg} / \mathrm{s}\) and a temperature of \(500^{\circ} \mathrm{C}\). The air initially at \(30^{\circ} \mathrm{C}\) and flowing at a rate of \(15 \mathrm{~kg} / \mathrm{s}\) is to be heated in the recuperator. The convective heat transfer coefficients on the exhaust gas and air side are \(750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(300 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Due to long term use of the gas turbine the recuperative heat exchanger is subject to fouling on both gas and air side that offers a resistance of \(0.0004 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) each. Take the properties of exhaust gas to be the same as that of air \(\left(c_{p}=1069 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\). If the exit temperature of the exhaust gas is \(320^{\circ} \mathrm{C}\) determine \((a)\) if the air could be heated to a temperature of \(150^{\circ} \mathrm{C}(b)\) the area of heat exchanger \((c)\) if the answer to part (a) is no, then determine what should be the air mass flow rate in order to attain the desired exit temperature of \(150^{\circ} \mathrm{C}\) and \((d)\) plot variation of the exit air temperature over a temperature range of \(75^{\circ} \mathrm{C}\) to \(300^{\circ} \mathrm{C}\) with air mass flow rate assuming all the other conditions remain the same.

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