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Chapter 11: Problem 83

In open heart surgery under hypothermic conditions, the patient's blood is cooled before the surgery and rewarmed afterward. It is proposed that a concentric tube, counterflow heat exchanger of length \(0.5 \mathrm{~m}\) be used for this purpose, with the thin-walled inner tube having a diameter of \(55 \mathrm{~mm}\). The specific heat of the blood is \(3500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). (a) If water at \(T_{h j}=60^{\circ} \mathrm{C}\) and \(\dot{m}_{h}=0.10 \mathrm{~kg} / \mathrm{s}\) is used to heat blood entering the exchanger at \(T_{c A}=18^{\circ} \mathrm{C}\) and \(\dot{m}_{c}=0.05 \mathrm{~kg} / \mathrm{s}\), what is the temperature of the blood leaving the exchanger? The overall heat transfer coefficient is \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) The surgeon may wish to control the heat rate \(q\) and the outlet temperature \(T_{c, 0}\) of the blood by altering the flow rate and/or inlet temperature of the water during the rewarming process. To assist in the development of an appropriate controller for the prescribed values of \(\hat{m}_{c}\) and \(T_{c \jmath}\), compute and plot \(q\) and \(T_{c, \rho}\) as a function of \(\dot{m}_{h}\) for \(0.05 \leq \dot{m}_{\mathrm{h}} \leq 0.20 \mathrm{~kg} / \mathrm{s}\) and values of \(T_{h, l}=50,60\), and \(70^{\circ} \mathrm{C}\). Since the dominant influence on the overall heat transfer coefficient is associated with the blood flow conditions, the value of \(U\) may be assumed to remain at \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Should certain operating conditions be excluded?

Short Answer

Expert verified
The temperature of the blood leaving the exchanger is \(T_{c,out}\) and can be calculated using the energy balance and heat transfer equations. For part (b), the results are computed and plotted as a function of the mass flow rate of water, \(\dot{m}_h\), for different values of \(T_{h,in}\). Based on the plotted graph, analyze the results and determine if any operating conditions should be excluded to prevent any safety issues or inefficiencies in the process.

Step by step solution

01

Energy Balance

For a counterflow heat exchanger, we will use the energy balance equation given by: \(q = \dot{m}_c c_{p,c} (T_{c,out} - T_{c,in}) = \dot{m}_h c_{p,h} (T_{h,in} - T_{h,out})\) where \(q\) is the heat transfer rate, \(\dot{m}_c\) and \(\dot{m}_h\) are the mass flow rates of blood and water respectively, \(c_{p,c}\) and \(c_{p,h}\) are their specific heats, and \(T_{c,in}\), \(T_{c,out}\), \(T_{h,in}\), \(T_{h,out}\) are the inlet and outlet temperatures of blood and water. Step 2: Using heat transfer equation
02

Heat Transfer Equation

The heat transfer equation for this heat exchanger is given by: \(q=UA\Delta T_{lm}\) where \(U\) is the overall heat transfer coefficient, \(A\) is the heat transfer area, and \(\Delta T_{lm}\) is the logarithmic mean temperature difference, which can be calculated as: \(\Delta T_{lm}=\frac{\Delta T_1 - \Delta T_2}{\ln(\frac{\Delta T_1}{\Delta T_2})}\) where \(\Delta T_1 = T_{h,in} - T_{c,in}\) and \(\Delta T_2 = T_{h,out} - T_{c,out}\) Step 3: Calculate the heat transfer area A
03

Calculate A

Since it is a concentric tube heat exchanger, we can calculate A using the following formula: \(A=2\pi r L\) where \(r=\frac{D_t}{2}\), \(D_t\) is the diameter of the tube, and \(L\) is the length of the exchanger. For the given problem, \(D_t = 0.055 \mathrm{m}\) and \(L= 0.5 \mathrm{m}\), so we can calculate A. Step 4: Calculate the temperature of blood leaving the exchanger
04

Calculate Tc,out

We can now find the temperature of the blood leaving the exchanger \(T_{c,out}\). First, find the heat transfer rate q using the energy balance equation. Then substitute q into the heat transfer equation and solve for \(T_{c,out}\) using the given input parameters. Part (b): Step 1: Calculate q and Tc,out as a function of mh
05

Calculate q and Tc,out for different mh

For this part, compute and plot \(q\) and \(T_{c,out}\) as a function of \(\dot{m}_{h}\) for \(0.05 \leq \dot{m}_{\mathrm{h}} \leq 0.20 \mathrm{~kg} / \mathrm{s}\) and values of \(T_{h,in}=50,60\), and \(70^{\circ} \mathrm{C}\). Repeat the steps from part (a) for each case and plot the results as a function of the mass flow rate of water, \(\dot{m}_h\). Step 2: Determine the operating conditions
06

Find the operating conditions

Based on the plotted graph, analyze the results and determine if any operating conditions should be excluded to prevent any safety issues or inefficiencies in the process.

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Key Concepts

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

Counterflow Heat Exchanger

Understanding the workings of a counterflow heat exchanger is fundamental to analyzing systems where heat is exchanged between two fluids moving in opposite directions. The counterflow design is instrumental in various applications due to its efficiency in transferring heat. Why is it so effective? Primarily because the temperature gradient between the two fluids remains high over the length of the exchanger, maximizing the potential for heat transfer.

In the given exercise, the heat exchanger plays a crucial role in controlling the temperature of the patient's blood during surgery. A concentric tube forms the core of this counterflow heat exchanger, with water flowing through the outer tube to heat or cool the blood running in the opposite direction inside the inner tube. The design specifications, such as the length and diameter of the tubes, directly impact the heat exchange efficacy.

  • The opposing flow direction ensures a high temperature differential between the fluids at each point along the exchanger.
  • Efficient heat exchange is critical, particularly in medical procedures like open-heart surgery where the patient's blood must be precisely cooled and rewarmed.
  • The concentric tube design aids in maintaining a compact form factor while ensuring effective heat transfer.

By adjusting the mass flow rates and inlet temperatures of the fluids, the surgeon can fine-tune the heat exchange process to achieve desired blood temperatures.

Logarithmic Mean Temperature Difference

The logarithmic mean temperature difference (LMTD) is a concept that lies at the heart of heat exchanger analysis. It offers a measure of the average temperature difference between the two fluids across the heat exchanger. LMTD is crucial because it reflects the varying temperature difference along the length of the exchanger in a way that a simple arithmetic difference cannot. It accounts for the logarithmic relationship between the heat transfer and temperature difference.

In the provided exercise, calculating the LMTD was essential to find the outlet temperature of the blood after it had been warmed. Since blood temperature has to be carefully controlled, understanding and applying LMTD ensures more accurate and effective thermal management.

  • LMTD incorporates the changing temperature difference between the hot and cold fluids, reflecting the true driving force for heat exchange.
  • It is calculated using the temperatures of both fluids at the inlet and outlet of the heat exchanger.
  • Understanding LMTD allows surgeons to manipulate the heat exchanger parameters more effectively, ensuring patient safety during surgeries.

In clinical settings, precise control over the patient's blood temperature is necessary, and the LMTD method provides a reliable foundation for such regulation.

Heat Transfer Rate

The heat transfer rate is a quantitative expression of the amount of thermal energy that is transferred per unit of time. In any heat exchanger, including the counterflow type described in the exercise, this rate is dictated by several factors, among which are the properties of the fluids, the flow rate, the surface area for heat exchange, and the overall heat transfer coefficient.

For the hypothetical medical scenario, the heat transfer rate is crucial. It determines how quickly the patient's blood can be cooled or warmed to the required temperature. Adequate control over this rate ensures safety and effectiveness of the surgical procedure. By applying the heat transfer rate equation, the surgeon or technician can attain optimal blood temperature conditions.

  • Both the mass flow rate and the specific heat of the blood are significant in calculating the heat transfer rate.
  • Adjusting the heat transfer rate leads to direct changes in blood's outlet temperature, beneficial for the patient's wellbeing during surgical procedures.
  • A precise heat transfer rate calculation allows for the fine-tuning required in designing an appropriate control system for the heat exchanger.

The plot of heat transfer rate against water mass flow rate and inlet temperature, as mentioned in the exercise, aids in visualizing the impact of these parameters and assists in determining optimal operating conditions to prevent safety issues or inefficiencies.

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Most popular questions from this chapter

As part of a senior project, a student was given the assignment to design a heat exchanger that meets the following specifications: \begin{tabular}{lccc} \hline & \(\dot{m}(\mathrm{~kg} / \mathrm{s})\) & \(T_{m, i}\left({ }^{\circ} \mathrm{C}\right)\) & \(T_{m, \theta}\left({ }^{\circ} \mathrm{C}\right)\) \\ \hline Hot water & 28 & 90 & \(-\) \\ Cold water & 27 & 34 & 60 \\ \hline \end{tabular} Like many real-world situations, the customer hasn't revealed, or doesn't know, additional requirements that would allow you to proceed directly to a final configuration. At the outset, it is helpful to make a first-cut design based upon simplifying assumptions, which can be evaluated to determine what additional requirements and trade-offs should be considered by the customer. (a) Design a heat exchanger to meet the foregoing specifications. List and explain your assumptions. Hint: Begin by finding the required value for \(U A\) and using representative values of \(U\) to determine \(A\). (b) Evaluate your design by identifying what features and configurations could be explored with your customer in order to develop more complete specifications.

A novel design for a condenser consists of a tube of thermal conductivity \(200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) with longitudinal fins snugly fitted into a larger tube. Condensing refrigerant at \(45^{\circ} \mathrm{C}\) flows axially through the inner tube, while water at a flow rate of \(0.012 \mathrm{~kg} / \mathrm{s}\) passes through the six channels around the inner tube. The pertinent diameters are \(D_{1}=10 \mathrm{~mm}, D_{2}=14 \mathrm{~mm}\), and \(D_{3}=50 \mathrm{~mm}\), while the fin thickness is \(t=2 \mathrm{~mm}\). Assume that the convection coefficient associated with the condensing refrigerant is extremely large. Determine the heat removal rate per unit tube length in a section of the tube for which the water is at \(15^{\circ} \mathrm{C}\).

A two-fluid heat exchanger has inlet and outlet temperatures of 65 and \(40^{\circ} \mathrm{C}\) for the hot fluid and 15 and \(30^{\circ} \mathrm{C}\) for the cold fluid. Can you tell whether this exchanger is operating under counterflow or parallelflow conditions? Determine the effectiveness of the heat exchanger.

Consider Problem 11.36. (a) For \(\dot{m}_{c \mathrm{CA}}=\dot{m}_{\mathrm{h}, \mathrm{B}}=10 \mathrm{~kg} / \mathrm{s}\), determine the outlet air and ammonia temperatures, as well as the heat transfer rate. (b) Plot the outlet air and outlet ammonia temperatures versus the water flow rate over the range \(5 \mathrm{~kg} / \mathrm{s} \leq \dot{m}_{c, \mathrm{~A}}=m_{h, \mathrm{~B}} \leq 50 \mathrm{~kg} / \mathrm{s}\).

The oil in an engine is cooled by air in a cross-flow heat exchanger where both fluids are unmixed. Atmospheric air enters at \(30^{\circ} \mathrm{C}\) and \(0.53 \mathrm{~kg} / \mathrm{s}\). Oil at \(0.026 \mathrm{~kg} / \mathrm{s}\) enters at \(75^{\circ} \mathrm{C}\) and flows through a tube of 10-mm diameter. Assuming fully developed flow and constant wall heat flux, estimate the oil- side heat transfer coefficient. If the overall convection coefficient is \(53 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the total heat transfer area is \(1 \mathrm{~m}^{2}\), determine the effectiveness. What is the exit temperature of the oil?
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