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Chapter 11: Problem 16

The hot and cold inlet temperatures to a concentric tube heat exchanger are \(T_{h i}=200^{\circ} \mathrm{C}, T_{c, i}=100^{\circ} \mathrm{C}\), respectively. The outlet temperatures are \(T_{k, o}=110^{\circ} \mathrm{C}\) and \(T_{\omega_{0}}=125^{\circ} \mathrm{C}\). Is the heat exchanger operating in a parallel flow or in a counterflow configuration? What is the heat exchanger effectiveness? What is the NTU? Phase change does not occur in either fluid.

Short Answer

Expert verified
The given temperature values indicate that the heat exchanger is operating in a counterflow configuration. The heat exchanger effectiveness is approximately 0.588, and the NTU (Number of Transfer Units) is approximately 0.883.

Step by step solution

01

Identify the flow configuration

The given temperatures are: - Hot fluid inlet temperature: \(T_{h,i} = 200^{\circ}C\) - Cold fluid inlet temperature: \(T_{c,i} = 100^{\circ}C\) - Hot fluid outlet temperature: \(T_{h,o} = 110^{\circ}C\) - Cold fluid outlet temperature: \(T_{c,o} = 125^{\circ}C\) Since the hot fluid temperature decreases from the inlet to the outlet, and the cold fluid temperature increases from the inlet to the outlet, we can deduce that it is a counterflow configuration. In a parallel flow configuration, the hot fluid temperature would decrease and the cold fluid temperature would increase, but they would always maintain an order where \(T_{h,i} > T_{h,o} > T_{c,o} > T_{c,i}\). In this case, that order is not maintained, so it must be a counterflow configuration.
02

Determine heat capacities of hot and cold fluids

To calculate the heat capacities, we will use the equation, \(Q = mc\Delta T\), where Q is the heat transferred, m is the mass flow rate, c is the specific heat capacity, and \(\Delta T\) is the temperature difference. Since there is no phase change in either fluid, the heat gained by the cold fluid equals the heat lost by the hot fluid. \(m_h c_h (T_{h,i} - T_{h,o}) = m_c c_c (T_{c,o} - T_{c,i})\) We will use this equation later, when calculating the effectiveness and NTU.
03

Calculate heat exchanger effectiveness

For a counterflow heat exchanger, the effectiveness can be calculated using the formula: \(\varepsilon = \frac{T_{h,o} - T_{h,i}}{T_{c,i} (1 - \frac{T_{h,i} - T_{h,o}}{T_{c,o} - T_{c,i}})}\) Plugging in the known values: \(\varepsilon = \frac{110 - 200}{100 (1 - \frac{200 - 110}{125 - 100})} \approx 0.588\) The heat exchanger effectiveness in this configuration is approximately 0.588.
04

Calculate NTU

To calculate the NTU, we use the following formula, which relates effectiveness, NTU, and the ratio of heat capacities, \(C_r = \frac{C_s}{C_L}\), where \(C_s\) is the smaller heat capacity and \(C_L\) is the larger heat capacity: \(\varepsilon = \frac{1 - e^{(-NTU(1 - C_r))}}{1 - C_re^{(-NTU(1 - C_r))}}\) First, we will calculate \(C_r\). From step 2, we know: \(C_s = m_s c_s = m_h c_h (T_{h,i} - T_{h,o}) = m_c c_c (T_{c,o} - T_{c,i}) = C_L\) \(C_r = \frac{C_s}{C_L} = 1\) Now, we can use the effectiveness formula and our known values to calculate NTU: \(0.588 = \frac{1 - e^{(-NTU(1 - 1))}}{1 - e^{(-NTU(1 - 1))}}\) Since \(C_r = 1\), the effectiveness equation simplifies to: \(\varepsilon = 1 - e^{-NTU}\) We can now solve for NTU: \(e^{-NTU} = 1 - 0.588\) \(-NTU = \ln(1 - 0.588)\) \(NTU \approx 0.883\) The NTU (Number of Transfer Units) for this heat exchanger is approximately 0.883.

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Key Concepts

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

Concentric Tube Heat Exchanger
A concentric tube heat exchanger is a type of heat exchanger where two fluids flow through concentric, or coaxial, tubes. One fluid flows through the inner tube, while the other fluid flows through the annular space between the inner and outer tubes. This design allows for efficient heat transfer between the two fluids.

In a concentric tube heat exchanger, the construction can be simple and cost-effective, making it a popular choice in industry. It is also flexible in handling various flow arrangements, such as parallel flow and counterflow. The temperature changes of both fluids provide a simple visual representation to determine the flow arrangement. If properly designed, these heat exchangers can have a high thermal efficiency.
  • The inner and outer tubes can differ in materials to withstand different pressure and temperature conditions.
  • Flow direction can be adjusted based on the required process, serving specific industrial needs.
By allowing the fluids to either move in the same direction (parallel flow) or opposite directions (counterflow), the concentric tube heat exchanger can be adapted to different heat transfer scenarios. In the context of the exercise, the heat exchanger operates in a counterflow configuration, where the fluids move in opposite directions.
Counterflow Heat Exchanger
Counterflow heat exchangers are a configuration where the two fluids move in opposite directions. This setup is particularly advantageous for maximizing the temperature difference across the exchanger's length, leading to more efficient heat transfer. By having the fluids flow counter to each other, the exit temperature of the cold fluid can be higher than the outlet temperature of the hot fluid, as seen in the problem scenario with outlet temperatures at 125掳C for cold fluid and 110掳C for hot fluid.

This arrangement creates the largest possible temperature difference between the incoming and outgoing fluids, optimizing the heat exchanged. The counterflow configuration can achieve higher efficiency compared to a parallel flow configuration. This is because the average temperature difference between the fluids across the length of the exchanger is greater.
  • Counterflow ensures the maximum possible heat transfer efficiency.
  • This configuration helps in achieving closer approach temperatures.
Certain industries prefer counterflow arrangements to achieve higher thermal efficiency, despite the slightly more complex design requirements. The exercise highlights the superior performance of counterflow, as evidenced by the calculated heat exchanger effectiveness.
Number of Transfer Units (NTU)
The Number of Transfer Units (NTU) is a dimensionless parameter in heat exchanger design. It is a critical measure of a heat exchanger's capacity to transfer heat relative to the heat capacity rate of the fluids involved. NTU is significant because it relates directly to the heat exchanger effectiveness, providing insight into how well the exchanger performs.

NTU can be defined mathematically as follows: NTU = \( \frac{UA}{C_{min}} \),
where
  • \(U\) is the overall heat transfer coefficient,
  • \(A\) is the heat transfer surface area,
  • \(C_{min}\) is the lower heat capacity rate of the two fluids.
This formula shows how NTU involves the effectiveness and the heat capacity ratio. For this specific exercise, the NTU was calculated to be approximately 0.883. This value indicates that the heat exchanger is moderately effective in transferring heat. The effectiveness and NTU have a direct relation; a higher NTU denotes a higher effectiveness in theory, suggesting a better thermal performance.

In practical applications, adjusting NTU determines the required size and design specifics of heat exchangers, significantly impacting cost and efficiency.

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Most popular questions from this chapter

A shell-and-tube heat exchanger is to heat \(10,000 \mathrm{~kg} / \mathrm{h}\) of water from 16 to \(84^{\circ} \mathrm{C}\) by hot engine oil flowing through the shell. The oil makes a single shell pass, entering at \(160^{\circ} \mathrm{C}\) and leaving at \(94^{\circ} \mathrm{C}\), with an average heat transfer coefficient of \(400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The water flows through 11 brass tubes of \(22.9-\mathrm{mm}\) inside diameter and 25.4-mm outside diameter, with each tube making four passes through the shell. (a) Assuming fully developed flow for the water, determine the required tube length per pass. (b) For the tube length found in part (a), plot the effectiveness, fluid outlet temperatures, and water-side convection coefficient as a function of the water flow rate for \(5000 \leq m_{c} \leq 15,000 \mathrm{~kg} / \mathrm{h}\), with all other conditions remaining the same.

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?

Water at a rate of \(45,500 \mathrm{~kg} / \mathrm{h}\) is heated from 80 to \(150^{\circ} \mathrm{C}\) in a heat exchanger having two shell passes and eight tube passes with a total surface area of \(925 \mathrm{~m}^{2}\). Hot exhaust gases having approximately the same thermophysical properties as air enter at \(350^{\circ} \mathrm{C}\) and exit at \(175^{\circ} \mathrm{C}\). Determine the overall heat transfer coefficient.

An automobile radiator may be viewed as a cross-flow heat exchanger with both fluids unmixed. Water, which has a flow rate of \(0.05 \mathrm{~kg} / \mathrm{s}\), enters the radiator at \(400 \mathrm{~K}\) and is to leave at \(330 \mathrm{~K}\). The water is cooled by air that enters at \(0.75 \mathrm{~kg} / \mathrm{s}\) and \(300 \mathrm{~K}\). (a) If the overall heat transfer coefficient is \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the required heat transfer surface area? (b) A manufacturing engineer claims ridges can be stamped on the finned surface of the exchanger, which could greatly increase the overall heat transfer coefficient. With all other conditions remaining the same and the heat transfer surface area determined from part (a), generate a plot of the air and water outlet temperatures as a function of \(U\) for \(200 \leq U \leq 400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What benefits result from increasing the overall convection coefficient for this application?

A shell-and-tube exchanger (two shells, four tube passes) is used to heat \(10,000 \mathrm{~kg} / \mathrm{h}\) of pressurized water from 35 to \(120^{\circ} \mathrm{C}\) with \(5000 \mathrm{~kg} / \mathrm{h}\) pressurized water entering the exchanger at \(300^{\circ} \mathrm{C}\). If the overall heat transfer coefficient is \(1500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the required heat exchanger area.
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