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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

In a fire-tube boiler, hot products of combustion flowing through an array of thin-walled tubes are used to boil water flowing over the tubes. At the time of installation, the overall heat transfer coefficient was \(400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). After 1 year of use, the inner and outer tube surfaces are fouled, with corresponding fouling factors of \(R_{f, i}^{N}=0.0015\) and \(R_{f, w}^{*}=0.0005 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), respectively. Should the boiler be scheduled for cleaning of the tube surfaces?

Saturated process steam at 1 atm is condensed in a shell-and-tube heat exchanger (one shell, two tube passes). Cooling water enters the tubes at \(15^{\circ} \mathrm{C}\) with an average velocity of \(3.5 \mathrm{~m} / \mathrm{s}\). The tubes are thin walled and made of copper with a diameter of \(14 \mathrm{~mm}\) and length of \(0.5 \mathrm{~m}\). The convective heat transfer coefficient for condensation on the outer surface of the tubes is \(21,800 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Find the number of tubes/pass required to condense \(2.3 \mathrm{~kg} / \mathrm{s}\) of steam. (b) Find the outlet water temperature. (c) Find the maximum possible condensation rate that could be achieved with this heat exchanger using the same water flow rate and inlet temperature. (d) Using the heat transfer surface area found in part (a), plot the water outlet temperature and steam condensation rate for water mean velocities in the range from 1 to \(5 \mathrm{~m} / \mathrm{s}\). Assume that the shell-side convection coefficient remains unchanged.

A single-pass, cross-flow heat exchanger with both fluids unmixed is being used to heat water \(\left(m_{c}=2 \mathrm{~kg} / \mathrm{s}\right.\), \(c_{p}=4200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) ) from \(20^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) with hot exhaust gases \(\left(c_{p}=1200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) entering at \(320^{\circ} \mathrm{C}\). What mass flow rate of exhaust gases is required? Assume that UA is equal to its design value of \(4700 \mathrm{~W} / \mathrm{K}\), independent of the gas mass flow rate.

A process fluid having a specific heat of \(3500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and flowing at \(2 \mathrm{~kg} / \mathrm{s}\) is to be cooled from \(80^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) with chilled water, which is supplied at a temperature of \(15^{\circ} \mathrm{C}\) and a flow rate of \(2.5 \mathrm{~kg} / \mathrm{s}\). Assuming an overall heat transfer coefficient of \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), calculate the required heat transfer areas for the following exchanger configurations: (a) parallel flow, (b) counterflow, (c) shell-and-tube, one shell pass and two tube passes, and (d) cross-flow, single pass, both fluids unmixed. Compare the results of your analysis. Your work can be reduced by using IHT.

The condenser of a steam power plant contains \(N=1000\) brass tubes \(\left(k_{\mathrm{t}}=110 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\), each of inner and outer diameters, \(D_{i}=25 \mathrm{~mm}\) and \(D_{o}=\) \(28 \mathrm{~mm}\), respectively. Steam condensation on the outer surfaces of the tubes is characterized by a convection coefficient of \(h_{o}=10,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If cooling water from a large lake is pumped through the condenser tubes at \(m_{c}=400 \mathrm{~kg} / \mathrm{s}\), what is the overall heat transfer coefficient \(U_{o}\) based on the outer surface area of a tube? Properties of the water may be approximated as \(\mu=9.60 \times\) \(10^{-4} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}, k=0.60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\mathrm{Pr}=6.6 .\) (b) If, after extended operation, fouling provides a resistance of \(R_{f, i}^{\prime}=10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), at the inner surface, what is the value of \(U_{o}\) ? (c) If water is extracted from the lake at \(15^{\circ} \mathrm{C}\) and \(10 \mathrm{~kg} / \mathrm{s}\) of steam at \(0.0622\) bars are to be condensed, what is the corresponding temperature of the water leaving the condenser? The specific heat of the water is \(4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).
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