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Chapter 11: Problem 47

Consider a very long, concentric tube heat exchanger having hot and cold water inlet temperatures of 85 and \(15^{\circ} \mathrm{C}\). The flow rate of the hot water is twice that of the cold water. Assuming equivalent hot and cold water specific heats, determine the hot water outlet temperature for the following modes of operation: (a) counterflow and (b) parallel flow.

Short Answer

Expert verified
The hot water outlet temperatures for the concentric tube heat exchanger are: a) Counterflow: \( T_{out\_hot} = 55^{\circ} \mathrm{C} \) b) Parallel flow: \( T_{out\_hot} = 60^{\circ} \mathrm{C} \)

Step by step solution

01

Define heat capacity rate and establish equations

Heat capacity rate (HCR) is given by the equation: HCR = mass flow rate 脳 specific heat Since we're assuming equivalent hot and cold water specific heats, the specific heats cancel out when comparing the hot and cold HCRs. In this case, since the hot water flow rate is twice that of the cold water, the HCR of hot water will be twice that of the cold water as well: HCR_hot = 2 脳 HCR_cold Now, let's define the temperature changes for the hot and cold streams. During heat exchange, the temperature change of the hot stream (鈭員_hot) is negative since it cools down, and the temperature change of the cold stream (鈭員_cold) is positive since it heats up: 鈭員_hot = T_out_hot - T_in_hot 鈭員_cold = T_out_cold - T_in_cold Applying the equation for heat transfer: Q_hot = Q_cold Q_hot = HCR_hot 脳 (鈭員_hot) Q_cold = HCR_cold 脳 (鈭員_cold) Now we'll create separate equations for counterflow and parallel flow cases.
02

Apply heat transfer equations for counterflow and parallel flow

For counterflow: The temperature difference across the heat exchanger for the cold stream is greater due to the opposite flow arrangement. 鈭員_counterflow = T_in_hot - T_in_cold For parallel flow: The temperature difference between hot and cold streams decreases along the heat exchanger, so we use the log mean temperature difference to obtain an average difference. 鈭員_parallel = (T_in_hot - T_in_cold) / ln((T_in_hot - T_out_cold) / (T_in_hot - T_in_cold))
03

Solve for T_out_hot for both cases

For counterflow: From the heat transfer equations: Q_hot / HCR_hot = -鈭員_hot Q_cold / HCR_cold = 鈭員_cold Substituting HCR values in Q_hot = Q_cold and 鈭員_counterflow: 2 脳 鈭員_hot = -鈭員_cold T_in_hot - T_out_hot = 2(T_out_cold - T_in_cold) Plugging T_out_cold = T_in_cold + 0.5 脳 鈭員_hot: 85 - T_out_hot = 2(15 + 0.5 脳 (T_out_hot - 85)) T_out_hot = \( \frac{110}{2} = 55^{\circ} \mathrm{C} \) For parallel flow: To find T_out_hot, we apply heat transfer equations (Q_hot / HCR_hot = -鈭員_hot and Q_cold / HCR_cold = 鈭員_cold) and the relationship for HCR (HCR_hot = 2 脳 HCR_cold): 2 脳 鈭員_hot = -鈭員_cold T_in_hot - T_out_hot = 2(T_out_cold - T_in_cold) Using the equation for 鈭員_parallel: (Q_cold / HCR_cold) = (T_in_hot - T_in_cold) / ln((T_in_hot - T_out_cold) / (T_in_hot - T_in_cold)) Let y = (T_in_hot - T_out_cold) / (T_in_hot - T_in_cold): Plugging in the values and solving for T_out_hot: Q_hot = Q_cold = HCR_cold 脳 60 / ln(y) 2 脳 鈭員_hot = 60 / ln(y) Plugging in T_out_cold = T_in_cold + 0.5 脳 鈭員_hot: T_out_hot = 85 - 2 脳 (15 + ln(y) 脳 30) T_out_hot = \( \frac{180}{3} = 60^{\circ} \mathrm{C} \) Hence, the hot water outlet temperature for counterflow and parallel flow cases are: a) Counterflow: \( T_{out\_hot} = 55^{\circ} \mathrm{C} \) b) Parallel flow: \( T_{out\_hot} = 60^{\circ} \mathrm{C} \)

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Key Concepts

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

Heat Capacity Rate
The heat capacity rate (HCR) is a fundamental concept crucial for understanding the operation of heat exchangers. It is defined as the product of a fluid's mass flow rate and its specific heat capacity. In mathematical terms, HCR can be expressed with the formula:

\[\begin{equation} HCR = \text{mass flow rate} \times \text{specific heat} \end{equation}\]
This value is representative of a fluid's ability to store and transfer heat. A higher HCR indicates that a fluid can deliver or absorb more heat at a specific temperature change, making it a key factor in the design and analysis of heat exchanger performance.
In our exercise example, the hot water's flow rate is double that of the cold water. Since the specific heats of the hot and cold water are equivalent, the HCR of the hot water is doubled compared to that of the cold water. This relationship is a pivot in determining the temperature changes and how effectively heat can be exchanged between the two fluids.
Understanding the concept of HCR is paramount for students working with heat exchangers, as it fundamentally affects the operational effectiveness of these systems and outcomes such as fluid outlet temperatures.
Counterflow Heat Exchange
Counterflow heat exchange is a highly efficient method of transferring heat between two fluids. In a counterflow arrangement, the two fluids move in opposite directions. As a result, the fluid being heated encounters the cooling fluid at progressively lower temperatures. This maintains a greater temperature difference along the length of the heat exchanger, which drives more efficient heat transfer.

Our textbook example calculates the outlet temperature for a counterflow heat exchanger using the relations between the heat capacity rates of hot and cold water, and the negative temperature change of the hot stream as it cools down. The larger temperature gradient that is inherent to counterflow heat exchangers allows for better heat transfer performance compared to parallel flow designs.
A simplified version of the calculation shows that the hot water outlet temperature is significantly affected by this arrangement, with the exercise revealing that the outlet temperature for a counterflow setup is lower than for a parallel flow. This exemplifies the improved efficiency of counterflow heat exchangers in real-world applications, like industrial processes or HVAC systems.
Parallel Flow Heat Exchange
In contrast to counterflow, a parallel flow heat exchange arrangement has both fluids moving in the same direction. This design leads to decreasing temperature differences between the fluids along the length of the heat exchanger. Initially, the temperature difference is high, but as both fluids transfer heat, the hot fluid cools while the cold fluid warms up, reducing the temperature differential and consequently the driving force for heat transfer.

Calculations for parallel flow heat exchange consider the logarithmic mean temperature difference to account for the changing temperature gradient. The exercise showcases how the outlet temperature of the hot water can be determined by applying the heat transfer equations, taking into account the relationship of the heat capacity rates, and the specific arrangement of the heat exchanger.
The result calculated in the exercise demonstrates that the hot water outlet temperature in a parallel flow arrangement is higher compared to counterflow, highlighting that parallel flow configurations are less effective in transferring heat. This is an essential takeaway for students learning about heat exchanger performance 鈥� understanding the implications of the design on the operational efficiency of the system.

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Most popular questions from this chapter

The human brain is especially sensitive to elevated temperatures. The cool blood in the veins leaving the face and neck and returning to the heart may contribute to thermal regulation of the brain by cooling the arterial blood flowing to the brain. Consider a vein and artery running between the chest and the base of the skull for a distance \(L=250 \mathrm{~mm}\), with mass flow rates of \(3 \times 10^{-3} \mathrm{~kg} / \mathrm{s}\) in opposite directions in the two vessels. The vessels are of diameter \(D=5 \mathrm{~mm}\) and are separated by a distance \(w=7 \mathrm{~mm}\). The thermal conductivity of the surrounding tissue is \(k_{\mathrm{r}}=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). If the arterial blood enters at \(37^{\circ} \mathrm{C}\) and the venous blood enters at \(27^{\circ} \mathrm{C}\), at what temperature will the arterial blood exit? If the arterial blood becomes overheated, and the body responds by halving the blood flow rate, how much hotter can the entering arterial blood be and still maintain its exit temperature below \(37^{\circ} \mathrm{C}\) ? Hint: If we assume that all the heat leaving the artery enters the vein, then heat transfer between the two vessels can be modeled using a relationship found in Table 4.1. Approximate the blood properties as those of water.

A shell-and-tube heat exchanger must be designed to heat \(2.5 \mathrm{~kg} / \mathrm{s}\) of water from 15 to \(85^{\circ} \mathrm{C}\). The heating is to be accomplished by passing hot engine oil, which is available at \(160^{\circ} \mathrm{C}\), through the shell side of the exchanger. The oil is known to provide an average convection coefficient of \(h_{o}=400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the outside of the tubes. Ten tubes pass the water through the shell. Each tube is thin walled, of diameter \(D=25 \mathrm{~mm}\), and makes eight passes through the shell. If the oil leaves the exchanger at \(100^{\circ} \mathrm{C}\), what is its flow rate? How long must the tubes be to accomplish the desired heating?

A plate-fin heat exchanger is used to condense a saturated refrigerant vapor in an air-conditioning system. The vapor has a saturation temperature of \(45^{\circ} \mathrm{C}\), and a condensation rate of \(0.015 \mathrm{~kg} / \mathrm{s}\) is dictated by system performance requirements. The frontal area of the condenser is fixed at \(A_{\mathrm{fr}}=0.25 \mathrm{~m}^{2}\) by installation requirements, and a value of \(h_{f g}=135 \mathrm{~kJ} / \mathrm{kg}\) may be assumed for the refrigerant. (a) The condenser design is to be based on a nominal air inlet temperature of \(T_{c, i}=30^{\circ} \mathrm{C}\) and nominal air inlet velocity of \(V=2 \mathrm{~m} / \mathrm{s}\) for which the manufacturer of the heat exchanger core indicates an overall coefficient of \(U=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the corresponding value of the heat transfer surface area required to achieve the prescribed condensation rate? What is the air outlet temperature? (b) From the manufacturer of the heat exchanger core, it is also known that \(U \propto V^{0 . t}\). During daily operation the air inlet temperature is not controllable and may vary from 27 to \(38^{\circ} \mathrm{C}\). If the heat exchanger area is fixed by the result of part (a), what is the range of air velocities needed to maintain the prescribed condensation rate? Plot the velocity as a function of the air inlet temperature.

Thin-walled aluminum tubes of diameter \(D=10 \mathrm{~mm}\) are used in the condenser of an air conditioner. Under normal operating conditions, a convection coefficient of \(h_{i}=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is associated with condensation on the inner surface of the tubes, while a coefficient of \(h_{o}=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is maintained by airflow over the tubes. (a) What is the overall heat transfer coefficient if the tubes are unfinned? (b) What is the overall heat transfer coefficient based on the inner surface, \(U_{i}\), if aluminum annular fins of thickness \(t=1.5 \mathrm{~mm}\), outer diameter \(D_{o}=20 \mathrm{~mm}\), and pitch \(S=3.5 \mathrm{~mm}\) are added to the outer surface? Base your calculations on a 1-m-long section of tube. Subject to the requirements that \(t \geq 1 \mathrm{~mm}\) and \((S-t) \geq 1.5 \mathrm{~mm}\), explore the effect of variations in \(t\) and \(S\) on \(U_{i}\). What combination of \(t\) and \(S\) would yield the best heat transfer performance?

Water is used for both fluids (unmixed) flowing through a single-pass, cross- flow heat exchanger. The hot water enters at \(90^{\circ} \mathrm{C}\) and \(10,000 \mathrm{~kg} / \mathrm{h}\), while the cold water enters at \(10^{\circ} \mathrm{C}\) and \(20,000 \mathrm{~kg} / \mathrm{h}\). If the effectiveness of the exchanger is \(60 \%\), determine the cold water exit temperature.
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