Under what conditions is the heat transfer relation $$ \dot{Q}=\dot{m}_{c} c_{p c}\left(T_{c, \text { out }}-T_{c, \text { in }}\right)=\dot{m}_{h} c_{p h}\left(T_{h, \text { in }}-T_{h, \text { out }}\right) $$ valid for a heat exchanger?

Short Answer

Expert verified
Answer: The heat transfer equation for a heat exchanger is valid under the conditions of steady-state operation, no heat losses to the surroundings, and satisfaction of the energy conservation principle.

Step by step solution

01

Describe the variables and terms

In the given equation, we have the following variables and terms: - \(\dot{Q}\): Heat transfer rate (W) - \(\dot{m}_c\): Mass flow rate of the cold fluid (kg/s) - \(c_{p c}\): Specific heat capacity of the cold fluid at constant pressure (J/kg K) - \(T_{c, \text { in }}\): Inlet temperature of the cold fluid (K) - \(T_{c, \text { out }}\): Outlet temperature of the cold fluid (K) - \(\dot{m}_h\): Mass flow rate of the hot fluid (kg/s) - \(c_{p h}\): Specific heat capacity of the hot fluid at constant pressure (J/kg K) - \(T_{h, \text { in }}\): Inlet temperature of the hot fluid (K) - \(T_{h, \text { out }}\): Outlet temperature of the hot fluid (K)
02

Identify energy balance and conservation laws

The given equation represents an energy balance in a heat exchanger. Since no work is involved in a heat exchanger process, the energy balance can be expressed as the heat transfer from the hot fluid equals the heat transfer to the cold fluid, as shown in the equation. The equation also has some implicit assumptions, such as steady-state operation (no time-varying process) and no heat losses to the surroundings. The energy conservation principle dictates that the energy input must equal the energy output. In the case of a heat exchanger, this means that the heat transfer from the hot fluid must equal the heat transfer to the cold fluid. $$ \dot{Q}_{\text{hot}} = \dot{Q}_{\text{cold}} $$
03

Deduce the necessary conditions

Now, we can deduce the conditions under which the given equation for heat transfer in a heat exchanger is valid: 1. Steady-state operation: The heat exchanger must be operating at a steady state, without any time-varying processes that would affect the heat transfer rate or temperatures. 2. No heat losses: There must be no heat loss to the surroundings. In other words, all the heat transfer from the hot fluid must be transferred to the cold fluid. 3. Energy conservation: The energy conservation principle, which states that the heat transfer from the hot fluid must equal the heat transfer to the cold fluid, must be satisfied. In summary, the given heat transfer equation for a heat exchanger is valid under the conditions of steady-state operation, no heat losses to the surroundings, and satisfaction of the energy conservation principle.

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Key Concepts

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

Steady-State Heat Transfer
Understanding the concept of steady-state heat transfer is essential when examining heat exchangers, as it is one of the primary conditions for the heat transfer relation \(\dot{Q}=\dot{m}_{c} c_{p c}\left(T_{c, \text { out }}-T_{c, \text { in }}\right)=\dot{m}_{h} c_{p h}\left(T_{h, \text { in }}-T_{h, \text { out }}\right)\) to be valid.

In a steady-state scenario, the temperatures, flow rates, and heat transfer rates remain constant over time. This means that there are no fluctuations in the heat exchange process, and the system has reached an equilibrium where the input and output are unchanging.

The importance of this condition cannot be overstated. Without it, calculations become complex, as time variations would have to be accounted for. For educational content that teaches this concept, visual aids such as diagrams showing a constant flow and temperature profile over time could be beneficial for students to visualize and comprehend steady-state operation.
Energy Balance
At the heart of every heat exchanger operation is the principle of energy balance. This concept is pivotal in ensuring the conservation of energy, a fundamental law of physics that dictates that energy cannot be created or destroyed within an isolated system.

When dealing with heat exchangers, the notion of energy balance is simplified to the idea that the amount of heat leaving one fluid must be equal to the amount of heat being absorbed by another fluid. Mathematically, this can be expressed as \(\dot{Q}_{\text{hot}} = \dot{Q}_{\text{cold}}\), where \(\dot{Q}_{\text{hot}}\) is the heat being lost by the hot fluid and \(\dot{Q}_{\text{cold}}\) is the heat being gained by the cold fluid.

In educational materials, it's advantageous to use real-world examples to illustrate energy balance, such as comparing it to a simple financial transaction where the amount given out is equal to the amount received. Such analogies make the concept more accessible and relatable to students.
Specific Heat Capacity
The specific heat capacity, denoted by \(c_p\), is a substance's heat capacity per unit mass. It represents the amount of energy required to raise the temperature of one kilogram of the substance by one Kelvin.

In the context of a heat exchanger, specific heat capacities of the fluids involved (hot and cold) are central to calculating the amount of heat transfer. The heat capacity dictates how much heat is needed to change the fluid's temperature, which directly affects the heat exchanger's efficiency. High specific heat capacities mean that the fluid can absorb or release a significant amount of heat without a large temperature change.

Educational content on this topic could benefit from incorporating interactive elements where students can simulate the effects of varying the specific heat capacities on the heat exchanger's performance. This hands-on approach encourages a deeper understanding of the role that specific heat capacity plays in thermal systems.

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Most popular questions from this chapter

The radiator in an automobile is a cross-flow heat exchanger \(\left(U A_{s}=10 \mathrm{~kW} / \mathrm{K}\right)\) that uses air \(\left(c_{p}=1.00 \mathrm{~kJ} /\right.\) \(\mathrm{kg} \cdot \mathrm{K})\) to \(\mathrm{cool}\) the engine coolant fluid \(\left(c_{p}=4.00 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\). The engine fan draws \(30^{\circ} \mathrm{C}\) air through this radiator at a rate of \(10 \mathrm{~kg} / \mathrm{s}\) while the coolant pump circulates the engine coolant at a rate of \(5 \mathrm{~kg} / \mathrm{s}\). The coolant enters this radiator at \(80^{\circ} \mathrm{C}\). Under these conditions, what is the number of transfer units (NTU) of this radiator? (a) 1 (b) 2 (c) 3 (d) 4 (e) 5

Cold water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the tubes of a heat exchanger with 2 -shell passes and 23 -tube passes at \(14^{\circ} \mathrm{C}\) at a rate of \(3 \mathrm{~kg} / \mathrm{s}\), while hot oil \(\left(c_{p}=2200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the shell at \(200^{\circ} \mathrm{C}\) at the same mass flow rate. The overall heat transfer coefficient based on the outer surface of the tube is \(300 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the heat transfer surface area on that side is \(20 \mathrm{~m}^{2}\). Determine the rate of heat transfer using \((a)\) the LMTD method and \((b)\) the \(\varepsilon-\mathrm{NTU}\) method.

A performance test is being conducted on a double pipe counter flow heat exchanger that carries engine oil and water at a flow rate of \(2.5 \mathrm{~kg} / \mathrm{s}\) and \(1.75 \mathrm{~kg} / \mathrm{s}\), respectively. Since the heat exchanger has been in service over a long period of time it is suspected that the fouling might have developed inside the heat exchanger that might have affected the overall heat transfer coefficient. The test to be carried out is such that, for a designed value of the overall heat transfer coefficient of \(450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and a surface area of \(7.5 \mathrm{~m}^{2}\), the oil must be heated from \(25^{\circ} \mathrm{C}\) to \(55^{\circ} \mathrm{C}\) by passing hot water at \(100^{\circ} \mathrm{C}\left(c_{p}=4206 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at the flow rates mentioned above. Determine if the fouling has affected the overall heat transfer coefficient. If yes, then what is the magnitude of the fouling resistance?

Cold water \(\left(c_{p}=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters a counterflow heat exchanger at \(18^{\circ} \mathrm{C}\) at a rate of \(0.7 \mathrm{~kg} / \mathrm{s}\) where it is heated by hot air \(\left(c_{p}=1.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters the heat exchanger at \(50^{\circ} \mathrm{C}\) at a rate of \(1.6 \mathrm{~kg} / \mathrm{s}\) and leaves at \(25^{\circ} \mathrm{C}\). The maximum possible outlet temperature of the cold water is (a) \(25.0^{\circ} \mathrm{C}\) (b) \(32.0^{\circ} \mathrm{C}\) (c) \(35.5^{\circ} \mathrm{C}\) (d) \(39.7^{\circ} \mathrm{C}\) (e) \(50.0^{\circ} \mathrm{C}\)

In a textile manufacturing plant, the waste dyeing water \(\left(c_{p}=4295 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(75^{\circ} \mathrm{C}\) is to be used to preheat fresh water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(15^{\circ} \mathrm{C}\) at the same flow rate in a double-pipe counter-flow heat exchanger. The heat transfer surface area of the heat exchanger is \(1.65 \mathrm{~m}^{2}\) and the overall heat transfer coefficient is \(625 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the rate of heat transfer in the heat exchanger is \(35 \mathrm{~kW}\), determine the outlet temperature and the mass flow rate of each fluid stream.

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