Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For a specified fluid pair, inlet temperatures, and mass flow rates, what kind of heat exchanger will have the highest effectiveness: double-pipe parallel- flow, double-pipe counterflow, cross-flow, or multipass shell-and-tube heat exchanger?

An air handler is a large unmixed heat exchanger used for comfort control in large buildings. In one such application, chilled water \(\left(c_{p}=4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters an air handler at \(5^{\circ} \mathrm{C}\) and leaves at \(12^{\circ} \mathrm{C}\) with a flow rate of \(1000 \mathrm{~kg} / \mathrm{h}\). This cold water cools \(5000 \mathrm{~kg} / \mathrm{h}\) of air \(\left(c_{p}=1.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) which enters the air handler at \(25^{\circ} \mathrm{C}\). If these streams are in counter-flow and the water-stream conditions remain fixed, the minimum temperature at the air outlet is (a) \(5^{\circ} \mathrm{C}\) (b) \(12^{\circ} \mathrm{C}\) (c) \(19^{\circ} \mathrm{C}\) (d) \(22^{\circ} \mathrm{C}\) (e) \(25^{\circ} \mathrm{C}\)

The condenser of a room air conditioner is designed to reject heat at a rate of \(15,000 \mathrm{~kJ} / \mathrm{h}\) from refrigerant-134a as the refrigerant is condensed at a temperature of \(40^{\circ} \mathrm{C}\). Air \(\left(c_{p}=1005 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) flows across the finned condenser coils, entering at \(25^{\circ} \mathrm{C}\) and leaving at \(35^{\circ} \mathrm{C}\). If the overall heat transfer coefficient based on the refrigerant side is \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the heat transfer area on the refrigerant side.

Consider a double-pipe heat exchanger with a tube diameter of \(10 \mathrm{~cm}\) and negligible tube thickness. The total thermal resistance of the heat exchanger was calculated to be \(0.025 \mathrm{~K} / \mathrm{W}\) when it was first constructed. After some prolonged use, fouling occurs at both the inner and outer surfaces with the fouling factors \(0.00045 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) and \(0.00015 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), respectively. The percentage decrease in the rate of heat transfer in this heat exchanger due to fouling is (a) \(2.3 \%\) (b) \(6.8 \%\) (c) \(7.1 \%\) (d) \(7.6 \%\) (e) \(8.5 \%\)

Consider a shell-and-tube water-to-water heat exchanger with identical mass flow rates for both the hotand cold-water streams. Now the mass flow rate of the cold water is reduced by half. Will the effectiveness of this heat exchanger increase, decrease, or remain the same as a result of this modification? Explain. Assume the overall heat transfer coefficient and the inlet temperatures remain the same.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free