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3 years ago

Explain the effect of heat transfer, reversibilities and irreversibilities on changes in entropy

Engineering

1 answer:

Natali5045456 [20]3 years ago

5 0

Answer is given below

Explanation:

The second law of thermodynamics states that in the reversible process, the entropy of the universe is constant, while in the unchanging process, such as the transfer of heat from a hot object to a cold object, the reflection of the universe increases.

ΔS = ΔQ ÷ T

The irreversible process increases the entropy of the universe. Since entropy is a state function, the change in entropy of the system is the same, the process is reversible or irreversible. The second law of thermodynamics can be used to determine whether a process is reversible.

for an irreversible process, ΔS is not same as reversible process, because more than 1 reversible process is required for 2 ΔS to be equal.

if the initial and final states are the same, ΔS for an irreversible process can be calculated as if it were a reversible process.

If the initial and final condition is same then ΔS can find out for irreversible process even if it were reversible process

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Another name for your computer, running the web browser program is: Web user The client The mainframe Browsing agent

timofeeve [1]

Answer:

Browsing agent

Explanation:

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6 0

3 years ago

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Fantom [35]

Summary

Students learn about the variety of materials used by engineers in the design and construction of modern bridges. They also find out about the material properties important to bridge construction and consider the advantages and disadvantages of steel and concrete as common bridge-building materials to handle compressive and tensile forces.

This engineering curriculum aligns to Next Generation Science Standards (NGSS).

Engineering Connection

When designing structures such as bridges, engineers carefully choose the materials by anticipating the forces the materials (the structural components) are expected to experience during their lifetimes. Usually, ductile materials such as steel, aluminum and other metals are used for components that experience tensile loads. Brittle materials such as concrete, ceramics and glass are used for components that experience compressive loads.

Learning Objectives

After this lesson, students should be able to:

List several common materials used the design and construction of structures.

Describe several factors that engineers consider when selecting materials for the design of a bridge.

Explain the advantages and disadvantages of common materials used in engineering structures (steel and concrete).

Educational Standards

NGSS: Next Generation Science Standards - Science

Common Core State Standards - Math

International Technology and Engineering Educators Association - Technology

State Standards

Suggest an alignment not listed above

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Worksheets and Attachments

Strength of Materials Worksheet (doc)

Strength of Materials Worksheet (pdf)

Strength of Materials Worksheet Answers (doc)

Strength of Materials Worksheet Answers (pdf)

Strength of Materials Math Worksheet (doc)

Strength of Materials Math Worksheet (pdf)

Strength of Materials Math Worksheet Answers (doc)

Strength of Materials Math Worksheet Answers (pdf)

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2 years ago

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2 years ago

Write a python program to apply this:

lakkis [162]

Answer:

# Initialize a dictionary with the keys

contestants = {"Darci Lynne":0, "Angelica Hale":0, "Angelina Green":0};

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if cName.lower() == "done":

break;

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8 0

3 years ago

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-m

kodGreya [7K]

Complete question is;

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100 mm thick plate (ρ = 7830 kg/m3, Cp = 550 J/kg K, k = 48 W/m K). The plate initially is at 200 °C and is to be heated to a minimum temperature of 550 °C. Heating is effected in a gas-fired furnace where the products of combustion at T∞ = 800 °C maintain a convection heat transfer coefficient of h = 250 W/m.K on both surfaces of the plate. How long should the plate be left in the furnace?

Answer:

860 seconds

Explanation:

We are given;

Initial Temperature; Ti = 200 °C

Minimum Temperature; T_i = 550 °C

T∞ = 800 °C

convection coefficient; h = 250 W/m².K

ρ = 7830 kg/m³

Cp = 550 J/kg K

k = 48 W/m K

Plate thickness = 100mm

Thus,L = 100/2 = 50mm = 0.05 m

Let's find the biot number from the formula;

Bi = hL/K

Bi = (250 × 0.05)/48

Bi = 0.2604

Now, lowest temperature in the slab is given as;

θ_o = (T_o - T∞)/(T_i - T∞)

θ_o = (550 - 800)/(200 - 800)

θ_o = 0.4167

Now, from online tables calculation, we can find the root of the biot number.

Thus, root of the biot number Bi = 0.2604 is;

ζ1 = 0.488 rad

Also, C1 is gotten as 1.0396

Now,formula for thermal diffusivity is;

α = k/ρc

α = 48/(7830 × 550)

α = 1.115 × 10^(-5) m²/s

Also, from online tables, f(ζ1) = 0.401

Thus, we can find the time the plate should the plate be left in the furnace from;

-(ζ1)²(αt/L²) = In 0.401

-(ζ1)²(αt/L²) = -0.9138

t = (-0.9138 × 0.05²)/-(0.488² × 1.115 × 10^(-5))

t ≈ 860 s

8 0

3 years ago