Skip to main content

Engineering Civil Engineering Sustainable Energy

a) The moment of inertia and rotational energy of the earth. The moment of inertia of the earth = 9.696×10 3 7 kg.m 2 The rotational energy of the earth = 2.562×10 2 9 J Given Information: The earth is a solid sphere with a uniform density. Explanation: The moment of inertia of the earth can be computed using the formula: I = k × m × r 2 Where, k is the constant related to geometry = 2/5 (For solid sphere) m is the mass of the earth = 5.972×10 2 4 kg r is the radius of the earth = 6.371×10 6 m Putting all the values, I = 2 5 × 5.972 × 10 24 × 6.371 × 10 6 2 I = 0.4 × 5.972 × 10 24 × 4.058 × 10 13 I = 9.696 × 10 37 kg ⋅ m 2 Hence, The moment of inertia of the earth = 9.696×10 3 7 kg.m 2 The rotational energy of the earth can be computed using the formula, E = 2 × π 2 × I × f 2 Where, I = 9.696×10 3 7 kg.m 2 f is the Frequency of the rotation = 1.157×10-5 s-1 Putting all the values, E = 2 × π 2 × 9.696 × 10 37 × 1.157 × 10 − 5 2 E = 1.9139 × 10 39 × 1.338 × 10 − 10 E = 2.562 × 10 29 J Hence, The rotational energy of the earth = 2.562×10 2 9 J b) The annual rotational energy loss of the earth and annual energy use by the society. The annual rotational energy loss of the earth = 1×10 2 0 J/y The annual energy use by the society = 16.4% Given Information: The earth’s rotational period is increasing at the rate = 1.7×10-5 s per year Explanation: The power generated with decreasing rotation frequency will be P = d E d t Where, d E is the change in energy d t is the change in time P = d 2 π 2 I f 2 d t P = 4 π 2 I f d f d t .... ( 1 ) Putting, f = 1 T Differentiating with respect to time, d f d t = − 1 T 2 d T d t d f d t = − 1 24 × 3600 2 × 1.7 × 10 − 5 365 × 24 × 3600 d f d t = − 7.23 × 10 − 23 s − 2 Putting all the values in equation (1) P = 4 × π 2 × 9.696 × 10 37 × 1.157 × 10 − 5 × − 7.23 × 10 − 23 P = 3.2 × 10 12 W Therefore, The earth’s rotational energy loss per year will be E L = 3.2 × 10 12 × 3.15 × 10 7 E L = 1 × 10 20 J y Hence, The annual energy loss in earth’s rotational energy = 1×10 2 0 J/y Also, The percent loss will be p = E L E P × 100 Where, E p = Annual power use by society = 6.1×10 2 0 J/y Putting all the values, p = 1 × 10 20 6.1 × 10 20 × 100 p = 16.4 % Hence, The annual rotational energy loss is 16.4% of the annual primary energy use by the society.

a) The moment of inertia and rotational energy of the earth. The moment of inertia of the earth = 9.696×10 3 7 kg.m 2 The rotational energy of the earth = 2.562×10 2 9 J Given Information: The earth is a solid sphere with a uniform density. Explanation: The moment of inertia of the earth can be computed using the formula: I = k × m × r 2 Where, k is the constant related to geometry = 2/5 (For solid sphere) m is the mass of the earth = 5.972×10 2 4 kg r is the radius of the earth = 6.371×10 6 m Putting all the values, I = 2 5 × 5.972 × 10 24 × 6.371 × 10 6 2 I = 0.4 × 5.972 × 10 24 × 4.058 × 10 13 I = 9.696 × 10 37 kg ⋅ m 2 Hence, The moment of inertia of the earth = 9.696×10 3 7 kg.m 2 The rotational energy of the earth can be computed using the formula, E = 2 × π 2 × I × f 2 Where, I = 9.696×10 3 7 kg.m 2 f is the Frequency of the rotation = 1.157×10-5 s-1 Putting all the values, E = 2 × π 2 × 9.696 × 10 37 × 1.157 × 10 − 5 2 E = 1.9139 × 10 39 × 1.338 × 10 − 10 E = 2.562 × 10 29 J Hence, The rotational energy of the earth = 2.562×10 2 9 J b) The annual rotational energy loss of the earth and annual energy use by the society. The annual rotational energy loss of the earth = 1×10 2 0 J/y The annual energy use by the society = 16.4% Given Information: The earth’s rotational period is increasing at the rate = 1.7×10-5 s per year Explanation: The power generated with decreasing rotation frequency will be P = d E d t Where, d E is the change in energy d t is the change in time P = d 2 π 2 I f 2 d t P = 4 π 2 I f d f d t .... ( 1 ) Putting, f = 1 T Differentiating with respect to time, d f d t = − 1 T 2 d T d t d f d t = − 1 24 × 3600 2 × 1.7 × 10 − 5 365 × 24 × 3600 d f d t = − 7.23 × 10 − 23 s − 2 Putting all the values in equation (1) P = 4 × π 2 × 9.696 × 10 37 × 1.157 × 10 − 5 × − 7.23 × 10 − 23 P = 3.2 × 10 12 W Therefore, The earth’s rotational energy loss per year will be E L = 3.2 × 10 12 × 3.15 × 10 7 E L = 1 × 10 20 J y Hence, The annual energy loss in earth’s rotational energy = 1×10 2 0 J/y Also, The percent loss will be p = E L E P × 100 Where, E p = Annual power use by society = 6.1×10 2 0 J/y Putting all the values, p = 1 × 10 20 6.1 × 10 20 × 100 p = 16.4 % Hence, The annual rotational energy loss is 16.4% of the annual primary energy use by the society.

Sustainable Energy

Sustainable Energy

2nd Edition

ISBN: 9781337551663

Author: DUNLAP, Richard A.

Publisher: Cengage,

See similar textbooks

Concept explainers

Fundamentals of World Energy Use

Population growth has a direct relation to the growing energy demand of mankind. A major emphasis has been laid on the usage of various energy supplies from both renewable and nonrenewable energy sources to meet the global energy demand. The energy supp…

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

Blurred answer

Chapter 18, Problem 12P

Chapter 18, Problem 14P

Students have asked these similar questions

3. A storage tank with a surface area of 35 m² and a wall thickness of 100 mm contains 2000 kg of hydrogen gas at a concentration of 110 kg/m³. The atmosphere outside the tank has a hydrogen concentration of 0.2 kg/m³. If D = 8×10-¹¹ m²/s, how long will it take (in years) for 4% of the hydrogen to diffuse out of the tank? Assume steady-state conditions.

A particle of mass m is situated somewhere in between planets X and Y. The particle's location is at a distance d from planet X and at a distance 1.5d from planet Y. If planet X has a mass of M, and planet Y has a mass of 3M, then which planet exerts greater gravitational force on the particle? By how much, in percent? (3

The radioactive isotope Carbon-14 has a half-life of 5,730 years. Given 300 grams of C-14, what is the remaining mass (in grams) of the isotope after 250 years have elapsed? A) 291.06g B) 8.94g C) 97.02g D) 202.98g

Chapter 18 Solutions

Sustainable Energy

Ch. 18 - Prob. 1P Ch. 18 - Prob. 8P Ch. 18 - Prob. 13P

Knowledge Booster

Background pattern image

Civil Engineering

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.

Similar questions

Recommended textbooks for you

Sustainable Energy
Civil Engineering
ISBN:9781337551663
Author:DUNLAP, Richard A.
Publisher:Cengage,

Text book image

Sustainable Energy

Civil Engineering

ISBN:9781337551663

Author:DUNLAP, Richard A.

Publisher:Cengage,

SEE MORE TEXTBOOKS