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Engineering Civil Engineering MindTap Engineering, 1 term (6 months) Printed Access Card for Dunlap's Sustainable Energy, SI Edition, 2nd

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.

MindTap Engineering, 1 term (6 months) Printed Access Card for Dunlap's Sustainable Energy, SI Edition, 2nd

MindTap Engineering, 1 term (6 months) Printed Access Card for Dunlap's Sustainable Energy, SI Edition, 2nd

2nd Edition

ISBN: 9781337551779

Author: DUNLAP, Richard A.

Publisher: Cengage Learning

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MindTap Engineering, 1 term (6 months) Printed Access Card for Dunlap's Sustainable Energy, SI Edition, 2nd

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