Skip to main content

Engineering Civil Engineering Sustainable Energy, Si Edition

a) The total energy available in MWh e The total energy available in MWh e = 1.04×10 4 MWh e Given Information: The area of the storage reservoir = 1 km 2 The average depth of the reservoir = 30 m The head of the water flowing to a lower reservoir = 150 m The efficiency of the generator = 85% Explanation: The volume of the storage reservoir will be V = A × h Here, V = Volume A = Cross-sectional area = 1 km 2 = 1000000 m 2 h = Depth = 30 m Putting all the values, V = 1000000 × 30 V = 3 × 10 7 m 3 The mass of the water in the reservoir will be m = ρ × V m = 1000 × 3 × 10 7 m = 3 × 10 10 k g The potential energy will be E = m × g × h Here, m = 3×10 1 0 kg g = Acceleration due to gravity = 9.81 m/sec 2 h = Head = 150 m So, E = 3 × 10 10 × 9.81 × 150 E = 4.414 × 10 13 J E = 1.225 × 10 7 k W h The total energy available will be E t o t a l = η × E Here, η = Efficiency= 85% = 0.85 Putting all the values, E total = 0.85 × 1.225 × 10 7 E total = 1.04 × 10 7 kWh e E total = 1.04 × 10 4 MWh e Hence, The total energy available = 1.04×10 4 MWh e b) The average power available The average power available = 866.67 MW e Given Information: The time period = 12 h Explanation: The average power available can be computed using the formula P a v g = E t o t a l t Here, E t o t a l = 1.04×10 4 MWh e t = 12 h Putting all the values, P a v g = 1.04 × 10 4 12 P a v g = 866.67 MW e Hence, The average power available = 866.67 MW e

a) The total energy available in MWh e The total energy available in MWh e = 1.04×10 4 MWh e Given Information: The area of the storage reservoir = 1 km 2 The average depth of the reservoir = 30 m The head of the water flowing to a lower reservoir = 150 m The efficiency of the generator = 85% Explanation: The volume of the storage reservoir will be V = A × h Here, V = Volume A = Cross-sectional area = 1 km 2 = 1000000 m 2 h = Depth = 30 m Putting all the values, V = 1000000 × 30 V = 3 × 10 7 m 3 The mass of the water in the reservoir will be m = ρ × V m = 1000 × 3 × 10 7 m = 3 × 10 10 k g The potential energy will be E = m × g × h Here, m = 3×10 1 0 kg g = Acceleration due to gravity = 9.81 m/sec 2 h = Head = 150 m So, E = 3 × 10 10 × 9.81 × 150 E = 4.414 × 10 13 J E = 1.225 × 10 7 k W h The total energy available will be E t o t a l = η × E Here, η = Efficiency= 85% = 0.85 Putting all the values, E total = 0.85 × 1.225 × 10 7 E total = 1.04 × 10 7 kWh e E total = 1.04 × 10 4 MWh e Hence, The total energy available = 1.04×10 4 MWh e b) The average power available The average power available = 866.67 MW e Given Information: The time period = 12 h Explanation: The average power available can be computed using the formula P a v g = E t o t a l t Here, E t o t a l = 1.04×10 4 MWh e t = 12 h Putting all the values, P a v g = 1.04 × 10 4 12 P a v g = 866.67 MW e Hence, The average power available = 866.67 MW e

Sustainable Energy, Si Edition

Sustainable Energy, Si Edition

2nd Edition

ISBN: 9781337551670

Author: DUNLAP, Richard A.

Publisher: Cengage Learning

See similar textbooks

Concept explainers

Components of Hydroelectric Power Plant

A hydro electric power plant derives the energy from the force of fast moving water to harness energy. The demand for energy needs has impacted the environment resulting in global warming. Extensive use of non-renewable energy sources like petrol and die…

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

Blurred answer

Chapter 17, Problem 20P

Chapter 18, Problem 2P

Students have asked these similar questions

Solve for forces on pin C and D

Borrow pit soil is being used to fill an 900,00 yd3 of depression. The properties of borrowpit and in-place fill soils obtained from laboratory test results are as follows:• Borrow pit soil: bulk density 105 pcf, moisture content = 8%, and specific gravity = 2.65• In-place fill soil: dry unit weight =120 pcf, and moisture content = 16%(a) How many yd3 of borrow soil is required?(b) What water mass is needed to achieve 16% moisture in the fill soil?(c) What is the in-place density after a long rain?

solve for dt/dx=f(t,x)=x+t^2

Chapter 18 Solutions

Sustainable Energy, Si Edition

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

SEE MORE QUESTIONS

Recommended textbooks for you

Sustainable Energy
Civil Engineering
ISBN:9781337551663
Author:DUNLAP, Richard A.
Publisher:Cengage,
Engineering Fundamentals: An Introduction to Engi...
Civil Engineering
ISBN:9781305084766
Author:Saeed Moaveni
Publisher:Cengage Learning
Sustainable Energy
Civil Engineering
ISBN:9781133108689
Author:Richard A. Dunlap
Publisher:Cengage Learning

Text book image

Sustainable Energy

Civil Engineering

ISBN:9781337551663

Author:DUNLAP, Richard A.

Publisher:Cengage,

Text book image

Engineering Fundamentals: An Introduction to Engi...

Civil Engineering

ISBN:9781305084766

Author:Saeed Moaveni

Publisher:Cengage Learning

Text book image

Sustainable Energy

Civil Engineering

ISBN:9781133108689

Author:Richard A. Dunlap

Publisher:Cengage Learning

SEE MORE TEXTBOOKS