ELEMENTARY PRINCIPLES OF CHEM. PROCESS.
ELEMENTARY PRINCIPLES OF CHEM. PROCESS.
4th Edition
ISBN: 9781119249214
Author: FELDER
Publisher: INTER WILE
Question
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Chapter 2, Problem 2.13P
Interpretation Introduction

(a)

Interpretation:

How many panels are required.

Concept introduction:

Kilowatt hour is a commercial unit of energy which is widely used in electrical application.

Kilowatt hour can be calculated by multiplying 1000 W with 3600 s. The consumption by home and other small business are measured in kilowatt hour while the consumption by medium and large-scale industries are measured in Megawatt hour.

Applying conversion factor to find the value of kWh as follows:

1kWh=1000×60×60=3600000J

Interpretation Introduction

(b)

Interpretation:  What would be total cost saving

Concept introduction:

Kilowatt hour is a commercial unit of energy which is widely used in electrical application.

Kilowatt hour can be calculated by multiplying 1000 W with 3600 s. The consumption by home and other small business are measured in kilowatt hour while the consumption by medium and large-scale industries are measured in Megawatt hour.

Applying conversion factor to find the value of kWh as follows:

1kWh=1000×60×60=3600000J

Interpretation Introduction

(c)

Interpretation:  Motivation to decide to install the solar panels

Concept introduction:

Kilowatt hour is a commercial unit of energy which is widely used in electrical application.

Kilowatt hour can be calculated by multiplying 1000 W with 3600 s. The consumption by home

and other small business are measured in kilowatt hour while the consumption by medium and

large-scale industries are measured in Megawatt hour.

Applying conversion factor to find the value of kWh as follows:

1kWh=1000×60×60=3600000J

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Consider a cloudless day in which the sun shines down across the united states. if 2659 KJ of energy reaches a square meter (m^2) of the united states in one hour, how much total solar energy reaches the entire united states per hour? the entire area of the united states is 9,158.960 km^2.
The total power used by humans worldwide is approximately 15 TW (terawatts). Sunlight striking Earth provides 1.336 kW per square meter (assuming no clouds). The surface area of Earth is approximately 197,000,000 square miles. How much of Earth's surface would we need to cover with solar energy collectors to power the planet for use by all humans? Assume that the solar energy collectors can only convert 10 % of the available sunlight into useful power.
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