EBK ENGINEERING FUNDAMENTALS: AN INTROD
EBK ENGINEERING FUNDAMENTALS: AN INTROD
5th Edition
ISBN: 9780100543409
Author: MOAVENI
Publisher: YUZU
bartleby

Concept explainers

Question
Book Icon
Chapter 13, Problem 16P
To determine

Convert the given sources data in the Table accompanied for the problems 13.14 to 13.20 in the textbook from Billions of kilowatt-hours to Billions of Btu.

Expert Solution & Answer
Check Mark

Answer to Problem 16P

The given sources data in the Table accompanied for the problems 13.14 to 13.20 in the textbook from Billions of kilowatt-hours to Billions of Btu are tabulated in Table 1.

Explanation of Solution

Given data:

Refer to the problem 13.16 in textbook for the accompanying table.

Formula used:

The value of the 1Btus is,

1Btus=1.055kW (1)

Convert 1 hr into seconds,

1hr=3600s

Rearrange the equation,

1s=13600hr (2)

Calculation:

Substitute equation (2) in equation (1),

1Btu13600hr=1.055kW1Btu=1.0553600kWhr1Btu=2.93×104kWhr

Rearrange the equation for the relation between kWhr to Btu.

1kWhr=12.93×104Btu=3412.96Btu (3)

For coal:

All the following calculations of conversion are in Billions.

Coal for the year 1980 is,

(Coal)1980=1161.562kWhr (4)

Substitute 3412.96Btu for 1kWhr in equation (4).

(Coal)1980=1161.562(3412.96Btu)=3964364.6Btu

Coal for the year 1990 is,

(Coal)1990=1594.011kWhr (5)

Substitute 3412.96Btu for 1kWhr in equation (5).

(Coal)1990=1594.011(3412.96Btu)=5440295Btu

Coal for the year 2000 is,

(Coal)2000=1966.265kWhr (6)

Substitute 3412.96Btu for 1kWhr in equation (6).

(Coal)2000=1966.265(3412.96Btu)=6710783.79Btu

Coal for the year 2005 is,

(Coal)2005=2040.913kWhr (7)

Substitute 3412.96Btu for 1kWhr in equation (7).

(Coal)2005=2040.913(3412.96Btu)=6965554.432 Btu

Coal for the year 2010 is,

(Coal)2010=2217.555kWhr (8)

Substitute 3412.96Btu for 1kWhr in equation (8).

(Coal)2010=2217.555(3412.96Btu)=7568426.5Btu

Coal for the year 2020 is,

(Coal)2020=2504.786kWhr (9)

Substitute 3412.96Btu for 1kWhr in equation (9).

(Coal)2020=2504.786(3412.96Btu)=8548734.42Btu

The coal for the year 2030 is,

(Coal)2030=3380.674kWhr (10)

Substitute 3412.96Btu for 1kWhr in equation (10).

(Coal)2030=3380.674(3412.96Btu)=11538105.14Btu

For petroleum:

All the following calculations of conversion are in Billions.

Petroleum for the year 1980 is,

(Petroleum)1980=245.9942kWhr (11)

Substitute 3412.96Btu for 1kWhr in equation (11).

(Petroleum)1980=245.9942(3412.96Btu)=839568.36Btu

Petroleum for the year 1990 is,

(Petroleum)1990=126.6211kWhr (12)

Substitute 3412.96Btu for 1kWhr in equation (12).

(Petroleum)1990=126.6211(3412.96Btu)=432152.74Btu

Petroleum for the year 2000 is,

(Petroleum)2000=111.221kWhr (13)

Substitute 3412.96Btu for 1kWhr in equation (13).

(Petroleum)2000=111.221(3412.96Btu)=379592.82Btu

Petroleum for the year 2005 is,

(Petroleum)2005=115.4264kWhr (14)

Substitute 3412.96Btu for 1kWhr in equation (14).

(Petroleum)2005=115.4264(3412.96Btu)=393945.68Btu

Petroleum for the year 2010 is,

(Petroleum)2010=104.8182BkWhr (15)

Substitute 3412.96Btu for 1kWhr in equation (15).

(Petroleum)2010=104.8182(3412.96Btu)=357740.32Btu

Petroleum for the year 2020 is,

(Petroleum)2020=106.6799kWhr (16)

Substitute 3412.96Btu for 1kWhr in equation (16).

(Petroleum)2020=106.6799(3412.96Btu)=364094.23Btu

Petroleum for the year 2030 is,

(Petroleum)2030=114.6741kWhr (17)

Substitute 3412.96Btu for 1kWhr in equation (17).

(Petroleum)2030=114.6741(3412.96Btu)=391378.11Btu

For Natural gas:

All the following calculations of conversion are in Billions.

Natural gas for the year 1980 is,

(Naturalgas)1980=346.2399kWhr (18)

Substitute 3412.96Btu for 1kWhr in equation (18).

(Naturalgas)1980=346.2399(3412.96Btu)=1181702.92Btu

Natural gas for the year 1990 is,

(Naturalgas)1990=372.7652BkWhr (19)

Substitute 3412.96Btu for 1kWhr in equation (19).

(Naturalgas)1990=372.7652(3412.96Btu)=1272232.71Btu

Natural gas for the year 2000 is,

(Naturalgas)2000=601.0382kWhr (20)

Substitute 3412.96Btu for 1kWhr in equation (20).

(Naturalgas)2000=601.0382(3412.96Btu)=2051319.33Btu

Natural gas for the year 2000 is,

(Naturalgas)2005=751.8189kWhr (21)

Substitute 3412.96Btu for 1kWhr in equation (21).

(Naturalgas)2005=751.8189(3412.96Btu)=2565927.833Btu

Natural gas for the year 2010 is,

(Naturalgas)2010=773.8234kWhr (22)

Substitute 3412.96Btu for 1kWhr in equation (22).

(Naturalgas)2010=773.8234(3412.96Btu)=2641028.31Btu

Natural gas for the year 2020 is,

(Naturalgas)2020=1102.762kWhr (23)

Substitute 3412.96Btu for 1kWhr in equation (23).

(Naturalgas)2020=1102.762(3412.96Btu)=3763682.59Btu

Natural gas for the year 2030 is,

(Naturalgas)2030=992.7706BkWhr (24)

Substitute 3412.96Btu for 1kWhr in equation (24).

(Naturalgas)2030=992.7706(3412.96Btu)=3388286.34Btu

For Nuclear:

All the following calculations of conversion are in Billions.

Nuclear of the year 1980 is,

(Nuclear)1980=251.1156kWhr (25)

Substitute 3412.96Btu for 1kWhr in equation (25).

(Nuclear)1980=251.1156(3412.96Btu)=857047.49Btu

Nuclear of the year 1990 is,

(Nuclear)1990=576.8617kWhr (26)

Substitute 3412.96Btu for 1kWhr in equation (26).

(Nuclear)1990=576.8617(3412.96Btu)=1968805.9Btu

Nuclear of the year 2000 is,

(Nuclear)2000=753.8929kWhr (27)

Substitute 3412.96Btu for 1kWhr in equation (27).

(Nuclear)2000=753.8929(3412.96Btu)=2573006.31Btu

Nuclear of the year 2005 is,

(Nuclear)2005=774.0726kWhr (28)

Substitute 3412.96Btu for 1kWhr in equation (28).

(Nuclear)2005=774.0726(3412.96Btu)=2641878.820Btu

Nuclear of the year 2010 is,

(Nuclear)2010=808.6948kWhr (29)

Substitute 3412.96Btu for 1kWhr in equation (29).

(Nuclear)2010=808.6948(3412.96Btu)=2760043Btu

Nuclear of the year 2020 is,

(Nuclear)2020=870.698kWhr (30)

Substitute 3412.96Btu for 1kWhr in equation (30).

(Nuclear)2020=870.698(3412.96Btu)=2971657.44Btu

Nuclear of the year 2030 is,

(Nuclear)2030=870.5909kWhr (31)

Substitute 3412.96Btu for 1kWhr in equation (31).

(Nuclear)2030=870.5909(3412.96Btu)=2971291.91Btu

For Renewable/Others:

All the following calculations of conversion are in Billions.

Renewable of the year 1980 is,

(Renewable)1980=284.6883kWhr . (32)

Substitute 3412.96Btu for 1kWhr in equation (32).

(Renewable)1980=284.6883(3412.96Btu)=971629.78Btu

Renewable of the year 1990 is,

(Renewable)1990=357.2381kWhr (33)

Substitute 3412.96Btu for 1kWhr in equation (33).

(Renewable)1990=357.2381(3412.96Btu)=1219239.46Btu

Renewable of the year 2000 is,

(Renewable)2000=356.4786kWhr (34)

Substitute 3412.96Btu for 1kWhr in equation (34).

(Renewable)2000=356.4786(3412.96Btu)=1216647.203Btu

Renewable of the year 2005 is,

(Renewable)2005=375.8663kWhr (35)

Substitute 3412.96Btu for 1kWhr in equation (35).

(Renewable)2005=375.8663(3412.96Btu)=1282816.647Btu

Renewable of the year 2010 is,

(Renewable)2010=475.7432kWhr (36)

Substitute 3412.96Btu for 1kWhr in equation (36).

(Renewable)2010=475.7432(3412.96Btu)=1623692.51Btu

Renewable of the year 2020 is,

(Renewable)2020=515.1523kWhr (37)

Substitute 3412.96Btu for 1kWhr in equation (37).

(Renewable)2020=515.1523(3412.96Btu)=1758194.19Btu

Renewable of the year 2030 is,

(Renewable)2030=559.1335kWhr (38)

Substitute 3412.96Btu for 1kWhr in equation (38).

(Renewable)2030=559.1335(3412.96Btu)=1908300.27Btu

Therefore, the conversion from the Billions of kilowatt-hours to Billions of Btu is tabulated in Table 1 as below,

Table 1

YearCoalPetroleumNatural GasNuclearRenewable/Other
19803964364.644839568.3651181702.929857047.498971629.780
19905440295.783432152.7491272232.7171968805.9081219239.346
20006710783.794379592.8242051319.3352573006.3121216647.203
20056965554.432393945.6862565927.8332641878.8211282816.647
20107568426.513357740.3242641028.3112760043.0051623692.512
20208548734.427364094.2323763682.5962971657.4461758194.194
203011538105.135391378.1163388286.3472971291.9181908300.270

Conclusion:

Hence, the conversion for the energy sources from billions kilowatt-hours to billion Btu has been explained.

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!
Students have asked these similar questions
Part 3: Problem-Solving. Solve the following problems. Show all calculations. 1. A retaining wall 5.80m high supports soil that has the following properties: Unit weight = 17.3 kN/m³ Angle of internal friction = 26 deg. Cohesion = 14.5 kPa a) Calculate the normal pressure acting at the back of the wall assuming no tensile crack occurs in the soil. b) Find the location of the tensile crack measured from the surface of horizontal backfill. c) Determine the active pressure acting on the wall in tensile crack occurs in the soil. 2. The soil material is supported by a retaining wall to a height of 6m. The unit weight of the soil is 16 kN/m³ and the angle of internal friction is 29 deg. Assume the soil is cohesionless. a) Determine the earth pressure on the wall. b) Find the total active pressure if surcharge of 14 kPa is applied on the surface of horizontal backfill. c) Locate the position of the total pressure from the bottom.
Question 3 (20 points): The traffic volume on a 2-lane highway is 1600 veh/hr in each direction Page 3 of 6 with a density of 20 veh/mi. A large dump truck enters the traffic stream from an adjacent construction site at 20 mph and carries on this way for 2 miles before turning off to the dump site. Because flow is so high in the opposite direction, no one can pass the truck. As a result, traffic back up behind the truck at four times the density (i.e., 4x20 = 80 veh/mi) at a volume of 1000 veh/hr. How many vehicles get caught in the traffic congestion before the truck exits the highway?
How can construction project managers find a balance between speeding up schedules and the risks of making more mistakes and needing rework, especially when using methods like fast tracking?
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
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:9781337551663
Author:DUNLAP, Richard A.
Publisher:Cengage,