Engineering Fundamentals: An Introduction to Engineering
Engineering Fundamentals: An Introduction to Engineering
6th Edition
ISBN: 9780357112311
Author: Saeed Moaveni
Publisher: Cengage Learning US
bartleby

Concept explainers

Question
Book Icon
Chapter 14, Problem 19P
To determine

Create a table that shows the windchill temperatures for the given range of ambient air temperature and wind speed.

Expert Solution & Answer
Check Mark

Answer to Problem 19P

A table is created for the windchill temperatures with the given range of ambient air temperature and wind speed by using excel.

Explanation of Solution

Given data:

The range of ambient air temperature is 30°C<Ta<10°C.

The range of wind speed is 20kmh<V<80kmh

Formula used:

Formula to calculate the more common equivalent windchill temperatures is,

Tequivalent=0.045(5.27V0.5+10.450.28V)(Ta33)+33

Here,

Ta is the ambient air temperature in °C, and

V is the wind speed in kmh.

Calculation:

Engineering Fundamentals: An Introduction to Engineering, Chapter 14, Problem 19P , additional homework tip  1

Refer to the Figure 1:

Column A shows the wind speed (V) with range from 20kmh<V<80kmh in increment of 10 and Row 8 shows the ambient air temperature with range from 30°C<Ta<10°C in increment of 5.

For the cell B9, the formula used to find the equivalent windchill temperatures as “=0.045(5.27V0.5+10.450.28V)(Ta33)+33”.

Written as “=(0.045*((5.27*($A$9)^0.5)+10.45-(0.28*$A$9))*(B8-33))+33”. Here, B8 cell represents the value of ambient air temperature is 30 and $A$9 is an absolute cell reference which represent the value of cell A9 as 20. The result obtained in the cell B9 would be 47.6. Use the Fill command to copy the formula into cell C9, only the B8 cell in the formula is automatically substituted by C8, resulting in a value 41.2. Then, by click and drag the bottom corner of the C9 cell extend through J9 in row-wise. The result is obtained as in Table 1.

For the cell B10, the formula used to find the equivalent windchill temperatures as “=0.045(5.27V0.5+10.450.28V)(Ta33)+33”.

Written as “=(0.045*((5.27*($A$10)^0.5)+10.45-(0.28*$A$10))*(B8-33))+33”. Here, B8 cell represents the value of ambient air temperature is 30 and $A$10 is an absolute cell reference which represent the value of cell A10 as 30. The result obtained in the cell B10 would be 54.6. Use the Fill command to copy the formula into cell C10, only the B8 cell in the formula is automatically substituted by C8, resulting in a value 47.7. Then, by click and drag the bottom corner of the C10 cell extend through J10 in row-wise. The result is obtained as in Table 1.

For the cell B11, the formula used to find the equivalent windchill temperatures as “=0.045(5.27V0.5+10.450.28V)(Ta33)+33”.

Written as “=(0.045*((5.27*($A$11)^0.5)+10.45-(0.28*$A$11))*(B8-33))+33”. Here, B8 cell represents the value of ambient air temperature is 30 and $A$11 is an absolute cell reference which represent the value of cell A11 as 40. The result obtained in the cell B11 would be 59.4. Use the Fill command to copy the formula into cell C11, only the B8 cell in the formula is automatically substituted by C8, resulting in a value 52.0. Then, by click and drag the bottom corner of the C11 cell extend through J11 in row-wise. The result is obtained as in Table 1.

For the cell B12, the formula used to find the equivalent windchill temperatures as “=0.045(5.27V0.5+10.450.28V)(Ta33)+33”.

Written as “=(0.045*((5.27*($A$12)^0.5)+10.45-(0.28*$A$12))*(B8-33))+33”. Here, B8 cell represents the value of ambient air temperature is 30 and $A$12 is an absolute cell reference which represent the value of cell A12 as 50. The result obtained in the cell B12 would be 62.6. Use the Fill command to copy the formula into cell C12, only the B8 cell in the formula is automatically substituted by C8, resulting in a value 55.0. Then, by click and drag the bottom corner of the C12 cell extend through J12 in row-wise. The result is obtained as in Table 1.

For the cell B13, the formula used to find the equivalent windchill temperatures as “=0.045(5.27V0.5+10.450.28V)(Ta33)+33”.

Written as “=(0.045*((5.27*($A$13)^0.5)+10.45-(0.28*$A$13))*(B8-33))+33”. Here, B8 cell represents the value of ambient air temperature is 30 and $A$13 is an absolute cell reference which represent the value of cell A13 as 60. The result obtained in the cell B13 would be 64.7. Use the Fill command to copy the formula into cell C13, only the B8 cell in the formula is automatically substituted by C8, resulting in a value 57.0. Then, by click and drag the bottom corner of the C13 cell extend through J13 in row-wise. The result is obtained as in Table 1.

For the cell B14, the formula used to find the equivalent windchill temperatures as “=0.045(5.27V0.5+10.450.28V)(Ta33)+33”.

Written as “=(0.045*((5.27*($A$14)^0.5)+10.45-(0.28*$A$14))*(B8-33))+33”. Here, B8 cell represents the value of ambient air temperature is 30 and $A$14 is an absolute cell reference which represent the value of cell A14 as 70. The result obtained in the cell B14 would be 66.1. Use the Fill command to copy the formula into cell C14, only the B8 cell in the formula is automatically substituted by C8, resulting in a value 58.2. Then, by click and drag the bottom corner of the C14 cell extend through J14 in row-wise. The result is obtained as in Table 1.

For the cell B15, the formula used to find the equivalent windchill temperatures as “=0.045(5.27V0.5+10.450.28V)(Ta33)+33”.

Written as “=(0.045*((5.27*($A$15)^0.5)+10.45-(0.28*$A$15))*(B8-33))+33”. Here, B8 cell represents the value of ambient air temperature is 30 and $A$15 is an absolute cell reference which represent the value of cell A15 as 80. The result obtained in the cell B15 would be 66.8. Use the Fill command to copy the formula into cell C15, only the B8 cell in the formula is automatically substituted by C8, resulting in a value 58.8. Then, by click and drag the bottom corner of the C15 cell extend through J15 in row-wise. The result is obtained as in Table 1.

Table 1 is created to shows a Wind Chill Temperature.

Table 1

Wind speed (Km/h)Ambient Temperature (°C)
-30-25-20-15-10-50510
20-47.6-41.2-34.8-28.4-22.0-15.6-9.2-2.83.6
30-54.6-47.7-40.7-33.8-26.8-19.9-12.9-6.01.0
40-59.4-52.0-44.7-37.4-30.0-22.7-15.4-8.1-0.7
50-62.6-55.0-47.4-39.8-32.2-24.7-17.1-9.5-1.9
60-64.7-57.0-49.2-41.5-33.7-25.9-18.2-10.4-2.7
70-66.1-58.2-50.3-42.5-34.6-26.8-18.9-11.0-3.2
80-66.8-58.8-50.9-43.0-35.1-27.2-19.3-11.3-3.4

Figure 2 shows a wind chill table in the excel sheet has obtained as similar to given Problem 14.19 in the textbook

Engineering Fundamentals: An Introduction to Engineering, Chapter 14, Problem 19P , additional homework tip  2

Conclusion:

Hence, a table is created for the windchill temperatures with the given range of an ambient air temperature, and wind speed have been explained using excel.

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
4th Order Rurka-Kutta Method This assignment will expand your skills on using Excel to perform some numerical analyses and plot your results. It is a means to familiarize you more with basic features in the spreadsheet. Preliminaries: a. Start a new file called R-K.xlsx (for this exercise, you may use the work you did in class but please be careful not to delete what you already have) b. Plan ahead Exercise 1: Using the 4" Order Ruga-Kutta Method that we recently learned, evaluate the following ordinary differential equation (ODE): dx f(t, x) = = x+t² dt At the initial position, X= 0, the time is = 0.1 minutes. Using a step size of, h = 0.05, what is the time spent to reach a position that is equal to 17 (Hint: use a O and b 1) Exercise 2 Using the 4th Order Ruga-Kutta Method, solve the following problem: The concentration of a certain non-reactive (conservative) chemical is given as a function of time by: de f(t,c) = = 30-3c Where c is the concentration and t is time. At the initial…
What percentage of all communication is screened out or changed by the receiver?”
In testing a certain kind or truck tire over rugged terrain, it is found that 20% of the trucks fail to complete the test run without a blowout. Of the next 13 trucks tested, find the probability that (a) from 2 to 6 have blowouts, (b) fewer than 4 have blowouts, and (c) more than 5 have blowouts. Click here to view page 1 of the table of binomial probability sums. Click here to view page 2 of the table of binomial probability sums. (a) The probability that from 2 to 6 trucks have blowouts is (Round to four decimal places as needed.)
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