When studying thermodynamics, preparing for exams using the Thermodynamics: An Engineering Approach (8th Edition, in SI units) test bank can greatly increase the probability of understanding the content. This particular test bank has been developed specifically to assist students with an extensive selection of illustrative questions, which are similar to the problems encountered in the textbook and classmates’ assignments as well as midterm and final exams.
Advantages of using the test bank
This test bank contains assorted multiple-choice questions, short answers, and even complete solutions covering all the important constraints in thermodynamics. These include:
Energy Transfer: The examination of the different forms of energy which include kinetic, potential, mechanical work, and thermal to name but a few.
Energy Analysis: Disaggregate the complex systems so that mass and energy analysis can be done at the single system level and not at the whole system.
Properties of Pure Substances: Study the quantitative and qualitative characteristics of the substances and their changes in state under different thermodynamic parameters.
Second Law of Thermodynamics: Significant variables i.e. exergy, entropy, and other critical concepts related to engineering thermodynamics.
Power and Refrigeration Cycles: Energy efficiency of power systems starting from gas turbine through to refrigeration systems.
Interpreting the Benefits of the Test Bank to Learning
This edition of the test bank becomes intelligible as it assumes a detailed scope along with easy explanations, From this perspective, practicing with answers to these questions centers on:
Improving Problem-Solving Skills: The questions ask you to solve actual engineering problems. In solving these, you gain hands-on experience.
Testing Your Knowledge: The exercise is intended to improve your understanding and revive your knowledge of an already covered material while pointing out the topics that require further study.
Mastering the Procedure of Numerical Calculations: The proceedings of many questions in the test bank are step-wise, thus ideal for the improvement of your calculations in the course of thermodynamics.
Boosting up Confidence for the Exams: If you are well equipped with such preparatory materials as This test bank, you will get more confidence to sit for your examinations.
Tips for Using the Test Bank Successfully
To maximize the benefits derived from Thermodynamics: An Engineering Approach 8th Edition Test Bank, you need to:
Practice Regularly: Aim to schedule time every week when you go through several questions with the view of building your understanding gradually.
Review Solutions Thoroughly: All the solutions provide step-by-step procedures. It is useful to know the reason for every step in order to enhance the understanding of the concept.
Real-World Questions: A generous number of questions in the test bank can be found from real-world experiences which is very useful to engineering students.
Why Buy This Test Bank?
Purchasing with this test bank becomes a good decision for any student of Thermodynamics: An Engineering Approach (8th Edition, SI Units). This not only assists you in working for the exam but deepens your knowledge of the subject through practice since it consists of numerous exercises meant to test your knowledge and analytical capabilities. Be it a midterm, or final, or you just want to do practice — this test bank has everything that you need to succeed.
Summary
To summarize, the Test Bank for Thermodynamics: An Engineering Approach, 8th Edition (SI Units) is a well-structured approach and is full of resources that will help you to improve the fundamental concepts in the area of thermodynamics. It is much possible with regular exercising that one will be able to increase the problem-solving skills needed for the exam and ensure that the content is known.
Test Bank For Thermodynamics An Engineering Approach 8Th edition By SI Units
Multiple-Choice Test Problems
Chapter 2: Energy, Energy Transfer, and General Energy Analysis
Çengel/Boles – Thermodynamics: An Engineering Approach, 8th Edition
(Numerical values for solutions can be obtained by copying the EES solutions given pasting them on a blank EES screen, and pressing the Solve command. Similar problems and their solutions can be obtained easily by modifying numerical values.)
Chap2-1 Heating by Resistance Heater
A 1.5-kW electric resistance heater in a room is turned on and kept on for 20 min. The amount of energy transferred to the room by the heater is
Solution Solved by EES Software. Solutions can be verified by copying and pasting the following lines on a blank EES screen.
We= 1.5 “kJ/s”
time=20*60 “s”
E_total=We*time “kJ”
“Some Wrong Solutions with Common Mistakes:”
W1_Etotal=We*time/60 “using minutes instead of s”
W2_Etotal=We “ignoring time”
Chap2-2 Heat Supplied by Vacuum Cleaner
A 200 W vacuum cleaner is powered by an electric motor whose efficiency is 70%. (Note that the electric motor delivers 200 W of net mechanical power to the fan of the cleaner). The rate at which this vacuum cleaner supplies energy to the room when running is
(a) 140 W(b) 200 W (c)286 W(d) 360 W (e) 86 W
Answer(c)286 W
Solution Solved by EES Software. Solutions can be verified by copying and pasting the following lines on a blank EES screen.
Eff=0.70
W_vac=0.2 “kW”
E=W_vac/Eff“kJ/s”
“Some Wrong Solutions with Common Mistakes:”
W1_E=W_vac*Eff “Multiplying by efficiency”
W2_E=W_vac “Ignoring efficiency”
W3_E=E-W_vac “Heat generated by the motor”
Chap2-3 Heat Convection
A 40-cm-long, 0.6-cm-diameter electric resistance wire is used to determine the convection heat transfer coefficient in air at 25∞C experimentally. The surface temperature of the wire is measured to be 150∞C when the electric power consumption is 90 W. If the radiation heat loss from the wire is calculated to be 30 W, the convection heat transfer coefficient is
Solution Solved by EES Software. Solutions can be verified by copying and pasting the following lines on a blank EES screen.
L=0.4 “m”
D=0.006 “m”
A=pi*D*L “m^2”
We=90 “W”
Ts=150 “C”
Tf=25 “C”
We-30= h*A*(Ts-Tf) “W”
“Some Wrong Solutions with Common Mistakes:”
We-30= W1_h*(Ts-Tf) “Not using area”
We-30= W2_h*(L*D)*(Ts-Tf) “Using D*L for area”
We 30= W3_h*A*(Ts-Tf) “Adding Q_rad instead of subtracting”
We= W4_h*A*(Ts-Tf) “Disregarding Q_rad”
Chap2-4 Heat Convection and Radiation
A 1.5-m2 black surface at 120∞C is losing heat to the surrounding air at 30∞C by convection with a convection heat transfer coefficient of 18 W/m2.∞C, and by radiation to the surrounding surfaces at 10∞C. The total rate of heat loss from the surface is
(a) 1483 W (b) 2430 W (c) 2448 W (d) 3913 W (e) 2609 W
Answer(d) 3913 W
Solution Solved by EES Software. Solutions can be verified by copying and pasting the following lines on a blank EES screen.
sigma=5.67E-8 “W/m^2.K^4”
eps=1
A=1.5 “m^2”
h_conv=18 “W/m^2.C”
Ts=120 “C”
Tf=30 “C”
Tsurr=10 “C”
Q_conv=h_conv*A*(Ts-Tf)“W”
Q_rad=eps*sigma*A*((Ts 273)^4-(Tsurr 273)^4) “W”
Q_total=Q_conv Q_rad “W”
“Some Wrong Solutions with Common Mistakes:”
W1_Ql=Q_conv “Ignoring radiation”
W2_Q=Q_rad “ignoring convection”
W3_Q=Q_conv eps*sigma*A*(Ts^4-Tsurr^4) “Using C in radiation calculations”
W4_Q=Q_total/A “not using area”
Chap2-5 Heat Conduction
Heat is transferred steadily through a 0.15-m thick 3 m by 5 m wall whose thermal conductivity is 1.2 W/m.∞C. The inner and outer surface temperatures of the wall are measured to be 18∞C to 4∞C. The rate of heat conduction through the wall is
(a) 112 W (b) 3360 W (c)2640 W (d) 38 W (e) 1680 W
Answer(e) 1680 W
Solution Solved by EES Software. Solutions can be verified by copying and pasting the following lines on a blank EES screen.
A=3*5 “m^2”
L=0.15 “m”
T1=18 “C”
T2=4 “C”
k=1.2“W/m.C”
Q=k*A*(T1-T2)/L “W”
“Some Wrong Solutions with Common Mistakes:”
W1_Q=k*(T1-T2)/L “Not using area”
W2_Q=k*2*A*(T1-T2)/L“Using areas of both surfaces”
W3_Q=k*A*(T1 T2)/L“Adding temperatures instead of subtracting”
W4_Q=k*A*L*(T1-T2) “Multiplying by thickness instead of dividing by it”
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