## Test Bank For Thermodynamics An Engineering Approach 8Th edition By SI Units

Test Bank for Thermodynamics: An Engineering Approach, 8th edition by Yunus A. Cengel and Michael A. Boles, provides students with a comprehensive and thorough understanding of thermodynamics, and how it is applied in engineering. The textbook covers the fundamental principles of thermodynamics, including energy, work, heat, and the laws of thermodynamics. It also explores topics such as thermodynamic cycles, thermodynamic property relations, and mixtures of gases and vapors. The Test Bank includes a range of questions and exercises that help students to develop their understanding of thermodynamics and to apply it to real-world engineering problems. The Test Bank is an essential resource for instructors teaching courses on thermodynamics at the undergraduate level.

Multiple-Choice Test Problems

Chapter 2: * Energy, **Energy Transfer, and General Energy Analysis*

Çengel/Boles – Thermodynamics: An Engineering Approach, 8^{th} Edition

(Numerical values for solutions can be obtained by copying the EES solutions given and 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

(a) 1.5 kJ (b) 60 kJ (c) 750 kJ (d) 1800 kJ (e) 3600 kJ

*Answer* (d) 1800 kJ

**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

(a) 0.48 W/m2.∞C (b) 127 W/m2.∞C (c) 63.7 W/m2.∞C (d) 95 W/m2.∞C (e) 200 W/m2.∞C “

*Answer* (c) 63.7 W/m2.∞C

**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-m^{2} 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/m^{2}.∞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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