2-Chapter 2-problems - slides notes PDF

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Solving Problems of Ch 2.

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Problem 1 Water is being heated in a closed pan on top of a range while being stirred by a paddle wheel. During the process, 30 kJ of heat is transferred to the water, and 5 kJ of heat is lost to the surrounding air. You move the paddle-wheel work by 500 N·m. Determine the final energy of the system if its initial energy is 10 kJ. 2

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Answer

Q = 30-5= 25 KJ W= - 500 N.m= 0.5 KJ E1=10 KJ E2  E1  Q  W

E2 - 10 = 25 – (-0.5) E2 = 35.5 KJ 4

Problem 2 Consider a room that is initially at the outdoor temperature of 20°C. The room contains a 100 W lightbulb, a 110 W TV set, a 200 W refrigerator, and a 1000 W iron. Electricity cross the wall. Assuming no heat transfer through the walls, determine the rate of increase of the energy content of the room when all of these electric devices are on. 5

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Answer Light= 100 W, TV= 110 W, Refrig.= 200 W, Iron= 1000 W, Q=0,  E=?

E 2  E1  Q  W since no energy is leaving the room in any form

Q  0

E 2  E1  W W  100  110  200 1000 1410 w

Electricity access to room as work

dE / dt  E  (1410)  1410 w 6

Problem 3

A fan is to accelerate air from rest to a velocity of 10 m/s at a rate of 4 m3/s. Determine the minimum power that must be supplied to the fan. Take the density of air to be 1.18 kg/m3. Consider the changes in only kinetic energy.

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Answer

E2  E1  Q W E2  E1  KE

W  KE 

Q0

1 2 mair (v2  v12 ) 2

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Problem 4

During steady-state operation, a gearbox receives power 60 kW through the input shaft and delivers power through the output shaft. For the gearbox as the system, the rate of energy transfer by convection to outside is 1.2 KJ/s. Calculate the power delivered through the output shaft of the gearbox in kW for steady state operation. 9

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Solution

Q= -1.2KJ/s

W 2

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Solution Steady state operation

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Problem 5

Example 1 Again

The initial pressure is 3 bar, the initial volume is 0.1 m3, and the final volume is 0.2 m3. Determine the work for the process, in kJ, if the work occurs as polytropic process with cases: (a) n=1.5, (b) n=0, (c) n=1. 13

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Solution   

P1= 3 bar, V1= 0.1 m3, V2= 0.2 m3. polytropic process: (a) n=1.5, (b) n=0, (c) n=1 W=? in kJ,

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  

P1= 3 bar, V1= 0.1 m3, V2= 0.2 m3. polytropic process: (a) n=1.5, (b) n=0, (c) n=1 W=? in kJ,

PV  const. n

P1V1  P2V2 n

n

V P 2  P1  1  V2

n

  

For n = 1.5

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For n = 0

W =30 kJ. 16

(c) n=1

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Problem 6 Four kilograms of a certain gas is contained within a piston–cylinder assembly. The gas undergoes a process for which the pressure–volume relationship is

The initial pressure is 3 bar, the initial volume is 0.1 m3, and the final volume is 0.2 m3. The change in specific internal energy of the gas in the process is u2 - u1= 4.6 kJ/kg. There are no significant changes in kinetic or potential energy. Determine the net heat transfer for the process, in kJ. 18

P1= 3 bar, V1= 0.1 m3, V2= 0.2 m3. u2 - u1= - 4.6 kJ/kg. KE=0, PE=0, Q=? kJ. 19

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Solution

W=17.6 KJ (from previous example)

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Credit Quiz Q: On a hot summer day, a student turns his fan on when he leaves his room in the morning. When he returns in the evening, will the room be warmer or cooler than the neighboring rooms? Why? Assume all the doors and windows are kept closed and room thermally isolated. A: Warmer. Because energy is added to the room air in the form of electrical work. 21

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Credit Quiz How to calculate the fist low of thermodynamics in adiabatic system.

Answer In adiabatic system there is no energy transfer by heat. Q=0

E2  E1  Q  W

E2  E1  ( W ) E2  E1  W

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Questions End of Ch 2- Problems Thank you

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