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Updated October 2012

SOLUTION MANUAL

CHAPTER 3

Fundamentals of

Thermodynamics

Borgnakke Sonntag

8e

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In-Text Concept Questions

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3.a

In a complete cycle what is the net change in energy and in volume?

For a complete cycle the substance has no change in energy and therefore no storage,

so the net change in energy is zero.

For a complete cycle the substance returns to its beginning state, so it has no

change in specific volume and therefore no change in total volume.

3.b

Explain in words what happens with the energy terms for the stone in Example 3.3.

What would happen if it were a bouncing ball falling to a hard surface?

In the beginning all the energy is potential energy associated with the

gravitational force. As the stone falls the potential energy is turned into kinetic energy and in the impact the kinetic energy is turned into internal energy of the stone

and the water. Finally the higher temperature of the stone and water causes a heat

transfer to the ambient until ambient temperature is reached.

With a hard ball instead of the stone the impact would be close to elastic

transforming the kinetic energy into potential energy (the material acts as a spring)

that is then turned into kinetic energy again as the ball bounces back up. Then the  ball rises up transforming the kinetic energy into potential energy (mgZ) until zero

velocity is reached and it starts to fall down again. The collision with the floor is not

perfectly elastic so the ball does not rise exactly up to the original height losing a

little energy into internal energy (higher temperature due to internal friction) with every bounce and finally the motion will die out. All the energy eventually is lost by

heat transfer to the ambient or sits in lasting deformation (internal energy) of the

substance.

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3.c

Make a list of at least 5 systems that store energy, explaining which form of energy.

A spring that is compressed. Potential energy (1/2) kx2

A battery that is charged. Electrical potential energy. V Amp h A raised mass (could be water pumped up higher) Potential energy mgH

A cylinder with compressed air. Potential (internal) energy like a spring.

A tank with hot water. Internal energy mu

A fly-wheel. Kinetic energy (rotation) (1/2) I2

A mass in motion. Kinetic energy (1/2) mV2

3.d

A constant mass goes through a process where 100 J of heat transfer comes in and

100 J of work leaves. Does the mass change state?

Yes it does.

As work leaves a control mass its volume must go up, v increases

As heat transfer comes in an amount equal to the work out means u is

constant if there are no changes in kinetic or potential energy.

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3.e

The electric company charges the customers per kW-hour. What is that in SI

units?

Solution:

The unit kW-hour is a rate

multiplied with time. For the standard SI units the rate of

energy is in W and the time is

in seconds. The integration in

Eq.4.21 becomes

1 kW- hour = 1000 W  60 min

hour   hour  60

s

min  = 3 600 000 Ws

= 3 600 000 J = 3.6 MJ

3.f

Torque and energy and work have the same units (N m). Explain the difference.

Solution:

Work = force   displacement, so units are N  m. Energy in transfer Energy is stored, could be from work input 1 J = 1 N m

Torque = force   arm, static, no displacement needed

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3.g

What is roughly the relative magnitude of the work in the process 1-2c versus the

process 1-2a shown in figure 3.15?

By visual inspection the area below the curve 1-2c is roughly 50% of the

rectangular area below the curve 1-2a. To see this better draw a straight line from state 1 to point f on the axis. This curve has exactly 50% of the area below it.

3.h

Helium gas expands from 125 kPa, 350 K and 0.25 m3 to 100 kPa in a polytropic  process with n = 1.667. Is the work positive, negative or zero?

The boundary work is: W =  P dV

P drops but does V go up or down?

The process equation is: PVn = C

so we can solve for P to show it in a P-V diagram

P = CV-n

as n = 1.667 the curve drops as V goes up we see

V2 > V1  giving dV > 0

and the work is then positive.

P

V W

1

2

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3.i

An ideal gas goes through an expansion process where the volume doubles.

Which process will lead to the larger work output: an isothermal process or a

polytropic process with n = 1.25?

The process equation is: PVn = C

The polytropic process with n = 1.25 drops the pressure faster than the isothermal

process with n = 1 and the area below the curve is then smaller.

P

V W

1

2

n = 1

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3.j

Water is heated from 100 kPa, 20oC to 1000 kPa, 200oC. In one case pressure is raised at T = C, then T is raised at P = C. In a second case the opposite order is

done. Does that make a difference for 1Q2 and 1W2?

Yes it does. Both 1Q2 and 1W2 are process dependent. We can illustrate

the work term in a P-v diagram.

P

T

V L

Cr.P.

S

1000 a

20 200

1

2

100

T C.P.

v

a

P

v

a

180 C

2 2

20 C 20

200

1 100

1000 200 C 100

1553 kPa 1000

1

b

b

In one case the process proceeds from 1 to state “a” along constant T then from “a” to state 2 al