Chapter 2: Sections 4 and 5

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Chapter 2: Sections 4 and 5

Lecture 03:

1
st

Law of Thermodynamics

and Introduction to Heat Transfer

Today’s Objectives:

Be able to recite the 1
st

Law of Thermodynamics

Be able indicate the sign conventions of the Work and Heat

Be able to distinguish between conduction, convection, and

Be able to calculate heat flow rate by conduction

Be able to calculate heat flow rate by convection

Bea able to calculate heat flow rate by radiation

Be able to solve Work
-
Energy system problems using the 1
st

Law.

Homework Assignment:

Read Chap 2. Sections 6 and 7

From Chap 2: Problems
49, 53,61, 68

Heat,
Q
: An interaction which causes a change in energy
due to differences in Temperature.

Heat Flow Rate, : the rate at which heat flows into or
out of a system,
dQ
/
dt
.

Heat flux, : the heat flow rate per unit surface area.

3

Sec

2.4: Energy Transfer by Heat

Q
Q Q dt

q
3 Types of Heat Transfer

Conduction

Convection

4

Sec

2.4.2: Heat Transfer Modes

Conduction
:

Heat transfer through a stationary media due to collision
of atomic particles passing momentum from molecule to
molecule.

Fourier’s Law

dT
Q A
dx

 
where
κ

is the thermal conductivity of the material.

Types of Heat Transfer

5

Sec

2.4.2: Heat Transfer Modes

Convection:

Heat transfer due to movement of
matter (fluids). Molecules
carry
away kinetic energy with them as
a fluid mixes.

Newton’s Law of Cooling
:

c b f
Q h A T T
 
h
c

= coefficient of convection

(
An empirical value, that depends on the material, the velocity, etc.)

Types of Heat Transfer

6

Sec

2.4.2: Heat Transfer Modes

:

Heat transfer which occurs as
matter exchanges
other
matter.

Stefan
-
Boltzmann Law:

4
b
Q AT


T
b

= absolute surface temperature

ε = emissivity of the surface

σ = Stefan
-
Boltzmann constant

Summary of Heat Transfer Methods

7

Sec

2.4.2: Heat Transfer Modes

dT
Q A
dx

 
where
A

is area

κ

is thermal conductivity

dT
/dx

Conduction
:

Convection
:

c b f
Q h A T T
 
where
A

is area

h
c

is the convection coefficient

T
b

-
T
f

is the difference

between th
e

body and the fluid temp.

4
b
Q AT


where
T
b

is absolute surface temperature

ε

is emissivity of the surface

σ

is Stefan
-
Boltzmann constant

A

is surface area

Example (2.45):
An oven wall consists of a 0.25” layer of steel

(

S
=8.7 BTU/(
h

ft

o
R
) )and a layer of brick
(

B
=0.42 BTU/(
h

ft

o
R
) ). At
steady state, a temperature decrease of 1.2oF occurs through the steel
layer. Inside the oven, the surface temperature of the steel is 540
o
F. If the
temperature of the outer wall of the brick must not exceed 105
o
F
,
determine the minimum thickness of brick needed.

8

9

Example Problem (2.50)

diameter of 0.5 m transfers energy by radiation from its outer surface at a rate
of 150 W. If the probe does not receive radiation from the sun or deep space,
what is the surface temperature in K? Let
ε
=0.8.

10

Recall from yesterday: by conventio
n

Heat, Q:

11

Sec

2.4: Energy Transfer by Heat

(This is reversed from the sign convention for work often used in
Physics. It is an artifact from engine

calculations.)

Q > 0
: Heat transferred
TO

the system

Q < 0
: Heat transferred
FROM

the system

W > 0
: Work done
BY

the system

W < 0
: Work done
ON

the system

Work, W:

+Q

+W

system

First Law of Thermodynamics

12

Sec

2.5: Energy Balance for Closed Systems

Energy is conserved

“The change in the internal energy of
a closed system is equal to the sum
of the amount of heat energy
supplied to the system and the
work done on the system”

E within

the system

net

Q

input

net

W
output

[ ]

=

[ ]

+

[ ]

system in out
E PE KE U Q W
        
W
Q
dE

where

denotes path dependent derivatives

The First Law of Thermodynamic
s

13

Sec

2.5: Energy Balance for Closed Systems

system in out
E Q W
  
Q
in

W
out

system

Δ
E

The 1st Law of Thermodynamics is an expanded form of the
Law of Conservation of Energy, also known by other name an
Energy Balance.

Example (2.55):
A mass of 10 kg undergoes a process during with there is
heat transfer from the mass at a rate of 5 kJ per kg, an elevation decrease
of 50 m, and an increase in velocity from 15 m/s to 30 m/s. The specific
internal energy decreases by 5 kJ/kg and the acceleration of gravity is
constant at 9.7 m/s
2
. Determine the work for the process, in kJ.

14

Example Problem (2.63)

A gas is compressed in a piston cylinder assembly form
p
1

= 2 bar to
p
2

= 8 bar,
V
2

= 0.02 m
3

in a process during which the relation between pressure and
volume is
pV
1.3
=
constant
. The mass of the gas is 0.2 kg . If the specific internal
energy of the gas increase by 50 kJ/kg during the process, deter the heat
transfer in kJ. KE and PE changes are negligible.

15

16

17

Example (2.70):
A gas is contained in a vertical piston
-
cylinder assembly
by a piston weighing 1000 lb
f

and having a face area of 12 in
2
. The
atmosphere exerts a pressure of 14.7 psi on the top of the piston. An
electrical resistor transfers energy to the

gas in the amount of 5 BTU as the elevation

of the piston increases by 2 ft. The piston and

cylinder are poor thermal conductors and

friction can be neglected. Determine the

change in internal energy of the gas, in BTU,

assuming it is the only significant internal

energy change of any component present.

P
atm
=14.7 psi

h

= 2 ft

A
piston

= 12 in
2

W
piston

= 1000
lb
f

W
elec
=
-

5 BTU

End of Lecture 03

Slides which follow show solutions to example
problems

18

Example (2.45):
An oven wall consists of a 0.25” layer of steel

(

S
=8.7 BTU/(
h

ft

o
R
) )and a layer of brick
(

B
=0.42 BTU/(
h

ft

o
R
) ). At
steady state, a temperature decrease of 1.2oF occurs through the steel
layer. Inside the oven, the surface temperature of the steel is 540
o
F. If the
temperature of the outer wall of the brick must not exceed 105
o
F
,
determine the minimum thickness of brick needed.

19

Solution to Example (2.45):

20

m i
Steel Steel Steel
Steel
T T
dT
Q A A
dx L
 
 

   
 
 
Steel Brick
Q Q
A A
   

   
   
Heat flow rate through steel:

0
m
Brick Brick
Brick
T T
Q A
L

 

 
 
 
Heat flow rate through steel:

Flow:

B
oth

materials have the same cross sectional area here

and the heat flow rate through each is the same.

Solution to Example (2.45):

21

0
m i m
Steel Brick
Steel Brick
T T T T
L L
 
   
 

   
   
Steel
i
m
m
Steel
Brick
Brick
L
T
T
T
T
L

0

Therefore:

and solving for
L
bric
k

with
κ
Brick

=
0.42
BTU/(
h

ft

o
R
)

κ
Steel

=
8.7
BTU/(
h

ft

o
R

T
i

=
540
o
F

T
m
=
538.8
o
F
T
0
= 105
o
F

and
L
Steel

= 0.25 in … solve for
L
brick

Solution to Example (2.45):

22

Therefore:

inches

36
.
4
25
.
0
2
.
1
105
8
.
538
7
.
8
42
.
0

Brick
L
Example Problem (2.50)

diameter of 0.5 m transfers energy by radiation from its outer surface at a rate
of 150 W. If the probe does not receive radiation from the sun or deep space,
what is the surface temperature in K? Let
ε
=0.8.

23

Solution:

4
b
Q AT


where:
ε

= 0.8

σ

= 5.67 x 10
-
8

W/m
2
•K
4

d

= 0.5m

dQ
/
dt

= 150 W

Q
out

3
1
6
sphere
A d

therefore:

3 3 3
1 1
(0.5 ) 0.6545
6 6
sphere
A d m m
 
  
4
b
Q AT


1/4
b
Q
T
A

 

 
 
0.25
1/4
8 2
2 4
150
266.6
0.8(5.67 10 )(.6545 )
b
Q W
T K
W
A
m
m K


 
 
 
  
 
 
 
 

 

 
Example (2.55):
A mass of 10 kg undergoes a process during with there is
heat transfer from the mass at a rate of 5 kJ per kg, an elevation decrease
of 50 m, and an increase in velocity from 15 m/s to 30 m/s. The specific
internal energy decreases by 5 kJ/kg and the acceleration of gravity is
constant at 9.7 m/s
2
. Determine the work for the process, in kJ.

24

system in out
E PE KE U Q W
        
Solution:

Principle: 1
st

Law of Thermodynamics

given:
m

= 10 kg

Q
in
/m

=
-
5 kJ/kg
Δ
h
=
-
50 m

v
1

= 15 m/s
v
2

= 30 m/s

Δ
U /
m=
-

5kJ/kg
g

= 9.7 m/s
2

also

PE mg h
  
2 2
2 1
1 1
2 2
KE mv mv
  
2
2
1
(10 )(9.7/)( 50 )
/
N
kg m s m
kg m s
 

2 2 2 2
2
1
(10 )(30 15 )/
2/
N
kg m s
kg m s
 

4850 4.85
1000
J kJ
N m kJ
N m J
    

3375 3.375
1000
J kJ
N m kJ
N m J
  

Example (2.55):
A mass of 10 kg undergoes a process during with there is
heat transfer from the mass at a rate of 5 kJ per kg, an elevation decrease
of 50 m, and an increase in velocity from 15 m/s to 30 m/s. The specific
internal energy decreases by 5 kJ/kg and the acceleration of gravity is
constant at 9.7 m/s
2
. Determine the work for the process, in kJ.

25

out in
W Q PE KE U
   
Solution continued:

(/)( ) ( 5/)(10 ) 50
U U m m kJ kg kg kJ
      
( 50 ) ( 4.85 ) (3.375 ) ( 50 )
kJ kJ kJ kJ
      
(/)( ) ( 5/)(10 ) 50
in in
Q Q m m kJ kg kg kJ
    
Solving for the Work done by the system:

1.475
kJ

Example Problem (2.63)

A gas is compressed in a piston cylinder assembly form
p
1

= 2 bar to
p
2

= 8 bar,
V
2

= 0.02 m
3

in a process during which the relation between pressure and
volume is
pV
1.3
=
constant
. The mass of the gas is 0.2 kg . If the specific internal
energy of the gas increase by 50 kJ/kg during the process, deter the heat
transfer in kJ. KE and PE changes are negligible.

26

Solution: starting with the 1
st

Law of Thermodynamics

in out
PE KE U Q W
      
where:
Δ
KE
=0
Δ
PE

= 0
Δ
U/
m = 50 kJ/kg
m

= 0.2 kg

p
1

= 2
bar
p
2

= 8
bar
V
1

=
?
V
2

= 0.02 m
3

also:
pV
1.3
=
constant

therefore:

1.3 1.3
1 1 2 2
pV p V

1/1.3
1/1.3
3 3
2
1 2
1
8
0.02 0.0581
2
p
bar
V V m m
p bar
 
 
  
 
 
 
 
Example Problem (2.63)

A gas is compressed in a piston cylinder assembly form
p
1

= 2 bar to
p
2

= 8 bar,
V
2

= 0.02 m
3

in a process during which the relation between pressure and
volume is
pV
1.3
=
constant
. The mass of the gas is 0.2 kg . If the specific internal
energy of the gas increase by 50 kJ/kg during the process, deter the heat
transfer in kJ. KE and PE changes are negligible.

27

Solution continued:

also:

therefore:

1.3 1.3
1 1 2 2
pV p V

1/1.3
1/1.3
3 3
2
1 2
1
8
0.02 0.0581
2
p
bar
V V m m
p bar
 
 
  
 
 
 
 
1.3 3.9 1.3
0.0495/(0.0495 )
p V bar m V

  
1.3 1.3 3 1.3 3.9
2 2
(8 )(0.02 ) 0.0495
pV p V bar m bar m
   
28

3.9
3 0.3 3 0.3
0.0495
(0.02 ) (0.581 )
0.3
bar m
m m
 

 
 
 

2
2 2
1 1
1
3.9
3.9 1.3 0.3
0.0495
((0.0495 ) )
0.3
V
V V
V V
V
bar m
W PdV bar m V dV V
 

   

 
so work done is:

2
3
100/1
0.338 33.8
1 1
kN m kJ
bar m kJ
bar kN m
    

3.9
0.9 0.9
0.0495
3.224 1.177
0.3
bar m
m m
 

 
 
 

3.9
0.9 0.9
0.0495
3.224 1.177
0.3
bar m
m m
 

 
 
 

29

Internal Energy is given as

Finally back at the 1
st

Law:

gives

in out
Q PE KE U W
      
0 0 (10 ) ( 33.78 )
kJ kJ
    
(/)( ) (50/)(0.2 ) 10
U U m m kJ kg kg kJ
    
23.78
kJ
 
in out
PE KE U Q W
    
30

Example (2.70):
A gas is contained in a vertical piston
-
cylinder assembly by a
piston weighing 1000 lb
f

and having a face area of 12 in
2
. The atmosphere exerts a
pressure of 14.7 psi on the top of the piston. An electrical resistor transfers energy
to the gas in the amount of 5 BTU as the elevation

of the piston increases by 2 ft. The piston and

cylinder are poor thermal conductors and friction

can be neglected. Determine the change in internal

energy of the gas, in BTU, assuming it is the only

significant internal energy change of any component

present
.

P
atm
=14.7 psi

h

= 2 ft

A
piston

= 12 in
2

W
piston

= 1000
lb
f

W
elec
=
-

5 BTU

Solution: Apply the 1
st

law of thermodynamics

in out
PE KE U Q W
    
in out
U Q W PE KE
    
31

where
mg

= 1000
lb
f

A

= 12 in
2

Δ
h

= 2
ft

W
elec_in

= 5 BTU

Because of the statement “poor
thermal
conductors”, it can be assumed that
this is an adiabatic process (Q = 0) and we will also assume that the process
occurs as a slow quasi
-
equilibrium process in which case the kinetic energy
terms will also be small (
Δ
KE = 0). Finally, since the piston floats on the
contained gas, the outside atmospheric pressure maintains a constant
pressure on the cylinder…so this is a constant pressure process (isobaric)

therefore:

0
in
Q

2
1
2 1
( )
V
PV
V
W pdV p V V
  

(1000 )(2 ) 2000
f f
PE mg h lb ft ft lb
     
0
KE
 
(for constant pressure)

5
elect
W BTU
 
(neg. since its put into the system)

32

for equilibrium:

2 1
V V A h
  
2
1
2 3
2 1
1
( ) (98.2/)(288 ) 2357
12
V
PV f f
V
ft
W pdV p V V lb in in lb ft
in
     

0
F
 
F
top
=
p
atm

A

W=1000lb
f

F
bottom
=p A

0
atm
pA p A W
  
0
bottom top
F F W
  
2
2
1000
14.7/98.03
12
atm
W lbf
p p lbf in psi
A in
    
and the increase in Volume:

2 3
2 1
12
12 (2 ) 288
1
in
V V in ft in
ft
  
therefore the work done by the gas was positive work by the system

33

Returning to the 1
st

law:

in out
U Q W PE KE
    
0 ( ) 0
PV elec
U W W PE
     
778
(2357 5 ) (2000 )
1
f
f
ft lb
U ft lbf BTU ft lb
BTU

      
467
U ft lbf
   
1
467 0.60
778
f
BTU
U ft lbf BTU
ft lb
     