Transcript Lecture 03: Heat and the 1 st Law of Thermodynamics

Chapter 2: Sections 4 and 5
Lecture 03:
1st Law of Thermodynamics
and Introduction to Heat Transfer
Today’s Objectives:
• Be able to recite the 1st Law of Thermodynamics
• Be able indicate the sign conventions of the Work and Heat
• Be able to distinguish between conduction, convection, and
radiation.
• 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 1st Law.
Reading Assignment:
• Read Chap 2. Sections 6 and 7
Homework Assignment:
From Chap 2: Problems 49, 53,61, 68
Sec 2.4: Energy Transfer by Heat
Heat, Q: An interaction which causes a change in energy
due to differences in Temperature.
Heat Flow Rate, Q: the rate at which heat flows into or
out of a system, dQ/dt.
Heat flux, q: the heat flow rate per unit surface area.
Q 
 Q dt
3
Sec 2.4.2: Heat Transfer Modes
4
3 Types of Heat Transfer
Conduction
Radiation
Convection
Conduction:
Heat transfer through a stationary media due to collision
of atomic particles passing momentum from molecule to
molecule.
Fourier’s Law
Q   A
dT
dx
where κ is the thermal conductivity of the material.
Sec 2.4.2: Heat Transfer Modes
5
Types of Heat Transfer
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:
Q  h c A  Tb  T f

hc = coefficient of convection
(An empirical value, that depends on the material, the velocity, etc.)
Sec 2.4.2: Heat Transfer Modes
6
Types of Heat Transfer
Radiation :
Heat transfer which occurs as
matter exchanges
Electromagnetic radiation with
other matter.
Stefan-Boltzmann Law:
Q   A Tb
4
Tb = absolute surface temperature
ε = emissivity of the surface
σ = Stefan-Boltzmann constant
Sec 2.4.2: Heat Transfer Modes
7
Summary of Heat Transfer Methods
Conduction:
Q   A
where A is area
κ is thermal conductivity
dT/dx is temperature gradient
dT
dx
Convection:
Q  h c A  Tb  T f
Radiation:
Q   A T
4
b

where A is area
hc is the convection coefficient
Tb -Tf is the difference between the
body and the fluid temp.
where Tb is absolute surface temperature
ε is emissivity of the surface
σ is Stefan-Boltzmann constant
A is surface area
8
Example (2.45): An oven wall consists of a 0.25” layer of steel
(S=8.7 BTU/(hftoR) )and a layer of brick (B=0.42 BTU/(hftoR) ). At
steady state, a temperature decrease of 1.2oF occurs through the steel
layer. Inside the oven, the surface temperature of the steel is 540oF. If the
temperature of the outer wall of the brick must not exceed 105 oF,
determine the minimum thickness of brick needed.
9
10
Example Problem (2.50)
At steady state, a spherical interplanetary electronics laden probe having a
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.
Sec 2.4: Energy Transfer by Heat
11
Recall from yesterday: by convention
Work, W:
W > 0 : Work done BY the system
W < 0 : Work done ON the system
(This is reversed from the sign convention for work often used in
Physics. It is an artifact from engine calculations.)
Heat, Q:
Q > 0 : Heat transferred TO the system
Q < 0 : Heat transferred FROM the system
system
+Q
+W
Sec 2.5: Energy Balance for Closed Systems
12
First Law of Thermodynamics
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
 E system   P E   K E   U  Q in  W o u t
dE   Q   W
where  denotes path dependent derivatives
Sec 2.5: Energy Balance for Closed Systems
13
The First Law of Thermodynamics
The 1st Law of Thermodynamics is an expanded form of the
Law of Conservation of Energy, also known by other name an
Energy Balance.
 E system  Q in  W o u t
ΔE
Qin
system
Wout
14
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/s2. Determine the work for the process, in kJ.
15
Example Problem (2.63)
A gas is compressed in a piston cylinder assembly form p1 = 2 bar to p2 = 8 bar,
V2 = 0.02 m3 in a process during which the relation between pressure and
volume is pV1.3 = 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.
16
17
Example (2.70): A gas is contained in a vertical piston-cylinder assembly
by a piston weighing 1000 lbf and having a face area of 12 in2. The
atmosphere exerts a pressure of 14.7 psi on the top of the piston. An
electrical resistor transfers energy to the
Patm=14.7 psi
gas in the amount of 5 BTU as the elevation
of the piston increases by 2 ft. The piston and
h = 2 ft
cylinder are poor thermal conductors and
Apiston = 12 in2
friction can be neglected. Determine the
Wpiston = 1000 lbf
change in internal energy of the gas, in BTU,
assuming it is the only significant internal
energy change of any component present.
Welec= - 5 BTU
18
End of Lecture 03
• Slides which follow show solutions to example
problems
19
Example (2.45): An oven wall consists of a 0.25” layer of steel
(S=8.7 BTU/(hftoR) )and a layer of brick (B=0.42 BTU/(hftoR) ). At
steady state, a temperature decrease of 1.2oF occurs through the steel
layer. Inside the oven, the surface temperature of the steel is 540oF. If the
temperature of the outer wall of the brick must not exceed 105 oF,
determine the minimum thickness of brick needed.
20
Solution to Example (2.45):
Heat flow rate through steel:
 T m  Ti 
   S teel A
   S teel A 

dx
L
 S teel 
dT
Q S teel
Heat flow rate through steel:
Q B rick
 T0  Tm 
   B rick A 

L
 B rick 
Steady State Heat Flow:
Both materials have the same cross sectional area here
and the heat flow rate through each is the same.
Q 
Q 




A
A

 S teel

 B rick
21
Solution to Example (2.45):
Therefore:
 S teel
 T m  Ti 
 T0  Tm 

   B rick 

L
L
 S teel 
 B rick 
and solving for Lbrick
L Brick 
 Brick  T 0  T m 
 Steel

 L Steel
 T m  Ti 
with κBrick = 0.42 BTU/(hftoR)
κSteel = 8.7 BTU/(hftoR
Ti = 540oF
Tm= 538.8oF
T0= 105oF
and LSteel = 0.25 in … solve for Lbrick
22
Solution to Example (2.45):
Therefore:
 0 . 42
L Brick  
 8 .7
  538 . 8  105 
0 . 25   4 . 36 inches


1 .2


23
Example Problem (2.50)
At steady state, a spherical interplanetary electronics laden probe having a
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.
Solution:
where:
Q   A Tb
4
A sp h ere 
ε
σ
d
dQ/dt
1
6
d
3
Qout
= 0.8
= 5.67 x 10-8 W/m2•K4
= 0.5m
= 150 W
1
1
3
3
3
therefore: A sp h ere   d   (0 .5 m )  0 .6 5 4 5 m
6
6
1/ 4
Q   A T
 Q 
Tb  


A


1/ 4
4
b

 Q 
Tb  


A






150W


W
8
2
 0.8(5.67  10
)(.6545 m ) 
2
4
m K


0.25
 266. 6 K
24
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/s2. Determine the work for the process, in kJ.
Solution:
Principle: 1st Law of Thermodynamics
 E system   P E   K E   U  Q in  W o u t
given:
m = 10 kg
Qin/m = -5 kJ/kg
Δh=-50 m
v1 = 15 m/s
v2 = 30 m/s
ΔU /m= - 5kJ/kg
g = 9.7 m/s2
also
PE  mg h
 (1 0 kg )(9 .7 m / s )(  5 0 m )
2
 4850 N  m
1
KE 
J
kJ
N  m 1000 J
2
1N
kg  m / s
2
  4 .8 5 kJ

1
m v2 
2
1
2
m v1
2
(1 0 kg )(3 0  1 5 ) m / s
2
2
2
2
 3375 N  m
J
kJ
N  m 1000 J
2
N
kg  m / s
2
 3 .3 7 5 kJ
25
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/s2. Determine the work for the process, in kJ.
Solution continued:
 U  (  U / m )( m )  (  5 kJ / kg )(1 0 kg )   5 0 kJ
Q in  ( Q in / m )( m )  (  5 kJ / kg )(1 0 kg )   5 0 kJ
Solving for the Work done by the system:
W o u t  Q in   P E   K E   U
 (  5 0 kJ )  (  4 .8 5 kJ )  (3 .3 7 5 kJ )  (  5 0 kJ )
 1 .4 7 5 kJ
26
Example Problem (2.63)
A gas is compressed in a piston cylinder assembly form p1 = 2 bar to p2 = 8 bar,
V2 = 0.02 m3 in a process during which the relation between pressure and
volume is pV1.3 = 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.
Solution: starting with the 1st Law of Thermodynamics
 P E   K E   U  Q in  W o u t
where: ΔKE=0
ΔPE = 0
p1 = 2 bar
also:
p2 = 8 bar
ΔU/m = 50 kJ/kg
V1 = ?
m = 0.2 kg
V2 = 0.02 m3
pV1.3 = constant
therefore:
1 .3
p1V1
 p 2V 2
1 .3
 p 
V1   2 
 p1 
1 / 1 .3
 8 bar 
V2  

 2 bar 
1 / 1 .3
0 .0 2 m  0 .0 5 8 1 m
3
3
27
Example Problem (2.63)
A gas is compressed in a piston cylinder assembly form p1 = 2 bar to p2 = 8 bar,
V2 = 0.02 m3 in a process during which the relation between pressure and
volume is pV1.3 = 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.
Solution continued:
also:p V
 p 2V 2
1 .3
1 .3
 (8 b a r )(0 .0 2 m )
3
p  0 .0 4 9 5 / V
1 .3
1 .3
 0 .0 4 9 5 b a r  m
 (0 .0 4 9 5 b a r  m
3 .9
)V
3 .9
 1 .3
therefore:
1 .3
p1V1
 p 2V 2
1 .3
 p 
V1   2 
 p1 
1 / 1 .3
 8 bar 
V2  

 2 bar 
1 / 1 .3
0 .0 2 m  0 .0 5 8 1 m
3
3
28
so work done is:
V2
W 

V1
V2
PdV 

((0 .0 4 9 5 b a r  m
3 .9
0.0495 bar  m
3.9
)V
 1 .3
)dV 
0 .0 4 9 5 b a r  m
 0 .3
V1


 0 .3


 V1
3.9
 3.224 m  0.9  1.177 m  0.9 


 0.3

V
 (0.02 m 3 )  0.3  (0.581 m 3 )  0.3 


 0.3
0.0495 bar  m
V2
3 .9
0.0495 bar  m
3.9
 3.224 m  0.9  1.177 m  0.9 


 0.3
  0.338 bar  m
3
100 kN / m
1 bar
2
1kJ
1kN  m
  33.8 kJ
29
Internal Energy is given as
 U  (  U / m )( m )  (5 0 kJ / kg )(0 .2 kg )  1 0 kJ
Finally back at the 1st Law:
 PE   K E   U  Q in  W out
gives
Q in   P E   K E   U  W o u t
 0  0  (1 0 kJ )  (  3 3 .7 8 kJ )
  2 3 .7 8 kJ
30
Example (2.70): A gas is contained in a vertical piston-cylinder assembly by a
piston weighing 1000 lbf and having a face area of 12 in2. 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
Patm=14.7 psi
of the piston increases by 2 ft. The piston and
cylinder are poor thermal conductors and friction
h = 2 ft
can be neglected. Determine the change in internal
Apiston = 12 in2
energy of the gas, in BTU, assuming it is the only
Wpiston = 1000 lbf
significant internal energy change of any component
present.
Solution: Apply the
1st
law of thermodynamics
 PE   K E   U  Q in  W out
 U  Q in  W o u t   P E   K E
Welec= - 5 BTU
31
where
mg = 1000 lbf
A = 12 in2
Δh = 2 ft
Welec_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 (ΔKE = 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:
KE  0
Q in  0
 P E  m g  h  (1 0 0 0 lb f )( 2 ft )  2 0 0 0 ft  lb f
W elect   5 B T U
(neg. since its put into the system)
V2
W PV 

V1
p d V  p (V 2  V1 )
(for constant pressure)
32
Ftop=patm A
for equilibrium:
W=1000lbf
F  0
Fb o tto m  Fto p  W  0
p A  p a tm A  W  0
p  p a tm 
W
Fbottom=p A
 1 4 .7 lb f / in 
2
A
1 0 0 0 lb f
1 2 in
2
 9 8 .0 3 p si
and the increase in Volume:
V 2  V1  A  h
V 2  V1  1 2 in ( 2 ft )
2
1 2 in
 2 8 8 in
3
1 ft
therefore the work done by the gas was positive work by the system
V2
W PV 

V1
p d V  p (V 2  V1 )  (9 8 .2 lb f / in )( 2 8 8 in )
2
3
1 ft
1 2 in
 2 3 5 7 lb f  ft
33
Returning to the 1st law:
 U  Q in  W out   PE   K E
 U  0  (W P V  W elec )   P E  0
 U   ( 2 3 5 7 ft  lb f  5 B T U
7 7 8 ft  lb f
1B T U
)  ( 2 0 0 0 ft  lb f )
 U   4 6 7 ft  lb f
 U   4 6 7 ft  lb f
1B T U
7 7 8 ft  lb f
  0 .6 0 B T U