chapter 2: sections 4 and 5 lecture 03: 1 st law of thermodynamics and introduction to heat transfer
TRANSCRIPT
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Chapter 2: Sections 4 and 5
Lecture 03: 1st Law of Thermodynamics and Introduction to Heat Transfer
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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:
Homework Assignment:
• Read Chap 2. Sections 6 and 7
From Chap 2: Problems 49, 53,61, 68
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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.
Sec 2.4: Energy Transfer by Heat
Q
Q Q dt
q
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3 Types of Heat TransferConductionRadiationConvection
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 LawdT
Q Adx
where κ is the thermal conductivity of the material.
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Types of Heat Transfer
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 fQ h A T T
hc = coefficient of convection(An empirical value, that depends on the material, the
velocity, etc.)
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Types of Heat Transfer
Sec 2.4.2: Heat Transfer Modes
Radiation : Heat transfer which occurs as
matter exchanges Electromagnetic radiation with other matter.
Stefan-Boltzmann Law:4bQ AT
Tb = absolute surface temperature
ε = emissivity of the surfaceσ = Stefan-Boltzmann constant
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Summary of Heat Transfer Methods
Radiation:
Sec 2.4.2: Heat Transfer Modes
dTQ A
dx
where A is area κ is thermal conductivity dT/dx is temperature gradient
Conduction:
Convection:
c b fQ h A T T
where A is area hc is the convection coefficient
Tb -Tf is the difference between the
body and the fluid temp.
4bQ AT
where Tb is absolute surface temperature
ε is emissivity of the surface σ is Stefan-Boltzmann constant A is surface area
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Example (2.45): An oven wall consists of a 0.25” layer of steel (S=8.7 BTU/(hftoR) )and a layer of brick (B=0.42 BTU/(hftoR) ). 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.
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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.
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Recall from yesterday: by convention
Heat, Q:
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 systemQ < 0 : Heat transferred FROM the system
W > 0 : Work done BY the systemW < 0 : Work done ON the system
Work, W:
+Q +W
system
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First Law of Thermodynamics
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 withinthe system
net Qinput
net W output[ ] =[ ] +[ ]
system in outE PE KE U Q W
WQdE where denotes path dependent derivatives
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The First Law of Thermodynamics
Sec 2.5: Energy Balance for Closed Systems
system in outE Q W
Qin Wout system
ΔE
The 1st Law of Thermodynamics is an expanded form of the Law of Conservation of Energy, also known by other name an Energy Balance.
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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.
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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.
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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 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.
Patm=14.7 psi
h = 2 ft
Apiston = 12 in2
Wpiston = 1000 lbf
Welec= - 5 BTU
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End of Lecture 03
•Slides which follow show solutions to example problems
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Example (2.45): An oven wall consists of a 0.25” layer of steel (S=8.7 BTU/(hftoR) )and a layer of brick (B=0.42 BTU/(hftoR) ). 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.
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Solution to Example (2.45):
m iSteel Steel Steel
Steel
T TdTQ A A
dx L
Steel Brick
Q Q
A A
Heat flow rate through steel:
0 mBrick Brick
Brick
T TQ A
L
Heat flow rate through steel:
Steady State Heat Flow: Both materials have the same cross sectional area here and the heat flow rate through each is the same.
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Solution to Example (2.45):
0m i mSteel Brick
Steel Brick
T T T T
L L
Steelim
m
Steel
BrickBrick L
TT
TTL
0
Therefore:
and solving for Lbrick
with κBrick = 0.42 BTU/(hftoR)
κSteel = 8.7 BTU/(hftoR
Ti = 540oF Tm= 538.8oF T0= 105oF
and LSteel = 0.25 in … solve for Lbrick
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Solution to Example (2.45):
Therefore:
inches 36.425.02.1
1058.538
7.8
42.0
BrickL
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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: 4bQ AT
where: ε = 0.8 σ = 5.67 x 10-8 W/m2•K4 d = 0.5m dQ/dt = 150 W
Qout
31
6sphereA d
therefore: 3 3 31 1(0.5 ) 0.6545
6 6sphereA d m m
4bQ AT
1/ 4
b
QT
A
0.25
1/ 4
8 22 4
150266. 6
0.8(5.67 10 )(.6545 )b
Q WT K
WA mm K
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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.
system in outE PE KE U Q W
Solution:Principle: 1st Law of Thermodynamics
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 2 2
2 1
1 1
2 2KE mv mv
22
1(10 )(9.7 / )( 50 )
/
Nkg m s m
kg m s
2 2 2 22
1(10 )(30 15 ) /
2 /
Nkg m s
kg m s
4850 4.85
1000
J kJN m kJ
N m J
3375 3.3751000
J kJN m kJ
N m J
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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.
out inW Q PE KE U
Solution continued:
( / )( ) ( 5 / )(10 ) 50U U m m kJ kg kg kJ
( 50 ) ( 4.85 ) (3.375 ) ( 50 )kJ kJ kJ kJ
( / )( ) ( 5 / )(10 ) 50in inQ Q m m kJ kg kg kJ
Solving for the Work done by the system:
1.475 kJ
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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
in outPE KE U Q W where: ΔKE=0 ΔPE = 0 ΔU/m = 50 kJ/kg m = 0.2 kg
p1 = 2 bar p2 = 8 bar V1 = ? V2 = 0.02 m3
also: pV1.3 = constant
therefore:1.3 1.3
1 1 2 2pV p V
1/1.3 1/1.3
3 321 2
1
80.02 0.0581
2
p barV V m m
p bar
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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:
therefore:1.3 1.3
1 1 2 2pV p V 1/1.3 1/1.3
3 321 2
1
80.02 0.0581
2
p barV V m m
p bar
1.3 3.9 1.30.0495 / (0.0495 )p V bar m V
1.3 1.3 3 1.3 3.92 2 (8 )(0.02 ) 0.0495pV p V bar m bar m
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3.93 0.3 3 0.30.0495
(0.02 ) (0.581 )0.3
bar mm m
22 2
1 1 1
3.93.9 1.3 0.30.0495
((0.0495 ) )0.3
VV V
V V V
bar mW PdV bar m V dV V
so work done is:
23 100 / 1
0.338 33.81 1
kN m kJbar m kJ
bar kN m
3.90.9 0.90.0495
3.224 1.1770.3
bar mm m
3.90.9 0.90.0495
3.224 1.1770.3
bar mm m
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Internal Energy is given as
Finally back at the 1st Law:
gives
in outQ PE KE U W
0 0 (10 ) ( 33.78 )kJ kJ
( / )( ) (50 / )(0.2 ) 10U U m m kJ kg kg kJ
23.78 kJ
in outPE KE U Q W
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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 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.
Patm=14.7 psi
h = 2 ft
Apiston = 12 in2
Wpiston = 1000 lbf
Welec= - 5 BTUSolution: Apply the 1st law of thermodynamics
in outPE KE U Q W
in outU Q W PE KE
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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:
0inQ
2
1
2 1( )V
PV
V
W pdV p V V
(1000 )(2 ) 2000f fPE mg h lb ft ft lb
0KE
(for constant pressure)
5electW BTU (neg. since its put into the system)
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for equilibrium:
2 1V V A h
2
1
2 32 1
1( ) (98.2 / )(288 ) 2357
12
V
PV f f
V
ftW pdV p V V lb in in lb ft
in
0F
Ftop=patm AW=1000lbf
Fbottom=p A0atmpA p A W
0bottom topF F W
22
100014.7 / 98.03
12atm
W lbfp p lbf in psi
A in
and the increase in Volume:
2 32 1
1212 (2 ) 288
1
inV V in ft in
ft
therefore the work done by the gas was positive work by the system
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Returning to the 1st law:
in outU Q W PE KE
0 ( ) 0PV elecU W W PE
778(2357 5 ) (2000 )
1f
f
ft lbU ft lbf BTU ft lb
BTU
467U ft lbf
1467 0.60
778 f
BTUU ft lbf BTU
ft lb