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:

Homework Assignment:

• Read Chap 2. Sections 6 and 7

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

3

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

4

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.

5

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.)

6

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

7

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

8

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.

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.

11

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

12

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

13

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.

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 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

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/(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.

20

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.

21

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

22

Solution to Example (2.45):

Therefore:

inches 36.425.02.1

1058.538

7.8

42.0

BrickL

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: 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

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.

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

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.

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

28

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

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 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)

32

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