1. Chapter 2: Sections 4 and 5 Lecture 03: 1st Law of Thermodynamics
and Introduction to Heat Transfer

2. Today’s Objectives: Reading Assignment: Homework Assignment:
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

3. 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, : the rate at which heat flows into or out of a system, dQ/dt. Heat flux, : the heat flow rate per unit surface area.

4. 3 Types of Heat Transfer Conduction Radiation Convection Conduction:
Sec 2.4.2: Heat Transfer Modes
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
where κ is the thermal conductivity of the material.

5. Types of Heat Transfer Convection:
Sec 2.4.2: Heat Transfer Modes
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:
hc = coefficient of convection
(An empirical value, that depends on the material, the velocity, etc.)

6. Types of Heat Transfer Radiation :
Sec 2.4.2: Heat Transfer Modes
Types of Heat Transfer
Radiation :
Heat transfer which occurs as matter exchanges Electromagnetic radiation with other matter.
Stefan-Boltzmann Law:
Tb = absolute surface temperature
ε = emissivity of the surface
σ = Stefan-Boltzmann constant

7. Summary of Heat Transfer Methods
Sec 2.4.2: Heat Transfer Modes
Summary of Heat Transfer Methods
Conduction:
where A is area
κ is thermal conductivity
dT/dx is temperature gradient
Convection:
where A is area
hc is the convection coefficient
Tb -Tf is the difference between the
body and the fluid temp.
Radiation:
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.

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
Sec 2.4: Energy Transfer by Heat
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

12. First Law of Thermodynamics
Sec 2.5: Energy Balance for Closed Systems
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 + =
where  denotes path dependent derivatives

13. The First Law of Thermodynamics
Sec 2.5: Energy Balance for Closed Systems
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.
Qin
Wout
system ΔE

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.

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/(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. Heat flow rate through steel:
Solution to Example (2.45):
Heat flow rate through steel:
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. with κBrick = 0.42 BTU/(hftoR) κSteel = 8.7 BTU/(hftoR
Solution to Example (2.45):
Therefore:
and solving for Lbrick
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:

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:
Qout
where: ε = 0.8
σ = 5.67 x 10-8 W/m2•K4
d = 0.5m
dQ/dt = 150 W
therefore: 

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
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

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:
Solving for the Work done by the system:

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
where: ΔKE= ΔPE = 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: 

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:
therefore: 

28. so work done is:

29. Internal Energy is given as
Finally back at the 1st Law:
gives

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 BTU
Solution: Apply the 1st law of thermodynamics 

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:
(neg. since its put into the system)
(for constant pressure)

32. for equilibrium:
Ftop=patm A
W=1000lbf
Fbottom=p A
and the increase in Volume:
therefore the work done by the gas was positive work by the system

33. Returning to the 1st law: