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

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Presentation on theme: "Chapter 2: Sections 4 and 5 Lecture 03: 1 st Law of Thermodynamics and Introduction to Heat Transfer."— Presentation transcript:

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

2 Todays 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 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 1 st 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. 3 Sec 2.4: Energy Transfer by Heat

4 3 Types of Heat Transfer Conduction Radiation 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. Fouriers Law where κ is the thermal conductivity of the material.

5 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. Newtons Law of Cooling: h c = coefficient of convection (An empirical value, that depends on the material, the velocity, etc.)

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

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

11 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

12 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 [ ] = + where denotes path dependent derivatives

13 The First Law of Thermodynamic s 13 Sec 2.5: Energy Balance for Closed Systems Q in W out system ΔEΔE 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/s 2. Determine the work for the process, in kJ. 14

15 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 16

17 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

18 End of Lecture 03 Slides which follow show solutions to example problems 18

19 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

20 Solution to Example (2.45): 20 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): 21 Therefore: and solving for L brick 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 = o F T 0 = 105 o F and L Steel = 0.25 in … solve for L brick

22 Solution to Example (2.45): 22 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= Solution: where: ε = 0.8 σ = 5.67 x W/m 2 K 4 d = 0.5m dQ/dt = 150 W Q out 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/s 2. Determine the work for the process, in kJ. 24 Solution: Principle: 1 st Law of Thermodynamics given: m = 10 kg Q in /m = -5 kJ/kg Δh=-50 m v 1 = 15 m/s v 2 = 30 m/s ΔU /m= - 5kJ/kg g = 9.7 m/s 2 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/s 2. Determine the work for the process, in kJ. 25 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 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 where: ΔKE=0 ΔPE = 0 ΔU/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:

27 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:

28 28 so work done is:

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

30 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

31 31 where mg = 1000 lb f A = 12 in 2 Δh = 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 (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: (for constant pressure) (neg. since its put into the system)

32 32 for equilibrium: F top =p atm A W=1000lb f F bottom =p A and the increase in Volume: therefore the work done by the gas was positive work by the system

33 33 Returning to the 1 st law:


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