Unit 1.3 - Thermodynamics
Notes
Thermodynamics
- the study of the effects of work, heat flow, and energy system - movement of thermal energy - engineers use thermodynamics in systems ranging from nuclear power plants to electrical components |
Thermal Energy versus Temperature
- Thermal Energy is kinetic energy in transit from object to another due to temperature. (Joules) - Temperature is the average kinetic energy of particles in an object; not the total amount of kinetic energy particles. (Degrees) |
Thermodynamic Equilibrium
- Thermal Equilibrium is obtained when touching objects within a system reach the same temperature. When thermal equilibrium is reached, the system loses its ability to do work. - Zeroth Law of Thermodynamics states that if two seperate systems are eqilibrium to a third system, then the first two systems are in equilibrium with each other. |
Examples of a thermodynamic system
- nuclear power plants - bike motors - electrical components |
Absolute Zero - ocurrs when all kinetic energy is removed from an object; 0K = -273*C
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1st Law of Thermodynamics
- Law of energy conservation applied to a thermal system. - Thermal energy can change form and location, but it cannot be created nor destroyed. Thermal energy can be increased within a system by adding thermal energy (heat) or by performing work in a system. |
2nd Law of Thermodynamics
- Thermal energy flows from hot to cold. - Entropy is the measure of how evenly distributed heat is within a system. A system tends to go from order to disorder. - The total amount of energy in the world does not change, but the availibility of that energy constantly decreases. |
Thermal Energy Transfer:
Convection, Conduction, and Radiation - Convection is the transfer of thermal energy by movement of fluid (liquid or gas). When fluid is heated, it expands, becomes less dense, and rises. - Conduction is the transfer of thermal energy within an object or between objects from molecule to molecule. - Radiation is the process by which energy is transmitted through a medium, including empty space, as electromagnetic waves. - Stefan's Law states that all objects lose and gain thermal energy by electromagnetic radiation. |
Thermal Energy Transfer Equations
A = area of thermal conductivity (m)
L = thickness of material (m) Pnet = radiated energy transfer (Stefan's constant) = 5.6696 x 10^-8 (W)/(m^2 x k^4) |
e = emissivity constant
T = temperature in Kelvin Q = energy transfer (Joules) m = mass of the material (kilograms) c = specific heat capacity of the material (J/kg x *C) |
(delta T) = change/difference in temperature (*C)
P = rate of energy transfer (Watts) (delta t) = change in time (seconds) k = thermal conductivity (J/s x m x C) |
U-Value - Coefficient of Heat Conductivity is the measure of a material's ability to conduct heat.
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R-Value - Thermal Resistance of a Material is the measure of a material's ability to resist heat. The higher the R-Value, the higher the resistance.
- Bulk R-Value = R-Value Object 1 + R-Value Object 2 + ... = Total R-Value |
Practice Problems
A 1.00kg piece of aluminum metal at 90*C is placed in a 4 liters (=4kg) of water at 25*C. Determine the final temperature (Tf).
List all known and unknown variables. mal = 1kg mwat = 4kg cal = 900 J/kg x *C Q = ? Tal = 90*C Twat = 25*C cwat = 4184 J/kg x *C Find equations and apply variables. Q = m x c x (delta T) Qwat = 4kg x (4184 J/kg x *C) x (Tf - 25*C) Qal = 1kg x (900J/kg x *C) x (90*C - Tf) Solve. 16736(Tf - 25*C) = 900(90*C - Tf) -> 16736Tf - 418,400 = 81,000 - 900Tf -> 17,636Tf = 499,400 -> Tf = 28.32*C |
A 3/16 in. thick acrylic testing box with dimensions of 10 in. x 10 in. is covered with an unknown 0.5 in. insulation material. Determine the thermal conductivity for the insulating material if a 25W bulb is used to heat the box. The bulb maintains the inside temperature at 10*C higher than the outside temperature.
List all known and unknown variables. L = .0127m A = .065m^2 P = 25W (delta T) = 10*C k = ? Find equations and apply variables. k = (P x L)/(A x (delta T)) k = (25W x .0127m)/(.065m^2 x 10*C) Solve. k = (.25 x .0127)/(.065 x 10) -> k = .3175/.65 -> k = .489 J/s x m x *C |
A student travels on a school bus in the middle of winter from home to school. The school bus temperature is 58*F. The student's temperature is 91.4*F. Determine the net energy transfer from the student's body during the 20 min. ride to school due to electromagnetic radiation. Note: Skin emissivity is .9, and the surface area of the student is 1.50m^2.
List all known and unknown variables.
A = 1.5m^2 e = .9 T1 = 58*F or 287.59K T2 = 91.4*F or 306.15K (delta t) = 1,200s Pnet = ? Q = ? |
Find equations and apply variables.
Pnet = (Stefan's constant) x A x e x (T2^4 - T1^4) P = Q/(delta t) Pnet = (5.6696 x 10^-8) x 1.5m^2 x .9 x (306.15^4 - 287.59^4) 148.815W = Q/1200s |
Solve.
Pnet = (7.65396 x 10^-8) x 1,944,290,107 ->Pnet = 148.815W |
Renewable Insulation Lab
Procedure
Your team will design a renewable composite insulation material. Equipment
- Computer - Heat Box Apparatus - LoggerPro Software - 2 -- Stainless Steel Temperature Probes - Insulation Materials - Tape - Standard and Metric Ruler |
Design Constraints
- Composite insulation material must produce minimum heat loss, representing good insulating value. - Composite insulation material must have overall uniform thickness less than or equal to one inch. - Composite insulation material must have consistent internal composition. - Composite insulation material dimensions must not exceed the overall dimensions of heat box apparatus top. - Individual insulation material(s) must be evironmentally friendly. - Individual insulation material(s) must be recyclable. - Individual insulation material(s) must be economical. - The change in temperature inside the box is directly related to the heat absorbed or released by the air in the box. |
Pre-Experiment Data
Inner dimensions (m) -> l = .15m w = .15m h = .175m Volume of air (m^3) -> V = .003938m^3 Description of insulation material -> Fabric Post-Experiment Data
Heat source light bulb wattage -> P = 25W Initial internal temperature (*C) -> TInitial 1 = 21.8*C Maximum internal temperature (*C) -> TMax 1 = 45.2*C Final internal temperature (*C) -> TFinal 1 = 35.9*C Initial room temperature (*C) -> TInitial 2 = 21.1*C Maximum room temperature (*C) -> TMax 2 = 27.1*C Final room temperature (*C) -> TFinal 2 = 25.4*C Heating times (s) -> t1 = 1200s Cooling times (s) -> t2 = 1200s |
Data Equations
Mass of the air inside the box: m = V x Dm V = 0.003938m^3 Dm = 1.20 kg/m^3 m = 0.003938(1.20) -> m = .004725kg Energy gained by the air in the box during heating: Q = m x c x (delta T) m = .004725kg c = 1,000 J/kg*C Delta T = 23.4*C (Tmax - Tinitial) Q = .004725(1000)23.4 -> Q = 111.4864J Energy lost by the air in the box during cooling: Q = m x c x (delta T) m = .004725kg c = 1,000J/kg*C Delta T = 9.3*C (Tmax - Tfinal) Q = .004725(1000)9.3 -> Q = 43.9332J Net energy retained within the box: Q = m x c x (delta T) m = .004725kg c = 1,000 J/kg*C Delta T = 14.1*C (Tfinal - Tinitial) Q = .004725(1000)14.1 -> Q = 66.6084J |