Energy Transfer and Thermodynamics - Article Example

Writer’s name] [Professor’s name] [Date] [Course title] Thermodynamics Q1. 1.1 First Law of Thermodynamics The first law of thermodynamics states that energy cannot be created or destroyed--in other words it has always been here. Let there be a system which absorb ∆Q amount of heat and as a consequence of this it performs ∆W amount of work. In this process the initial equilibrium states “i” of the system changes to a final equilibrium state “f” in a particular way and ∆Q-∆W is computed. Now this system is changed from the same initial state “i” to the final state “f” but along a different path. This procedure is repeated many times. It is observed that ∆Q-∆W comes out the same in all cases inspite of the fact that ∆Q and ∆W separately depend on the path taken. ∆Q-∆W depends only on the initial and final states. ∆Q is the energy added to the system and W is equal to the energy that has been extracted, from the system by the performance of work. The difference ∆Q-∆W which is retained within the system is the change in the energy of the system. It follows that the initial energy change of a system is independent of the path and is therefore equal to Ui internal energy of the system in state “f” minus the internal energy in state “i” or Uf – Ui therefore it follows that ∆Q-∆W=Uf-Ui=∆U The change in internal energy of a system in any process is equal to net heat flows into system minus the total work W done by the system Application of the First Law of Thermodynamics Isobaric Process: Isobaric process is that process which takes place at constant pressure. In such a process the heat transferred and the works done are both non-zero. ∆Q=P (V2-V1) +∆U This is the form of 1st law of thermodynamics in an isobaric process. Isochoric process: Isochoric process is defined as that process in which the volume of the system remains constant. U2-U1=∆Q (Isochoric process) Isothermal Process: If the temperature of the system remain constant throughout the process. It is called an isothermal process. ∆Q=∆W (Isothermal process) Adiabatic Process: The process in which no heat flows into or out of the system is called an adiabatic process. ∆U= -∆W 1.2 Second Law of Thermodynamics. Any device which converts heat into mechanical energy is called heat engine .The essentials of a heat engine are the furnace, or hot body, the working substance and a condenser or cold body. In a steam engine, the steam absorbs heat from the furnace, converts some of it into work by applying pressure to the piston and rejects the rests to the condenser. The point to note is that the furnace is at a higher temperature than the condenser and this conversion of heat into work is possible when the working substance falls in temperature as shown in figure in appendix. We can generalise it by saying that the two bodies must be maintained at different temperature for the working of a heat engine. A continuous supply of work has never yet been obtained from a single supply of heat otherwise we could build a ship which would use far more heat in the ocean water without needing any fuel. This lead to the first way of stating the second law of thermodynamics due to Kelvin. Let us consider the fact that no one has ever built a refrigerator which will work without a supply of energy. A refrigerator is essentially a machine for conveying heat from one body at a lower temperature to another at a higher temperature. In other words, it is only possible to make heat flow from a cold body to a hot body by using up work. The statement of second law of thermodynamics due to clausius comes from the consideration of this fact of refrigerator and it is stated. Application of Second Law of Thermodynamics. The Carnot Engine. It consists of a hot body of infinite thermal capacity, a similar cold body, a perfect heat insulator, and a cylinder fitted with a piston enclosing any working substance. The cylinder has a heat conducting base and non conducting walls and piston. The working substance is taken though the following cycle of operations known as the Carnot cycle. Part 4 Answer 1. First Law of Thermodynamics: When a system undergoes a complete thermodynamics cycle the intrinsic energy of the system is the same at the beginning and end of the cycle. Since the intrinsic energy of the system is unchanged the First Law of Thermodynamics states that “When a system undergoes a thermodynamic cycle then the net heat supplied to the system from its surrounding plus the net work input to the system from its surrounding must equal to zero.”(Eastop p18) Equation: ∑ Q+ ∑W=0 Where: ∑Q= the algebraic sum of the heat supplied to (+) or rejected from (-) the system. ∑W= the algebraic sum of the work done by surroundings on the system (+) or by the system on surroundings (-). Second Law Of thermodynamics: The Second Law of Thermodynamics which is also a natural law indicates that in any complete cycle the gross heat supplied plus the net work output must be greater than zero. Thus for any cycle in which there is a net work output (i.e. W –ve), heat must always be rejected at a higher temperature there must always be a positive work input. (Eastop p92) Equation: Q1 > -W Answer No 2. Open System: Open systems can exchange both matter and energy with an outside system. They are portions of larger systems and in intimate contact with the larger system. Your body is an open system. (http://www.bluffton.edu/~bergerd/NSC_111/thermo2.html). Air compressor is an example of open system. Closed System: A system can be defined as a collection of matter in a specific boundary. The boundaries can also be flexible; like the fluid which is present in the cylinder of a reciprocating engine in the expansion stroke ‘s boundaries are the cylinder walls as well as the piston crown. When piston moves, boundaries will also move ..This type of system is known as a closed system. In cylinder, fluid is being compressed by the piston is an example of closed system. Isolated System: Isolated systems are capable of exchanging neither energy nor matter with an outside system. Whilst they might be a part of larger systems, they do not have any contact with the outside system. The physical world is a good example of an isolated system. Heat cannot be transferred between isolated systems; however it can be transferred. A good example of this aymptotically flat space time Adiabatic System: In thermodynamics, an adiabatic system is a closed system, which does not allow the heat to be exchanged. Thus it can called a "prefect " system, in which it is presumed that the work which is required to change an adiabatic system from one distinct condition to another defined state remains the same the work is done. Thermos Container or dewar flask is the examples of adiabatic system. Answer No 3: Formulae for Kinetic Energy: K.E =½mv² Where, KE = Kinetic Energy, M = Mass of object, V = Velocity or Speed of object Formulae for Potential Energy: PE=mgh Where, PE = Potential Energy, m = Mass of object, g = Acceleration of Gravity, h = Height of object Formulae For Work: W=Fd Where, W = Work, F = Force, d = Distance Formulae For Heat Transfer: q = m (∆T) Cp Where q = heat transferred, ∆T = the change in temperature and Cp = the specific heat. Answer No.4 Problem Solving: 1. t=98.6°f We know that °C=5/9 (°F-32) =5/9 (98.6-32) =37 Normal body temperature is 37°C We know that R=°F +459.67 = 98.6+459.67= 558.27 Normal body temperature is 558.27R 2. Zero on Rankine scale is called as absolute zero. °F=R-459.60 = -459.67 3. We know that K=°C+273=30+273=303 Rise in temperature =303K 4. We know that R=°F+459.67=60+459.67 = 519.67 We know that °C=5/9 (°F-32) = 5/9(60-32) = 15.56 We know that K=°C+273=15.56+273=288.56 Rise in temperature: R=519.67, °C=15.56 and K=288.56 Answer No 5. Mechanism of Heat Transfer In general there are three ways in which heat may be transferred, given under the headings below. Conduction: Conduction is the transfer of heat from one part of a substance to another part of the same substance, or from one substance to another in physical contact withit, without appreciable displacement of the molecules forming the substances. (Rajput p715) Convection: Convection is the transfer of heat within a fluid by the mixing of one portion of the fluid with another (Rajput p716). The movement of the fluid may be caused by differences in density resulting from the temperature differences as in natural convection (or free convection), or the motion may be produced by mechanical means, as in forced convection. Radiation: All matter continuously emits electromagnetic radiation unless its temperature is absolute zero. It is found that the higher the temperature then the greater is the amount of energy radiated. If therefore two bodies at different temperature are so placed that the radiation from each body is intercepted by the other, then the body at the lower temperature will receive more energy than it is radiating, and hence its internal energy will increase, similarly the internal energy of the body at the higher temperature will decrease. (Dincer&Rosen p42). In any particular example in practice heat may be transferred by a combination of conduction, convection, and radiation and it’s usually possible to assess the effects of each mode of heat transfer separately and then to sum up the results. Answer No.6 Transient Heat Conduction Differ From Steady State Conduction: Steady state conduction is a kind of conduction that takes place when the difference in temperature remains constant thus the equilibration time, the spatial distribution of temperatures does not change. There are many situations in which the temperature drastically rises and falls like when we place a hot copper ball into oil at a considerably low temperature, with the intention to is in analyze the extraordinary change of temperature in the hot copper ball over time. The above described kind of heat conduction may be considered as transient conduction. Answer No 7 Natural Convection Differ From Forced Convection: The movement of the fluid may be caused by differences in density resulting from the temperature differences as in natural convection (or free convection), or the motion may be produced by mechanical means, as in forced convection. Answer No.8 Problem Solving M= mass flow rate =2kg/sec ∆HC= Heat of combustion for C3H8=46450 KJ/Kg qHc=heat release rate =? We know that qHc=∆Hc *m =46450*2=92900 KJ/sec Heat release rate = 92900 KJ/sec. Answer No.9 Problem Solving q= rate of heat transfer=? We know that q= σ K4A If T=20°C=293K and A=1 σ =Stefen-boltzman constant = 5.67*10¯8, w/m²-k4 So, q/A= 5.67*10¯8 *(293)4 = 5.67*10¯8* 7.37*10¯9 = 417.88 J Rate of heat transfer is 417.88 J Answer No.10 Fourier’s Law of Conduction Fourier’s law stats that the rate of flow of heat through a single homogeneous solid is directly proportional to the area A of the section at right angles to the direction of heat flow, and to the exchange of temperature with respect to the length of the path of the heat flow, dt/dx.This is an empirical law based on observation. Dq/dt = −λ A dT/dx Thermal Conductivity: The term “λ” is called the thermal conductivity. The thermal conductivity of a substance can be defined as the heat flow per unit area per unit time when the temperature decreases by one degree in unit distance. The units of “λ” are usually written as W/Mk or KW/m K. Values of Thermal Conductivity: 1. Metals: 1 Btu/(hr °F ft2/ft) = 1 Btu/(hr °F ft) = 1.731 W/(m K) = 1.488 kcal/(h m °C) Metal Temperature - t - (oF) Thermal Conductivity - k - (Btu/(hr oF ft)) Aluminium, pure 68 118 200 124 400 144 Carbon Steel, max 0.5% C 68 31 Carbon Steel, max 1.5% C 68 21 752 19 2192 17 Copper, pure 68 223 572 213 1112 204 Gold 68 182 T(°C) = 5/9[T(°F) - 32] 1 Btu/(hr °F ft²/ft) = 1 Btu/(hr °F ft) = 1.731 W/(m K) = 1.488 kcal/(h m °C) 2. Insulating Material Temperature limits of some common insulation materials are as under. Insulation Material Low Temperature Range High Temperature Range (oC) (oF) (oC) (oF) Calcium Silicate -18 0 650 1200 Cellular Glass -260 -450 480 900 Elastomeric foam -55 -70 120 250 Fiberglass -30 -20 540 1000 Mineral Wool 0 32 1000 1800 Polyisocyanurate or polyiso -180 -290 150 300 Answer No 11 Stefan-Boltzmann Law: It was found experimentally by Stefan, and proved theoretically by Boltzmann, that the emissive power of black body is directly proportional to the fourth power of its absolute tempreture, and this is known as the Stefan-Boltzmann law. i.e. EB = σ T4 The value of σ is 5.67*10¯8 W/m² (K) 4 Emissivity: Most surfaces are not blackbody emitters, and emit some fraction of the amount of thermal radiation that a blackbody would. This fraction is known as emissivity. If a surface emits ½ or 0.5 as much radiation at a given wavelength and temperature as a blackbody, it is said to have an emissivity of 0.5. If it emits 1/10 or 0.1 as much as a blackbody, it has an emissivity of 0.1 and so on. Obviously, a blackbody has an emissivity of 1.0 at all temperatures and wavelengths. (http://www.temperatureconsultant.com/EMISSIVITY.htm) Answer No 12 Combustion: Source of heat in engines as well as power plants is chemical energy of substances called fuels. This energy is released during the chemical reaction of the fuel with oxygen . This process is called combustion. Combustion occurs in a controlled way in some type of combustion chamber after the starting of combustion via a number of ways .The most appropriate source of the supply of oxygen is that of the environment which consists of nitrogen, oxygen as well as trace of other gases . Usually no effort is made to extract the oxygen from the environment, and the nitrogen, etc accompanies the oxygen into the combustion chamber. Heat Of Combustion: The energy released during the process of combustion is called the heat of combustion. In an oxidation process fuel elements combine with oxygen which is rapid and is accompanied by the evolution of heat. Heat Release Rate: The rate at which heat is generated by fire. The HRR can be viewed as the engine driving the fire. This tends to occur in a positive-feedback way: heat makes more heat. HRR is measured in Joules per second (also called Watts). Since a fire puts out much more than 1 Watt, HRR is usually quantified in kilowatts (1000 W) or megawatts (a million watts). (http://www.interfire.org/termoftheweek.asp?term=2116) Types of Combustion Rapid In Rapid combustion a huge amount of heat along with light energy is released, this at times ignites a fire. This type of combustion is utilized to machines such as internal combustion engines and in thermo baric weapons. Apart from the production of heat and energy at times a huge amount of gas is released Slow This kind of combustion is occurs in places where there is low temperature. Complete Only a few products are production in this kind of combustion, suppose a hydrocarbon will burn inside oxygen, has been known to generate carbon dioxide and water. If any kind of fuel will burn in the air, the end product of combustion consist of nitrogen as well. Carbon tends to produce carbon dioxide. Nitrogen tends to produce nitrogen dioxide. Sulfur tends to produce sulfur; Iron will tend to produce iron (III) oxide. This helps in concluding that complete combustion is practically impossible to achieve. Turbulent This kind of combustion is identified by turbulent flows. This kind of combustion is mostly utilized by industries as the turbulence assists in the mixing procedure between the fuel and oxidizer. Microgravity Almost all flames act in a different way in a microgravity environment; for instance, a candle's flame may appear in a shape of a spear. Studies based on this type of combustion helps in improving the fire safety procedures of spacecraft and various concepts regarding combustion physics. Incomplete Incomplete combustion takes place when oxygen is less then the required amount to permit the fuel (to react entirely with the oxygen to generate carbon dioxide and water, this kind of combustion is can be cooled down by a heat sink like a solid surface Definitions Specific Heat Capacity Specific heat is the measure of heat capacity. Latent Heat Latent heat is the heat energy involved in the phase change of water (www.uwsp.edu/gEo/faculty/ritter/glossary/l_n/latent_heat.html) Calorimetry Calorimetry is the quantitative measurement of the heats which is essential or evolves while chemical process is taking place. Combustion Temperature Combustion temperature is kind of temperature that is measured in Kelvin degrees. This temperature is the temperature which measured in combustion chambers. Chemical Equilibrium Acids and bases neutralize each other and form a salt as a by product. This reaction reaches what is called equilibrium. Answer No.13Problem Solving Thermal efficiency = 1- T²/T¹=? T2= 10°C = 283 K T1= 300°C= 573 K We know that Efficiency = 1-T²/T¹= 1- 283/ 573 = efficiency = 50.6% Thermal Efficiency is 50.6% Answer No.14 Problem Solving Mw1 = 1.25 kg Mw2 =1.25 kg = mw1 Tw1 = 17°C = 290 K Tw2 = 85°C = 358 K Tf =? We know that Mw1Cw∆T=mw2Cw∆T Mw1Cw (Tf –Tw1) = mw2Cw (Tw2 –Tf) Tf – Tw1= Tw2 –Tf 2 Tf = Tw2 +Tw1= 358+290= 648 Tf = 648/2 = 324 K Final Temperature is 324 K or 51°C Answer No 15. Problem Solving Xice = 1mm=.001m = thickness A= 0.75 m² = area T= 2 minutes = 120 sec P (watts) = P (J/S) =? We know that L ice = 80 Kcal / kg = 80* 4.18 J/kg = 334400 J/kg = Latent heat of fusion. Dice = 0.93 *10³ kg /m³ = density Mice =Dice*Vice = density*volume=density (area*thickness) = Dice*A*Xice = 0.93*10³*0.75*0.001= 0.6975 kg We know that ∆ Qice = miceL= 0.6975*33440 = 233244 J Power = P= ∆Q/t = 233244/120= 1943.7 W Answer No.16 Problem solving Xbrick =? Kbrick = 0.6 w/mk= thermal conductivity of brick. Xair = 10cm = 0.1 m ∆Q/∆t (brick) = ∆Q/∆t (air) Kair = 0.026 w/mk Kbrick A ∆T/Xbrick = Kair A ∆T/Xair 0.6 (A) ∆T/ Xbrick = .026 (A) ∆T/0.1 Xbrick = 2.3 m (thickness of brick is 2.3 m) Answer No.17 Problem solving Rsun = 7.108 (power) m P= 4 * 1026 W Stefen-Boltzman constant =σ = 5.67* 10¯8 J/S-m²-k 4 Tsun =? P= σ T4 sun 4π R² sun Tsun = (P/ σ 4π R²sun) ¼ = {4 *1026/ (5.67 *10¯8) (4π) (7*108)²}¼ = {4 *1026 / 3.49 *1011}¼ = (1.146 * 1015)¼ = 5817.93 K Surface Temperature is 5817.92 K Answer No.18 Problem solving: K Plywood = 2.5 cm = 0.025 m X concrete = 10 cm = 0.1 m T1 = 30°C = 303 K T2 = 10°C = 283 K K plywood = 0.12 w/mk K concrete = 1.3 w/m-k a) For plywood ∆Q/ ∆T plywood = K plywood A ∆T/ X (eq. 1) = 0.12 * 1* (303-T2) / 0.025 = 4. 8 (303-T2) = 1454.4 – 4.8 T2 b) For Concrete ∆Q/ ∆T = K concrete A * ∆T/ X (eq. 2) = 1.3 *1 * (T2 -283)/ 0.1 = 13T2 – 3679 Now put in equation No.1 a) ∆Q/ ∆T = K A ∆T/ X = (0.12) (1) (303-288.4)/ 0.025 = 70.08 J/S b) Put in equation 2 because heat rate is same so ∆Q/ ∆T = K A ∆T/ X = 1.3 *1 (T2 -283)/0.1 = 13T2 – 3679 13T2 = 70.08 + 3679 = 3749.08 = T2 = 3749.08 / 13 = 288.4 K = 15.4°C Hence Proved Answer No.19 Problem Solving X= 5mm = 0.005m A= 370 cm² = 0.037m M coffee = 0.5 kg Coffee = 4200 J/kg-k T1 = 80°C = 353 K T2 = 20°C = 293K K Styrofoam = 0.033 J/mk We know that 1) ∆Q/ ∆T = K Styrofoam A ∆T/ X = 0.033 (0.037) (353-293)/0.005 = 14.65 J/S 2) ∆Q coffee = mc ∆T = (0.5) (4200) (80-70) = 21000 J ∆t = ∆Q coffee / (∆Q/ ∆T) = 21000/ 14.65 = 1433.45 sec = 23.89 min Time required to cool from 80°C to 70°C = 23.89 min Answer 20 Laminar and Turbulent Flow: For flow through a tube it can be assumed that the flow is turbulent when Re> 2100 approximately. For flow through a tube that the flow is laminar when Re< 2100 approximately. Reynolds’s Number: The Reynolds’s number is a dimensionless factor that describes the flow behaviour of liquids. Liquids are not only described by their velocity but also by how chaotic the flow patterns are. (www.8d.com/Final%208pad%20%20Why%20the%20Hole%20Reynolds.html) The Reynolds number is a dimensionless group given by Re=ρCl /η OR Cl / v Where ρ is the fluid density, C is the fluid mean velocity, l the characteristics linear dimension, η the dynamic velocity of the fluid, and v the kinematics viscosity of the fluid η/ρ Answer No.22 Problem solving: L= 12m T1 = -5°C = 268 K ∆L =? T2 = 35°C = 308K ∞ = 1*10¯5 °C¯1 We know that ∆L = ∞ L ∆T = 1 *10¯5 * 12 * {35- (-5)} = 4.8 *10¯³ m Change in length is 4.8 *10¯³ m Answer No.23 Problem solving Tice = -10°C = 263 K T tea = 90°C = 363 K T f = 10°C = 283 K M tea = 1 L = 1kg M ice =? ∆Q tea = ∆Q ice melts + ∆Q ice Cm tea ∆T = m ice L + Cm ice ∆T 4200*1*(363-283) = m (334000) + 2100*m*(283-263) 336000 =m {334000 + 42000) 336000 =m (376000) m= 0.8936 kg = 893.6 g 20g ice cube = m= 893.6/20 = 44.6 44 ice cube of 20g Answer No: 24 Problem solving L= 3.66m T1 = -28°C = 245 K T2 = 39°C = 312K ∞aluminium= 2.34 *10¯5 /°C ∆L =? We know that ∆L = ∞ L ∆T = 2.34*10¯5 *3.66 * {39-(-28)} = 5.74 *10¯³ m Answer No 25. Problem solving L1 = 11.2m L2 = 11.7m T1 = 22°C T2 =? ∞ Steel = 1.24 *10¯5 /°C We know that ∆L = ∞ L ∆T (L2-L1)= 1.24*10¯5 * 11.2*(T2 – 22) = T2 – 22 – 22 = 0.5/1.388 * 10¯4 T2 = 3600 +22 = 3622°C Work cited Dincer Ibrahim & Rosen A. Marc; Thermal Energy Storage: Systems and Applications Wiley 2002 p42 http://www.interfire.org/termoftheweek.asp?term=2116 retrieved on 26th February 2009 http://www.temperatureconsultant.com/EMISSIVITY.htm retrieved on 26th February 2009 R.K. Rajput; A Textbook of Engineering Thermodynamics, Laxmi Publication 2005 pp715-716 T. D Eastop; Applied thermodynamics for engineering technologists Wiley; 5th edition, 1993 pp18-92 www.8pad.com/Final%208pad%20-%20Why%20the%20Hole%20Reynolds.html retrieved on 26th February 2009 www.uwsp.edu/gEo/faculty/ritter/glossary/l_n/latent_heat.html retrieved on 26th February 2009 Read More