Laws of Thermodynamics - Article Example

Q1. a) First Law of Thermodynamics: The first law of thermodynamics is also called the law of conservation of energy and it states that the energy is always conserved; it can neither be created nor destroyed though energy can be converted from one form into another. ∆E System = 0 b) Second Law of Thermodynamics: The second law of thermodynamics is also called the law of entropy. This law predicts that the degree of randomness (entropy) of any isolated system always increases with time. There are several statements which explain second law of thermodynamics. ∆S System ≥ 0 Kelvin-Planck statement: It is impossible for any heat engine to produce net work in a complete cycle if it exchanges heat only with bodies at a single fixed temperature. Thus to produce the work the cycle should exchange heat with two reservoirs which are a different temperatures. The reservoir at high temperature is the source and other one at low temperature reservoir is the sink. Clausius statement: It is impossible to construct a device which, operating in a cycle, will produce no effect other than the transfer of heat from a colder to a hotter body. Thus in order to transfer heat from low temperature reservoir to high temperature reservoir, external work must be done on the cycle. Q2. a) An open system has no restriction on energy and matter transfer through the boundary of the system. Example is an open steel tumbler. Both water vapour and energy in for of heat can flow in and out of the system. dU ≠ 0 and dV ≠ 0 b) A closed system has restriction on flow of matter across its boundaries though energy can flow across. Example is a closed metallic jar. No material content flows out or in but cooling or heating takes place. dU ≠ 0 and dV = 0 c) An isolated system has restriction on transfer of both matter and energy across its boundaries. Example is a insulated flask from which neither content nor energy comes out in any form. dU = 0 and dV = 0 d) An adiabatic system is an insulated system from which no heat is transferred in or out. Example is a thermos container. dQ = 0 Q3. a) Kinetic Energy, KE= (1/2) mV2 Where m is the mass and V is the velocity b) Potential Energy, PE= mgz Where m is mass, g is acceleration due to gravity and z is the datum. c) Work, W=Wext(external work)+Wflow(flow work) Wflow= m (p/r) Where m is mass, p is pressure and r is the density d) Change in Heat content, dQ = dE + dW Q4. Formula for conversion of temperature scale is: TK K = (TK - 273.15) °C = [1.80 * (TK - 273.15) + 32] °F = 1.80 TK °Rankine i. Normal body temperature is 98.6°F = 37°C = 558 °Rankine ii. Absolute zero = 0 K = -459.4°F iii. Temperature rises by 30°C = Temperature rise 30 K iv. Temperature rise by 60°F = Temperature rise by 33.33°C = Temperature rise by 33.33K = Temperature rise by 59.4 °Rankine Q5. There are basically there mechanisms of heat transfer. They are namely, Conduction, convection and radiation. a) Conduction: It is a mode of direct heat flow through any matter and it is a result of actual physical contact between two parts of a same body or two different bodies. Heat is always conducted from hot to cold part or body but is never the other way round. Heat flows through the shortest and easiest route. Example is heating of a metallic container kept on heater. b) Convection: It is a mode of heat transfer in gases and liquids which is caused by the actual flow of the material itself i.e. the movement of the matter. Generally the movement of the matter is turbulently upward after getting heated though with a component of sideways motion also. Example is boiling of water. c) Radiation: This is a mode of heat transfer in vacuum or in space without and help of matter. The transfer takes place through infrared or electromagnetic waves. Example is the heat og sun coming to earth. Q6. Heat transfer is often classified as being steady or transient (also called unsteady). The term steady implies no change with time at any point within the medium, while transient implies variation with time or time dependence. Therefore, the temperature or heat flux remains unchanged with time during steady heat transfer through a medium at any location, although both quantities may vary from one location to another. During transient heat transfer process, the temperature normally varies with time but not with position. Most heat transfer problem encountered in practice is transient in nature, but they are usually analyses under some presumed steady condition since they are easy to analyse. Transient heat conduction Steady heat conduction Q7. In natural convection, the driving force for heat transfer to take place is buoyancy. This buoyancy is because of difference in fluid density because of acceleration in the system. The fluid around the source of heat gets less dense after being heated and rises above. This gives way to cooler and denser fluid to go down near the source. This process keeps on continuing and the heating takes place and it is said that convection current is formed. Forced convection, on the other hand, takes place when forced draft of the fluid through external means like mechanical pumps or fans is used to artificially form the convection current. Many times, the heating process is a combination of both natural and forced convection. Q8. Given data is, Mass Flow Rate: 2kg/s Heat of Combustion for C3H8: 46450kJ/kg Heat release rate is to be determined. Heat release rate = Flow rate X Heat of combustion = 2kg/s X 46450kJ/kg = 92900 kJ/s Q9. Given data is Temperature of black body T: 20 °C = 293.15 K Rate of heat transfer per unit is to be determined. The formula for heat transfer for black body is: Q = σ A T4 σ is the Stefan-Boltzmann Constant = 5.67 x 10-8 Watts-m-2-°K-4 And A = 1 for unit area. Thus Q = 5.67 x 10-8 Watts-m-2-°K-4 X 1 X (293.15)4 = 418 Watts-m-2 Q10. Fourier's law is the law of heat conduction and states that the time rate of heat transfer through a material is proportional to the negative temperature gradient and to the area at right angles, to that gradient, through which the heat is flowing. Where, is the local heat flux, [W·m−2] is the material's conductivity, [W·m−1·K−1], is the temperature gradient, [K·m−1]. Thermal conductivity is a material property represented by notation ‘k’ and indicates material’s ability to conduct heat. High value of thermal conductivity allows heat conduction to occur readily. It can be defined by the help of Fourier’s law as the quantity of heat, ΔQ, transmitted during given time (Δt) through a thickness x, in a direction normal to a surface of area A, induced by a temperature difference ΔT. Since thermal conductivity is a material property, it varies from material to material. In general metals exhibit very high thermal conductivity followed by insulating materials which have moderately low thermal conductivity and last comes gases including air which are poor conductors and have very low value of thermal conductivity. Q11. The Stefan–Boltzmann law is also known as Stefan's law and it highlights the heat transfer related to black bodies. This law states that the total energy radiated per unit surface area of a black body in unit time which is know as the emissive power, is directly proportional to the fourth power of the black body's absolute temperature T. The formula for heat transfer per unit surface area of black body is: Q = σ T4 σ is the Stefan-Boltzmann Constant = 5.67 x 10-8 Watts-m-2-°K-4 Emissivity of any material is a measure of its ability to radiate the absorbed energy. It can be defined as the ratio of energy radiated by a particular material to energy radiated by a black body under the same conditions. It is expressed by ‘ε’ or ‘e’. Since it is a ratio, it has got no unit and is dimensionless. For a true black body, ε =1 For real objects, ε m1 * c1 * ΔT1 = m2 * c2 * ΔT2 => ΔT1 = ΔT2 => T - T1 = T2 – T => T = (T1 + T2) / 2 = (17 + 85)/2 = 510C Q15. Given data is Surface area of the windshield = 0.75 m2 Thickness of ice layer = 1 mm = 0.001 m Air temperature = 0 0C Time to melt Ice = 2 minute = 120 seconds Volume of Ice to be melted = 0.75 X 0.001 = 0.00075 m3 Mass of ice to be melted = Volume X Density= 0.00075 X 916.7 kg/m³ at 0 °C = 0.6875 Kg Now heat required to melt this amount of ice at 0 °C = mass x specific latent heat of fusion = 0.6875 kg X 334720 J/kg = 230130 J Thus the power required = 230130/ 120 = 1917.75 Watt= 1.92 KW Q16. The heat flow per second through a cross section area (A) through conduction is given by: H = k * A * ΔT/x Where, H = the heat flowed per second k = Coefficient of thermal conductivity A = Total cross-sectional area of the conducting surface ΔT = Difference between the temperature x = Thickness of conducting surface separating the two temperatures. Since it’s given that for same condition for brick and air, Heat conducted through brick = Heat conducted through air k1 * A1 * ΔT1/x1 = k2 * A2 * ΔT2/x2 Where, k1, k2 = Coefficient of thermal conductivity of brick and air respectively A1 = A2 = Total cross-sectional area of the conducting surface of brick and air (same condition for brick and air) ΔT1 = ΔT2 = Difference between the temperature (same condition for brick and air) x1, x2 = Thickness of brick and air respectively Hence, k1 / x1 = k2 / x2 => k1 * x2 = k2 * x1 Given, Coefficient of thermal conductivity of brick k1= .9 W/mK Coefficient of thermal conductivity of air k2= 0.023 W/mK Thickness of air = x2 = 10 cm = .01 m X1 = (k2 * x2) / k1 = (.023 * .01)/ .9 = 2.56 x 10-4 m = .0256 cm Q17. Incomplete question (data not sufficient) Energy Radiated from a black surface is given by: W = σ * A * T4 Where, σ = Stefan-constant T = Absolute temperature of the radiating surface A = Area of radiating surface of sun = 4πr2 Given, W = Not Given σ = 5.67 x 10-8 Wm-2K-4 (assume) r = 6.96 * 108 meters (assume) Hence, A = 4 * (22 / 7) * (6.96 * 108)2 = 6.08 x 1018 Hence, Temperature of sun (T) = [W / (σ * A)](1/4) = Q18. The heat flow per second through a cross section area (A) through conduction is given by: H = k * A * ΔT/x Where, H = the heat flowed per second k = Coefficient of thermal conductivity A = Total cross-sectional area of the conducting surface ΔT = Difference between the temperature x = Thickness of conducting surface separating the two temperatures. For plywood it’s given, k = .17 W/mK A = 1 m2 (per unit area) ΔT = 30 – 10 = 200C = 293.15K Q19. The heat flow per second through a cross section area (A) through conduction is given by: H = k * A * ΔT/x Where, H = the heat flowed per second k = Coefficient of thermal conductivity A = Total cross-sectional area of the conducting surface ΔT = Difference between the temperature x = Thickness of conducting surface separating the two temperatures Given, Conductivity of Styrofoam k = .1 W/m-K A = 370 cm2 = .037 m2 ΔT = 800C – 200C = 600C =333.15 K x = 5 mm = .005 m i) Hence heat transfer per second (H) = (.1 * .037 * 333.15) / .005 = 646.13 W ii) Total thermal content of coffee (Q) = m * c * ΔT Given, m = .5 kg c = Specific heat capacity of coffee = 2.08 KJ kg−1 K−1 ΔT = 800C – 700C = 100C = 283.15K Hence, Q = .5 * 2.08 * 283.15 = 294.48 KJ = 294.48 KW = 294480 W Hence total time required for the coffee to cool from 800C to 700C: t = Q / H = 294480 / 646.13 = 455.76 seconds Q20. In laminar flow the motion of the particles of fluid is very orderly with all particles moving in straight lines parallel to the pipe walls. Turbulent flow is regarded by disordered and stochastic property changes. Turbulence has low momentum diffusion, high momentum convection, and rapid variation of pressure and velocity in space and time. Reynolds number is defined as the ratio of inertial forces to viscous forces and it quantifies the importance of these forces in fluid flow. Reynolds number, Re, which is a dimensionless number is given by Where р = density, u = mean velocity, d = diameter and µ = viscosity Laminar flow: Re < 2000 Transitional flow: 2000 < Re < 4000 Turbulent flow: Re > 4000 Q21. A compressible fluid can be defined as the type of fluid whose density changes with the application of high pressure-gradients. In a compressible fluid, the application of a force at one point in a system never results in flow through the system. Rather, the fluid gets compressed at the point of application of the force. On the other hand an incompressible fluid is not reduced in volume by an increase in pressure. The pressure applied at the one point is propagated throughout the system without any gradient. Thus the application of force results in flow through the system. Q22. Change in length of concrete slab due to thermal expansion is given by: ΔL = L0 * αL * ΔT Where, ΔL = Change in length L0 = Original length αL = Thermal expansion co-efficient ΔT = Change in temperature Given, L0 = 12 m αL = 1 X 10-5 ºC-1 ΔT = 35°C – (-5°C) = 40°C Hence, ΔL = 12 * 1 X 10-5 * 40 = 4.8 x 10-3 m Q23. Let the weight of ice be m2 Thermal content of tea at temperature T1 = 900C = 363.15K is given by: Q1 = m1 * c1 * ΔT1 Where, m1= mass of the tea = 1 L = 1 Kg, C1 = specific heat capacity of the tea (water) ΔT1 = change in temperature by mixing water of other temperature = T1 - T T = Final temperature of mix Thermal content of ice at temperature T2 = -100C = 263.15K is given by: Q2 = m2 * c2 * ΔT2 Where, m2= mass of the ice c2 = specific heat capacity of the ice ΔT2 = change in temperature by mixing ice and tea = T – T2 After mixing both the water we get: Q1 = Q2 => m1 * c1 * ΔT1 = m2 * c2 * ΔT2 m2 = (m1 * c1 * ΔT1)/ ( c2 * ΔT2) Given, m1 = 1 kg T = 100C = 283.15K c1 = 2.08 KJ kg−1 K−1 ΔT1 = 80K c2 = 2.05 KJ kg−1 K−1 ΔT2 = 20K Hence, m2 = (1 * 2.08 * 80) / (2.05 * 20) = 4.06 Kg Hence number of ice cubes of weight 20 g (.2 Kg) = 4.06 / .2 = 20.29 = 21 ice cubes Q24. Change in length of aluminium siding due to thermal expansion is given by: ΔL = L0 * αL * ΔT Where, ΔL = Change in length L0 = Original length αL = Thermal expansion co-efficient ΔT = Change in temperature Given, L0 = 3.66 m αL = 2.3 X 10-5 ºC-1 (for aluminium) ΔT = 39°C – (-28°C) = 67°C Hence, ΔL = 3.66 * 2.3 X 10-5 * 67 = 5.64 x 10-3 m Hence changed length of aluminium siding = L0 + ΔL = 3.66 + 5.64 x 10-3 = 3.6656 m = 36.66 cm Q25. Change in length of steel due to thermal expansion is given by: ΔL = L0 * αL * ΔT Where, ΔL = Change in length L0 = Original length αL = Thermal expansion co-efficient ΔT = Change in temperature Given, ΔL = Final length – original length = 11.7 m – 11.2 m = .05 m L0 = 11.2 m αL = 1.1 X 10-5 ºC-1 (for steel) Initial temperature of steel (T) = 220C Hence, Change in temperature ΔT = ΔL / (L0 * αL) = .05 / (11.2 * 1.1 X 10-5) = 405.840C So, Final temperature of steel = T + ΔT = 22 + 405.84 = 427.840C Read More