Thermodynamics Formula Sheet First law of thermodynamics for a system: (Qin - Cout) + (Win - Wout) + (Emass,in - Emass,out) = AU + AKE + APEB Steady Flow Energy Equation (SFEE) for a single stream control volume: Qcv - Wev = m (hz - hy) + (52 20:) + g(zz -zz)) Rates of Conductive & Convectiont Transfer: Qcond = -kcond A ax Qconv = -hconvAAT Rate of Radiative Heat Transfer: Qrad = Azo (TB* - TA* ) where o = 5.67x10-8 W/m2.K4 Entropy balance for a system: at (Ssystem) = Sin - Sout + Sgen where Sinlout = Linjout Tinlout Gibbs equations: TdS = du + PdV = dH - Vdp Gibbs Energy: G =H -TS Boltzmann distribution: s = k In w where k = 1.38x 10-23 J/K For an ideal gas: PV = nRT where R = 8.314 J/(mol.K) and R = - Rspec where an over dot indicates rate and A area K Boltzmann's Cy, cp specific heat constant AU = mcvAT capacity at constant Kcond thermal conductivity V velocity V. P m mass V volume Emass energy associated n number of moles W work with mass transfer PE potential energy Z height above datum KE kinetic energy P pressure e emissivity g gravitational Q heat transfer Stefan Boltzmann constant Rspec gas constant, constant G Gibbs energy Rspec = Cp - CV w number of viable h, H specific enthalpy, s, S specific entropy, arrangements enthalpy, entropy AH = mcpAT time hconv heat transfer temperature coefficient internal energy,Q2 Two 400W fans are switched on in a perfectly-insulated, closed room at atmospheric pressure and a temperature of 18 C. If the fans operate continuously and the room is 5m x 6m with 3.5m high ceiling: (a) Calculate the temperature of the air in the room after 2 hours, stating all of the assumptions made. [15] (b) What would be the temperature of the room if there was a heat loss through the walls of 750 W throughout the two hours? [5] Assume that the specific gas constant and specific heat capacity of air at a constant volume are 287 J/kg.K and 717 J/kg.K respectivelyDr D Litskevich 45228 Q3 For a university building, steam at 150 C for heating is supplied from a remote boiler house through a 35m length of insulated pipe over the road at first floor level. The pipe has an inside diameter of 0.15m with a wall thickness of 10mm and conductivity of 48 W/(m.K). The insulation is 50 mm thick with a conductivity of 0.87 W/(m.K). State Fourier's law of conduction for an infinitely long cylinder and define the terminology used. [5] Calculate the heat loss when the atmospheric temperature is -5 C and the heat transfer coefficients between the steam and pipe and from the insulation to the atmosphere are 2.84 and 34.1 kW/(m2.K) respectively. [15]PV = MRT - + 1 0 1 3 x 10 5 X 105 = mx ( 28 7 X 16 ) X ( 201.15 ) = > m = 12 7 3 2 3 / ky This is the mass of all present in the room. Now, time for which fan are ON time = 2 hours = 2 X 6 0 X6 0 seconds . eq ( 1 ) => Win = 80 0 *6 0 x6 0x2 Win - 576 OX103 ,Joules First law of twomodynamics ( Sin Pout ) + ( Win , Wock ) + Fman,in = Eman, out) = AUT AKE + APEBO . Win = 4U As. twain is inside the room. So- temperature change with Constant volume. AU = Cvm ( 47 ) = Win = mcv (AT ). of the room If T, in two final temperature" then Win = ( 12 7 . 3 2 31 ) ( 717 ) ( TS - Ti )2) a ) Given, pressure of air = 1 atm = 1013x10p. Intial temperature of the room Ti= 18 => T' = 18+ 27315 K > T; = 291 15 Kelvin volume of air = volume of room = 5mx6 mx 3 5 m ->V = 105m 3 Two Jan of grow are on . . Win 2 X 400 time Win = (80 0 x time) Joceles - (1 No wonk is done by allo ." Wool= 0- (2) Room is perfectly - insulated . Heat trans for on mass transfer with phen octaide on Surrounding is Zero . Bin = Boces = Emand,in = Emann, out = 0.- Assumption : ") ain is ideal gas. So potential Energy= 11) Kinetic energy of ain Consider to be negligible. Now, by idea yao equation7 5 760 X 10 = ( 127 . 3 231) ( 717) (75- 291.15) > TF = 354' 2452 k If = 81 09 51 7 13 ' C This in two temperature of the room. of room b) Now, walls are not insulted, heat Is ocet percent time is 750 Jocele. gout - 750 time 7 Bout = (7 50 x time ) J; Hime in second, total heat loss on oct from boom in 2 hours in Socel = ( 7 5 0 X 2 X 60 *60 ) Joule => Bout = 5400x103 Jocelen .'. From lot law of teenmodynamics in - Pout ) + ( Win = Wool ) + 0 = Autoto * ( - 5400 * 10 3 / 4 ( 5 160 * 105 ) = MCV AT 3 60 X 10 3 - (12 - . 3 2 31 ) ( =117 ) ( T - 291.15) If = 295 . 09 3 4513 Kelvin Tf - 21 94 3 4 5 1 3 0 c /This i tw temperature of room after * 2 hours when walls are not insulted X please understand Joule work 7 watt = [ heat Excalla time [ time ] [ time ]