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,Q6 A vertical cylindrical tank, 2m diameter, has a sharp-edged orifice with diameter d = 0.05m near the bottom. The orifice has a discharge coefficient of 0.6, see Figure Q6. (a) Derive the orifice discharge rate in terms of depth of water in the tank above the orifice, h. [7] (b) If water enters the tank at a constant rate Qin of 0.0095m'/s, find the depth of water above the orifice when the level in the tank becomes stable. [8] (c) If the tank is emptied and water then runs into the tank at 0.02m /s, the orifice remaining open, find the rate of rise in water level in the tank when the level has reached a depth of 1.7m above the orifice. [5] Qin = 0.0095m /s d = 0.05m