m = P 1V 1 RT 1 P 2 = P 1. P 2 V 2 T 2 = P 1V 1 T 1 V 1 2 V 1 T 2 = T 1. {z} T 2 = T 1 1W 2 = PdV = P 1 (V 2 V 1 ). Z T2 (c vo + αt)dt.

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1 NAME: SOLUTION AME 0 Thermodynamic Examination Prof J M Power March 00 Happy 56th birthday, Sir Dugald Clerk, inventor of the two-troke engine, b March 85 (5) A calorically imperfect ideal ga, with ga contant R and initially at P, T, V, fill a cylinder which i capped by a frictionle mobile piton The ga i heated until V = V The pecific heat i given by c v (T) = c vo + αt, where c vo and α are contant Find the final temperature and the heat tranferred to the ga in term of given quantitie The ma of the ga, m, i The proce i iobaric o From the ideal ga law, we have m = P V RT P = P P V T = P V T P V T = T P V {z} = T = T V V The work i Z W = PdV = P (V V ) The change in internal energy i Z u u = c v(t) = Z T (c vo + αt) T T u u = c vo(t T ) + α T T u u = c vo(t T ) + α (T T ) u u = c vo(t T ) + α (T T ) V u u = c vot + αt V! V V

2 So U U = P V V c vot + αt RT V U U = P V R From the firt law, U U = Q W, o Q = P V R c vo V V + αt V!! V V!! V Q = U U + W V c vo + αt V!! V V V cvo Q = P V V R + + αt V + R V V + P V V Overall performance on thi problem wa not very good The bigget problem wa imple; many tudent forgot calculate the total energy U via multiplication by the ytem ma Many tudent alo had difficulty in integrating the pecific heat to get the internal energy Alo, many tudent neglected to account for the work, which influenced the heat tranfer via the firt law Latly, a few tudent failed to realize that the proce wa iobaric (5) A phere of aluminum with radiu of 00 m i initially at 500 K It i uddenly immered in a very large tub of water at 00 K The heat tranfer coefficient i h = 0 kw/m /K Auming the phere ha a patially uniform temperature and contant material propertie, find the time when the phere temperature i 00 K The firt law hold that dt = Q Ẇ There i no work for thi problem, o Ẇ = 0 Thu dt = Q Now we neglect kinetic and potential energy change of the aluminum, o With m, c and E o contant, we thu have Now we know that So the firt law reduce to E = mct + E o dt = mc dt Q = ha(t T) mc = ha(t T) dt dt = ha (T T) mc dt = ha (T T) ρv c dt = h(πr ) ρ (T T) πr c dt = h (T T) ρrc T T = h ρrc dt ln(t T) = h ρrc t + C

3 At t = 0, we have T = T o, o T T = C exp h ρrc t T T o = C T T = (T T o)exp ρrc t h T T = exp h T T o ρrc t ln T T T T o = h ρrc t t = ρrc T T ln h T T o From Table A, we find for aluminum ρ = 700 /m and c = 090 //K So we look for the time when T = 00 K and get 700 m t = (00 m ) 090 K (00 K) (00 K) ln 0 kw (00 K) (500 K) m K t = 0 (50) Conider the Rankine cycle below Find P = 5 MPa T = 00 C heat lo from non-adiabatic turbine = 5000 kw turbine boiler m = 00 / P = 0 kpa x = condener P = 5 MPa T = 60 C pump P = 0 kpa x = 0 warm lake water out, ΔT = 0 C cold lake water in (a) the heat tranfer rate to the boiler (kw), (b) the power output of the turbine (kw), (c) the overall thermal efficiency, (d) the thermal efficiency of a Carnot cycle operating between the ame temperature limit, (e) an accurate ketch of the cycle on a T diagram, (f) the ma flow rate of external lake cooling water to exchange heat with the condener if the lake cooling water temperature rie i deigned to be 0 C At tate, we have two propertie From the table, we learn h = 98, = 069 K, T = 58 C

4 At tate, the compreed liquid table give h = 665, = 08 After the boiler, we know two propertie and find h = 975, = 5880 After the turbine, we know two propertie and find So the heat tranfer rate to the boiler i Q = ṁ(h h ) = 00 h = 586, = K K K 665 = 779 kw For the turbine we have /dt = Q W + ṁ(h h ) Now /dt = 0, o the power output of the turbine i W = ṁ(h h ) Q = (5000 kw) W = 08 kw The power requirement of the pump, aumed to be adiabatic, i W = ṁ(h h ) = The cycle efficiency i 98 η = Ẇnet (08 kw) (78 kw) = = Q 779 kw = 78 kw The Carnot efficiency for an engine operating between the ame temperature limit i η = T L = T H = 056 T ( C) (//K) The heat lo in the condener i Q = ṁ(h h ) = = 98 kw

5 Now for the lake water we need ṁ w = Q = ṁ wc( T) Q ṁ w = c T 98 kw (0 K) 86 K = 858 Overall performance on thi problem wa good but not outtanding Some had fundamental problem identifying the numerical value of the tate variable; thi wa intended to be an eay part of thi problem a there wa no interpolation involved A few failed to realize that tate wa a compreed liquid Mot got the turbine power; a few forgot to multiply by the ma flow rate Many did not correctly account for the heat tranfer in the turbine and had a ign error Mot did not account for the pump work in calculation of thermal efficiency Surpriingly many people ued Celiu and not Kelvin to calculate the ideal efficiency A few calculated the efficiency for a heat pump, not a power cycle T diagram were generally bad, with a few very good Some problem with T diagram include: ) Not howing that the entropy increaed in the pump, ) Not howing that the entropy increaed in the turbine, ) Not howing the vapor dome, ) Not howing the turbine temperature wa greater than the critical temperature, 5) Not howing the iobar wa an iotherm under the vapor dome, 6) Not placing tate and at the edge of the vapor dome Mot people got the cooling water flow rate correct 5