Name: SID: Dscusson Sesson: Chemcal Engneerng Thermodynamcs 141 -- Fall 007 Thursday, November 15, 007 Mdterm II SOLUTIONS - 70 mnutes 110 Ponts Total Closed Book and Notes (0 ponts) 1. Evaluate whether the statement s true or false. a) The effcency of the Carnot cycle s dependent on the workng flud. False - The effcency of the Carnot cycle depends only on the temperature of the hot and cold reservor. b) An deal Rankne cycle conssts of two adabatc and two sobarc processes. True - The deal Rankne cycle conssts of an adabatc turbne/pump and an sobarc boler/condenser. c) A vapor compresson cycle utlzng a turbne has a hgher COP than one wth a throttle valve operatng under the same evaporator and condenser pressures. True Work s recovered by usng a turbne. d) When usng Amagat s law, water has the same partal molar volume when added to ether pure ethanol or pure water. True - Amagat s law states that the partal molar volume of speces n mxture s equal to the molar volume of the pure speces. e) The chemcal potental of ce at 0 C and 1 atm s less than the chemcal potental of lqud water at 0 C and 1 atm. False - At 0 C and 1 atm, ce and lqud water coexst n equlbrum and ther chemcal potentals are therefore equal. f) For a pure substance, the chemcal potental s equal to the total Gbbs energy. False - For a pure substance, the chemcal potental s equal to the molar Gbbs energy. g) As pressure approaches zero, fugacty approaches negatve nfnty. False - Fugacty approaches pressure; chemcal potental approaches negatve nfnty. h) The natural log of the fugacty coeffcent of speces n a mxture, ln( φ ), s an example of a partal molar property. True - ln( φ ) n ln = n ( φ) T,P,n j, whch satsfes the defnton of a partal molar property. ) An deal soluton s one n whch the components do not nteract wth each other. False An deal soluton s one n whch nteractons between dfferent components s the same as nteractons between smlar components and all components are of smlar sze and shape. j) Actvty coeffcent s greater than 1 for a mxture exhbtng postve devaton from Raoult s law. True
. It s freezng n your hut n Sbera. You are out of frewood, but have plenty of electrcty. Desperate for warmth, you decded to construct a heat pump usng ammona: Sbera Q A W B 1 Evaporator (A) Compressor (B) Throttle (D) 4 Condenser (C) Hut Q C Saturated vapor leaves the evaporator at 18 F and saturated lqud leaves the condenser at 56 F. The effcency of the compressor s 0.67. a) Draw the cycle on a T-S dagram usng an deal compressor. On the dagram, label the equpment (A, B, C, D) and streams (1,,, 4). b) Calculate the coeffcent of performance for the ammona cycle when the real compressor s used. c) Unsatsfed wth your current setup, you consder the alternatve cycles. What s the hghest possble COP when operatng between 18 F and 56 F? Soluton a) Draw the cycle on a T-S dagram assumng deal equpment. On the dagram, label the equpment (A, B, C, D) and streams (1,,, 4). T 4 C B D 1 A S b) Calculate the coeffcent of performance for the ammona cycle when a real compressor s used. By the defnton of COP, COP = Q n / (-W) COP = Q A / (-W) From an energy balance on the cycle,
H = Q W = 0 -W = -Q = Q C Q A Combnng these relatonshps, COP = Q A / (-W) = Q A / (Q C Q A ) From an energy balance on the evaporator (A), H A = Q A W A = Q A Q A = H H 1 From an energy balance on the condenser (C), H C = -Q C W C = -Q C Q C = H H 4 Therefore, the COP for an deal compressor s ( H H1) COP = Q A / (Q C Q A ) = H H H H ( ) ( ) 4 1 Saturated vapor leaves the evaporator at 18 F, mplyng T = 18 F. From the saturated ammona table at T = 18 F for pure vapor, P = 46.1 psa, H = 617. Btu/lb m R, and S = 1.006 Btu/lb m R Saturated lqud leaves the condenser at 56 F, mplyng T 4 = 56 F. From the saturated ammona table at T 4 = 56 F for pure lqud, P 4 = 99.91 psa, and H 4 = 104.7 Btu/lb m R. Snce an deal compressor s sentropc, S = S = 1.006 Btu/lb m R by the ext condtons of the evaporator. Snce the heat exchanger operates at constant pressure, P = P 4 = 99.91 psa by the ext condtons of the condenser. From the superheated ammona graph, H = 660 Btu/lb m R and T = 80 F at S = 1. Btu/ lbm.r and P = 99.91 psa. Snce a throttle s senthalpc, H 1 = H 4 = 104.7 Btu/lb m R by the ext condtons of the condenser. The effcency of the compressor s 0.67. ( H H ) ( 660 617.) η = = = 0.67 H ' ' ' 681. 1 H H H 617. = Btu/lb m R ( ) ( ) Therefore, the COP for the cycle usng the real compressor s ( H H1) ( 617. 104.7) COP = = ' H H H H 681.1 104.7 617. 104.7 ( ) ( ) 4 1 ( ) ( ) = 8.0 c) Unsatsfed wth your current setup, you consder alternatve cycles. What s the best possble COP when operatng between 18 F and 56 F? The best possble COP can be acheved wth the Carnot refrgeraton cycle. Processes A and C would be replaced by sothermal heatng/coolng. Processes B and D would reman as an adabatc compresson/expanson. Startng from COP = Q A / (Q C Q A ) By the Second Law on an sothermal process,
Q A = T low S A -Q C = T hgh S C Substtuton gves COP = Q A / (Q C Q A ) = T low S A / (-T hgh S C T low S A ) From an entropy balance on a reversble cycle, S total = S A + S B + S C + S D = 0 By the Second Law on an reversble adabatc process, S B = 0 S D = 0 Substtutng S B and S D nto the entropy balance gves S C = - S A Therefore, the COP for the Carnot refrgeraton cycle s Tlow ( 18 + 460) COP = = = 1.58 T T 56 + 460 18 + 460 hgh low ( ) ( ) (0 ponts). A bnary mxture obeys the Margules relatonshps: ln γ1 = x [A1 + (A 1 A1 )x1] ln γ = x1 [A 1 + (A1 A 1 )x ] a) Keepng temperature constant, sketch ln γ 1 and ln γ over the entre range of composton ( x 1 = 0 to x 1 = 1) where A 1 > A 1, and the mxture exhbts postve devatons from Raoult s Law. Label the plot axes. Clearly ndcate how ln γ 1 and ln γ approach the ln γ = 0 horzontal axs. Label the ntercepts of ln γ 1 at x 1 = 0 and G ln γ at x 1 = 1 wth the proper Margules parameters. Sketch E on the same plot. RT b) At 50 K, we are gven ln γ 1 = 0.50 and ln γ = 1. 0 and for pure components: P1 sat = 50kPa and P sat = 0kPa Assumng low pressure and the vapor phase to be an deal gas, determne:. the total pressure (n kpa) exerted by a soluton of composton x 1 = 0. 40.. the vapor phase compostons at ths pressure for x 1 = 0. 40.
Soluton: lnγ 1 lnγ Intersecton at max of G E /RT lnγ =A 1 lnγ 1 =A 1 lnγ 1 G E /RT lnγ Slope = 0 x 1 Slope = 0 Margules Equatons: ln γ1 = A1 = 0.50 ln γ = A1 = 1.0 ln γ1 = x [A1 + (A 1 A1 )x1] ln γ = x1 [A1 + (A1 A1)x ] γ1 = exp(0.6 [0.50 + (1.0 0.50)0.40]) = 1.4646 γ = exp(0.4 [1.0 + (0.50 1.0)0.60]) = 1.059 sat sat P = y1 P + y P = x1γ1p1 + xγ P = 0.4(1.4646)(50kPa) + 0.6(1.059)(0kPa) = 4.00kPa sat x1γ1p1 0.4(1.4646)(50kPa) y1 = = = 0.697 P 4.00kPa y = 1 y1 = 0.0 (0 ponts) 4. For ths problem show ALL work and clearly show how smplfcatons and substtutons occur. Analogous to the conventonal partal property M, we can defne the partal property M ~ at a constant T, V, n j : a) Show that: M ~ Recall: = M M ~ + (nm) n M (V V ) V T,V,n j x x x w y = y + w z w y y z Hnts: Try w = V and note that all dervatves hold T and b) Show that M ~ satsfes the summablty relaton: M = x M~ Note: ths part can be done ndependently of part (a). n constant. j
Soluton: a) Usng the relaton x y z x x w = y + w w y y z where x = nm, y = n, w = V, z = P and Tandn j areheld cons tan t for all dervatves M (nm) n T,P,n j (nm) = n T,V,n j (nm) + V = n,n j (V) n T,P, n j obtaned by applyng the relaton to the defnton of the conventonal partal property. M (M) (V) M ~ = + n obtaned by applyng the defnton of the new partal property and V n T,P,n j seeng that f n j and n are held constant, n s held constant and can be pulled out of the dervatve. We need to get rd of the volume term somehow, so we use the defnton of the partal molar volume: V (nv) n T,P,n j V = n n T,P,n j n + V n T,P,n j V = n n T,P,n j + V = V V n T,P,n j = V V n Pluggng ths nto M (M) (V) M ~ + n V n T,P,n j = gves the fnal relaton: M ~ M = M + (V V ) V b) M = x M ~ M = M + V = x M M + (V V ) V M xv xv = M + V = xm V x M + x (V V ) V M V = M + V M = M + V ( V V) = M x (V V )