Ministry of high education& scientific Research University of echnology Materials engineering department Class: Second Year Subject: Chemical Metallurgy Examiner: Final Examination (0-0) وزارة التعليم العالي و البحث العلمي الجامعت التكنولوجيت قسم هندست المواد Date: 7/05/0 Allowed time: 3 hrs Note: Attempt only Five Questions. Q / (A) Determine two parameters used to predict spontaneity of the process, then determine when each one from them is useful in dealing with different conditions of the process. State their relationship with enthalpy change. (B) Is the reaction spontaneous under standard conditions? 4KClO 3 (s) 3KClO 4 (s) + KCl (s) i H f (kj/mol) S o KClO 3-397.7 43. KClO 4-43.8 5.0 KCl -436.7 8.6 (0 Marks) Q / (A) Explain in details the Reversible and Irreversible processes, give an example. (B) Water at 33 C is pumped from storage tank at the rate of 0. m 3 /sec. he motor for the pump supplied work at the rate of 60 KJ/Kg. he water passes through heat exchanger of 000 KJ/Kg, and is delivered to elevation of 0 m above the first tank at the velocity 5 m/sec. What is the temperature of the water delivered to this elevation? (0.94 Btu = 0 3 J). Density of water=000 Kg/m 3. Sketch Flow Diagram. Enthalpy, Btu/kg emperature C Enthalpy, Btu/kg emperature C 60 50 0 90 79 70 6 0 (0 Marks) Q 3 / (A) For closed system adiabatic processes for an ideal gas, by law of hermodynamic prove that: beginning from the differential first P P (B) One Kilogram of air is heated reversibly at constant pressure from an initial state of 300 K and bar until its volume triple. Calculate Q, W, U and H for the process. Assume that air obeys the relation PV/=83.4 bar cm 3 mol K, Cp=9 J mol K, and Molecular Weight of air=9 Kg/Kmol. (0 Marks)
Pressure Q 4 / For the following Figure, calculate for each process U, H, Q & W. Given that: = =600 K P = 0 bar, P = 3 bar, P 3 = bar Cp=3.5R, Cv=.5R R=8.48 J/mol. K Volume (0 Marks) Q 5 / Sulfur dioxide gas is oxidized in 90 percent excess air with 80 % conversion to Sulfur trioxide. he gases enter the reactor at 40 o C and leave at 460 o C. How much heat must be transfer from the reactor on the basis mole of entering gas? 4 3 SO + O SO 3 Given data; Standard heat of formation at 5 C, and Constants. i H 98 J/mole A 0 3 B 0 6 C 0-5 D SO -96830 5.699 0.80 --- -.05 O ---- 3.639 0.506 ---- -0.7 SO 3-39570 8.060.056 ---- -.08 N ----- 3.80 0.593 ---- 0.040 (0 Marks) Q 6 / (A) One mole of an ideal gas, Cp= (7/)R, Cv=(5/)R, is compressed adiabatically in a piston/cylinder device from bar and 40 C to 5 bar. he process is irreversible and requires 35 percent more work than a reversible adiabatic compression from the initial state to the same final pressure. What is the final temperature and entropy change of the gas? (B) For an ideal gas with constant heat capacities show that: For a pressure change from P to P, the sign of S of an isothermal change is opposite that for a constant volume change. (0 Marks) Good Luck
Q / (A) Question Solutions of the Final Examination (0-0)-Chemical Metallurgy University of echnology,materials engineering department he major parameters to predict spontaneity of the process are Entropy change ( S) and Free Energy ( G), the first one is useful in dealing with different conditions of the process, but usually the second ( G) be more useful for certain conditions of temperature and pressure. Note that ΔG is composite of both ΔH and ΔS: If ΔH < 0 and ΔS > 0.spontaneous at all A reaction is spontaneous if ΔG < 0. Such that: If ΔH > 0 and ΔS < 0.not spontaneous at any If ΔH < 0 and ΔS < 0.spontaneous at low If ΔH > 0 and ΔS > 0.spontaneous at high Q / (B) ΔG = ΔH -ΔS ΔH rxn = ᵧi ΔH i = 3( 43.8kJ) + ( 436.7kJ) 4( 397.7kJ) = 44kJ ΔS rxn = ᵧi S o i = 3(5.0J/K) + (8.6 J/K) 4(43.J/K) = 36.8J/K ΔG rxn = ΔH rxn ΔS rxn = 44kJ (98K)( 36.8J/K)(kJ/000J)= 33kJ ΔG rxn < 0; therefore, reaction is spontaneous under standard conditions. Q / (A) A process is reversible when its direction can be reversed at any point by an infinite change in external conditions. Once the process is initiated, no infinite change in external conditions can reverse its direction; the process is irreversible. he apparatus is shown in figure (a gas in piston/cylinder). Expansion processes result when mass is removed from the piston (we assume the piston without friction), then the piston will rise and new balance of the piston level will be achieved. he oscillation of the peiton level due to that mass m is suddenly removed to a shelf. If we change the mass by powder the piston gradually rise and the other hand will return to initial level in the same path. m Figure : Gas Expasion l
Question Solutions of the Final Examination (0-0)-Chemical Metallurgy University of echnology,materials engineering department Q / (B) H + E p + E k = Q W s Mass flow rate= Density x Volumetric flow rate= 0. m 3 /sec X 000 Kg/m 3 =00 Kg/sec. E p = m Z g= 00 Kg/sec x 0 m x 9.8 m/s = 980 kj/sec. E k =m x u /= 00 Kg/sec (5-0)/= 50 kj/sec. (u could be neglected; tank). Q = 000 kj/kg x 00 Kg/sec = 00000 kj/sec. (positive sign assume heat absorbed by the system). W s = 60 kj/kg x 00 Kg/sec = 6000 kj/sec. hen H= Q W s E p E k = = 00000 kj/sec 6000 kj/sec 980 kj/sec 50 kj/sec = 8940 kj/sec. = H H By extrapolation method H (at 33 C) = 44 Btu/Kg x ( kj/0.94 Btu) x 00 Kg/sec= 5300 kj/sec. hen H = H + H = 8940 kj/sec. + 5300 kj/sec. = 9840 kj/sec. H = 9840 kj/sec. x (0.94 Btu/ kj) / 00 Kg/sec = 7.5 Btu/kg. By extrapolation method (have H=93.5 Btu/kg ) = 66.8 C. Q 3 / (A) First law of thermodynamic, U = Q W (Adaibatic Q =0) C v d = P dv, P=R/V hen C v d = R(dV/V) d/= R/C v (dv/v) R/C v = (C p C v )/ C v = C p / C v =ᵞ d/ = (ᵞ ) dv/v (Integration) ln( / )=ln(v /V ) -(ᵞ-) V V ( ) P V / = P V /, then V /V = ( / )(P /P ) substitute in the last equation hen: P P
Question Solutions of the Final Examination (0-0)-Chemical Metallurgy University of echnology,materials engineering department Q 3 / (B) No. of moles = /9 Kg/Kmol =0.0 34 mole P V / = P V / = 83.4 bar cm 3 mol K, constant pressure P= P V / = V /, and V / = 3V /, =3 =3 x 300= 900 V = 300 X 83.4 bar cm 3 mol K = 494 cm 3 mol and V = 848 cm 3 V = 3V = 544 cm 3 At constant pressure Q = H=nC p ( )= 0.034 x 9 (900-300) = 600 J W=P V= X 0 5 (544 cm 3-848 cm 3 ) = 696 x 0 5 N/m. cm 3 =69.6 J U = Q W = 600 J 69.6 J = 430.4 J. Q 4 / = (Isothermal process), H=0, U=0 P Q=W= R ln 0 =8.34 x 600 ln = 605 J. P 3 3 Constant Volume (W=0) P P 3 3, constant volume. 3 = 600 x /3= 066.6 K H= C p ( 3 )= 3.5x8.34(066.6 600)= 559.4 J U= C v ( 3 )=.5x8.34(066.6 600)= 085.3 J Q = U = 085.3 J 4 Adiabatic Process (Q=0), and P 4 =P 3 = 3 bar 4 P P 4, 600 0 4 3, and =.4 hen 4 = 34.3 K H= C p ( 4 )= 3.5x8.34(600 34.3)= 355.4 J U= C v ( 4 )=.5x8.34(600 34.3)= 9679.6 J W= - U= 9679.6 J 3 4 Constant pressure (Isoparic), Q= H H= C p ( 4 3 )= 3.5x8.34(34.3 066.6)= 970.0 J U= C v ( 4 3 )=.5x8.34(34.3 066.6)= 407. J W = Q U = 970.0 407.= 56.8 J
Question Solutions of the Final Examination (0-0)-Chemical Metallurgy University of echnology,materials engineering department Q 5 / Basis: mole of SO entering., =40+73.5= 683.5 K, = 460+73.5= 733.5 K. O required (computed on base completely conversion) According reaction equation ( mole SO Vs 0.5 mole O required) Excess O = 0.5 x 0.9= 0.45 mole otal O entering = 0.5 + 0.45= 0.95 mole. 79 N entering= 0.95 x = 3.57 mole. Leaving the reactor: SO = x ( 80% conversion) = 0. mole O reacted = x 80% x 0.5= 0.4 mole. O output = input-reacted = 0.95 0.4 = 0.55 mole. SO 3 produced = No. of mole SO reacted = 0.8 mole. in =40 o C SO mol O 0.95 mol N 3.57 mol H R Q= H H P out =460 o C SO 0. mol SO 3 0.8 mol O 0.55 mol N 3.57 mol H= H R + H 98 + H P H 98 H 98 = ᵧi ΔH i = - 39570 + (- ) (- 96830)=- 98890 J am =( + )/=708.5 K. H R = n i Cp m,hi (98.5 in ) H P = n i Cp m,hi ( out - 98.5) n i Cp m,hi =R( n i A i + ( n i B i ) am + ni Di ) n i A i = x 5.699 + 0.95 x 3.639 + 3.57 x 3.80= 0.86 n i B i = 3.4, n i D i = -.088.088 n i Cp m,hi = 8.34(0.86+3.4 x 708.5 + )= 73.5 683x733 he: H R = n i Cp m,hi (98.5 683)= 6677.5 J And the same way H P = 78665.3 J Finally H= 6677.5 J + ( 98890 J) + 78665.3 J = 86996. J
Question Solutions of the Final Examination (0-0)-Chemical Metallurgy University of echnology,materials engineering department Q 6 / (A) Compressed adiabatically, Q=0 U = W he first step we find reversible and adiabatic work (i.e S=0) S=0=C p ln P R ln P 7/Rln 4 =Rln, = 465.3 K 33.5 W= U= C v ( )= 5/R(465.3-33.5)= W rev. = 363. J W irrev =.3 x W rev (More 30 %).= 4. J = C v ( irrv. )=5/R( irrv 33.5), irrv. = 5 K 5 4 S=7/R ln R ln 33.5 =.7 J/K > 0 spontaneous process. Q 6 / (B) S=C p ln P R ln, P. Isothermal process = P hen ( S) iso = R ln P. Constant volume P P, and P P P And S=C p ln P R ln P P P = (C p -R)ln P = +Cv ln P P herefore ( S) opposite sign of ( S) V