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Solution HW-1 CHEE310 Spring 2014 2 .9. A system comprised of chloroform, 1,4-dioxane, and ethanol exists as a two-phase vapor-liquid system at 50°C and 55 kPa. It is found, after the addition of some pure ethanol, that the system can be returned to two-phase equilibrium at the initial T and P. In what respect has the system changed, and in what respect has it not changed? Components – Chloroform, 1,4-dioxane, ethanol Number of components = N = 3 Number of phases = = 2 Degrees of freedom = F = N - + 2 = 3 – 3 + 2 = 3 Since temperature and pressure are specified, this leaves one more variable to be defined. So if some pure ethanol is added to the three-component system, it will change the compositions of the two phases; T and P can still be maintained at their srcinal values. As a result of addition of ethanol to the system and returning the system to two-phase equilibrium at the initial T and P, the temperature and pressure of the system remain unchanged, but the compositions of the two phases will change. 2.10. For the system described in Pb. 2.9: (a) How many phase-rule variables in addition to T and P must be chosen so as to fix the compositions of both phases? (b) If the temperature and pressure are to remain the same, can the overall composition of the system be changed (by adding or removing material) without affecting the compositions of the liquid and vapor phases? (a) In problem 2.9, there are three degrees of freedom. So after specifying T and P, one more variable needs to be specified. (b) Addition or removal of one or two components to the system at specified T and P would result in redistribution of the components in the two phase system and thus changing the mole fractions of the components in the two phases. However, if the materials are added or removed such that the mole-fraction of any phase does not change (i.e. the whole mixture is added or removed in the srcinal proportions), then the only the amounts of the two phases will change. Hence, the answer to this question is yes. 2.27. Nitrogen flows at steady state through a horizontal, insulated pipe with inside diameter of from the valve the pressure is 100 (psia), the temperature is 120 ( o F), and the average velocity is 20 (ft)(s -1 ). If the pressure just downstream from the valve is 20 (psia), what is the temperature? Assume for nitrogen that P V/ T is constant, C V = (5/2) R, and Cp = (7/2) R. (Values for R are given in App. A.) Pipe diameter = 1.5 inch = 0.125 ft P 1 = 100 psia = 14400 lb f /ft 2 P 2 = 20 psia = 2880 lb f /ft 2 T 1 = 120 o F = 48.9 o C = 580 R Gas constant for nitrogen = R = 1545 ft. lb f /(lb mol .R) u 1 = 20 ft/s Molecular weight of N 2 = 28 lb m /lb mol R = 1545 [(ft.lb f )/(lb mol .R)]x[1/28 lb mol /lb m ]x[32.2 (lb m .ft)/(lb f .s 2 )] = 1545x(1/28)x(32.2) ft 2 /(s 2 .R) = 1776.75 ft 2 /(s 2 .R) For flow past a partially opened valve: H 1 + u 12 /2 = H 2 + u 22 /2 From gas law: P 1 u 1 / T 1 = P 2 u 2 / T 2 u 2 = u 1 (P 1 /P 2 )(T 2 /T 1 ) u 2 = (20 ft/s)(100/20)(T 2 /580) = T 2 /5.8 ft/s H 2 – H 1 = C P (T 2 – T 1 ) = (7/2 R) (T 2 – T 1 ) (7/2 R)(T 2 – 580) + (1/2) (T 2 /5.8) 2 – (1/2) (20) 2 = 0 Check units: [ft 2 /(s 2 .R)]x(R) + (ft/s) 2 - (ft/s) 2 units are consistent. (0.0149) T 22 + 6218.6 T 2 – 3607003 = 0 T 2 = 579.2 R = 119.2 o F 2.31. In the following take Cv = 5 and Cp = 7 (Btu)(lb mole) -1 ( o F) -1 for nitrogen gas: (a) Three pound moles of nitrogen at 70 (°F) contained in a rigid vessel, is heated to 350 ( o F). How much heat is required if the vessel has a negligible heat capacity? If it weighs 200 (lb m ) and has a heat capacity of 0.12 (Btu)(lb mole) -1 ( o F) -1 , how much heat is required? (b) Four pound moles of nitrogen at 400 (°F) is contained in a piston/cylinder arrangement. How much heat must be extracted from this system, which is kept at constant pressure, to cool it to 150 (°F) if the heat capacity of the piston and cylinder is neglected? Fluid: Nitrogen gas Molecular weight = 28 lb m /lb mol Cp = 7 (Btu)(lb mole) -1 ( o F) -1 C V = 5 (Btu)(lb mole) -1 ( o F) -1 (a) Rigid vessel constant volume system m gas = 3 lb mol T 1 = 70 o F = 530 R T 2 = 350 o F = 810 R (i) Negligible heat capacity of vessel. Heat required = C V (T 2 – T 1 ) = [3 lb mol ] [5 (Btu)(lb mole) -1 ( o F) -1 ] [(810-530)R] = 4200 (Btu) (ii) Mass of vessel = 200 lb m {Comment: The problem statement says weight, but the units suggest mass} Heat capacity of the vessel = 0.12 (Btu)(lb m ) -1 ( o F) -1 {Comment: The problem states (Btu)(lb mole) -1 ( o F) -1 , but heat capacity of vessel should be in terms of mass of vessel. Hence mass units have been used.} Heat required to raise the temperature of vessel = (200 lb m )[0.12 (Btu)(lb m ) -1 ( o F) -1 ][(810-530) R] = 6720 Btu Total heat required = 10920 Btu (b) Piston-cylinder system constant pressure system assuming the piston is free to move. Mass of gas = m = 4 lb mol T 1 = 400 o F = 860 R T 2 = 150 o F = 610 R Heat capacity of piston/cylinder is negligible. Heat required = (4 lb mol ) [7 (Btu)(lb mole) -1 ( o F) -1 ] [(610-860)R] = - 7000 Btu Since heat required is negative, +7000 Btu must be removed from the system.
