Question

You are working for a venture capital firm and are tasked with determining if an invention should be considered for financial feasibility studies. The invention consists of a tube into which compressed air is admitted. The tube separates the compressed air at an intermediate temperature into two air flows, one hot and one cold. A schematic of the invention is shown below in fig. 1. The compressed air admitted into the device is divided such that 60% of it exits at the cold outlet and 40% at the hot outlet. The operating parameters for the device as given by the designer are given in table 1. Note that the device has two modes of operation. The compressed air in the device performs no external mechanical work and no significant heat transfer occurs within the device. Table 1: Operating Parameters for Device. The mass flow rate is not required for the initial analysis (a) For each mode of operation (Mode 1 or Mode 2) determine if the device violates the law of conservation of energy. Ignore any changes in macroscopic kinetic or potential energy in the flow. Derive all necessary equations based on the assumption that air behaves as an ideal gas with y =1.4 (b) For a closed system consisting of n moles of an ideal gas, show that AS for a reversible process that takes the gas from state 1(n, Pi, Vi, Ti) to state 2(n, Pf, Vf, Tf) is: delta S - nR[gamma/gamma - 1 In (T_l/P_i)] Your derivation should start from the Is' law of thermodynamics, the Claussius definition for change of Entropy (dS = dQ_rev/T) and the ideal gas equations: PV = nRT = (y-l)U. c) For each mode of operation (Mode 1 or Mode 2) determine if the device violates the 2nd Law of Thermodynamics. (Assume air behaves as an ideal gas with gamma = 1.4) d) Comment on the feasibility of the device and whether you would recommend your company to start considering financial feasibility

Answer 1

Data given, Draw the given data table. Mode 1 5 atm 1 atm (20+273) K (-18 +273) K (70+273) K Mode 2 5 atm 5 atm Pi (20+273) K (-18 +273) K (70+273) K Consider the first law of thermodynamics Substitute 0 for Q, 0 for W

Write the expression for change in internal energy for cold gas Substitute 0.6m for m, Write the expression for change in internal energy for hot gas. Substitute 0.4m for m Calculate the total change in internal energy. Substitute the values.

Calculation for mode1 Substitute the values in equation (1). dU0.6mc, (-18-20)+0.4mc, (70-20) For mode 1, the value of dU is non-zero. Hence,mode 1 violates lanw of comservation of energy Calculation for mode 2. Substitute the values in equation (1). dU0.6mc, (-18-20)+0.4mc, (70-20) For mode 2, the value of dU is non-zero. Hence, mode 2 violates law of conservation of energy

Consider the first law of thermodynamics Substitute Tds for Q. pdV for W Tas-pdV = dU Consider the expression for enthalpy. Differentiate the above expression Hence

RT Substitute c dT for dh, _- for T Integrate the above equation. yR Substitute for c For n moles Pi Hence, the required expression of change in entropy is obtained

For mode1 For cold gas Write the expression for change in entropy for cold gas Substitute the values 1.4 .(-18+273 1.4-1 20 +273 AS- (0.6x8.314)1n - 6.603 J/K For hot gas Write the expression for change in entropy for hot gas Pi Substitute the values 1.470+273 1.4-1 20+273 AS (0.4x8.314)In = 7.186 J/K

Calculate the total change in entropy. AS AS+AS Substitute the values. AS-6.603+7.186 613.8 J/K Since AS > 0, it does not violate 2nd law For mode 2: For cold gas: Write the expression for change in entropy for cold gas Pi Substitute the values. AE: (0.6 x 8.3 14). 1.4 -2.425J/K I n(. .18.:.273) 1.4-1 20 +273

For hot gas: Write the expression for change in entropy for hot gas Pi Substitute the values. AS, (0.4x8.314 14-1 20 +273)5 1.833 J/K Calculate the total change in entropy AS AS+AS Substitute the values. AS--2.4251.833 Since AS <0, it violates 2nd law. There is no feasibility because it violates 1st law of thermodynamics.