Question

7.27 One kg of helium initially at 20°C and 1 bar is contained within a rigid, insulated tank. The helium is stirred by a paddle wheel until its pressure is 1.45 bar. Employing the ideal gas model with k = 1.67, determine the work and the exergy destruction for the helium, each in kJ. Ignore the effects of motion and gravity and let To = 20°C, po = 1 bar.

I need those answers ===>  W = -411.2 kJ, Ed = 339.2 kJ

Answer 1

Soli: riven T, = 20°C P = ibor P2 = 1045 bar 5o = 20°c Po=1 boo Insulation Y llega Energy balance lavation; त su =a w =) w=-m[U (T2)-U(T) ] - o ideal gas lavation; PV = MRT 7: T, = T (M 2 ) = (20+273x) ( 1:45) -425 Pev = MRT The valve et Cp is constant for monoatomic gases 1. To = 5 Ř & Ev = 3/8 R = 105 (8.314 KJ 7 Lu.003 KgK 0 > WC-MCV (To-T) = -x)(1.5) [4:30 tot het ons [495-293] K W = -411.2 kJ]

The exeggy destruction is F. = Too [o is amt of entropy produced] Now, as = s( 50) + 5 =) = m(89-9) [reg) uny- (iun do] wzo [cos) o-les) un deš Jue 5 m (15 p.) ima = (1kg) (1:5)( 8.314 Mode) in (1.45) = 1.158 KI 4.003 KgK 1.158 KU So; E = Too = (293k ) (1,158 km) EJ = 339.2 KJ