Solution cont. Alternatively, Closed system Imaginary piston' Boundary work: P;1 MPa !(constant) Pidv=-Pi(V2 -Vi)=-P[Vank-(Vink +Vi)]=PV; 1MPa ;=12 V--volume occupied by the steam before it enters tank Energy balance for the closed system P2=1 MPa △E=￠-Won (b)The closed-system Wbin=△U equivalence miPiVi=mu2-miui ->w2=u;+PV=h; 上游充通大 Mar/29,Wed,2017 6 SHANGHAI JIAO TONG UNIVERSITY
Mar/29, Wed, 2017 6 Solution cont. Alternatively, Closed system 1MPa Boundary work: -- volume occupied by the steam before it enters tank Energy balance for the closed system E Q Wb,in
Example 22.2 Cooking with a Pressure Cooker Known: pressure regulator (petcock) ·Initially1 kg water 30 min Qin 1atm>p=75kPa(gage) 1atm 100 kPa ● unsteady System boundary. H20 m=I kg V=6L Determine: CV p=75 kPa(gage) Vapor e>Higher p, ·T2,m2 T(133C Liquid Assumptions: ·Uniform flow 0m=500W ·△PE,△KE→0 one exit and no inlets ·p(thus T)constant Steam leaves as a saturated vapor ·Wcy→0 上游充通大学 Mar/29,Wed,2017 7 SHANGHAI JLAO TONG UNIVERSITY
Mar/29, Wed, 2017 7 Example 22.2 Cooking with a Pressure Cooker Higher p, T (133˚C) pressure regulator (petcock) Known: • Initially 1kg water • 30 min Qin 1atmp=75kPa(gage) • 1atm = 100 kPa • unsteady Determine: • T2 , m2 Assumptions: • Uniform flow • ∆PE, ∆KE 0 • p (thus T) constant • Steam leaves as a saturated vapor • WCV0 CV one exit and no inlets