adrian bejan advanced engineering thermodynamics pdf

This search leads to this portion of the table: T P = 0.5 MPa P = 0.6 MPa v u v u ⋯ ⋯ ⋯ ⋯ ⋯ 800☌ 0.9896 3662.1 0.8245 3661.8 ⋯ ⋯ ⋯ ⋯ ⋯ Fitting v 2 between 0.9896 and 0.8245, we interpolate linearly for pressure and find 2 P 0. This is the equivalent of traveling along the u = u 2 line and looking for the v value that comes closest to v 2. Advanced Engineering Thermodynamics Adrian Bejan An advanced. At state ( 2 ), we know u 2 and v 2 therefore, one way to proceed is to look at the table of superheated steam properties and find the u values of order 3662 kJ /kg. A TEXTBOOK OF CHEMICAL ENGINEERING THERMODYNAMICS K.

The first one is the volume 1 1 2 1 f ,T 1 fg,T 3 v v v x v 0.001044 0.5(1.6729 0.001044) 0.837 m / kg         The second property is the internal energy: this comes from the first law 1 2 1 2 2 1 Q W m(u u )      ( 1 ) where W 1−2 = 0 and 1 1 1 f,T 1 fg,T u u x u 418.94 (0.5)(2087.6) 1462.74 kJ/kg      Equation (1) yields 2 1 1 2 1 u u Q 3662 kJ/kg m     (c) To find T 2 and P 2, we must first locate state (2) on the P(v, t) surface (or tables).

(b) To pinpoint state (2), we must determine two properties at the final state.

Adrian bejan advanced engineering thermodynamics pdf