Elementary Principles 256 - 7.24(a H E k E p = Q Ws E k E p...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
7- 11 7.24 (a) ΔΔ Δ ±± ± HE EQ W kp s ++= ; Δ ± E k , Δ ± E p , ± WH Q s =⇒ = 0 Δ ² H 400 3278 °= C, 1 atm kJ kg bg (Table B.7) ² H 100 2676 °⇒ = C, sat'd 1 atm kJ kg (Table B.5) 22 oo 100 kg H O(v)/s 100 kg H O(v)/s 100 C, saturated 400 C, 1 atm (kW) Q ± ± . . Q = 100 kg kJ 10 J s kg 1 kJ Js 3 3278 2676 0 602 10 7 (b) Δ UE W ; Δ E k , Δ E p , WU Q = = 0 Δ () 3 final m ˆˆ ˆ Table B.5 100 C, 1 atm 2507 , 1.673 400 C, UV V P ⇒° = ° = = ° Interpolate in Table B.7 to find P at which ˆ V =1.673 at 400 o C, and then interpolate again to find ˆ U at 400 o C and that pressure: 3 o final 3.11 1.673 1.673 m /g 1.0 4.0 3.3 bar , (400 C, 3.3 bar) = 2966 kJ/kg 3.11 0.617 VP U = + = ⎛⎞ ⎜⎟ ⎝⎠ ( ) [ ] ( ) 37 ˆ 2966 2507 kJ kg 10 J kJ 4.59 10 J QU m U ⇒= Δ=Δ= = × The difference is the net energy needed to move the fluid through the system (flow work). (The energy change associated with the pressure change in Part (b) is insignificant.) 7.25 ² ,. Hl H O C kJ kg 2 ch 20 839 (Table B.5) ² . H steam, sat'd 20 bars, 2797 2 = (Table B.6) o [kg H O(l)/h] [kg H O(v)/h] 20 C 20 bar (sat'd) =0.65(813 kW) 528 kW mm Q = ± (a)
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

Ask a homework question - tutors are online