Hw 8 solutions

Hw 8 solutions - Solution of Homework 8 7-113 The work...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Solution of Homework 8 7-113 The work produced for the process 1-3 shown in the figure is to be determined. Assumptions Kinetic and potential energy changes are negligible. Analysis The work integral represents the area to the left of the reversible process line. Then, kJ/kg 0 = + + = + + = + = 100)kPa - /kg)(400 m (1.0 kPa ) 500 100 ( 2 /kg m ) 0 . 1 5 . 0 ( ) ( ) ( 2 3 3 2 3 2 1 2 2 1 3 2 2 1 3 - in,1 P P P P dP dP w v v v v v 7-127E R-134a is expanded in an adiabatic process with an isentropic efficiency of 0.95. The final volume is to be determined. Assumptions 1 Kinetic and potential energy changes are negligible. 2 The device is adiabatic and thus heat transfer is negligible. Analysis We take the R-134a as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as ) ( 1 2 out energies etc. potential, kinetic, internal, in Change system mass and work, heat, by nsfer energy tra Net out in u u m U W E E E = = = ±² ±³ ´ ´ From the R-134a tables (Tables A-11E through A-13E), Btu/lbm 445 . 93 ) 898 . 82 )( 9897 . 0 ( 401 . 11 9897 . 0 19962 . 0 02605 . 0 22362 . 0 psia 20 R Btu/lbm 22362 . 0 Btu/lbm 48 . 108 F 100 psia 120 2 2 2 2 1 2 2 1 1 1 1 = + = + = = = = = = = = ° = = fg s f s fg f s s s u x u u s s s x s s P s u T P The actual work input is Btu/lbm 14.28 Btu/lbm ) 448 . 93 48 . 108 )( 95 . 0 ( ) ( 2 1 out , out , = = = = s T s T a u u w w η The actual internal energy at the end of the expansion process is P (kPa) v (m 3 /kg) 2 1 500 400 100 0.5 3 T s 1 2 s 120 psia 20 psia 2a
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Btu/lbm 20 . 94 28 . 14 48 . 108 ) ( out , 1 2 2 1 out , = = = ⎯→ = a a a a w u u u u w The specific volume at the final state is (Table A-12E) /lbm ft 2745 . 2 ) 01182 . 0 2772 . 2 )( 9988 . 0 ( 01182 . 0 9988 . 0 898 . 82 401 . 11 20 . 94 Btu/lbm 20 . 94 psia 20 3 2 2 2 2 2 2 = + = + = = = = = = fg f fg f a a x u u u x u P v v v The final volume is then 3 ft 22.75 = = = /lbm) ft 5 lbm)(2.274 10 ( 3 2 2 v V m 7-137 Air enters an adiabatic compressor with an isentropic efficiency of 84% at a specified state, and leaves at a specified temperature. The exit pressure of air and the power input to the compressor are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Air is an ideal gas with variable specific heats.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This homework help was uploaded on 04/16/2008 for the course ME 2322 taught by Professor Oler during the Spring '06 term at Texas Tech.

Page1 / 6

Hw 8 solutions - Solution of Homework 8 7-113 The work...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online