FTFS Chap20 P032

# FTFS Chap20 P032 - Chapter 20 Natural Convection 20-32 A...

This preview shows pages 1–2. Sign up to view the full content.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 20 Natural Convection 20-32 A fluid flows through a pipe in calm ambient air. The pipe is heated electrically. The thickness of the insulation needed to reduce the losses by 85% and the money saved during 10-h are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The local atmospheric pressure is 1 atm. Properties Insulation will drop the outer surface temperature to a value close to the ambient temperature, and possible below it because of the very low sky temperature for radiation heat loss. For convenience, we use the properties of air at 1 atm and 5 C (the anticipated film temperature) (Table A-22), 1- 2 5 K 003597 . K ) 273 5 ( 1 1 7350 . Pr /s m 10 382 . 1 C W/m. 02401 . = + = = = = =- f T k Analysis The rate of heat loss in the previous problem was obtained to be 29,094 W. Noting that insulation will cut down the heat losses by 85%, the rate of heat loss will be W 4364 W 094 , 29 15 . ) 85 . 1 ( insulation no = =- = Q Q The amount of energy and money insulation will save during a 10-h period is simply determined from kWh 3 . 247 h) kW)(10 094 . 29 85 . ( , = = = t Q Q saved total saved \$22.26 = = ) kWh / 09 . )(\$ kWh 3 . 247 ( = energy) of cost t saved)(Uni Energy ( saved Money The characteristic length in this case is the outer diameter of the insulated pipe, insul insul c t t D L 2 3 . 2 + = + = where t insul is the thickness of insulation in m. Then the problem can be formulated for T s and t insul as follows: ) 7350 . ( ) /s m 10 382 . 1 ( ) 2 3 . ( K] ) 273 )[( K 003597 . )( m/s 81 . 9 ( Pr ) ( 2 2 5 3-1 2 2 3- +- =- = insul s c s t T L T T g Ra ( 29 [ ] ( 29 [ ] 2 27 / 8 16 / 9 6 / 1 2 27 / 8 16 / 9 6 / 1 7350 . / 559 . 1 387 . 6 . Pr / 559 . 1 387 . 6 . + + = + + = Ra Ra Nu m) )(100 2 3 . ( C W/m. 02401 . insul s c c t L D A Nu L Nu L k h + = = = = The total rate of heat loss from the outer surface of the insulated pipe by convection and radiation becomes ] ) K 273 30 ( )[ .K W/m 10 67 . 5 ( ) 1 . ( + ) 273 ( 4364 ) ( ) ( 4 4 4 2 8 4 4 +-- - =- +- = + =- s s s s surr s s s s rad conv T A T hA T T A T T hA Q Q Q In steady operation, the heat lost by the side surfaces of the pipe must be equal to the heat lost from the exposed surface of the insulation by convection and radiation, which must be equal to the heat conducted through the insulation. Therefore, ] 3 . / ) 2 3 . ln[( K ) m)(298 C)(100 W/m. 035 . ( 2 W 4364 ) / ln( ) ( 2 tank insul s o s insulation t T D D T T kL Q Q +- = - = = The solution of all of the equations above simultaneously using an equation solver gives T s = 281.5 K = 8.5 C and t insul = 0.013 m = 1.3 cm ....
View Full Document

## This homework help was uploaded on 04/18/2008 for the course EML 3007 taught by Professor Chung during the Spring '08 term at University of Florida.

### Page1 / 23

FTFS Chap20 P032 - Chapter 20 Natural Convection 20-32 A...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online