Heat_Transfer__Convection examples tmb rev

Heat_Transfer__Convection examples tmb rev - Heat Transfer...

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

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

Unformatted text preview: Heat Transfer Convection ENES 101 / 101H / 101Y Introductory Engineering Science Modes of Heat Transfer Heat transfer due to convection involves the energy exchange between a surface and an adjacent fluid (gas or liquid). If a solid body is exposed to a moving fluid having a different temperature, energy is carried or convected away from the fluid. Forced convection when a fluid is made to flow past the solid surface by an external agent (pump/fan) Free or natural convection where a warmer or cooler fluid next to the solid boundary causes circulation because of the density difference resulting from the temperature variation throughout a region of the fluid. Newton's Law of Cooling q = h A T = h A(Ts - T ) q rate of heat transfer (W or Btu/hr) A surface area (m2 or ft2) T temperature difference between T surface Ts and the fluid (K or oF) W m 2K h convective heat transfer coefficient Btu hr ft o F Example #3 A tent has a surface area of 30m . Suppose the tent contains a heater that generates 1800 Watts. a) If the outside temperature is 0 C and there is no wind, what will be the interior temperature? We are convecting 1800 W of energy from the air inside the tent to the tent walls, TWi = TWo = TS Then from the tent walls to the outside air. Assume tent wall is thin and it will not TWi=Two=TS Ti provide significant insulation To To o 2 Ti Example #3 (cont.) From Table 6.4 h is between 5 30 W/m K So assume that h = 7 W/m K Q = ho A(TS - To ) 1800W = 10 o TS =6 C o 2 inside air to inside tent surface h = 10 W/m K i 2 W 30m 2 (TS - To ) 2 m 2K ( outside tent surface to outside air on a calm night Q = hi A(Ti - TS ) W 1800W = 7 2 ( 30m 2 )(Ti - TS ) mK Ti = 14.57 o C ) Example #3b a) What is the heat output (W) required in order to maintain an interior temperature of 68 F? How many Btu/hr is this? o Convert 68oF to oC (6832)5/9 = 20 oC Q = hi A(Ti - TS ) W Q = 7 2 30m 2 ( 20 - 6 ) oC mK Btu / hr Btu Q = 2940W 3.412 = 10,031 1 W hr ( ) Example #3c a) Suppose the outside temperature is 0 C and the heat output is 2940W, but the wind velocity is 40 mph. What will be the temperature inside the tent in F? mi 1hr Given h = 5 + 10 v.8 1m -4 40 o o m = 17.9 hr 3600s 6.21x10 mi s W .8 .8 h = 5 + 10v = 5 + 10(17.9 ) = 105.44 2 mK Example #3c (cont) Tent surface temperature: Q = ho A(TS - To ) 2940W = 105.44 TS = 0.93 o C W 30m 2 TS - 0o C m 2K ( )( ) Inside temperature: Q = hi A(Ti - TS ) W 2940W = 7 2 30m 2 Ti - 0.93 o C mK Ti = 14.93 o C ( )( ) Combined Conduction/Convection Heat Transfer Most Heat Transfer problems involve more than one mode of heat transfer. Many involve Convection from fluid to a solid surface Conduction through homogeneous material Convection from solid to fluid Example #4 Wooden cabin with 2cm thick walls kwood = 0.1 W/mK A = 30 m2 Q = 1800 W hi = 5W/m2K ho = 20 W/m2K Example #4a a) Sketch temperature profile A=30m2 T i T si T so T =0C o o Wood Wall x = 2cm k = 0.1W/mK Example #4b a) T = 0 C, what is T ? convection (outside air to outside wood wall) o o i Q = ho A(TSo - To ) TSo = Q 1800W + To = + 0 =3o C W ho A 2 20 2 30m m K ( ) conduction (outside wood wall to inside wood wall) dT kA (TSi - TSo ) = dx x Qx 1800W ( 0.02m ) TSi = + TSo = +3 o C W kA 2 0 .1 30m mK TSi = 15 o C Q = -kA ( ) Example #4b (cont) convection (inside wood wall to inside Q = hi A(Ti - TSi ) Q air) Ti = + TSi = hi A Ti = 27 o C 1800W + 15 o C W 2 5 2 30m m K ( ) T =27 C i o T =15 C si o T =3 C so o T =0C o o 2 cm Example #4c a) What heat output (W) is required to maintain inside temp of ~68 F (20 C)? o kAo (TSi - TSo ) = hi A(Ti - TSi ) - To ) = x Q = ho A(TSo Ti - simplify:To Q= 1 x 1 + + ho A kA hi A 1 20W / m 2K 30m 2 Q = 1333.3W Q= ( )( ) 20 o C 0.02m 1 + + 2 ( 0.1W / mK ) 30m 5W / m 2K 30m 2 ( ) ( )( ) Example #4d a) Suppose Q=1800W, Ti = 72oF(22.2oC), To = 12oF( 11.1oC). What is x? Ti -To T Q= = = 1800W 1 x 1 R + + ho A kA hi A solving for x=0.03m or 3cm Radiation The energy is transferred via electromagnetic wave propagation. No medium is required. Works most efficiently in a vacuum, but also occurs in a medium. StefanBoltzman Law Radiant heat transfer is defined via the StefanBoltzman Law q = A T4 q rate of radiant energy emission (W or Btu/hr) A area of emitting surface (m2 or ft2) T absolute temperature (K or oR) Stefan Boltzman constant W -8 5.676 x10 m 2K 4 Btu 0.1714 x10 -8 hr ft 2 o R 4 Heat Transfer Temperature profile Ti T1 T2 T3 T4 To For steady state heat transfer (1) Q T Ti - To = = 1 t t t 1 A R + + + + hi k 1 k 2 k 3 ho Modes of heat transfer between Ti and T1 T1 and T2 T2 and T3 T3 and T4 T4 and To (2) (3) convection conduction conduction conduction convection h convective heat transfer coeff. t material thickness k material thermal conductivity ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online