One-Dimensional, Steady-State. State Conduction without Thermal Energy Generation

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1 One-Dimensional, Steady-State State Conduction without Thermal Energy Generation

2 Methodology of a Conduction Analysis Specify appropriate form of the heat equation. Solve for the temperature distribution. Apply Fourier s Law to determine the heat flux. Simplest Case: One-Dimensional, Steady-State Conduction with No Thermal Energy Generation Alternative conduction analysis Common Geometries: The Plane Wall: Described in rectangular (x) coordinate. Area perpendicular to direction of heat transfer is constant (independent of x). The Tube Wall: Radial conduction through tube wall. The Spherical Shell: Radial conduction through shell wall.

The Plane Wall Consider a plane wall between two fluids of different temperature: Heat Equation: d dt k = 0 dx dx (3.) Implications: Heat flux q is independent of x.

3 The Plane Wall Consider a plane wall between two fluids of different temperature: Heat Equation: d dt k = 0 dx dx (3.) Implications: Heat flux q is independent of x. ( x ) ( q ) Heat rate is independent of x. Boundary Conditions: ( 0 ), ( ) x T = T T L = T s, s,2 Temperature Distribution for Constant k : x T( x) = Ts, + ( Ts,2 Ts,) (3.3) L

4 Heat Flux and Heat Rate: dt k q x = k = ( Ts, Ts,2 ) (3.5) dx L dt ka qx = ka = ( Ts, Ts,2 ) (3.4) dx L ΔT Thermal Resistances Rt = and Thermal Circuits: q L Conduction in a plane wall: Rtcond, ka Convection:, R t conv ha = (3.6) = (3.9) Thermal circuit for plane wall with adjoining fluids: R tot L = + + ha ka h A q 2 T T (3.2),,2 x = (3.) Rtot

Thermal Resistance for Unit Surface Area: L Rtcond, = R tconv, = k h 2 Units: Rt W/K R t m K/W Radiation Resistance: Rtrad, ha Contact Resistance: = = Rtrad, r hr 2 2 r εσ s sur s sur

5 Thermal Resistance for Unit Surface Area: L Rtcond, = R tconv, = k h 2 Units: Rt W/K R t m K/W Radiation Resistance: Rtrad, ha Contact Resistance: = = Rtrad, r hr 2 2 r εσ s sur s sur ( )( ) h = T + T T + T (.9) R = tc, T A T q x B R R = tc, tc, Ac Values depend on: Materials A and B, surface finishes, interstitial conditions, and contact pressure (Tables 3. and 3.2)

6 Composite Wall with Negligible Contact Resistance: q x T T,,4 = (3.4) Rtot R tot LA LB LC R = = A h ka kb kc h4 A tot Overall Heat Transfer Coefficient (U) : A modified form of Newton s Law of Cooling to encompass multiple resistances to heat transfer. q = UAΔ T (3.7) x R tot = UA overall (3.9)

7 Series Parallel Composite Wall: Note departure from one-dimensional conditions for k k. F G Circuits based on assumption of isothermal surfaces normal to x direction or adiabatic surfaces parallel to x direction provide approximations for. q x

8 ALTERNATIVE CONDUCTION ANALYSIS: STEADY STATE NO HEAT GENERATION NO HEAT LOSS FROM THE SIDES A(x) and k(t) IS TEMPERATURE DISTRIBUTION ONE-DIMENSIONAL? q x = q x + dx IS IT REASONABLE TO ASSUME ONE-DIMENSIONAL TEMPERATURE DISTRIBUTION IN x? FROM THE FOURIER S LAW: q x x dx A( x) T = 0 T 0 k( T ) dt q x = A( x) k( T ) dt dx

9 Tube Wall The Tube Wall Heat Equation: d dt kr = 0 (3.23) rdr dr What does the form of the heat equation tell us about the variation of q r with r in the wall? Is the foregoing conclusion consistent with the energy conservation requirement? How does q vary with r? r Temperature Distribution for Constant k: Ts, T s,2 r T( r) = ln + Ts ln ( r/ r2) r 2,2 (3.26)

10 Heat Flux and Heat Rate: dt k q = k = T T ( ) ( ),,2 r s s dr r ln r2/ r 2π k q = 2π rq = T T ( ) ( ),,2 r r r r s s ln 2/ 2π Lk q 2π rlq = T T (3.27) ( ) ( ),,2 r r r= r s s ln 2/ Conduction Resistance: ln ( r2/ r) Rtcond, = Units K/W 2π Lk ln ( r2/ r) Rtcond, = Units m K/W 2π k (3.28) Why is it inappropriate to base the thermal resistance on a unit surface area?

11 Composite Wall with Negligible Contact Resistance T, T,4 qr = = UA T T R tot (,,4 ) (3.30) Note that UA = R tot is a constant independent of radius. But, U itself is tied to specification of an interface. U = ( AR ) i i tot (3.32)

12 Spherical Shell Heat Equation d 2 r dr dt 2 r = dr 0 What does the form of the heat equation tell us about the variation of q with r? Is this result consistent with conservation of energy? r How does q vary with r? r Temperature Distribution for Constant k : ( ) = s, ( s, s,2 ) T r T T T ( r/ r) ( r r 2) /

13 Heat flux, Heat Rate and Thermal Resistance: dt k q = k = T T r 2 s s dr r r r2 ( ) ( ) ( ),,2 / / 4π k q r q T T R 2 r = 4π r = s s 2 tcond, = ( ) ( ) ( ),,2 / r / r ( / r ) ( / r ) 2 4π k (3.35) (3.36) Composite Shell: ΔToverall qr = = UAΔT R tot UA = R Constant tot overall ( ) Depends on U = AR A i i tot i

14 Problem 3.23: Assessment of thermal barrier coating (TBC) for protection of turbine blades. Determine maximum blade temperature with and without TBC. Schematic: ASSUMPTIONS: () One-dimensional, steady-state conduction in a composite plane wall, (2) Constant properties, (3) Negligible radiation

15 ANALYSIS: For a unit area, the total thermal resistance with the TBC is R tot,w h o ( L k ) Zr R t,c ( L k ) In h = i ( ) R tot,w = m K W = m K W With a heat flux of T,o T,i 300 K 5 2 q w = R = = W m 3 2 tot,w m K W the inner and outer surface temperatures of the Inconel are = + ( ) ( ) Ts,i(w) T,i qw hi = 400 K W m 500 W m K = 04 K ( ) ( ) Ts,o(w) = T,i + hi + L k q In w ( ) ( ) = 400 K m K W W m = 74 K

16 Without the TBC, ( ) R 3 2 tot, wo h o L k In h i = + + = m K W ( ) q wo T,o T,i R tot, wo = = W/m 2 The inner and outer surface temperatures of the Inconel are then ( ) Ts,i(wo) = T,i + q wo hi = 22 K [( ) ( ) ] Ts,o(wo) = T,i + hi + L k q In wo = 293 K Use of the TBC facilitates operation of the Inconel below T max = 250 K. COMMENTS: Since the durability of the TBC decreases with increasing temperature, which increases with increasing thickness, limits to its thickness are associated with reliability considerations.

17 Problem 3.62: Suitability of a composite spherical shell for storing radioactive wastes in oceanic waters. SCHEMATIC: ASSUMPTIONS: () One-dimensional conduction, (2) Steady-state conditions, (3) Constant properties at 300K, (4) Negligible contact resistance. PROPERTIES: Table A-, Lead: k = 35.3 W/m K, MP = 60K; St.St.: 5. W/m K. ANALYSIS: From the thermal circuit, it follows that T T 4 q= q r 3 = & R π tot 3

18 The thermal resistances are: RPb = / ( 4π 35.3 W/m K ) = K/W 0.25m 0.30m RSt.St. = / ( 4π 5. W/m K ) = K/W 0.30m 0.3m ( π ) Rconv = / m 500 W/m K = K/W Rtot = K/W. The heat rate is then 5 3 ( )( ) 3 q=5 0 W/m 4 π / m = 32,725 W and the inner surface temperature is ( ) T = T + R tot q=283k k/w 32,725 W = 405 K < MP = 60K. Hence, from the thermal standpoint, the proposal is adequate. COMMENTS: In fabrication, attention should be given to maintaining a good thermal contact. A protective outer coating should be applied to prevent long term corrosion of the stainless steel.

SOFTbank E-Book Center Tehran, Phone: , For Educational Use. PROBLEM 1.45

SOFTbank E-Book Center Tehran, Phone: , For Educational Use. PROBLEM 1.45 PROBLEM 1.45 KNOWN: Rod of prescribed diameter experiencing electrical dissipation from passage of electrical current and convection under different air velocity conditions. See Example 1.3. FIND: Rod