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DRAFT ME 470 – Heat Transfer Exam 1 Formulae First Law of Thermodynamics ˙ E st = ˙ E in - ˙ E out + ˙ E g (1) Heat Diffusion Equation ρc ∂T ∂t = ∂x k ∂T ∂x + ∂y k ∂T ∂y + ∂z k ∂T ∂z g ′′′ (2) Plane Wall without Thermal Energy Generation (0 x L) T (x)=(T s,2 - T s,1 ) x L + T s,1 and ˙ q ′′ x = k L (T s,1 - T s,2 ) (3) Plane Wall with Thermal Energy Generation (-L x L) T (x)= ˙ g ′′′ L 2 2k 1 - x 2 L 2 + T s and ˙ q ′′ x (x)=˙ g ′′′ x (4) Hollow Cylinder without Thermal Energy Generation (r 1 r r 2 ) T (r)=(T s,1 - T s,2 ) ln(r/r 2 ) ln(r 1 /r 2 ) + T s,2 and ˙ q ′′ r (r)= k(T s,1 - T s,2 ) r ln(r 2 /r 1 ) (5) Solid Cylinder with Thermal Energy Generation (0 r r o ) T (r)= ˙ g ′′′ r 2 o 4k 1 - r 2 r 2 o + T s and ˙ q ′′ r (r)= 1 2 ˙ g ′′′ r (6) Hollow Sphere without Thermal Energy Generation (r 1 r r 2 ) T (r)= T s,1 - (T s,1 - T s,2 ) 1 - (r 1 /r) 1 - (r 1 /r 2 ) and ˙ q ′′ r (r)= k(T s,1 - T s,2 ) r 2 [(1/r 1 ) - (1/r 2 )] (7) Steady-State Thermal Resistances R wall L kA R cyl ln(r 2 /r 1 ) 2πLk R sph r 2 - r 1 4πr 1 r 2 k (8) R conv 1 hA R rad 1 h rad A R fin θ b ˙ q fin (9) Fins ε fin ˙ q fin ˙ q no fin = ˙ q fin hA c,b θ b and η fin ˙ q fin ˙ q max = ˙ q fin hA s,fin θ b (10) Convection heat loss from fin tip θ θ b = cosh m(L - x)+( h/mk) sinh m(L - x) cosh mL +( h/mk) sinh mL and ˙ q fin = hPkA c θ b sinh mL +( h/mk) cosh mL cosh mL +( h/mk) sinh mL (11) 5

Heat Transfer equations

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Heat Transfer equations

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Page 1: Heat Transfer equations

DR

AFT

ME 470 – Heat TransferExam 1 Formulae

First Law of Thermodynamics

Est = Ein − Eout + Eg (1)

Heat Diffusion Equation

ρc∂T

∂t=

∂x

[

k∂T

∂x

]

+∂

∂y

[

k∂T

∂y

]

+∂

∂z

[

k∂T

∂z

]

+ g′′′ (2)

Plane Wall without Thermal Energy Generation (0 ≤ x ≤ L)

T (x) = (Ts,2 − Ts,1)x

L+ Ts,1 and q′′x =

k

L(Ts,1 − Ts,2) (3)

Plane Wall with Thermal Energy Generation (−L ≤ x ≤ L)

T (x) =g′′′L2

2k

(

1−x2

L2

)

+ Ts and q′′x(x) = g′′′x (4)

Hollow Cylinder without Thermal Energy Generation (r1 ≤ r ≤ r2)

T (r) = (Ts,1 − Ts,2)ln(r/r2)

ln(r1/r2)+ Ts,2 and q′′r (r) =

k(Ts,1 − Ts,2)

r ln(r2/r1)(5)

Solid Cylinder with Thermal Energy Generation (0 ≤ r ≤ ro)

T (r) =g′′′r2o4k

(

1−r2

r2o

)

+ Ts and q′′r (r) =1

2g′′′r (6)

Hollow Sphere without Thermal Energy Generation (r1 ≤ r ≤ r2)

T (r) = Ts,1 − (Ts,1 − Ts,2)

[

1− (r1/r)

1− (r1/r2)

]

and q′′r (r) =k(Ts,1 − Ts,2)

r2[(1/r1)− (1/r2)](7)

Steady-State Thermal Resistances

Rwall ≡L

kARcyl ≡

ln(r2/r1)

2πLkRsph ≡

r2 − r14πr1r2k

(8)

Rconv ≡1

hARrad ≡

1

hradARfin ≡

θbqfin

(9)

Fins

εfin ≡qfin

qno fin

=qfin

hAc,bθband ηfin ≡

qfin

qmax

=qfin

hAs,finθb(10)

Convection heat loss from fin tip

θ

θb=

coshm(L− x) + (h/mk) sinhm(L− x)

coshmL+ (h/mk) sinhmLand qfin =

hPkAc θbsinhmL+ (h/mk) coshmL

coshmL+ (h/mk) sinhmL(11)

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Page 2: Heat Transfer equations

DR

AFT

Negligible heat loss from fin tip

θ

θb=

coshm(L− x)

coshmLand qfin =

hPkAc θb tanhmL (12)

Lc ≡ L+Ac

PLc,rectangular fin ≈ L+

t

2Lc,cylindrical fin = L+

D

4(13)

Prescribed temperature at the fin tip

θ

θb=

(θL/θb) sinhmx+ sinhm(L− x)

sinhmLand qfin =

hPkAc θbcoshmL− θL/θb

sinhmL(14)

Infinitely long finθ

θb= e−mx and qfin =

hPkAc θb (15)

where

θ ≡ T − T∞ and m ≡

hP

kAc

(16)

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