Tsur 0K:=J.P. Holman: 3.09)

Use table 3-1 to determinethe shape factor for this problem.

D 2.4m:=

r 0.5m:=

S4π r⋅

1r

2 D⋅−

:= Tsphere 30K:=

S 7.014m= k 1.7Wm K⋅

:=

Use eq. 3-23 to calculate the heat loss.

q k S⋅ Tsphere Tsur−( )⋅:=

q 357.703W=

Temp

487.5

362.5

437.5

412.5

K=Temp0 487.5K=Temp Find T1 T2, T3, T4,( ):=

T4T1 TB+ TA+ T2+

4=

T3TC T1+ T2+ TD+

4=

T2T3 T4+ TA+ TD+

4=

T1TC TB+ T4+ T3+

4=

Given

T4 T3:=T3 T2:=T2 T1:=T1TA TB+ TC+ TD+

4:=

Set a guess value for all temperatures

TD 500K:=

So, Fourier´s law becommes asystem of linear equations.

Ti j,

Ti 1+ j, Ti j 1+,+ Ti 1− j,+ Ti j 1−,+

4=

and henceTA 100K:= TC 400K:=∆x ∆y=Here,

dTdx

dTdy

+ 0=

1

2 3

3

Generally, 2-D with no internalheat gereration and steady state:

TB 700K:=

J.P. Holman: 3.45)

τ0_cylinder 4.344 103× s=

τ0_surfρ cp⋅ V⋅

hsurf Asurf⋅:= τ0_surf 2.78 103× s=

Now, according to the handbook, page 11, and the textbook, section 4-5.

T T00−

Ti T00−e

τ−

τ0_cylinder e

τ−

τ0_surf⋅= which can be simplified toT T00−

Ti T00−e

τ−

τ0_tot=

where τ0_tot1

τ0_cylinder

1τ0_surf

+

1−:= τ0_tot 1.695 103× s=

which also could be expressed as

τ0_totρ cp⋅ V⋅

Σ h A⋅( )= τ0_tot

ρ cp⋅ V⋅

hl Al⋅ hsurf Asurf⋅+:= τ0_tot 1.695 103× s=

So, solving for the time, for which T Tend:=

τ τ0_tot− lnT T00−

Ti T00−

⋅:= Thus, τ 687.313 s=

τ 11.455min=

dl 0.1liter:=Exam problem P.36) Asurf 0.006m2:= hsurf 50W

m2 K⋅:=

Ti 65K:=

Tend 50K:=

T00 20K:=

V 2dl:= hl 8W

m2 K⋅:=

mass 0.2kg:=

Al 0.024m2:=ρmassV

:= ρ 1000kg

m3=

cp 4170J

kg K⋅:=

In the text, it is assumed that inner thermal resistance is neglected!

=> Lumped Capacity!!!!, Eq. 4-5.

T T00−

Ti T00−e

hA⋅

ρ c⋅ V⋅

− τ⋅=

Since we have heat transfer at several interfaces, the superposition principal is to be used, see page 11 in Handbook by Granryd.

τ0_cylinderρ cp⋅ V⋅

hl Al⋅:=

α 1.381 10 7−×m2

s=

In fig 2.88 c, the mean temperature of a infinite cylinder is already calculated.

ϑmϑ0

T T00−

T0 T00−=

T T00−

T0 T00−0.6= Figure gives Fo 1 10 1−⋅:=

Thus the time for this is

τFo r2⋅

α:= τ 764.871 s= τ 12.748min=

b)If consideration is taken to the edges, what is the temperature?

See the edges as the special case of plates Bihout

L2

⋅

kcoke:= Bi 9.914=

Foα τ⋅

L2

2:= Fo 0.032=

Figure 2.78-c in handbook.

ϑmϑ0

plates0.88= Hence, the average temperature in the can is

ϑmϑ0

can0.6 0.88⋅= Tcan T00 T0 T00−( ) 0.6⋅ 0.88⋅+:=

Tcan 10.56 °C=

°C K:=Exam problem P.37)dCoke 65mm:=

T0 20°C:=r

dCoke2

:=T00 0°C:=

L 115mm:=hout 100

W

m2 K⋅:=

Heat transfer by conduction inside the can!

kcoke 0.58Wm K⋅

:= cp 4200J

kg K⋅:= ρ 1000

kg

m3:=

a) Calculate how long tim it takes to cool the Coke to T 12°C:= if the can is considered to be long.

b) Calculate the temperature the coke will have at the calculated instance if the edges are considered.

a)

Bihout r⋅

kcoke:= Bi 5.603=

Foα τ⋅

r2:=

τ where αkcokeρ cp⋅

:=

Rex 3.392 104×= ν 2.211 10 5−×m2

s=

The boundary layer becomes turbulent at the critical Reynoolds number Recr 5 105⋅:=

Rex Recr> 0= => No, the boundary layer is still laminar.

So, eq. 5-21a is used: δx

5.0

Rex=

and δ5.0 xδ⋅

Rex:= δ 0.679mm=

How does the entire layer look like? xcrRecr ν⋅

uoo:= xcr 0.369m=

δ x( )5 x⋅

uoo xδ⋅

ν

x xcr<if

x 0.381uoo xδ⋅

ν

1−

5

⋅ 10.256uoo xδ⋅

ν

1−

⋅−

⋅

otherwise

:=

0 0.1 0.2 0.3 0.4 0.50

0.01

0.02

0.03

δ x( )

x

°C K:=bar 105Pa:=J.P. Holman, 5.14)Too 90°C:=

p 1bar:= uoo 30ms

:=

xδ 2.5cm:= δ

Check if the boundary layer is laminar or turbulent at the location.

ν a b T⋅+=Rex

uoo xδ⋅

ν:= where

1

1

350

400

1− 20.76

25.90

⋅1

363.15

⋅ 22.112=90 273.15+ 363.15=

ν1

1

350

400

1− 20.76

25.90

⋅ 10 6−⋅1

363.15

⋅m2

s⋅≡

(Table A5)

cp 2.006 103×J

kg K⋅=

ν1

1

40

60

1− 0.00024

0.839 10 4−⋅

⋅

1

50

⋅m2

s⋅:= ν 1.619 10 4−×

m2

s=

k1

1

40

60

1− 0.144

0.140

⋅1

50

⋅Wm K⋅

⋅:= k 0.142Wm K⋅

=

µ ν ρ⋅:= µ 0.141Pa s⋅=

Prµ cp⋅

k:= Pr 1.99 103×=

αk

ρ cp⋅:= α 8.138 10 8−×

m2

s=

Recalculate the properties if necessary when the outlet temperature is known!

Reu di⋅

ν:= Re 23.155= LAMINAR FLOW

Constant wall temperature, use eq. 6-9.

Nu 3.660.0668

diL

⋅ Re⋅ Pr⋅

1 0.04diL

Re⋅ Pr⋅

2

3

⋅+

+:= Nu 9.162=

°C K:=J.P. Holman, 6.09)Engine Oil as Fluid

di 1.25cm:= Tin 38°C:= u 30cms

:=

L 3m:= Twall 65°C:=

Estimate the total heat transfer and the outgoing oil temperature!

Assume the average bul temperature of the oil for properties! TbulkTin Twall+

2:= Tbulk 51.5 °C=

Set Tbulk 50°C:= for simplicity!

Table A-4, page 657:

ρ1

1

40

60

1− 876.05

864.04

⋅1

50

⋅kg

m3⋅:= ρ 870.045

kg

m3=

cp1

1

40

60

1− 1964

2047

⋅1

50

⋅J

kg K⋅⋅:=

ϑmTwall Tin−( ) Twall Tout−( )+

2=

∆T Tout Tin−=

A π di⋅ L⋅:= A 0.118m2=

Qm uπ di

2⋅

4⋅ ρ⋅:= Qm 0.032

kgs

=

So, inserted in energy balance

h π⋅ di⋅ L⋅Twall Tin−( ) Twall Tout−( )+

2⋅ u

π di2⋅

4⋅ ρ⋅ cp⋅ Tout Tin−( )⋅=

Solving for

Tout4 h⋅ L⋅ Twall⋅ 2 h⋅ L⋅ Tin⋅− u di⋅ ρ⋅ cp⋅ Tin⋅+( )

2 h⋅ L⋅ u di⋅ ρ⋅ cp⋅+( ):= Tout 44.014K=

and q Qm cp⋅ Tout Tin−( )⋅:= q 386.302W=

Controll the assumed Bulk temperature TbulkTin Tout+

2:= Tbulk 41.007 °C=

Redo all the calculations with new thermal properties of the OIL !!!!!!!!!!

a simpler eq. is 6-10

νw1

1

60

80

1− 0.839 10 4−⋅

0.375 10 4−⋅

⋅

1

65

⋅m2

s⋅:= νw 7.23 10 5−×

m2

s=

ρw1

1

60

80

1− 864.04

852.02

⋅1

65

⋅kg

m3⋅:= ρw 861.035

kg

m3=

µw νw ρw⋅:= µw 0.062Pa s⋅=Re Pr⋅diL

⋅ 191.998= > 10, and can be used

Nu 1.86 Re Pr⋅( )

1

3⋅diL

1

3

⋅µ

µw

0.14⋅:=

Nu 12.03=

So, the heat transfer coefficient is hNu k⋅di

:= h 136.665W

m2 K⋅=

Energy balance over the tube

q h A⋅ ϑm⋅= Qm cp⋅ ∆T⋅=

where

Reu DH⋅

ν:= Re 2.87 105×= Highly turbulent !

Dittus Boelter, eq. 6-4

Nu 0.023 Re0.8⋅ Pr0.4⋅:= Nu 465.508=

hNu k⋅DH

:= h 20.358W

m2 K⋅=

Use Petukhov equation, eq. 6-7 & 6-8. n 0:= for gases! µw µ:=

f 1.82 log Re( )⋅ 1.64−( ) 2−:= f 0.015=

Nu

f8

Re⋅ Pr⋅

1.07 12.7f8

⋅ Pr

2

3 1−

⋅+

µµw

n⋅:= Nu 385.037=

hNu k⋅DH

:= h 16.839W

m2 K⋅=

∆p fLDH

⋅ ρ⋅u2

2⋅:=

∆pL

0.802Pam

=

Figure 6-4 gives f 0.0145:=

∆p fLDH

⋅ ρ⋅u2

2⋅:=

∆pL

0.8Pam

=

J.P. Holman, 6.21)Air duct H 45cm:= u 7.5

ms

:= T 300K:=

B 90cm:= p 1atm:= L 1m:=

A H B⋅:= A 0.405m2= P 2 H B+( ):= P 2.7m=

DH4 A⋅P

:= DH 0.6m=

Table A-5

ρ 1.1774kg

m3:= cp 1005.7

Jkg K⋅

:= µ 1.8462 10 5−⋅ Pa s⋅:= νµρ

:= ν 1.568 10 5−×m2

s=

k 0.02624Wm K⋅

:= αk

ρ cp⋅:= α 2.216 10 5−×

m2

s= Pr

µ cp⋅

k:= Pr 0.708=

q h A⋅ T00 Tice−( )⋅= => q'' h T00 Tice−( )⋅=

Thus, τε

h T00 Tice−( )⋅=

b)a) ha 8W

m2 K⋅:= L 5m:=

hb hb_conv 1 0.5+( )⋅= uoo 5ms

:=

τaε

ha T00 Tice−( )⋅:= ReL

uoo L⋅

νair:= ReL 1.88 106×=

τb 66.268min=

τb 3.976 103× s=τbε

hb T00 Tice−( )⋅:=

hb 18.957W

m2 K⋅=hb hb_conv 1 0.5+( )⋅:=

hb_conv 12.638W

m2 K⋅=hb_conv

NuL kair⋅

L:=

NuL 2.622 103×=NuL 0.036 ReL0.8⋅ 835−

Pr

1

3⋅:=F.S.

τa 2.617hr=

NuL 2.683 103×=NuL Pr

1

3 0.037 ReL0.8⋅ 871−

⋅:=τa 157.031min=

τa 9.422 103× s=Use eq. 5-85

The flow is first laminar andat the end of the plate it becomes turbulent!

°C K:=Exam problem P.34)Given

rice 335000Jkg

:= Tice 0°C:=

T00 4°C:=ρice 900kg

m3:=

Formelsamling:To melt ice ∆x 1mm:= it takes

E M rice⋅= EA

ρice A⋅ ∆x⋅ rice⋅

A= ρice ∆x⋅ rice⋅=

ρair 1.293kg

m3:=

Heat transfer to the ice

Pr 0.718=Prµair cp_air⋅

kair:=

τεq''

=ε q'' τ⋅==>E0

tτq

⌠⌡

d= q τ⋅=kair 0.0241

Wm K⋅

:=cp_air 1006J

kg K⋅:=

The time required is given byµair 1.72 10 5−× Pa s⋅=µair νair ρair⋅:=

ε 3.015 105×J

m2=ε ρice ∆x⋅ rice⋅:=ε

EA

=

νair 13.3 10 6−⋅m2

s:=

Prµ cp⋅

k:= ν

µρ

:= α 5.279 10 5−×m2

s=

Pr 0.682= ν 3.599 10 5−×m2

s= β

1Tf

:= β 2.059 10 3−×1K

=

Grg β⋅ Ts Tair−( )⋅ H3⋅

ν2

:= Gr 5.848 109×=

Gr Pr⋅ 3.986 109×= Tab (7-1) => C 0.10:= n13

:=

Equation 7-25, ==> Nu C Gr Pr⋅( )n⋅:= Nu 158.558=

hNu k⋅H

:= h 6.252W

m2 K⋅=

q h H⋅ L⋅ Ts Tair−( )⋅:= q 2.344 103× W=

°C K:=J.P. Holman, 7.09) L 1m:=

Ts 400°C 273.15°C+:=

Tair 25°C 273.15°C+:=

TfTs Tair+

2:= H 1m:=

Tf 485.65K=

Tf 273.15°C− 212.5K=

Table A-5:

ρ1

1

450

500

1− .7833

0.7048

⋅

1

TfK

⋅kg

m3⋅:= ρ 0.727kgm-3=

cp1

1

450

500

1− 1020.7

1029.5

⋅

1

TfK

⋅J

kg K⋅⋅:= cp 1.027 103×

Jkg K⋅

=

µ1

1

450

500

1− 2.484

2.671

⋅ 10 5−⋅

1

TfK

⋅ Pa⋅ s⋅:= µ 2.617 10 5−× Pa s⋅=

k1

1

450

500

1− 0.03707

0.04038

⋅

1

TfK

⋅Wm K⋅

⋅:=k 0.039

Wm K⋅

=

αk

ρ cp⋅:=

β1Tf

:= β 3.304 10 3−×1K

= Grg β⋅ Ts Tair−( )⋅ H3⋅

ν2

:= Gr 2.239 1012×=

Gr Pr⋅ 1.583 1012×= Tab (7-1) => C 0.10:= n13

:=

Equation 7-25, ==> Nu C Gr Pr⋅( )n⋅:= Nu 1.165 103×=

hNu k⋅H

:= h 4.402W

m2 K⋅=

q h 2⋅ H⋅ L⋅ Ts Tair−( )⋅:= q 4.086 103× W=

OR Use eq. 7-29

RaL Gr Pr⋅:= Nu 0.8250.387 RaL

1

6⋅

10.492Pr

9

16+

8

27

+

2

:= Nu 1.284 103×= hNu k⋅H

:= h 4.851W

m2 K⋅=

q h 2⋅ H⋅ L⋅ Ts Tair−( )⋅:= q 4.502 103× W=

°C K:=J.P. Holman, 7.17) L 1.3m:=

Ts 55°C 273.15°C+:=

Tair 4°C 273.15°C+:=

TfTs Tair+

2:= Tf 302.65K=

Table A-5:

ρ1

1

300

350

1− 1.1774

0.9980

⋅

1

TfK

⋅kg

m3⋅:= H 7m:=

cp1

1

300

350

1− 1005.7

1009.0

⋅

1

TfK

⋅J

kg K⋅⋅:=

µ1

1

300

350

1− 1.8462

2.075

⋅ 10 5−⋅

1

TfK

⋅ Pa⋅ s⋅:=

k1

1

300

350

1− 0.02624

0.03003

⋅

1

TfK

⋅Wm K⋅

⋅:=

αk

ρ cp⋅:= Pr

µ cp⋅

k:= ν

µρ

:= ρ 1.168kgm-3= cp 1.006 103×J

kg K⋅= µ 1.858 10 5−× Pa s⋅=

k 0.026Wm K⋅

= α 2.251 10 5−×m2

s= Pr 0.707= ν 1.591 10 5−×

m2

s=