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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=