Heat Loss From a Buried Oil Pipe

Heat loss from a buried oil line Conduction shape factor

[12], page 144, Eq. (3.32)

Example 3.3 [12]

Heat loss from a buried oil line P =

P: heat flow rate [W]

An oil pipeline whith an outside diameter "d1", is buried with its k: thermal conductivity of media [W/(m K)]

centerline "h" below ground level in damp soil. S: conduction shape factor [m]

The length of the line is "L" and the mass flow rate of the oil is "m". Dt: difference in surface temperatures [K]

Required are: Oil outlet temperature for an uninsulated pipe "tout_UP" Table 3.2 [12], pages 142 and 143 presents

shape factors for a series of situations.

For a buried cylinder, case 8 of Table 3.2

0.3 m

h = 1 m

L = 5000 m

m = 2.5 kg/s

120 °C

23 °C

s = 150 mm

0.03 W/(m K)

1.5 W(m K)

2000 J/(kg K)

For the Uninsulated Pipe case (UP) and for the Insulated Pipe (IP)

the outlet temperatures and heat losses are required

Assumptions:

Steady state

Deep ground temperature same as surface temperatureNegligible resistance of the pipe wall and for convection from de oilIsothermal surface exposed to oil

k * S * Dt

The oil inlet temperature is "tin" and the ground level soil is at "tGLS"

and for an insulated pipe "tout_IP" with an insulation layer with a

thickness "s" and a thermal conductivity "kins". The soil thermal

conductivity is "kSoil". The oil specific heat is "cp_oil"

d1 =

tin =

tGLS =

kins =

kSoil =

cp_oil =

h

t1

L

Ground level soil at t2 = tGLS

r1

S=2⋅π⋅L

cosh−1 (hr1)

S=2⋅π⋅L

Ln(2⋅hr1)

for h>3⋅r1

Microsoft Editor de ecuaciones 3.0

A.- Non insulated pipe

Outlet temperature

h = 1 m

0.15 m

k: thermal conductivity of media [W/(m K)] 6.7

S: conduction shape factor [m] h>3

Dt: difference in surface temperatures [K] 120 °C

S = 0.974

Table 3.2 [12], pages 142 and 143 presents L = 5,000 m 23 °C

shape factors for a series of situations. h = 1.0 m 25.6 °C

0.15 m

For a buried cylinder, case 8 of Table 3.2 S = 12,128 m Heat flow rate

Q =

m = 2.5 kg/s

2 kJ/(kg K)

120 °C

25.6 °C

Q = 472.2 kW

1.5 W(m K)

S = 12,128 m

m = 2.5 kg/s

2000 J/(kg K)

3.64

Effectivness

3.640.974

h / r1

r1 =

h / r1 =

h / r1 = tout = tin - e * (tin - tGLS)

tin =

2 * p * L / Ln(2*h/r1) e =

tGLS =

tout =

r1 =

m * cp_oil * (tin - tout)

cp_oil =

tin =

tout =

Ntu = kSoil *S / (m * cp_oil)

kSoil =

cp_oil =

Ntu =

e = 1 - Exp(-Ntu)

Ntu =e =

L

Ground level soil at t2 = tGLS N tu=k⋅Sm⋅c p

(Eq . e )

ε=1−e−N tu (1 .59 )

ε=t in−touttin−tGLS

(1 .57 )

Rev. cjc. 21.02.2014

From Annex A 1289 W/K

m = 2.5 kg/s

2000 J/(kg K)

0.258

Insulated radius Effectivness

0.15 m

s = 0.15 m

0.30 m

0.26

with 0.227

S =

L = 5,000 m Outlet temperature

h = 1.0 m

0.30 m

S = 16,560 m

0.30 m 120 °C

0.15 m 0.227

L = 5000 m 23 °C

0.03 W/(m K) 98.0 °C

1.5 W/(m K)S = 16,560 m Heat flow rate

0.00077571 Q =

1289 W/K m = 2.5 kg/s

2 kJ/(kg K)

Ntu 120 °C

98.0 °CQ = 110.2 kW

(Note)The uninsulated case cannot be solved exactly using the shape factor concept,since when the insulation is added to the pipe, the outer surface of the insulationwill not be isothermal. However, to get some idea of the effect of the insulation,an isothermal surface will be assumed. [12]

B.- Insulated pipe (note) Ntu = Pere*L*U / (m * cp )

Pere*L*U =

cp_oil =

Ntu =

r2 = r1 + s

r1 =

r2 = e = 1 - Exp(-Ntu)

Ntu =

e =

2 * p * L / Ln(2*h/r1)

r2 =

1/(Pere*L*U)=Ln(re/ri) /(2*p*L*kinsul)+1/(ksoil * S) tout = tin - e * (tin - tGLS)

re = tin =

ri = e =

tGLS =

kinsul = tout =

ksoill =

1/(Pere*L*U)= m * cp_oil * (tin - tout)

Pere*L*U =

cp_oil =

tin =

tout =

1Per e⋅L⋅U

=

Ln( reri )2⋅π⋅L⋅k insul

+ 1k soil⋅S

(g )

N tu=U⋅Per⋅Lm⋅c p

(Eq . e ' )

ε=1−e−N tu (1 .59 )


(1 .57 )

Este link muestra la parte del libro que contiene el ejercicio

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Heat loss from a buried oil line

An oil pipeline whith an outside diameter "d1", is buried with itscenterline "h" below ground level in damp soil.The length of the line is "L" and the mass flow rate of the oil is "m".

Required are: Oil outlet temperature for an uninsulated pipe "tout_UP"

h =L =m =

s =

h =

S =L =h =

S =

S =m =

Effectivness





d1 =

tin =

tGLS =

kins =

kSoil =

cp_oil =

h / r1

r1 =

h / r1 =

h / r1 =

r1 =

Ntu =

kSoil =

cp_oil =

Ntu =

N tu=k⋅Sm⋅c p

(Eq . e )

Heat loss from a buried pipe . [12 ] page 145

Effectiveness of the heat exchanger ε

ε=1−e−N tu (1 .59 )

ε=t in−toutt in−tGLS

(1 .57 )

Number of transfer units N rSub { size 8{ ital tu } }


(Eq . a)

Heat transfer Δ dot ital {Q}} for an element ¿of exchanger of length Δx andperimeter Per with a global heattransfer coefficient U and atemperature differential Δ ` ital tΔQ̇=U⋅Per⋅Δx⋅Δt (Eq . b ).Heat transfer rate for the pipelineΔQ̇=k⋅ΔS⋅Δt (Eq . c )

k: soil thermal conductivity [Wm⋅K ]ΔS: shape factor for given geometryΔt: temperature differential betweenfluid and soil

From equations (b ) and (c )ΔQ̇=U⋅Per⋅Δx⋅Δt (Eq . b )ΔQ̇=k⋅ΔS⋅Δt (Eq . c )U⋅Per⋅Δx=k⋅ΔSAnd for the pipeU⋅Per⋅L=k⋅S (Eq . d )Replacing (Eq . d ) in (Eq . d )


=k⋅Sm⋅c p

(Eq . e )


=k⋅Sm⋅c p


(Eq . e ' )

N tu=k⋅Sm⋅c p

(Eq . e )

.

Outlet temperature

e =

Ntu =

e =

tout =

tin =e =

tGLS =

tout =


N tu=k⋅Sm⋅c p

(Eq . e )

ε=1−e−N tu (1 .59 )


(1 .57 )

Heat loss from a buried pipe . [12 ] page 145

Effectiveness of the heat exchanger ε

ε=1−e−N tu (1 .59 )

ε=t in−toutt in−tGLS

(1 .57 )

Number of transfer units N rSub { size 8{ ital tu } }


(Eq . a)

Heat transfer Δ dot ital {Q}} for an element ¿of exchanger of length Δx andperimeter Per with a global heattransfer coefficient U and atemperature differential Δ ` ital tΔQ̇=U⋅Per⋅Δx⋅Δt (Eq . b ).Heat transfer rate for the pipelineΔQ̇=k⋅ΔS⋅Δt (Eq . c )

k: soil thermal conductivity [Wm⋅K ]ΔS: shape factor for given geometryΔt: temperature differential betweenfluid and soil

From equations (b ) and (c )ΔQ̇=U⋅Per⋅Δx⋅Δt (Eq . b )ΔQ̇=k⋅ΔS⋅Δt (Eq . c )U⋅Per⋅Δx=k⋅ΔSAnd for the pipeU⋅Per⋅L=k⋅S (Eq . d )Replacing (Eq . d ) in (Eq . d )


=k⋅Sm⋅c p

(Eq . e )


=k⋅Sm⋅c p


(Eq . e ' )

N tu=k⋅Sm⋅c p

(Eq . e )

.


(Eq . e ' )

Heat loss from a buried oil line

An oil pipeline whith an outside diameter "d1", is buried with itscenterline "h" below ground level in damp soil.The length of the line is "L" and the mass flow rate of the oil is "m".

Required are: Oil outlet temperature for an uninsulated pipe "tout_UP"

0.3 m1 m

5000 m2.5 kg/s

120 °C

23 °C150 mm

0.03 W/(m K)

1.5 W(m K)

2000 J/(kg K)

1 m

0.15 m

6.7

h>3

5,000 m1.0 m

0.15 m12,128 m

1.5 W(m K)12,128 m

2.5 kg/s

2000 J/(kg K)

3.64





2 * p * L / Ln(2*h/r1)

kSoil *S / (m * cp_oil)

N tu=k⋅Sm⋅c p

(Eq . e )


3.64

0.974

Outlet temperature

120 °C0.974

23 °C

25.6 °C

1 - Exp(-Ntu)

tin - e * (tin - tGLS)

ε=1−e−N tu (1 .59 )


(1 .57 )

Annex A

Heat flow across a cylinder[ 12 ] , page 69 , equation (2. 14 )

Q=2⋅π⋅k⋅L⋅(t1−t2 )

Ln(r2

r1)

(2. 14 )

Qinsul=2⋅π⋅re⋅L⋅(t i−t e )rekinsul

⋅Ln (reri )Pere=2⋅π⋅re

Qinsul=Per e⋅L⋅(t i−te )rek insul

⋅Ln(reri )Qinsul⋅

rek insul

⋅Ln(reri )⋅1Per e⋅L

=t i−t e (e )

Q soil=k soil⋅S⋅(t e−t soil)Q soil⋅

1ksoil⋅S

⋅¿ t e−t soil ( f )

Adding equations (e ) and ( f )

Qinsul⋅rek insul

⋅Ln(reri )⋅1Per e⋅L

+Q soil⋅1ksoil⋅S

=t i−t e+t e−t soil

with Qinsul =Q soil=Q

Q⋅1Pere⋅L

⋅[rek insul⋅Ln(reri )+Pere⋅Lk soil⋅S ]=t i−t soil

Q=ti−t soil

1Pere⋅L

⋅[rek insul⋅Ln(reri )+Pere⋅Lksoil⋅S ]Pere⋅L⋅U=Pere⋅L⋅{1rek insul⋅Ln(reri )+Pere⋅Lk soil⋅S }

Q=Pere⋅L⋅{1rek insul⋅Ln(reri )+Pere⋅Lksoil⋅S }⋅(t i−t soil )

Q=Pere⋅L⋅U⋅( ti−t soil )

Pere⋅L⋅U=Pere⋅L⋅{1rek insul⋅Ln(reri )+Pere⋅Lk soil⋅S }Pere⋅L⋅U=¿

Pere⋅L

rek insul

⋅Ln(reri )+Pere⋅Lk soil⋅S

1Pere⋅L⋅U

=1Pere⋅L [r ek insul⋅Ln(reri )+Pere⋅Lksoil⋅S ]

1Pere⋅L⋅U

=12⋅π⋅re⋅L [rek insul⋅Ln(reri )+Pere⋅Lk soil⋅S ]

1Pere⋅L⋅U

=

Ln(reri )2⋅π⋅L⋅k insul

+1ksoil⋅S

(g )

1Per e⋅L⋅U

=


+ 1k soil⋅S

(g )

Q=Pere⋅L⋅{1rek insul⋅Ln(reri )+Pere⋅Lksoil⋅S }⋅(t i−t soil )

Q=Pere⋅L⋅U⋅( ti−t soil )

Pere⋅L⋅U=Pere⋅L⋅{1rek insul⋅Ln(reri )+Pere⋅Lk soil⋅S }Pere⋅L⋅U=¿

Pere⋅L

rek insul

⋅Ln(reri )+Pere⋅Lk soil⋅S

1Pere⋅L⋅U

=1Pere⋅L [r ek insul⋅Ln(reri )+Pere⋅Lksoil⋅S ]

1Pere⋅L⋅U

=12⋅π⋅re⋅L [rek insul⋅Ln(reri )+Pere⋅Lk soil⋅S ]

1Pere⋅L⋅U

=

Ln(reri )2⋅π⋅L⋅k insul

+1ksoil⋅S

(g )

1Per e⋅L⋅U

=


+ 1k soil⋅S

(g )

[12] Heat and mass transferAnthony F. MillsIrwin 1995

Heat Loss From a Buried Oil Pipe

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