Vol.9,No.9,61/69 (2010)

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***

Temperature Drop of Steam Flowing Through Control ValvesMasahiro URATA, Toshiharu KAGAWA

Abstract: The steam temperature and pressure are important operating factors in many chemical processes that use thermalenergy of the steam. Solvent desorption processes require precise estimate of temperature drop of steam at a control valvethrough which the steam is supplied to the processes. This paper proposes a procedure for estimate of the temperature drop insteam flow through a control valve. Theoretical estimate was made assuming a one-dimensional compressible fluid flowingthrough a throttle. In control valves in practice, heat radiation and flow irregularities induced by wall geometry causecondensation and re-evaporation, which are not included in the theory. Experiments were carried out for steam of about 0.53MPa (abs) pressure with about 430 K temperature. While the theoretically estimated temperature drop by a pressure dropfrom 0.5 MPa (abs) to 0.1 MPa (abs) is about 15 K, deviation of experimental values from the theoretical values was within 1K when the flow rate is not very small. The result of the paper can be effectively used for planing of plants that use steam asthe medium for heating and solvent desorption.

Key words: Control valve, Steam flow, Temperature drop, Joule-Thomsons effect, Heat transfer

1)

****Production Engng. Center, Fuji Film Co. ltd.**Precision and Intelligence Laboratory, TokyoInstitute of Technology(Received November 24, 2009)TRIA009/10/0909 (C) 2009SICE

1020%Joule-Thomson 2) 4)

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Joule-Thomson

160 0.6 MPa 5k

0.6 MPa 1600.6 MPa-0.1 MPa 160-100

A [m2]cp [J/kg/K]Da [m]Db [m]Dc [m]E [J]e [J/kg]GrGrashof [-]g [m]h [J/kg]L [m]m [kg/s]NuNuselt [-]PrPrandtl [-]p [Pa]Q [W]

R 1, ~4, H [W/K]ReReynoldsRH [W/K]S [m2]T [K, ]T [K, ]T0 [K, ]TM [K, ]TV [K, ]u [J/kg]v [m3/kg]w [m/s] [W/m2/K] [K-1] [W/m/K]a [W/m/K]s [W/m/K] [Pa.s] [m2/s] [kg/m3]12M

2.1 Fig. 1Fig. 1Fig. 1Fig. 1

Fig. 1 Flow through a throttle

Fig. 1 4),5)

T1p1w1

T2p2w2

S1 S2

Q

Throat

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w wh q h1

2

2

2

1 220

+ + = (1)

q Q m= / (2)

v = 1 / (3)

h u pv= + (4)

10 m/s 5102 m2/s20.5 kJ/kg 1500.5MPa(abs) 2753kJ/kg

h q h1 2 0+ (5)

h h2 1

h h w w2 1 22

1

2 2 >> / (6)

(4)

h h1 2= (7)

h c Tp= (8)

cp

T T1 2= (9)

Joule-Thomson Joule-Thomson

2.2 Joule-Thomson

Joule-Thomson h p T 6), 7)

dh hT

dT hp

dpp T

=

FHGIKJ +

FHGIKJ (10)

ch

Tp p=

FHIK (11)

dh=0 (10), (11)

FHGIKJ =

FHGIKJ =

FHGIKJ +

FHGIKJ

LNMM

OQPP

Tp c

hp c

up

pvp

T p T p T T

1 1 b g

(12)(12) 0 0 Joule-Thomson 1 MPa 200(p1, T1)

(p2, T2)

( , ) ( , ) ( , )p T p T p T1 1 1 2 2 2 (13)

(11)(10)

h h c T p dT v Tv

pdppT

T

Tp

p

2 1 1 21

2

2

1

2

= + z z FHG IKJ

LNM

OQP

( , )

(14)

v v T p= ( , )2 (15)

(14) 2

1p1, p2, T1,T2p1, p2, T1 T2p=f(v, T)Van derWaals cph p T 4) 5), 6) 7)

F

p=250 kPa 2716.8 kJ/kg 127.4 (101.3 kPa) h=cpT cp1T Joule-ThomsonJoule-Thomson p1=250 kPa, h1=2740 kJ/kg- 64-

xFig. 2Fig. 2Fig. 2Fig. 2

ig. 2 h-T-p diagram for steam

()

k2

F

100

110

120

130

140

150

160

2680 2700 2720 2740 2760h [kJ/kg]

p=600 kPap=550 kPa

p=500 kPap=450 kPa

p=400 kPap=350 kPap=300 kPap=250 kPa

p=200 kPa

p=150 kPa

p=101.3 kPa Saturation limit

Superheated steam

Wet steam

T []T1=138.2 p2=p=101.3

Pa 131.7 7.1 Joule-Thomson

.3 8) Fig. 3Fig. 3Fig. 3Fig. 3

ig.3 Piping model

DaDbDc

InsulatorPipe

L1 L2

TM,pM

T

T1,p1

T2,p2

TV

Insulator

8

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Fig.1Fig.1

Q Q Qt = +1 2 (16)

Q1, Q2Q 1, 2Fig.3 (1) R1(2) R2(3) R3(4) R4 RH

R R R R RH = + + +1 2 3 4 (17)

9, 10)

Q T T RV H= a f / (18) R1~R4

(1) R1R1=0

Q A T Ti V a= ( ) (19)

R A D Li A a1 1= =/ ( ),

i [W/m2/K]

Nu Di a= / (20)

Nu Re Pr= 0 023 0 8 0 4. . . (21)

10) 13)(21) 10), 11)

R1 R1(21)

Pr a a cp= = / , / ( ) (22)

Fig. 4Fig. 4Fig. 4Fig. 4 0.4 1.04 0.99

Fig. 4 Prandtl number of steam

mw=mv/S

RewD wD mD S m

Da a a

a

= = = =

/ 4(23)

R1=0 RH1%(2) R2R3 9), 10)

RDb

2

1= log (24)

(3

P

0.96

0.98

1

1.02

1.04

1.06

1.08

1.1

1.12

100 110 120 130 140 150 160

Pr

T []

Superheated steam

Water

Saturation limit

p=100 kPa

p=200 kPa

p=300 kPa

p=400 kPap=500 kPa

p=600 kPaL Ds a2

RL

D

Dc

b

3

1

2=

log (25)

) R4 Nu Gr

r

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10)13)

Nu GrPr= ( )0 13 1 3. / GrPr=108~1012 (26)

Nu GrPr= ( )0 56 1 4. / GrPr=104~108 (27) NuGr Pr Pr=0.71

Grg D Tc

=

3

2

(28)

T

T T T= 0

=

FHIK

1

T p (29)

=

=

=

=

FHIKFHIK

FH

IK

1 1

1 1

Tv

v

T

RT

p

p

R

T

T T

p p

p

( / )

( / )(30)

T0Gr (4) R1, R2, R3

Q A T T T T R= =

0 0 0 0 4( ) ( ) / (31)

RA

D

N Ac

u a

4

0 0 0

1= =

(32)

T T0

R417

T T Q R R RV = + +0 1 2 3( ) (33)

T T Q R R R RV = + + + ( )1 2 3 4 (34)

(17),(18),(27),(28),(31)

T TR

R R R RT T

RA

D

Nu A

Nu GrPr

Grg D T T

v

c

a

c

04

1 2 3 4

4

0 0 0

1 4

3

0

2

1

0 56

=

+ + +

= =

=

=

( )

U

V

||||

W

||||

a f

.

( )

/

(35)

(35)

[ , , , ]T R Nu Gr0 4

4 (18) Q Qt(16)(2) q

3333

3.1 Fig.5Fig.5Fig.5Fig.5 PC PC

Fig. 5Experimental setup

QFS

D/A converter Signal processor

PC

p2p1

T1 T2

Control valve

Steam main line

Weighing balance

Variable throttle

Table 1

Table 1 Specifications of the test rig

Table 2 3.2

Table 3Table 3Table 3Table 3

= =q h c Tt p (36)

Table 2 Test parameters

0.2 KTable 3

Table 3 Enthalpy comparison

Table 3 1200 kg/h 0.8 K2400 kg/h 0.1 K 4Table 4

Table 4 Temperature drop by Joule-Thomsons effect

Unit Steel Glass wool [W/m] 46 0.0450

D a [m] 0.2D b [m] 0.22 0.22D c [m] 0.36L 1 [m] 0.9 0.9L 2 [m] 0.9 0.9

1/ /A [K/W] 0.00147 1.94 m [kg/h] 1200 2400p 1 [kPa] 530 530T 1 [C] 158.4 162

h 1 2761.2 2769.6p 2 [kPa] 113 124T 2 [C] 143.6 147.5

h 2 2762.8 2769.8 h

[kJ/kg] -1.6 -0.2

m [kg/h] 1200 2400p 1 kPa] 530 530T [C] 158.4 162

h 1 [kJ/kg] 2761.2 2769.6p 2 [kPa] 113 124T 2 [C],Theory 142.8 147.4T 2 [C],

Experiment 143.6 147.5

m (kg/h) 1200 2400T 1 [C] 158 162

p 1 [kPa] 530 530Pr [-] 1.079 1.079

[Pa.s] 1.43E-05 1.43E-05 [J/m/K] 0.0315 0.0315

Re 1.49E+05 2.97E+05Nu [-] 325 566

[W/m2/K] 52.4 91.21/ /A [K/W] 0.0337 0.0194

m (kg/h) 1200 2400T 2 [C] 144 147.5

p 2 [kPa] 113 123Pr [-] 0.985 0.985

[Pa.s] 1.39E-05 1.41E-05- 67-

[J/m/K] 0.0283 0.0287Re 1.53E+05 3.01E+05

Nu [-] 321 552 [W/m2/K] 45.2 78.91/ /A [K/W] 0.0391 0.0224 [1 1200 kg/h 0.8 K2400 kg/h 0.1 K(15.6 K 14.6 K)

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Joule-Thomson Joule-Thomson 1)

4444

Joule-Thomson Joule-Thomson 9Da

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Simulation Model for Compressible Fluid PassingThrough a Pressure Control Valve, CD-ROM Proc. of2008 Asia Simulation Conference, pp.547-550, IEEECatalog N: CP0858D-CDR, Beijing: 10-12 October(2008)

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Transfer, MacGraw-Hill(1959)

2002

()19741976,()- 69-