Preventing exposed water pipes from freezing

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Preventing exposed water pipes from freezing

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PREVENTING

EXPOSED

WATER

PIPES FROM

FREEZING

by

D.G.

Stephenson

A

water pipe that is exposed to an environment at temperatures

lower than the freezing point of water will not necessarily freeze eyen

without insulation if there is a continuaus flow through it, but when there

is no flow i t will freeze regardless of insulation. The Lequired minimum

flow rate depend8 on the temperature of the water entering the exposed section of pipe and the resistance to heat transfer from the water to

the environment. This note presents an equation relating these parameters

.

It can be u s e d to solve for any one of the three variables when the other

two are knawn. Idormation is also presented which facilitates the

calculation of the thermal resistance between water in the pipe

and

the environment, for conditions that could cause freezing.

B a s i c Eauation for Heat Loss

The rate of heat transfer from fluid flowing through a pipe is:

L ' 8

mean

Q =

R

Total

where

R

is the resistance t o heat flow per unit

Total

length of pipe,

L is the length of the exposed eection of pipe,

0 is the difference between the fluid temperature and the ambient air temperature.

Let T

=

temperature of fluid entering pipe, in

T

= temperature of fluid leaving pipe, out

T

= temperature of the environment.

(4)

Then

0.

= T

- T

~n in ambient

8

= T

- T

out out ambient

0 - 8

in out

T

- T

0 - -

-

- in out

mean In@.

18:

1

In out trip. m

/e

out

p

The heat loss can also be related to the fluid flow rate and the temperature drop between inlet and outlet.

where W is mass flow rate through the pipe,

C is the specific heat of the fluid (- 1.0 for watek).

Combining the two independent expressions for Q gives:

J n ( 8 .

/B

=

L

In o u t

W C R ~ o t a l

Thermal R e s i s t a n c e s

The total resistance to heat flow between fluid flawing thraugh

a pipe and the outside environment is the sum of the following four components :

(a) Rinside' which depends on the rate of flow and the inside diameter of the pipe. Figure 1 gives the relaticrn~hip for cold water flowing

through a long pipe.

where 0. D. and I. D. a r e the outside and inside diameters of the pipe

respectively and

K

i s the thermal conductivity of the pipe material.

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P R

( o . D . / I . D . )

- - in6 ulation Rinsu~ation ~ I T K

insulation

Material

For most pipe insulations the conductivity has a value of about

0.025 ~ t u / h r f t OF.

Copper

ALurninurn S tee1

Plastic

where

H

and H are khe conductances per lineal foot of pipe due to

3: R

convection and radiation respectively. H depends on the outside

C

diameter of the cylinder and the velocity of the air blowing across it.

The relationship is shown graphically in Figure 2, H depends on

R

220

120 28

0.1

the diameter of the cylinder and the emissivity of the outer surface. F o r surfaces with a high value of emissivity,

HR

is approximately equal to twice the outside diameter expressed in feet. This is

usually small compared to H and a lower ernissivity surface makes

C

i t smaller still.

Minimum W a t e r Temperature

If a pipe is to remain completely free of ice, its inside surface

temperature must not fall below 32

'F,

which means that the water

temperaturemustbeabove32"F. T h e r n i n i m w n v a l u e o f T i s g i v e n b y : out

T

= 3 2 + Rinside

out

(6)

R

or Total

out

R

+ a

pipe insulation

+

R

out side

Exarnnle Problem

Find the required inlet w a t e r temperature to prevent freezing

in a 500-ft length

of 6-in.

schedule 40 steel pipe covered by a l -in. layer

of insulation. The minimum water flow rate will be 3 gallons/rnin. (1 800

lb/hr) and the ambient conditions are -10°F with a 30-mile-per-hour wind.

Data:

6-in.

schedule 40 steel pipe

-

I.

D.

pipe = 6 . 0 6 5 in. 0. D. pipe = 6.625 in. 0. D. insulation = 8 . 6 2 5 in.

Solution:

A. Calculation of Thermal Resistances

From Figure 1,

R

= 0 . 2 5 inside

R

=

I n ( 6 . 6 2 5 / 6 . 0 6 5 )

/

[ZIT x 2 8 ) = 0 . 0 0 0 5 i. e . , negligible pip=

R

= In

( 8 . 6 2 5 / 6 . 6 2 5 ) /' (Zn x 0 . 0 2 5 )

= 1,68

insulation A

R

=

outside

_{+ %}

C Wind Speed x

0,

D. = 30 x 8 . 6 2 5 12 = 2 1 . 6 insulation

.:

From Figure 2, H = 19.8 C

(7)

B.

Calculation of Water Outlet Temperature R ~ o h a l

o ut

R

32

-

T

pipe Rinsnlation

t R

outs ide

- -

- 9 8 _{x 42} ₌ _{4 8}_degrees

1 . 7 3

T

= T

+

8 = -10

+

48 = 3 8 ° F

out ambient out . =

C.

Calculation of W a t e r Inlet Temperature

The problem might have been to determine the minimum

allowable flow rate for a given inlet temperature, say

T.

= 48

"F.

In

Solution:

A. as in first example as suming that flow will be laminar and

R

= . 2 5 regardless of flow rate

inside

B.

as in dirst example

C.

Calculation of Minimum Flow Rate

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F I G U R E

1 T H E R M A L

R E S I S T A N C E B E T W E E N

P1

PE A N D

W A T E R

F L O W I N G

T H R O U G H

I T

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1.0 10 W I N D S P E E D x 0 . _0._{I N S U L A T I O N}

F I G U R E 2