Preventing Exposed Water Pipes From Freezing

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Technical Note (National Research Council of Canada. Division of Building Research), 1964-06-01

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Preventing Exposed Water Pipes From Freezing

Stephenson, D. G.

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DIVISION OF BUILDING RESEARCH

NATIONAL RESEARCH COUNCIL OF CANADA

'Jr

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1HIN ][ CAlL

NOTlE

ｾ

ｾｉ

NOT FOR PUBLICATION

PREPARED BY _{D. G. Stephenson} CHECKED BY _A.G.W. APPROVED BY N. B. H.

DATE June 1964

PREPARED FOR Inquiry Reply

SUBJECT _{PREVENTING EXPOSED WATER PIPES FROM FREEZING}

A water pipe that is exposed to an environment at temperatures lower than the freezing point of water will not necessarily freeze even without insulation if there is a continuous flow through it, but when there is no flow it will freeze regardless of insulation. The required minimum flow rate depends 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 used to solve for anyone of the three variables when the other two are known. Information 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.

Basic Equation for Heat Loss

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

Q

=

L·a

mean R Total where R

TotaI is the resistance to heat flow per unit length of pipe,

L is the length of the exposed section of pipe,

a

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

....

,

2

-Let T. = temperature of fluid entering pipe, 1n

T = temperature of fluid leaving pipe, out

T

=

temperature of the environment. ambient

Then

e.

_1n

=

T._1n - T_{am lent}b'

e

= T - T

out out ambient

T -T in out = 1n

(e. /e

t) In ou

e.

- e

In out

e

mean

=

1n

(e. /e. )

, 1n out

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

Q = W C (T. - T )

. In out

where W is mass flow rate through the pipe,

C is the specific heat of the fluid (= 1. 0 for water). Combining the two independent expressions for Q gives:

L 1. n (e.

Ie

t) =

-W-C-R---In ou Total

or

e.

_In

=

e

_out eL/WCRTotal

Thermal Resistances

The total resistance to heat flow between fluid flowing through a pipe and the outside environment is the sum of the following four

components:

21TK . pipe

R. ' d ' which depends on the rate of flow and the inside diameter of 1nSI e

the pipe. Figure 1 gives the relationship for cold water flowing through a long pipe.

In(O.D·II.D.} . p1pe R. = p1pe (a) (b)

( c)

3

-where O. D. and I. D. are the outside and inside diameters of the pipe respectively and K. is the thermal conductivity of the pipe material.

pipe

Btu Material _{K (it hr of)}

Copper 220 Aluminum 120 Steel 28 Plastic O. I In(O.D./I.D.). It. R

=

lnsu a ion insulation 21T K. I . lnsu abon

For most pipe insulations the conductivity has a value of about 0.025 Btu/hr ft of.

(d) R = I

outside H c

+

H R where H

c and HR are the conductances per lineal foot of pipe due to 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

R depends on the diameter of the cylinder and the emissivity of the outer surface. For sudaces with a high value of emissivity, H

R is approximately equal to twice the outside diameter expres sed in feet. This is

usually small compared to H_{. c} and a lower emissivity surface makes it smaller still.

Minimum Water Temperature

If a pipe is to remain completely free of ice, its inside surface temperature must not fall below 32of, which means that the water

temperature must be above 32°F. The minimum value of T is given by: out

T

=

32

+

out 4 -( R. 'd ) - 1ns1 e

R ._pIpe + R._{1nsu ahon}l ' + R_{ou Sl e}t ' d (32 - Tambient)

or ( R ) ( ) Total

e

out = R--.ＭＭＫＭｒＭＧＭｾｬＢＺＢＢＺＢＺＢＢＢ + R -d - 32 - T ambient

pIpe 1nsu ahon outsi e

Example Problem

Find the required inlet water temperature to prevent freezing in a 500-ft length of 6-in. schedule 40 steel pipe covered by a I-in. layer

of insulation. The minimum water flow rate will be 3 gallons/min. (1800

1b/hr) and the ambient conditions are -10 of with a 30-mile-per-hour wind.

Data: 6-in. schedule 40 steel pipe

Solution: I.D. I?ipe O. D. pipe O. D. insulation = 6.065 in. = 6.625 in. = 8.625 in.

A. Calculation of Thermal Resistances

w

11' X 1. D. 1800 =1I'x6.065!12= 1134 From Figure 1, _{R. 'd = 0.25} 1nS1 e

=

0.05 R. = 1n (6.625/6.065) / (211' x 28) = 0.0005 i. e., negligible pIpe R. l ' = 1n (8.625/6.625) / (211' x 0.025) = 1. 68 1nsu ahon . R

=

1 outside H c + HR Wind Speed x O. D. . l ' = 30 x 8'16: 5 = 21. 6 1nsu ahon • From Figure 2, H = 19.8 c H_R = 2 x₁₂8. 625 = 1 4_. R = 1 outside 19. 8 + 1.4 RTota1

=

0.25 + 1. 68 + 0.05 = 1. 98

ＭＭＭＭＭＭＭＭＭＭＭＭＭＭＭｾＭＭ

5

-B. Calculatio:t;l of Water Outlet Temperature

8

o (32 - Tambient)\

=

ｾＺｾｾ

x 42 = 48 degrees

T - T

+

8

=

-10

+

48

=

3BoF

out - ambient out ======

C. Calculation of Water Inlet Temperature

8. 1 L/WCRTotal =8 e out L/WCRTotal 500 = 1800 x 1 x 1. 98 = 0.140 8. = 48 eO. 140 = 55 degrees ln T. = T b'

+

8. = -10

+

55 = 45°F 10 am lent ln

-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 assuming that flow will be laminar and R. 'd =.25 regardless of flow rate

lnSl e

B. as in fir s t example

C. Calculatio of Minimum Flow Rate L/wCR_T t 1

=

1n (8.

18

t) o a ln ou 8. = T. - T . = 48

+

10 = 58 ln ln amblent 1n(8./8 .t} = 1n(58/48} = 0.190 ln ou W = L

=

CRT t 11n(8._{o a ln}

18

_out

}

500 lx1.98xO.190 = l330lb/hr.

1.0

·

=>

·

.10

l

-·

co

"-

L.L. 0

--

.

ｾ .c:

-Q)

e

"0

.-V) c

.01

0::: H , _tl H - ｦｾ -L _{! P} r ,

I;

1=

-I

iff

t-t-H-H •

--

...

-

-

-

-laminar Flow

Turbulent Flow

ｾＱＱ

-....

III

-

·M

Win Ib/hr

,

I. D. in

ft

-

ｾ

MIl

II

ｾ _-

m

--

-m

ｾｮｭ

I

m

H-H.

W

JT

I.

D.

FIGURE

1

THERMAL RESISTANCE BETWEEN PIPE AND WATER FLOW1NG

THROUGH IT

" 100

..

, 1111 li'UIl! m l-' IIi ,I-f-+++--H+I H ＱｩｦＱｬｾＱＭＭｦＭＫｉＭＧｈＭ

1

I'

I

i , I i II II: I I I I i

'1"

1 .. -, I JI

I_

III:

J

!l

I

--

ｾ 6 ＵＱｾＬ 4 3 1.0 10

WIND SPEED x O. D. INSULATION

2 R4FR-H++,'++++-IRfI¥.,j'

,q

ffi

I I' 'i iI' I f1 i , '; "

ｦｬＮＱＭＧｾＧｉｾＭｉｌｉｌＧｊｬｾｉｉｊｾＭＧｾＱｊｾｭｩｉｔｭｭｲｬｬｬｾＢｩｩｬｬｾｉｉｉｉＧ

ｾｩｭｭｾｾＺ

I

ＡｾｉＬ

m

1 I I +tIl 60

rH-

1=__ I 11" 11 '1 lIill!;ll I II

50 f=t::

=

Wind Speed in MPH I

III -

I

- Itm

U lill

40 ｴＭｾｾｉｉＭｈ O. D. Insulation In Feet 1 JII L

:-t j t! IJ j I

30 E . ［ｾ , Il; fR'

I

I

n

r

ｾ

U: ffFR-I.rilfHtfHffl:!tIHfi!ffilt'-i-f'-+-;li-i+l-Lt++-tH+ttt I

20

｟ｾ

I I

I

j'111f I II ,... , I I' r=

+:1-I

jll

ｾｉ ｾｬｩｾｬｬＫ

ｾ

10 u.. o

..

-. u :c FIGURE 2