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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NOT FOR PUBLICATION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 RTotaI 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. ambientThen
e.
1n=
T.1n - Tam lentb'e
= T - Tout out ambient
T -T in out = 1n
(e. /e
t) In oue.
- e
In oute
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/WCRTotalThermal 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 abonFor 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 Hc 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 eR .pIpe + R.1nsu ahonl ' + Rou Sl et ' d (32 - Tambient)
or ( R ) ( ) Total
e
out = R--.MMKMrMGMセャBZBBZBZBBB + R -d - 32 - T ambientpIpe 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 HR = 2 x128. 625 = 1 4. R = 1 outside 19. 8 + 1.4 RTota1=
0.25 + 1. 68 + 0.05 = 1. 98MMMMMMMMMMMMMMMセMM
5
-B. Calculatio:t;l of Water Outlet Temperature
8
o (32 - Tambient)\
=
セZセセ
x 42 = 48 degreesT - T
+
8=
-10+
48=
3BoFout - 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/wCRT 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 ln18
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MKiMGhM1
I'
I
i , I i II II: I I I I i'1"
1 .. -, I JII_
III:
J!l
I
--
セ 6 UQセL 4 3 1.0 10WIND 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
IAセiL
m
1 I I +tIl 60rH-
1=__ I 11" 11 '1 lIill!;ll I II50 f=t::
=
Wind Speed in MPH IIII -
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 I20