TECHNICAL SCIENCES
ACCOUNT FOR THE INFLUENCE OF THE ECCENTRICITY OF SPHERICAL INSALLATIONS FOR MEASURING THERMAL CONDUCTIVITY OF SUBSTANCES
Naziyev J.
doctor of technical sciences, professor of «Physics» department of Azerbaijan State University of Oil and Industry
Baku, Azerbaijan DOI: 10.5281/zenodo.7049648
Abstract
To calculate thermal conductivity using spherical measuring instruments, it is necessary to know the heat flux and temperature difference between symmetrical spherical working surfaces. For the case of symmetric spherical surfaces, the calculation equation is well known. In this work, the influence of the eccentricity of the spherical surfaces of devices in calculating the thermal conductivity is theoretically determined.
Keywords: heat transfer, thermal conductivity, spherical calorimeter, eccentricity.
To increase the efficiency and strength of equipment used in various production processes where heat transfer processes take place, it is necessary to have data on the thermophysical properties of the materials from which the equipment and the working fluid are made. Thermal conductivity is the most important of them.
There are different methods for determining thermal conductivities, both stationary and non-stationary [1, p. 19-72]. Recently, non-stationary methods have been used more [2, p. 347-355; 3, p. 12-22]. According to the shape of the working surfaces, cylindrical, spherical and flat bicalorimeters are distinguished [1, p. 7781].
Let us determine the exact solution of the problem of the influence of eccentricity for eccentric spherical
surfaces. Consider the differential equation of heat conduction [4, p. 30]
d 2t d2t A ■ + —- = 0
dx2 dyz
(1)
We solve this equation using the method of point heat sources, where the resulting temperature field is found by adding the fields from separate symmetrically located sources A and B (Fig. 1).
For the case of point sources, the temperature difference
0 =
Q
1 1
V r r ;
(2)
We compose the following system of equations
Y= 0
1 1 1
1 1
r d — r d + r
m m m m
— r = 2(h — r )
m
r + r = 27n
mm
a — a = a H — h = a
(3)
Here Y = h - a'.
Solving this system of equations with respect to h and H we get
2h = a 2H = a' +
— + r — a r a
R
— + R2 — a" R a
(4)
For the joint solution of the last three equations, we have
R6
(a)2 (a)
R3 r
+ 2
(R5
— + — a a
I —(r 4 — - )— 2(a" R3 + a r3 )+[(a)2 R2 +(a)2 r2 ]—
— 2a 2| —+ — I a a
2a2 (r2 + r2 )+ 2(a R + a r)+ 2Rr(ar + a r)+
(5)
y
+ 2Rr — r2 + — Rz
a
a
— 2R2r2(4 + - | —2Rr I a a )
Under certain assumptions, it is possible to obtain an analytical solution to the problem of the eccentricity of spherical bodies.
For the surfaces temperature difference, we write
At =
Q
4kX
1111
—---r + —---r
V rm rm rM rM y
. (6)
Taking into account the inequality (Fig.)
1 1 1 + — rM 1,
r m r m rM
and taking into account that rm )) rm and rM )) rM
1 1
, and fractions —- , —— have opposite signs, we can
rm rM
make the first assumption:
( R2 r2 -
—— — + a a
V a a y
r — r
n m
rN — rM
— 2 R2 r2 + a 4 = 0.
Where
r — r
n m
rN rM
r
R
(13)
(14)
From (11) and (12)
r„ — r
r„ — r,
r — r
n m
r\T —
rm rM rn rN
rmrM
rnrN
. (15)
'N 'M N 'M
Then the following inequalities can be written:
........ f "' ' \2 f
-)
V rmrM y
r,
M
V 'm y
V AR2
(
Vry
. (16)
At =
Q
4kX
1 1
V rm
(8)
' M y
Since the temperatures over the entire surface of the first and second spheres are the same
(9)
----- = ----- , (10)
or
1 1 1 1
r r r r
m m n n
1 1 1 1
rM rM rN 'N
r — r = (r — r
n m \ n m )
(r — r )•
nm
rr
mn
rr
mn
rN rM = —
(rN rM^
rMrN
rMrN
It is also seen from the figure that
(11)
(12)
Let's make the following second assumption:
rmrM
rmrM
rnrN
rnrN
( R12
Vr y
(17)
Comparing relations (14), (15) and (17) we obtain
( R13 Vr y
r„ — r
= 1.
rN rM
From figure 1
rm + a = r
rM + a" = R
r — a = r
n
-n— a = R
a —a = a
Using equation (19) from (18) we find
(18)
3
r
y
/
6
r
1
rM
m
rm rM
rm rM
)
'mrM
rn-N
J
a = ■
ar
R - r3
aR
a =
d3 3
R - r
(20)
Taking into account (40), the (8) can be written as
x = -
_Q_
4 nAt
(R - r 1 -
R - r
(21)
Rr
1-
aR1
n ! 3
R - r
1-
ar
n ! 3
R - r
From (21) for limiting cases we obtain:
At =
At = 0 at a = R - r Q R - r
at a = 0
4nX Rr
The resulting expression (21) is very convenient for calculating the thermal conductivity coefficient, taking into account the eccentricity of the working surfaces of equipment.
.References
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