Heat Transfer Media

Water like air is available in abundance, relatively cheaper price.

Heat Transfer Fluids
Water is a very good candidate for heat transfer media and it
can convert into steam and again, steam is again posed a very
good heat transfer media.
Water/Steam
The main problem with the water steam HTF is the scarcity of
water especially in desert regions.
Thermal oils
High-temperature steam corrosion in several different alloys has
been reported in the literature.
Organics
Water is commonly used as a heat transfer media. The best
example, boiler where steam produced. The only limitation with
Molten Salts the water is that it has some pressure-sensitive or temperature-
sensitive approaches.

Liquid metals In that case some more ranges of heat transfer fluid is required
like thermal oils.
 Heat Transfer Media

Heat Transfer Fluids • These offer again a long-range of

temperature spectrum.
Water/Steam • Mineral oil (petrolatum and paraffin
oil), silicone oil, and synthetic oils
have been tested and used as HTFs
Thermal oils mostly in solar system.
• These oils can be thermally stable
Organics only up to 400°C and beyond this
may have the tendency to dissociate.
Molten Salts • Another issue with the thermal oils is
that these oils are very expensive
because of the processing cost.
Liquid metals
 Heat Transfer Media

Heat Transfer Fluids • Organic materials are used as HTFs mostly in solar systems.
For example, biphenyl and diphenyl oxide are commonly used.

Water/Steam • This Biphenyl/Diphenyl oxide is a eutectic mixture of two very

stable organic compounds; Biphenyl (C12H10) and Diphenyl
oxide (C12 H10O).
Thermal oils • Operating temperature range of this biphenyl/diphenyl oxide is
usually within 12-393°C.

Organics • The first solar thermal power plant with these organic materials
as the heat transfer fluid was commissioned in 2009 at
Badajoz, Spain.
Molten Salts • Another similar type of plant named Megha Solar plant is
located at Anantapur, Andhra Pradesh in India, started in year
2014 with a production capacity of 50-megawatt electricity.
Liquid metals
 Heat Transfer Media

Heat Transfer Fluids • There are certain molten salts, which are having a large spectrum of
operating temperature range. These molten salts make excellent HTFs
due to their thermal stability at high temperatures (generally >500°C).
Water/Steam
• Molten salts also have properties comparable to water at high
temperatures, including similar viscosity and low vapor pressure.
Thermal oils • Another important advantage of utilizing molten salts in the power
system is their thermal energy storage capability.

Organics • While the salts are limited by their own thermal properties when it
comes to high temperature stability, the stability of piping and
container materials must also be taken into consideration when it
Molten Salts comes to the temperature range at which these salts are handled.
• Examples of some commercially available are given in the following
table.
Liquid metals
 Heat Transfer Media
Name Melting Stability Viscosity(Pa Thermal Heat Corrosion
point(°C) temperature( -s) Conductivity Capacity(k rate(µm/y
°C) (W/mK) J/kg K) ear)
NaNO3(7%)-KNO3(53%) 142 535 0.00316@300 0.2@300°C 1.56@300°C 2
NaNO2(40%) °C
NaNO3(7%)-KNO3(45%)- 120 500 0.00637@300 0.52@300°C 1.45@300°C 6-10
Ca(NO4)2(48%) °C
NaNO3(28%)-KNO3(52%)- 130 600 0.03 - 1.4-1.5 -
Li(NO3)(20%)
Li2CO3(32.1%)-Na2CO3(33.4%)- 400 800-850 0.0043@800° - 1.4-1.5 -
K2CO3(34.5%) C
KNO3(50-80%)-Li(NO3)(0-25%)- 99 430 - - 1.66 -
Ca(NO4)2(10-45%)
Li-Na-K flourides/carbonates 400 900 1.17@400°C 8-12
NaCl(7.5%)-KCl(23.9)-ZnCl2 204 850 0.004@600- 0.325@300°C 0.81 110-200
(68.6%) 800°C
 Heat Transfer Media

Heat Transfer Fluids

• Liquid metals has been used in various nuclear industries since 1940s
and are currently being studied for use in the solar thermal systems as
Water/Steam HTF and thermal storage media.
• So far, they have several promising properties including extensive
operating temperature range, low viscosity, and efficient heat transfer
Thermal oils characteristics.
• For example, liquid sodium has an operating temperature range from
98-883°C.
Organics • The cost of these liquid metals are relatively higher than that of molten
salt or steam/water.
Molten Salts • Also heat capacities of these liquid metals are relatively lower than
various commercially available nitrates/nitrite-based salts and hence,
they are less favorable to be used as a thermal energy storage media.
Liquid metals
 Heat Transfer Media
Liquid metals - Liquid Sodium

• Liquid sodium was first used as HTF and storage medium in 1981 in a test plant (500kW) located at
Almeria, Spain.
• Liquid sodium has an operating temperature range of 98-883°C and viscosity is reported to 0.00021
Pascal second. The thermal conductivity is 1.25 watt per meter kelvin at 600 degrees Celsius.
• The main disadvantage is its high combustibility when in contact with water and that hydrogen is one
of the products from this liquid sodium water fire.
• Another issue is that the high cost of liquid sodium which is almost four times higher than solar salt.
• As far as corrosion issues are concerned, generally liquid sodium is less aggressive than the other
liquid metals with stainless steel.
• The ceramics such as SiC and stainless steel are highly compatible to be used as piping container
material with liquid sodium.
 Heat Transfer Media
Liquid metals - Liquid Na-K(22.2-77.8 wt%) eutectic mixture
• One of the special features of this Na-K mixture is in liquid state even at room temperature. Its melting point is
(-12°C) and the boiling point is 785°C.
• The cost of this mixture is high almost four times higher than that of the solar salt.
• Corrosion data on the piping container alloy with this particular mixture is not available in the literature.

Liquid metals - Liquid Pb-Bi(44.5-55.5 wt%) eutectic mixture

• Lead-Bismuth eutectic mixture has a high boiling point i.e around 1533°C and its melting point is relatively high
i.e. around 125°C compared to other liquid metals.
• Since LBE is chemically inert with both air and water, there is no hazardous fire risk as in the case of liquid Na.
But the cost of this mixture is also extremely high, almost 26 times higher than that of the solar salt.
• Besides lead mixtures are toxic to humans. Stainless steel and nickel alloys showed corrosion with LBE (>250
m/year). But ceramic materials such as SiC showed good corrosion resistance with almost zero weight loss
even after 1000 hour of immersion at the temperature up to 750 °C.
 Heat Transfer Media
Liquid metals - Liquid Cd-Bi, Sn-Bi, Bi-Zn and Ca-Cu eutectic mixtures
• There binary mixtures available of liquid metals are being investigated for use in HTFs in solar systems.
• Combinational material synthesis and high-throughput characterization techniques are used to effectively
identify which are less corrosive composition with piping or container alloys.
• Although the exact molar compositions have not been reported, the above mentioned binary mixtures are
suggested as promising candidates in their ongoing work. Corrosion tests with the metallic alloys are also
ongoing.

Name Melting Stability Viscosity Thermal Heat Corrosion

point(°C) temperature (Pa-s) Conductivi Capacity rate
(°C) ty (W/mK) (kJ/kg K) (µm/year)
Na 98 883 0.00021@600° 46@600°C 1.25@600° -
C C
Na(22.2%)-K(77.8%) -12 785 0.00018@600° 26.2@600°C 0.87@600° -
C C
NaNO3(28%)-KNO3(52%)- 125 1533 0.001@600°C 12.8@600°C 0.15@600° 25-250
Li(NO3)(20%) C
 Heat Transfer in Insulation Materials

Cylindrical Insulating Materials Spherical Insulating Materials

Critical thickness of insulation

for cylinder and for spheres
 Heat Transfer in Insulation Materials
Heat transfer in cylindrical insulating materials

• Let us consider a long cylinder of length L, inner radius r1 and r2 which is

exposed to the temperature difference of T1 – T2 as shown in the figure.

• The length of the cylinder is very large compared with the diameter of the
cylinder.
• So, assume that the heat will flow only in the radial direction that is the
space coordinates r need to specify the system.

• So, the area of heat flow in the cylinder will be Ar= 2πrL
Now according to the Fourier's law of heat transfer through conduction in small element dr will be:

dT dT
qr = −kAr = - 2rLk
dr dr
 Heat Transfer in Insulation Materials
Heat transfer in cylindrical insulating materials
• Boundary conditions: T= Ti at r= ri and T= To at r= ro

ro
dr
To 2kL(Ti − To )
qr  = - 2Lk  dT q=
ro
ri
r Ti ln( )
ri
• The thermal resistance in this particular case:
ro
ln( )
ri
Rth =
2kL
• This thermal resistance concept can be used for multilayer
cylindrical walls:
 Heat Transfer in Insulation Materials
Problem-1: A thick wall tube made up of stainless steel with inner diameter of 4 centimeter and outer
diameter of 8 centimeter. The thermal conductivity of the stainless steel is 19 W/mK. It is covered within
insulation of a thickness 4 centimeter and that is k = 0.17 W/mK. The temperature at outside of the
insulation is around the 200°C. The inside wall of the pipe is maintained at 550°C temperature. calculate
the heat loss per unit length of the pipe and tube insulation interface temperature.
Solution:
Given; Inner radius of the tube (ri)= 2 cm = 0.02 m
Outer radius of the thick wall tube (ro) = 8cm = 0.04 m
The pipe is covered with 4 cm thick insulation materials then r3= ro + t = 4+4= 8 cm
There is only 1-D radial heat transfer in the pipe the heat loss per unit length of the pipe is given as;

2L(Ti − To )
q=
ro r3
ln( ) ln( )
ri ro
+
ks ki
 Heat Transfer in Insulation Materials
Ti = 550°C= 823K ks = 19 W/mK
To = 200°C= 473K ki= 0.17 W/mK

q 2 (823 − 473 )
=
On putting all these values in the previous equation we have, L 0.04 ln( 0.08 )
ln( )
0.02 + 0.04
19 0.17
q/L= 54.163 W/m
q 2 (T − 473 )
Now, for interface temperature = 54.163 =
L 0.08
ln( )
0.04
0.17

Solving this equation, we get T= 508.148K

 Heat Transfer in Insulation Materials
Problem-2: What will be the thickness of insulation, if the temperature of inner side is 300 °C and
temperature at outer side is 38 °C? The thermal conductivity of the insulation is k = 0.057 W/mK and
the inner radius is given as r1 = 0.035 m. If heat transfer per unit length of the cylinder that that is q/L
is 60 W/m.
Solution: Given; Inner radius of the tube (r1)= 2 cm = 0.02 m
Ti = 300°C= 573K, To= 38°C= 311K
q/L= 60 W/m
q 2 (Ti − T0 ) 2 (573 − 311 )
Now, use formula = 60 = =
L r r
ln( 2 ) ln( 2 )
r1 0.035
k 0.057

Solving, this we get r2= 0.167 m

And thickness of insulation = r2- r1 = 0.167-0.035= 0.132m
 Heat Transfer in Insulation Materials
Heat transfer in spherical insulating materials
Let us consider a sphere of inside radius ri and outside radius ro and take a small element of any
radius dr
T1 be the temperature at the inner part of this sphere and T2 be the
temperature of the outside surface where T1 > T2

Rate of heat flux according to the Fourier law is given as

r2 To
dT 2 dT
dr
Q = −kAr = - 4r k Q  2 = - 4k  dT
dr dr r1
r Ti

Solving, previous equation with boundaries conditions T=T1 at r=r1 and T=T2 at r=r2

4k (T1 − T2 ) 4r1r2 k (T1 − T2 )

Q= Q=
1 1 (r2 − r1 )
( − )
r1 r2
 Heat Transfer in Insulation Materials
Problem-3: If a spherical ball of steel of inner radius 2 cm and outer radius of 4 cm is filled
with hot matter and the temperature of it was 275°C. The thermal conductivity of the steel is
7.68 W/mK. If it is exposed to ambient temperature 25°C calculate the heat loss. The
convective heat transfer coefficient is given as 0.20 W/m2K.
Solution: Given r1= 0.02m, r2= 0.04m, ks=7.68 W/mK, h0= 0.20 W/m2K, T1= 275°C =548K, Ta=
25°C =298K
Now, outside area of the sphere, A0 =4*3.14*0.04*0.04= 0.02011 m2
Now, use the following equation

(T1 − Ta ) (548 − 298)

Q= =
1 1 1 1 1 1 1 1
( − )+ ( − )+
4k s r1 r2 Ao ho 4 (7.68) 0.02 0.04 0.02011* 0.2

Solve, this we get Q= 0.998W

 Heat Transfer in Insulation Materials
Optimum thickness of insulation

▪ It can be obtained by economical approach.

On increasing the thickness of insulation,
there is a reduction in the loss of heat, thus
saving in the operating cost but at the same
time the cost of insulation will be increased
with increase in the thickness.

▪ The optimum thickness of insulation is one

on which the total annual cost of insulation
is minimum.
 Heat Transfer in Insulation Materials
Critical thickness of insulation for cylinder
• The insulation on the external surface of the pipeline and the vessel to reduce the heat loss to the
ambient atmosphere.
• The greater insulation will result in less in heat loss which always not be true.
• Let us consider a long cylinder of length L, inner radius r1 and r2 which is carrying steam at temperature
of Ti and wrapped with insulation of thermal conductivity (outside temperature To) k2 to a radius r3. Let hi
and ho be the heat transfer coefficients in inner and outer side.
• Then, rate of heat transfer is given by

2L(Ti − To )
q=
r2 r3
ln( ) ln( )
1 r1 r2 1
+ + +
hi r1 k1 k2 ho r3
 Heat Transfer in Insulation Materials
Critical thickness of insulation for cylinder

• Let q be the function of r3 and the other parameter being kept constant it will rise up to a maxima of a certain
value of r3 and the value is called the critical radius of insulation.
• On simplifying the above equation, we have
(Ti − To ) 𝑟2 𝑟3
q= 1 1 ln( ) ln( ) 1
r r 𝑟1 𝑟2
ln( 2 ) ln( 3 ) 𝑅𝑡𝑜𝑡𝑎𝑙 = ( + + + )
1 1 r1 r2 1 2𝜋𝐿 ℎ𝑖 𝑟1 𝑘1 𝑘2 ℎ𝑜 𝑟3
( + + + )
2L hi r1 k1 k2 ho r3
• All the parameters in this equation are constant except r3, which is depending upon the thickness of insulation
required for minimum amount of heat transfer.

dRtotal 1 1 1 1 Solving, this equation we get

=0= (0 + 0 + + (− 2 )) r3 = (k3/h0)= Critical radius of insulation
dr3 2L r3 k 2 ho r3
 Heat Transfer in Insulation Materials
Critical thickness of insulation for cylinder

The thickness up to which heat flow

increases and after which heat flow
decreases is termed critical thickness.

• If the critical radius of insulation is

greater than the outer radius of the pipe
then adding insulation up to rC will
increase the heat loss from the pipe.
• The addition of the insulation thereafter
reduces the heat loss from the pipe.
• When the critical radius of insulation is
less than or equal to the outer radius of
the pipe or container then addition of
insulation will immediately reduce the
heat loss from the pipe.
 Heat Transfer in Insulation Materials
Problem-4: What will be the critical radius of insulation for material (asbestos where k = 0.20 W/mK)
surrounding the pipe and exposed to the room temperature at 25 °C with ho 2.5 W/m2K. Calculate the heat
loss from a 230°C of a 60 mm diameter pipe when covered with the critical radius of insulation and without
insulation. Would any fiber glass insulation having the thermal conductivity of 0.03 W/mK cause decrease in
heat transfer.
Solution: Given; Inner radius of the tube (r1)= 30 mm = 0.03 m, kas= 0.20 W/mK, ho = 2.5 W/m2K
Ts = 230°C= 503K, To= 25°C= 298K
rc= (k/ ho)=0.08m
Use formula, heat loss with insulation can be calculated as

q 2 (Ti − To ) 2 (503 − 298)

= = = 130.05W / m
L ln( rc ) ln(
0.08
)
r1 1 0.03 + 1
+
k as ho r0 0.20 0.08 * 2.5
 Heat Transfer in Insulation Materials
Now, heat loss without insulation can be calculated as

q
= 2rh(To − Tamb ) = 2 (0.03)(2.5)(503 − 298) = 96.606W / m
L
Percentage increase in heat transfer given by
% Increase = (Heat transfer without insulation- Heat transfer with insulation)/ Heat transfer without insulation
= (130.05-96.606)/96.606= 0.3462= 34.62%

Also, given that k (for glass insulation)= 0.03 W/mK

Thus, rc= (k/ ho)=0.03/2.5= 0.012m
 Heat Transfer in Insulation Materials
Critical thickness of insulation for sphere

For spheres the heat loss in terms of area and inner and outer heat transfer coefficients can be written as

(T1 − T2 )
q=
1 1 1 1
+ (r2 − r1 ) + (r3 − r2 ) +
hi 4r1 hi 4r1r2 k1
2
hi 4r2 r3 k 2 ho 4r32

From here

1 1 1 1
Rtotal = + (r2 − r1 ) + (r3 − r2 ) +
hi 4r1 hi 4r1r2 k1
2
hi 4r2 r3 k 2 ho 4r32
 Heat Transfer in Insulation Materials
Here, let r1 and r2 are the inner and outside diameter of the sphere and r3 be the outer radius of the insulation
and r2 can be considered as inner radius for insulating material.

Differentiate Rtotal w.r.t and r3 and solve

dRtotal 1 −2
= 0+0+ +
dr3 4r3 k 2 ho 4r33
2

On solving, we get r3 = (2k2 / ho) = rc

Above, equation gives the critical radius of thickness for the sphere, where k2 is the thermal conductivity of the
insulating material and ho is the convective heat transfer coefficient for the insulating material.
 Heat Transfer in Insulation Materials
Problem-5: If a sphere of inner diameter 40 mm having the inner temperature Ti= 200°C is surrounded by
insulation. The thermal conductivity of the insulating material k is 0.017 W/mK and the sphere is exposed to
ambient atmosphere at 30°C temperature, with ho = 0.20 W/m^2K. Calculate the heat loss with and without the
insulation and the percentage increase/decrease in the heat loss and the critical thickness of the insulation.
Solution:
Given: Radius of sphere (r) = 40mm/2 = 20 mm = 0.02 m
Ti= 200°C = 200+273 = 473K
rc = 2k/ho = 2x0.017 W/mK/ (0.20 W/m2K)
rc = 0.17 m
Tambient= 30°C = 30+273 = 303K
On putting these values in the heat transfer equation heat transfer with insulation is given by
(473 − 303)
Q= = 0.7717W
1 1
(0.17 − 0.02) +
4  0.02  0.017  0.17 4 (0.17) 2  0.2
 Heat Transfer in Insulation Materials
Without insulation
Q= ho A(Ti -Tambient)= 0.2(4π*0.02*0.02)(473-303)= 0.1709W

Now, percentage change in heat transfer is given by

(Qinsulation − Qwithoutinsulation ) 0.7717 − 0.1709

%Change =  100 =  100 = 35.55%
Qwithoutinsulation 0.1709

Thus, percentage change in heat transfer= 35.55%