Heat Transfer Analysis in Annulus Flow

This assignment delves into the analysis of heat transfer within a system involving turbulent flow through an annulus. It calculates key parameters such as mean velocity, Reynolds number, and Nusselt number using established equations and correlations. The Dittus-Boelter equation is employed to determine the convective heat transfer coefficient (h) for the turbulent flow. Furthermore, it investigates the overall heat transfer coefficient (U) considering the specific conditions of the annulus system.

HEAT TRANSFER AND COMBUSTION
[Document subtitle]
FEBRUARY 10, 2018
[Company name]
[Company address]

HEAT TRANSFER AND COMBUSTION
Introduction to Heat Exchanger and Determining Heat Transfer Coefficient
Example
The return condensate from a steam trapping assembly is to be used to warm boiler make up
water, utilizing a double tube counter-flow heat exchanger (Khartchenko, 2014). The condensate
has an average temperature of 800C and is available at a rate of 10kg/s. The boiler makes up
water has an average temperature of 250C and has a mass flow rate of 15 kg/s. The inner tube of
heat exchanger has a diameter of 40mm, the outer shell is 60 mm in diameter.
Assumptions:
 The thermal resistance of the inner tube is negligible as the tube is highly thermally
conductive and its thickness is negligible.
 Both the oil flows are fully developed.
 The properties of the water are constant.
Hot water 10 kg/s
Cold water 15 kg/s
Given: Using fluid properties table(table A-9)for water, properties are obtained.
1
Hot water
 Taverage = 80oC
  = 971.8 kg/m3
 Pr = 2.22
 K =0.670 W/m-K
 =0.355*10-3 kg/m-s
 Cp=4197 J/kg-K
Cold water
 Taverage =25oC
  = 997 kg/m3
 Pr = 6.14
 K =0.607 W/m-K
 =0.891*10-3 kg/m-s
 Cp=4180 J/kg-K

For Cold water, prandtl number can be calculated by-
Pr= c p
k
Pr= 4180∗0 .891∗10−3
.607 =6.1357
For Hot water, prandtl number can be calculated by-
Pr= c p
k
Pr= 4197∗0 .355∗10−3
.670 =6.2238
So, we can see that the Prandtl numbers for Hot and Cold Water are approximately same as given
in table A-9.
In Heat Exchanger, Heat transfer is mainly due to Convection. So overall Heat transfer
coefficient “U” will be -
1
U = 1
hi + 1
ho
Where:
hi =heat transfer coefficient for inner tube
ho =heat transfer coefficient for outer annulus.
Calculation for hi:
Diameter of tube=40 mm=0.040m
So hydraulic diameter of circular tube=0.040m
Mass flow rate (mo)=15 kg/s
So Mean velocity (Vm) is,
2

V m = mo
Ac = mo
( 1
4 D2
)
V m = 15
997∗1
4 ∗¿ 0.042
=11.97 m/ s
Where: =Density
Ac=Cross Sectional Area
Reynold’s number (Corda, 2017) (Re)-
 Reynold’s number is used to determine the type of flow of fluid.
 It is the ratio of Inertia forces to Viscous Forces.
ℜ= Inertia Force
Viscous Force =V m DH
❑ = DH V m
❑
ℜ= 0.04∗11.97
8.937∗10−7 =5,35,750.25
 For,
Re< 2300–Laminar Flow
2300<Re<4000 –Transition to turbulent
Re>4000 –Turbulent flow
Hence Flow inside the tube is Turbulent.
Prandtl Number (Pr)-
 The thermal Boundary layer is best described by the dimensionless parameter Prandtl
number (Pr) (Maleshwar, 2009).
3

Pr= molecular diffuivity of momentum
molecular diffusivity of heat
Pr=❑
α = cp
k
Pr= 4180∗0 .891∗10−3
.607 =6.14
 The prandtl number of fluids range from less than 0.01 for liquid metals to more than
1,00,000 for heavy oils.
 Heat diffuses more quickly in liquid metals (Pr<<1) and very slowly in oils (Pr>>1)
relative to momentum.
Nusselt number (Janna, 2000)(Nu)-
 Nusselt number represents the enhancement of heat transfer through a fluid layer as a
result of convection relative to conduction across the same fluid layer.
 Large the Nusselt number, the more effective is the convection.
Nu= qconviction
qconduction
= h Lc
k
Where: h=coefficient of convective heat transfer
k=thermal conductivity
Lc=Characteristic length
 For Nu>>>1, Thermal Conviction>>Thermal conduction.
Due to flow being fully turbulent, The Nusselt number (NU) is derived using the ‘Dittus and
Boelter’ equation which is:
4

NU =0.023 ℜ0.8 Pr0.4
NU =0.023(535761.61)0.8 (6.14)0.4 =1820.42
NU = hi Lc
k
hi = k
Lc
NU
hi =.607
0.04 ∗1820.42=27624.92 W /m2 K
Calculation for ho:
Outer diameter of annulus (Do) =60 mm=0.060 m
Mass flow rate (mo) =10 kg/s
Inner Diameter of annulus (Di) =40 mm=0.040 m
So hydraulic diameter of circular tube (DH=Do-Di) =0.060-0.040=0.020 m
So mean Velocity will be,
Vm= mo
Ac = mo
1
4 ( Do¿¿ 2−Di2)¿
V m = 10
1
4 ∗971.8∗π (.062−0.042)
=6.55 m/ s
So Reynold’s no. will be,
ℜ= DH V m
❑
ℜ= 0.02∗6.55
3.653∗10−7 =358607.88
5

So Flow in Annulus is also Turbulent.
Due to flow being fully turbulent, The Nusselt number (NU) is derived using the ‘Dittus and
Boelter’ equation (Incropera, 2006) which is:
NU =0.023 ℜ0.8 Pr0.3
NU =0.023(358607.88)0.8 (2.22)0.3=811.57
So, ho will be,
NU = ho Lc
k
ho= k
Lc
N U
ho= 0.670
0.02 ∗811.57=27187.595W /m2 K
Calculation for Overall Heat Transfer Coefficiant(U):
1
U = 1
hi + 1
ho
1
U = 1
27624.92 + 1
27187.595
U =13702.256W / m2 K
Variations-
To increase Heat transfer,
6

Q=UA (T Hot−T cold )
So we can Change-
1) Contact area (A) - We can increase contact area by increasing Length of Heat Exchanger
or we can increase it by passing same tube no. of times.
2) By increasing the temperature of hot Fluid and decreasing the Temperature of cold fluid.
3) Overall heat transfer coefficient (U)- U can be increased by increasing heat transfer
coefficient of tube and annulus (h). To increase heat transfer coefficient of tube or
annulus, we can-
 Use liquids of higher Prandtl number.i.e. liquids with Higher value of dynamic
viscosity(μ) and specific Heat (cp) and lower value of thermal conductivity.
 Increase the mass flow rate of liquids.
Now to demonstrate the increase in Heat Transfer Coefficient, we are increasing the mass flow
rate.
Example 2
The return condensate from a steam trapping assembly is to be used to warm boiler make
up water, utilizing a double tube counter-flow heat exchanger. The condensate has an
average temperature of 800C and is available at a rate of 20kg/s. The boiler make up water
has an average temperature of 250C and has a mass flow rate of 30 kg/s. The inner tube of
heat exchanger has a diameter of 40 mm, the outer shell is 60 mm in diameter.
Assumptions:
 The thermal resistance of the inner tube is negligible as the tube is highly thermally
conductive and its thickness is negligible.
 Both the oil flows are fully developed.
 The properties of the water are constant.
7

Given: Using fluid properties table(table A-9)for water, properties are obtained (Fox, 2004).
For Cold water, prandtl number can be calculated by-
Pr= c p
k
Pr= 4180∗0 .891∗10−3
.607 =6.1357
For Hot water, prandtl number can be calculated by-
Pr= c p
k
Pr= 4197∗0 .355∗10−3
.670 =6.2238
So we can see that the Prandtl numbers for Hot and Cold Water are approximately same as given
in table A-9.
Calculation for hi :
Diameter of tube= 40 mm=0.040m
So hydraulic diameter of circular tube=0.040 m
8
Hot water
 Taverage = 80oC
  = 971.8 kg/m3
 Pr = 2.22
 K =0.670 W/m-K
 =0.355*10-3 kg/m-s
 Cp=4197 J/kg-K
Cold water
 Taverage =25oC
  = 997 kg/m3
 Pr = 6.14
 K =0.607 W/m-K
 =0.891*10-3 kg/m-s
 Cp=4180 J/kg-K

Mass flow rate (mo)=30 kg/s
So Mean velocity (Vm) is,
V m = mo
Ac = mo
( 1
4 D2
)
V m = 30
997∗1
4 ∗¿ 0.042
=23.94 m/s
Reynold’s no, will be,
ℜ=V m DH
❑
ℜ= 997∗0.04∗23.94
0.891∗10−3 =1071523.232
Due to flow being fully turbulent, The Nusselt number (NU) is derived using the ‘Dittus and
Boelter’ equation (Glicksman, 2016) which is:
NU =0.023 ℜ0.8 Pr0.4
NU =0.023(1071523.232)0.8(6.14)0.4=3169.5416
NU = hi Lc
k
hi = k
Lc
NU
hi =.607
0.04 ∗3169.5416=48097.79 W / m2 K
9

Calculation for ho:
Outer diameter of annulus (Do) =60 mm=0.060 m
Mass flow rate (mo) =20 kg/s
Inner Diameter of annulus (Di) =40 mm=0.040 m
So hydraulic diameter of circular tube (DH=Do-Di) =0.060-0.040=0.020 m
So mean Velocity will be,
Vm= mo
Ac = mo
1
4 ( Do¿¿ 2−Di2)¿
V m = 20
1
4 ∗971.8∗π (.062−0.042)
=13.1 m/s
So Reynold’s no. will be,
ℜ= DH V m
❑
ℜ= 0.02∗13.1
3.653∗10−7 =717218.724
So Flow in Annulus is also Turbulent.
Due to flow being fully turbulent, The Nusselt number (NU) is derived using the ‘Dittus and
Boelter’ equation which is:
NU =0.023 ℜ0.8 Pr0.3
NU =0.023(717218.724)0.8 (2.22)0.3=1413.0418
So, ho will be,
10

NU = ho Lc
k
ho= k
Lc
N U
ho= 0.670
0.02 ∗1413.0418=47336.90W /m2 K
Calculation for Overall Heat Transfer Coefficient (U) (Gavhane, 2008):
1
U = 1
hi + 1
ho
1
U = 1
48097.79 + 1
47336.90
U =23857.1558W /m2 K
U Previous=13702.256 W /m2 K
So, we can see that the overall Heat Transfer Coefficient has been increased to approximately
1.74 time of previous value by doubling the mass flow rate of both the fluids. We can also change
the fluids and increase the heat transfer coefficient.
References
Corda, S., 2017. Introduction to Aerospace Engineering With A Flight Test Perspective.
Chichester: John Wiley.
Fox, R. W., 2004. Introduction to Mechanics. 5th ed. New Delhi: Wiley India (p.) Ltd.
Gavhane, K. A., 2008. Heat Transfer. 8 ed. Pune: Nirali Prakashan.
Glicksman, L. R. L. V. J. H., 2016. Modeling and Approximation in Heat Transfer. New York:
Cambridge University Press.
Incropera, F. P., 2006. Fundamentals Of Heat And Mass Transfer. 5th ed. New Delhi: John
Willey & Sons (Asia) Pte. Ltd.
11

Janna, W. S., 2000. Engineering Heat Transfer. 2nd ed. Boco Raton: CRC Pess.
Khartchenko, N. V. K. V. V., 2014. Advance Energy System. 2nd ed. Boca Raton: CRC Press.
Maleshwar, M. T., 2009. Fundamentals of Heat and Mass Transfer. 2nd ed. Delhi: Dorling
Kindersley(India) Pvt. Ltd.
12