Uploaded by Mustapha Melhaoui

Assignment Week 3

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Convective Heat Transfer
Assignment 3: Solution
Q. 1
A.
C.
Sol.
A fluid is flowing inside a circular tube of 5 cm radius. The maximum velocity of fluid
inside tubeis 0.3 m/s and velocity profile at any section along the axis of the tube is
parabolic. Determine the average fluid temperature in the tube if temperature profile is
given as: T(r) = 300+150(r/R)3. Consider temperature scale in Kelvin.
212.5 K
B. 106.25 K
167.14 K
D. 334.28 K
Given
Radius, R = 5 cm = 0.05 m
Maximum velocity, Umax = 0.3 m/s
Assume parabolic velocity profile.
Temperature profile, 𝑇 π‘Ÿ = 300 + 150
Determine
Average temperature
Solution
Given assume parabolic velocity profile,
π‘Ÿ 3
𝑅
So velocity profile in pipe flow, π‘ˆ π‘Ÿ = π‘ˆπ‘šπ‘Žπ‘₯ × 1 −
π‘ˆ
π‘Ÿ 2
𝑅
We know, average velocity, π‘ˆπ‘Žπ‘£π‘” = π‘šπ‘Žπ‘₯
2
𝑅
2
π‘‡π‘Žπ‘£π‘” =
𝑇 π‘Ÿ × π‘ˆ π‘Ÿ × π‘Ÿ × π‘‘π‘Ÿ
π‘ˆπ‘Žπ‘£π‘” × π‘… 2 0
𝑅=0.05
2
π‘Ÿ 3
π‘Ÿ
π‘‡π‘Žπ‘£π‘” =
300 + 150
× 0.3 × 1 −
2
0.15 × π‘… 0
𝑅
𝑅
π‘‡π‘Žπ‘£π‘” = 334.28 𝐾
Q. 2
A.
C.
Sol.
2
× π‘Ÿ × π‘‘π‘Ÿ
Water enters inside a tube of closed feed water heater having 5 cm diameter at 30 °C at a
rate of 0.25kg/s and is heated by the saturated steam withdrawn from high pressure
turbine at 150 °C. Determine the length of the tube required to heat the water to 130 °C
if the average heat transfer coefficient is 1200 W/m2-K. Properties of water at mean
temperature are: ρ = 970.87 kg/m3, μ = 343x10-6 N-s/m2, k = 0.671 W/m-K, Pr = 2.14.
9.95 m
B. 7.93 m
4.63 m
D. 2.76 m
Given
Diameter, D = 5 cm = 0.05 m
Inlet temperature, Ti = 30 ˚C = 303 K
Surface temperature, TS = 150 ˚C = 423 K
Mass flow rate, m = 0.25 kg/s
Average heat transfer coefficient, havg = 1200 W/m2K
Properties at mean temperature are given.
Determine
Length of tube.
Solution
For constant temperature cases
−β„Žπ‘Žπ‘£π‘” 𝐴𝑠
πœƒπ‘’
= 𝑒π‘₯𝑝
πœƒπ‘–
π‘šπ‘π‘
−β„Žπ‘Žπ‘£π‘” 𝐴𝑠
𝑇𝑠 − 𝑇𝑒
= 𝑒π‘₯𝑝
𝑇𝑠 − 𝑇𝑖
π‘šπ‘π‘
423 − 403
−1200 × πœ‹ × 0.05 × πΏ
= exp
0.671 ×2.14
423 − 303
0.25 × 343 × 10 −6
𝐿 = 9.95 π‘š
Q. 3
A.
C.
Sol.
Water is heated inside the circular tube of 5 cm internal diameter and 7 m length from
20 °C to 75 °C. An electric resistance heater is employed around the periphery of the
tube that provides uniform heating. The outer surface of the heater is well insulated such
that all the heat is transferred to water. Determine the power rating of resistance heater if
the water flows inside the tube at a rate of 0.1 kg/s. Properties of water at mean
temperature are: ρ = 989.12 kg/m3, μ = 577x10-6 N-s/m2, k = 0.640 W/m-K, Pr = 3.77.
2.3 kW
B. 23 kW
2300 kW
D. 230 kW
Given
Diameter, D = 5 cm = 0.05 m
Length, L = 7 m
Inlet temperature, Ti = 20 ˚C
Exit temperature, Te = 75 ˚C
Mass flow rate, m = 0.1 kg/s
Properties are given at mean temperature.
Determine
Pumping rating.
Solution
Heat transfer rate,
π‘ž = π‘š 𝐢𝑝 βˆ†π‘‡
0.64 × 3.77
π‘ž = 0.1 ×
75 − 20
577 × 10−6
π‘ž = 23 π‘˜π‘Š
Q. 4
A.
C.
Sol.
Also, calculate the surface temperature of the tube at exit given in Q. 3 if the Nusselt
number is estimated using the Dittus-Boelter correlation as Nu = 0.023Re0.8Pr0.4.
B. 125.72 °C
398.72 °C
D.
80.07 °C
98.22 °C
Given
Same as above in Q3.
Determine
Surface temperature at exit.
Solution
4π‘š
= 4413.3086
πœ‹πœ‡π·
𝑁𝑒 = 0.023 𝑅𝑒 0.8 π‘ƒπ‘Ÿ 0.4
𝑅𝑒 =
β„Ž × 0.05
= 0.023 4413.3086 0.8 3.77 0.4
0.640
β„Ž = 412.36878 π‘Š/π‘š2 𝐾
π‘ž = β„Žπ΄π‘  𝑇𝑠 − 𝑇𝑒 𝑒π‘₯𝑖𝑑 = 412.36878 × πœ‹ × 0.05 × 7 × π‘‡π‘  − 75
𝑇𝑠 = 125.72 °πΆ
𝑁𝑒 =
Q. 5
A.
C.
Sol.
Air flows inside a 10 m long square duct of 0.25 m side at 70 °C and atmospheric
pressure. The duct passes through a room which is maintained at 40 °C. Consider the
surface of duct attains isothermal conditions at 40 °C when air flows at a rate of 0.12
m3/s. Determine the temperature of air at the exit of duct. Estimate average Nusselt
number using Dittus-Boelter correlation as Nu = 0.023Re0.8Pr0.3. Use hydraulic diameter
4𝐴
of duct, π·β„Ž = 𝑝 𝑐 as length scale to calculate Reynolds number. Here, AC and p are the
cross-sectional area and perimeter of duct cross-section, respectively. Use following
properties of air: ρ = 1.023 kg/m3, μ = 200.2x10-7 N-s/m2, k = 0.0297 W/m-K, Pr =
0.701.
B. 61.32 °C
62.5 °C
D. 56.06 °C
57.47 °C
Given
Side, a = 0.25 m
Length, L = 10 m
Inlet temperature, Ti = 70 ˚C = 343 K
Constant Surface temperature, TS = 40 ˚C = 313 K
Volumetric flow rate, 𝑉 = 0.12 m3/s
Determine
Temperature at exit of duct.
Solution
𝑉 = π΄π‘ˆπ‘Žπ‘£π‘”
0.12
π‘ˆπ‘Žπ‘£π‘” =
= 1.92 π‘š/𝑠
(0.25 × 0.25)
4𝐴
4(0.25×0.25)
Hydraulic Diameter, π·β„Ž = 𝑐 =
= 0.25 π‘š
𝑃
2(0.25+0.25)
πœŒπ‘ˆπ‘Žπ‘£π‘” π·β„Ž 1.023 × 1.92 × 0.25
𝑅𝑒 =
=
= 24527.47253
πœ‡
200.2 × 10−7
𝑁𝑒 = 0.023 𝑅𝑒 0.8 π‘ƒπ‘Ÿ 0.3
β„Ž × 0.25
𝑁𝑒 =
= 0.023 24527.47253 0.8 0.701 0.3
0.0297
β„Ž = 7.98 π‘Š/π‘š2 𝐾
For constant temperature cases:
πœƒπ‘’
−β„Ž 𝐴𝑠
= 𝑒π‘₯𝑝
πœƒπ‘–
π‘šπ‘π‘
𝑇𝑠 − 𝑇𝑒
−β„Ž 𝐴𝑠
= 𝑒π‘₯𝑝
𝑇𝑠 − 𝑇𝑖
π‘šπ‘π‘
40 − 𝑇𝑒
= exp
40 − 70
−7.98 × 2(0.25 + 0.25) × 10
1.023 × 0.12 ×
𝑇𝑒 = 56.06 °πΆ
0.0297 ×0.701
200.2 × 10 −7
Q. 6
A.
C.
Sol.
Air flows through an unisulated circular duct of 12 cm diameter at a mass flow rate of
0.03 kg/s. Air enters the duct at 95 °C and leaves the duct at a temperature of 70 °C.
Length of the duct is 7 m and it is exposed to ambient at 3 °C. Determine the surface
temperature of the duct at exit if heat transfer coefficient between duct surface and
ambient is 5 W/m2-K. Use Dittus-Boelter correlation as Nu = 0.023Re0.8Pr0.3 to calculate
the heat transfer coefficient between duct surface and inside air. Properties of air at
mean temperature are: ρ = 0.993 kg/m3, μ = 210x10-7 N-s/m2, k = 0.0304 W/m-K, Pr =
0.699.
B. 56.7 °C
82.5 °C
D. 42.75 °C
49.78 °C
Given
Diameter, D = 12 cm = 0.12 m
Length, L = 7 m
Inlet temperature, Ti = 95 ˚C
Exit temperature, Te = 70 ˚C
Mass flow rate, π‘š = 0.03 kg/s
Determine
Surface temperature at exit.
Solution
4π‘š
4 × 0.03
𝑅𝑒 =
=
= 15157.6
πœ‹ πœ‡ 𝐷 210 × 10−7 × πœ‹ × 0.12
𝑁𝑒 = 0.023 𝑅𝑒 0.8 π‘ƒπ‘Ÿ 0.3
β„Žπ‘– × 0.12
𝑁𝑒 =
= 0.023 15157.6 0.8 0.699 0.3
0.0304
β„Žπ‘– = 11.57 π‘Š/π‘š2 𝐾
Apply energy balance at wall,
Energy taken by wall from hot fluid = Energy released by wall to surrounding
β„Žπ‘– 𝐴𝑠 𝑇𝑒 − 𝑇𝑠 = β„Žπ‘œ 𝐴𝑠 (𝑇𝑠 − 𝑇∞ )
11.57 343 − 𝑇𝑠 = 5 𝑇𝑠 − 276
𝑇𝑠 = 322.78 𝐾 = 49.78 °πΆ
Q. 7
Water enters inside a circular duct of 12 cm diameter and 7 m length at a temperature of
20 °C. The initial 2 m long section of the duct is maintained at ambient temperature of
20 °C. Remaining section of the duct is wrapped with and electric resistance heater
which maintains the duct surface at constant temperature of 60 °C. Flow is assumed to
be hydrodynamically fully developed before it enters the heated section of the duct.
Determine the exit temperature of water, neglecting the axial conduction between heated
and unheated portion of the tube if mass flow rate of water is 0.1 kg/s. Use Edwards et
al. (1979) correlation to calculate the heat transfer coefficient: 𝑁𝑒 = 3.66 +
0.065
1+0.04
A.
C.
Sol.
𝐷
π‘…π‘’π‘ƒπ‘Ÿ
𝐿
2 3
𝐷
π‘…π‘’π‘ƒπ‘Ÿ
𝐿
. Use following properties of water: ρ = 997 kg/m3, μ = 855x10-6 N-
s/m2, k = 0.613 W/m-K, Pr = 5.83.
28.79 °C
26.50 °C
Given
Diameter, D = 12 cm = 0.12 m
Length, L = 7 m
B. 27.26 °C
D. 29.78 °C
Inlet temperature, Ti = 20 ˚C
Mass flow rate, m = 0.1 kg/s
Determine
Exit temperature.
Solution
π‘š = πœŒπ΄π‘‰
0.1
𝑉=
πœ‹
997 × 4 × (0.12)2
𝑉 = 8.87 × 10−3 π‘š/𝑠
4π‘š
𝑅𝑒 =
= 1240.974
πœ‹πœ‡π·
𝐷
0.065 𝐿 𝑅𝑒 π‘ƒπ‘Ÿ
𝑁𝑒 = 3.66 +
𝐷
1 + 0.04 𝐿 𝑅𝑒 π‘ƒπ‘Ÿ
0.12
5
0.065
𝑁𝑒 = 3.66 +
0.12
5
1 + 0.04
π‘π‘’π‘˜
𝐷
2
3
× 1240.974 × 5.83
× 1240.974 × 5.83
2
3
= 8.69
 β„Ž=
= 44.39 π‘Š/π‘š2 𝐾
For constant temperature cases
πœƒπ‘’
−β„Ž 𝐴𝑠
= 𝑒π‘₯𝑝
πœƒπ‘–
π‘šπ‘π‘
𝑇𝑠 − 𝑇𝑒
−β„Ž 𝐴𝑠
= 𝑒π‘₯𝑝
𝑇𝑠 − 𝑇𝑖
π‘šπ‘π‘
60 − 𝑇𝑒
−44.39 × πœ‹ × 0.12 × 5
= exp
0.613 ×5.83
60 − 20
0.1 × 855 × 10 −6
𝑇𝑒 = 27.26 ˚𝐢
Q. 8
A.
C.
Sol.
Engine oil flows through a circular tube of diameter 100 mm and 10 m length whose
surface is maintained at uniform heat flux of 200 W/m2. Determine the temperature of
the tube surface at exit if the inlet temperature and mass flow rate of engine oil are 25 °C
and 0.1 kg/s. Use following properties of engine oil: ρ = 868.8 kg/m3, μ = 11.36x10-2 Ns/m2, k = 0.142 W/m-K, Pr = 1585.
B. 28.17 °C
52.86 °C
D. 60.48 °C
66.65 °C
Given
Diameter, D = 100 mm = 0.1 m
Heat flux, q = 200 W/m2
Inlet temperature, Ti = 25 ˚C = 298 ˚K
Determine
Exit temperature.
Solution
Mass flow rate, m = 0.3 Kg/s
π‘š = πœŒπ΄π‘‰
𝑉=
0.1
868.8 ×
πœ‹
4
× 0.1
2
= 0.0146 π‘š 𝑠
π‘ž = π‘šπ‘π‘ (𝑇𝑒 − 𝑇𝑖 )
0.142 × 1585
200 × πœ‹ × 0.1 × 10 = 0.1 ×
× (𝑇𝑒 − 25)
11.36 × 10−2
𝑇𝑒 = 28.17 ˚𝐢
πœŒπ‘‰π·
868.8×0.0146 ×.1
Reynolds number, 𝑅𝑒 =
=
= 11.16
−2
πœ‡
11.36 × 10
οƒ  Flow is laminar, therefore Nusselt number, Nu = 4.36 (Constant heat flux condition)
π‘π‘’π‘˜
and β„Ž = 𝐷 = 6.19 π‘Š/π‘š2 𝐾
Applying energy balance at tube exit, π‘ž = β„Ž(𝑇𝑠 − 𝑇𝑒 ) οƒ  200 = 6.19 × (𝑇𝑠 − 28.17)
οƒ  Temperature of tube surface at exit, 𝑇𝑠 = 60.48 ˚𝐢
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