A slab of Aluminium 5 cm thick initially at200°C is suddenly immersed in a liquid at70°C for which the convection heat transferco-efficient is 525 W/m2K. Determine thetemperature at a depth of 12.5 mm from oneof the faces 1 minute after the immersion.Also calculate the energy removed per unitarea from the plate during 1 minute ofimmersion. Take P= 2700 bar,с. 0.9 kJ/kg. k 215 W/mK,a' = 8.4 x 10^-5 m/s.
Answers
Answer:
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Answer:
A long aluminium cylinder 5.0 cm in diameter and initially at 200°C is suddenly exposed to a convection environment at 70°C with heat transfer coefficient of 525 W/m^2.K.525W/m
2
.K. Calculate the temperature at the radius of 1.25 cm 1 minute after the cylinder exposed to the environment.
Step-by-Step
Solution
Report Solution
Given : A long cylinder
D = 5.0 cm, T_i= 200°cm,D=5.0cm,T
i
=200°cm,
T_∞ = 70°C, h = 525 W/m^2.K,T
∞
=70°C,h=525W/m
2
.K,
r = 1.25 cm, t = 1 min = 60 s.
To find : The temperature at the radius of 1.25 cm in the cylinder.
Assumptions :
(i) No radiation exchange
(ii) The physical properties for aluminium cylinder as
ρ = 2700 kg/m^3, C = 900 J/kg.K, k = 210 W/m.Kρ=2700kg/m
3
,C=900J/kg.K,k=210W/m.K
Analysis : Since the position temperature is to determine, thus using Heisler charts. The radius of cylinder
r_o = \frac{D}{2}= \frac{0.05 m}{2}= 0.025 mr
o
=
2
D
=
2
0.05m
=0.025m
Biot number
Bi = \frac{hr_o}{k} = \frac{525 × 0.025}{210} = 0.0625Bi=
k
hr
o
=
210
525×0.025
=0.0625
\frac{1}{Bi} = 16
Bi
1
=16
Fourier number
Fo =\frac{αt}{r_o^2} =\left ( \frac{k}{\rho C} \right )\frac{t}{r_o^2}Fo=
r
o
2
αt
=(
ρC
k
)
r
o
2
t
\left ( \frac{210}{2700 ×900} \right )\frac{60}{(0.025)^2}= 8.29(
2700×900
210
)
(0.025)
2
60
=8.29
The dimensionless centre temperature from Heisler chart, Fig. 6.32 (a)
\frac{T_{c} – T_{\infty }}{T_{i} – T_{\infty }} = 0.35
T
i
–T
∞
T
c
–T
∞
=0.35
∴ T_c= 70 + 0.35 × (200 – 70) = 115.5°CT
c
=70+0.35×(200–70)=115.5°C
The dimensionless position
\frac{r}{r_o} = \frac{1.25 cm}{2.5 cm} = 0.5
r
o
r
=
2.5cm
1.25cm
=0.5
\frac{1}{Bi}= 16
Bi
1
=16
From Fig. 6.32 (b)
\frac{T_{c} – T_{\infty }}{T_{i} – T_{\infty }}= 0.98
T
i
–T
∞
T
c
–T
∞
=0.98
T = 70 + 0.98 × (115.5 – 70)
= 114.59°C. A
Alternatively
Since Biot number is much less than 0.1, thus this problem can also be solved by using the lumped system analysis, eqn. (6.10)
δ = \frac{r_o}{2}= 0.0125 mδ=
2
r
o
=0.0125m
\frac{T_{c} – T_{\infty }}{T_{i} – T_{\infty }}=exp\left [ -\frac{ht}{ρ δ C} \right ]
T
i
–T
∞
T
c
–T
∞
=exp[−
ρδC
ht
]
T = 70 + (200 – 70) × exp ×\left [ -\frac{525 ×60}{2700 ×900 ×0.0125} \right ]T=70+(200–70)×exp×[−
2700×900×0.0125
525×60
]
= 70 + 130 × 0.354 = 116°C.
It is temperature in the cylinder with error of 1.2% only.