Answer:
a) Time it will taken for the column surface temperature to rise to 27°C is
17.1 hours
b) Amount of heat transfer is 5320 kJ
c) Amount of heat transfer until the surface temperature reaches 27°C is 4660 kJ
Explanation:
Given that;
Diameter D = 30 cm
Height H = 4m
heat transfer coeff h = 14 W/m².°C
thermal conductivity k = 0.79 W/m.°C
thermal diffusivity α = 5.94 × 10⁻⁷ m²/s
Density p = 1600 kh/m³
specific heat Cp = 0.84 Kj/kg.°C
a)
the Biot number is
Bi = hr₀ / k
we substitute
Bi = (14 W/m².°C × 0.15m) / 0.79 W/m.°C
Bi = 2.658
From the coefficient for one term approximate of transient one dimensional heat conduction The constants λ₁ and A₁ corresponding to this Biot number are,
λ₁ = 1.7240
A₁ = 1.3915
Once the constant J₀ = 0.3841 is determined from corresponding to the constant λ₁
the Fourier number is determined to be
[ T(r₀, t) -T∞ ] / [ Ti - T∞] = A₁e^(-λ₁²t') J₀ (λ₁r₀ / r₀)
(27 - 28) / (14 - 28) = (1.3915)e^-(17240)²t (0.3841)
t' = 0.6771
Which is above the value of 0.2. Therefore, the one-term approximate solution (or the transient temperature charts) can be used. Then the time it will take for the column surface temperature to rise to 27°C becomes
t = t'r₀² / ₐ
= (0.6771 × 0.15 m)² / (5.94 x 10⁻⁷ m²/s)
= 23,650 s
= 7.1 hours
Time it will taken for the column surface temperature to rise to 27°C is
17.1 hours
b)
The heat transfer to the column will stop when the center temperature of column reaches to the ambient temperature, which is 28°C.
Maximum heat transfer between the ambient air and the column is
m = pV
= pπr₀²L
= (1600 kg/m³ × π × (0.15 m)² × (4 m)
= 452.389 kg
Qin = mCp [T∞ - Ti ]
= (452.389 kg) (0.84 kJ/kg.°C) (28 - 14)°C
= 5320 kJ
Amount of heat transfer is 5320 kJ
(c)
the amount of heat transfer until the surface temperature reaches to 27°C is
(T(0,t) - T∞) / Ti - T∞ = A₁e^(-λ₁²t')
= (1.3915)e^-(1.7240)² (0.6771)
= 0.1860
Once the constant J₁ = 0.5787 is determined from Table corresponding to the constant λ₁, the amount of heat transfer becomes
(Q/Qmax)cyl = 1 - 2((T₀ - T∞) / ( Ti - T∞)) ((J₁(λ₁)) / λ₁)
= 1 - 2 × 0.1860 × (0.5787 / 1.7240)
= 0.875
Q = 0.875Qmax
Q = 0.875(5320 kJ)
Q = 4660 kJ
Amount of heat transfer until the surface temperature reaches 27°C is 4660 kJ
