A thin shell of aluminum of diameter 5mm and thickness 0.3mm drops off a conveyor vertically down to the ground. The temperature of the shell is initially 75^◦C. While it is falling to the ground that is some 15m below, the shell cools by losing heat to the ambient air at 20^◦C via a constant avera

Question

A thin shell of aluminum of diameter 5mm and thickness 0.3mm drops off a conveyor vertically down to the ground. The temperature of the shell is initially 75°C. While it is falling to the ground that is some 15m below, the shell cools by losing heat to the ambient air at 20°C via a constant average heat transfer coefficient of [latex]90 W/m^{2} °C[/latex]. Determine the temperature of the shell as it hits the ground.

Accepted Answer

With the usual notation, the given data is given in the table. The Biot number is then calculated as

[latex] au =\frac{ ho cL_{ch} }{h} =\frac{2700 imes 896 imes 0.0003}{90} =8.064s[/latex]

[latex]Bi=\frac{hL_{ch}}{k} =\frac{90 imes 0.0003}{236}=1.14 imes 10^{-4}[/latex]

Since the Biot number is less than 0.1 it is reasonable to use lumped formulation to solve the problem.

The first-order time constant may now be calculated using Eq. 3.3, to describe the temperature transient in the aluminum shell, as it falls to the ground.

[latex] au =\frac{ ho c}{h} imes \frac{V}{S} =\frac{ ho c}{h} imes L_{ch}[/latex] (3.3)

Characteristic length(see Table 3.1):[latex] L_{ch} = 0.3 mm = 0.0003 m[/latex]
Thermal conductivity of aluminum: [latex]k = 236 W/m°C[/latex]
Density of aluminum: [latex]ρ = 2700 kg/m^{3}[/latex]
Specific heat of aluminum: [latex]c = 896 J/kg°C[/latex]
Heat transfer coefficient: [latex]h = 90 W/m^{2}°C[/latex]
Height of fall of shell: [latex]s = 15 m[/latex]
Acceleration due to gravity: [latex]g = 9.8 m/s^{2}[/latex]
Note: Properties of aluminum are taken from Table I.4

Table 3.1 Characteristic length scale for a first-order system

Geometry	[latex]L_{ch}[/latex]
Large slab of thickness δ losing heat from both sides	δ/2
Long solid cylinder of diameter D	D/4
Solid sphere of diameter D	D/6
Long tube of small wall thickness δ << D losing heat from outside	δ
Spherical shell of small wall thickness δ << D losing heat from outside	δ

Table I.4 Thermal properties of metallic solids*

Metal	ρ [latex]kg/m^{3}[/latex]	C J/kg°C	k W/m°C	[latex]\alpha imes 10^{5} m^{2} /s[/latex]
Aluminum	2702	903	237	9.71
Alumel	8574	464	29.2	0.734
Brass	8470	377	116	3.63
Bronze	8830	377	52	1.36
Cast Iron (4% C)	7272	420	51	1.67
Chromel	8730	448	17.3	0.493
Copper	8900	385	21.7	6.33
Constantan	8900	390	22.7	0.654
Duralumin	8933	385	401	11.7
Gold	19300	129	317	12.7
Iron	7870	447	80.3	2.28
Nickel	8900	444	90.7	2.3
Platinum	21450	133	71.6	2.51
Silver	10500	235	429	17.4
Stainless Steel 304	7900	477	14.9	0.395
Tin	7310	227	66.6	4.03
Titanium	4500	522	21.9	0.932
Tungsten	19300	132	174	6.83

*Tabulated values are at room temperature and representative

Time taken by the shell to fall to the ground is now calculated. Motion of the shell is under gravity and it is assumed that the shell starts to fall with zero initial velocity. The time t of travel before the shell hits the ground is

[latex]t=\sqrt{\frac{2s}{g} }=\sqrt{\frac{2 imes 15}{9.8}}=1.75s[/latex]

The temperature of the shell at the end of this time is determined using the response of a first-order system to a step input.With initial temperature of shell of [latex]T_{0} = 75°C[/latex] and ambient temperature of [latex]T_{∞} = 20°C[/latex], temperature of the shell as it hits the ground is obtained using Eq. 3.7 as

[latex]\underbrace{\frac{T-T_{\infty } }{T_{0} -T_{\infty }}}_{ heta } =e^{-\frac{t}{ au } }[/latex] (3.7)

[latex]T=T_{\infty }+(T_{0} -T_{\infty })e^{-\frac{t}{ au } }=20+(75-20)e^{-\frac{1.75}{8.064 } }=64.3°C[/latex]

Question 3.1: A thin shell of aluminum of diameter 5mm and thickness 0.3mm...

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