A copper sheet of dimensions 50 × 100 × 2 mm is exposed on both sides to an ambient fluid at 25^◦C with an average heat transfer coefficient of 7W/m^2^◦C. Is it proper to consider the copper sheet as a lumped system for transient analysis? If the answer is yes, determine the time constant of the fi

Question

A copper sheet of dimensions 50 × 100 × 2 mm is exposed on both sides to an ambient fluid at 25°C with an average heat transfer coefficient of [latex]7W/m^{2}°C[/latex]. Is it proper to consider the copper sheet as a lumped system for transient analysis? If the answer is yes, determine the time constant of the first-order system. If the initial temperature of the plate is 100°C how long does one have to wait for the temperature to become 50°C?

Accepted Answer

Properties of copper are taken from Table I.4. They are

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

Density: [latex]ρ = 8950 kg/m^{3}[/latex]
Specific heat: [latex]c = 385 J/kg°C[/latex]
Thermal conductivity: [latex]k = 386 W/m°C[/latex]

The characteristic length scale in the problem is semi plate thickness (see Table 3.1) given by [latex]L_{ch} = 0.001m[/latex]. The Biot number is calculated as

[latex]Bi=\frac{hL_{ch}}{k} =\frac{7 imes 0.001}{386}=1.813 imes 10^{-5}[/latex]

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

Since the Biot number is smaller than 0.1 it is indeed reasonable to treat the cooling process as a lumped first-order process. The time constant for the cooling process is obtained as

[latex] au =\frac{ ho cL_{ch} }{h} =\frac{8950 imes 385 imes 0.001}{7} =492.3s [/latex]

Other appropriate data specified in the problem are given below:
Initial temperature of copper plate: [latex]T_{0} = 100°C[/latex]
Ambient temperature: [latex]T_{∞ } = 25°C[/latex]
Final temperature of copper plate: [latex]T_{f} = 50°C[/latex]

We solve for t using Eq. 3.7 to get

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

[latex]t=- au imes \ln \left(\frac{T_{f}-T_{\infty } }{T_{0}-T_{\infty }} ight)=-492.3 imes \ln \left(\frac{50-25}{100-25} ight) =540.8s \approx 9 min[/latex]

Question 3.2: A copper sheet of dimensions 50 × 100 × 2 mm is exposed on b...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions