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Question

A spherical thermocouple junction of diameter 0.706 mm is to be used for the measurement of temperature of a gas stream. The convective heat transfer coefficient on bead surface is 400 W/m2K. Thermophysical properties of thermocouple material are k=20 W/mK, c=400 J/kgK and ρ=8500 kg/m3. If the thermocouple initially at 30 C is placed in a hot stream of 300 C, the time taken by the bead to reach 298 C, is

A

2.35 s
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B

4.9 s
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C

14.7 s
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D

29.4 s
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Solution

The correct option is B
4.9 s
Given data:

h=400 W/m2K: k=20 W/mK: c=400 J/kgKρ=8500 kg/m3: T(t)=298C
Now, L=VA=1/6πD3πD2=D6=0.706×1036=1.76×104m

Biot number,
Bi=hLk=400×1.76×10420 =0.0023Since Bi<0.1

Lumped system analysis can be used
TTTiT=ehAtρcV
29830030300=e400×6×t8500×0.706×103×400

7.407×103=et
Taking loge both sides, we get
loge 7.407×1=t
or t=4.90 seconds

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