Please verify the performance by doing the calculations • Material: AISI 304 Stainless Steel• Shape: Spherical (2 mm diameter)• Heat transfer coefficient: h = 2.2(V/D)0.5• Time to 0.99 A T: 18 seconds• Minimum flow velocity: 43 m/sVerify the performance of the existing thermocouple• Check that for a minimum velocity of 43 m/s you can reach 99% of the T. value at the center of thethermocouple in 18 seconds

Answered: • Material: AISI 304 Stainless Steel •…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

Please verify the performance by doing the calculations

• Material: AISI 304 Stainless Steel
• Shape: Spherical (2 mm diameter)
• Heat transfer coefficient: h = 2.2(V/D)0.5
• Time to 0.99 A T: 18 seconds
• Minimum flow velocity: 43 m/s
Verify the performance of the existing thermocouple
• Check that for a minimum velocity of 43 m/s you can reach 99% of the T. value at the center of the
thermocouple in 18 seconds

Expert Solution

Step 1

Calculate the heat transfer coefficient h using the given relation.

$h = 2.2 {(\frac{V}{D})}^{0.5}$

Here V is the velocity of flow, and D is the diameter of the spherical thermocouple.

Substitute 43 m/s for V and 2 mm for D.

$h = 2.2 {(\frac{43 m / s}{2 mm})}^{0.5} = 2.2 {(\frac{43 m / s}{2 mm (\frac{0.001 m}{1 mm})})}^{0.5} = 2.2 {(\frac{43 m / s}{0.002 m})}^{0.5} = 322.583 W / m^{2} \cdot ° C$

Calculate the characteristic length of the thermocouple by using the formula for characteristic length of the sphere.

$L_{c} = \frac{D}{6}$

Here, L_c is the characteristic length of the thermocouple.

Substitute 2 mm for D.

$L_{c} = \frac{2 mm}{6} = 0.33 mm = 0.33 mm (\frac{0.001 m}{1 mm}) = 0.33 \times 10^{- 3} m$

Write the required properties of AISI 304 stainless steel.

$k = 16.2 W / m \cdot ° C ρ_{g} = 8000 kg / m^{3}$

Here, k is the conduction heat transfer coefficient and $ρ_{g}$ is the density of the thermocouple.

Calculate the Biot number.

$B i = \frac{h L_{c}}{k}$

Here, Bi is the Biot number.

Substitute $322.583 W / m^{2} \cdot ° C$ for h, $0.33 \times 10^{- 3} m$ for L_c, and 16.2 W/m-^oC for k.

$B i = \frac{(322.583 W / m \cdot ° C) (0.33 \times 10^{- 3} m^{3})}{16.2 W / m \cdot ° C} = 0.006$

Since the Biot number is less than 0.1, the lumped heat analysis is valid.

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