In a planned experiment, a thermocouple is to be exposed to a step change in temperature. The response characteristics of the thermocouple must be such that the thermocouple's output reaches 98% of the final temperature within 5 s. Assume that the thermocouple's bead (its sensing element) is spherical with a density equal to 8000 kg/m3, a specific heat at constant volume equal to 380 J/(kg.K), and a convective heat transfer coefficient equal to 210 W/(m2.K). Determine the maximum diameter [] that the thermocouple can have and still meet the desired response characteristics. The unit is millimeter.
Answers
Answer:
max Diameter = 0.530 mm
Explanation:
Calculate the maximum Diameter that the thermocouple should have
applying this formula : e = [tex]\frac{SvCv}{hA}[/tex] ------ ( 1 )
mass = density * volume
Time constant = mc / hA
attached below is the detailed solution
r ( diameter ) = 0.530 mm
Related Questions
If it is struck by a rigid block having a weight of 550 lblb and traveling at 2 ft/sft/s , determine the maximum stress in the cylinder. Neglect the mass of the cylinder. Express your answer to three significant figures and include appropriate units.
Answers
This question is incomplete, The missing image is uploaded along this answer below;
Answer:
the maximum stress in the cylinder is 3.23 ksi
Explanation:
Given the data in the question and the diagram below;
First we determine the initial Kinetic Energy;
T = [tex]\frac{1}{2}[/tex]mv²
we substitute
⇒ T = [tex]\frac{1}{2}[/tex] × (550/32.2) × (2)²
T = 34.16149 lb.ft
T = ( 34.16149 × 12 ) lb.in
T = 409.93788 lb.in
Now, the volume will be;
V = [tex]\frac{\pi }{4}[/tex]d²L
from the diagram; d = 0.5 ft and L = 1.5 ft
so we substitute
V = [tex]\frac{\pi }{4}[/tex] × ( 0.5 × 12 in )² × ( 1.5 × 12 in )
V = 508.938 in³
So by conservation of energy;
Initial energy per unit volume = Strain energy per volume
⇒ T/V = σ²/2E
from the image; E = 6.48(10⁶) kip
so we substitute
⇒ 409.93788 / 508.938 = σ²/2[6.48(10⁶)]
508.938σ² = 5,312,794,924.8
σ² = 10,438,982.5967
σ = √10,438,982.5967
σ = 3230.9414
σ = 3.2309 ksi ≈ 3.23 ksi { three significant figures }
Therefore, the maximum stress in the cylinder is 3.23 ksi
