Answer:
Ceramics
Explanation:
Ceramics are used to make cooktops if you look it up and according to my engineering and design class ceramics have high heat resistance and melting points, and make good electrical insulators.
Well this question is though because we have never seen such a thing ! and to be quite frank when that happens , nothing good comes from it. Black Holes
Answer:
How must used oil storage containers be marked? Containers and aboveground tanks used to store used oil at generator facilities must be labeled or marked clearly with the words “Used Oil" (40 CFR Section 279.22(c)).
Explanation:
i think it will help you
Answer:
The average thickness of the blubber is<u> 0.077 m</u>
Explanation:
Here, we want to calculate the average thickness of the Walrus blubber.
We employ a mathematical formula to calculate this;
The rate of heat transfer(H) through the Walrus blubber = dQ/dT = KA(T2-T1)/L
Where dQ is the change in amount of heat transferred
dT is the temperature gradient(change in temperature) i.e T2-T1
dQ/dT = 220 W
K is the conductivity of fatty tissue without blood = 0.20 (J/s · m · °C)
A is the surface area which is 2.23 m^2
T2 = 37.0 °C
T1 = -1.0 °C
L is ?
We can rewrite the equation in terms of L as follows;
L × dQ/dT = KA(T2-T1)
L = KA(T2-T1) ÷ dQ/dT
Imputing the values listed above;
L = (0.2 * 2.23)(37-(-1))/220
L = (0.2 * 2.23 * 38)/220 = 16.948/220 = 0.077 m
Answer:
maximum allowable electrical power=4.51W/m
critical radius of the insulation=13mm
Explanation:
Hello!
To solve this heat transfer problem we must initially draw the wire and interpret the whole problem (see attached image)
Subsequently, consider the heat transfer equation from the internal part of the tube to the external air, taking into account the resistance by convection, and conduction as shown in the attached image
to find the critical insulation radius we must divide the conductivity of the material by the external convective coefficient
