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Stels [109]
3 years ago
11

Define the terms (a) thermal conductivity, (b) heat capacity and (c) thermal diffusivity

Engineering
1 answer:
IceJOKER [234]3 years ago
6 0

Explanation:

<u>(a)</u>

<u>The measure of material's ability to conduct thermal energy (heat) is known as thermal conductivity.</u> For examples, metals have high thermal conductivity, it means that they are very efficient at conducting heat.<u> The SI unit of heat capacity is W/m.K.</u>

The expression for thermal conductivity is:

q=-\kappa \bigtriangledown T

Where,

q is the heat flux

\kappa is the thermal conductivity

\bigtriangledown T is the temperature gradient.

<u>(b)</u>

<u>Heat capacity for a substance is defined as the ratio of the amount of energy required to change the temperature of the substance and the magnitude of temperature change. The SI unit of heat capacity is J/K.</u>

The expression for Heat capacity is:

C=\frac{E}{\Delta T}

Where,

C is the Heat capacity

E is the energy absorbed/released

\Delta T is the change in temperature

<u>(c)</u>

<u>Thermal diffusivity is defined as the thermal conductivity divided by specific heat capacity at constant pressure and its density. The Si unit of thermal diffusivity is m²/s.</u>

The expression for thermal diffusivity is:

\alpha=\frac{\kappa}{C_p \times \rho}

Where,

\alpha is thermal diffusivity

\kappa is the thermal conductivity

C_p is specific heat capacity at constant pressure

\rho is density

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Section lines represent surfaces exposed by a cutting plane<br> a. True b. False
oksian1 [2.3K]

Answer:

these is very interesting question but I still know the answer is a. True

7 0
3 years ago
2. Using an RLC circuit to design a band-reject filter, with the two cut off frequencies 40k and 90k Hz. Design the values of R,
nadya68 [22]

Answer:

C = 59.17 nF

Q = 2.6

Explanation:

given data

frequencies = 40k Hz

frequencies  = 90k Hz

solution

we take here R, L C take in series

so cut off frequency is express as

Wc1 = -\frac{R}{2L}+ \sqrt{(\frac{R}{2L})^2+\frac{1}{LC}}   =  40000

wc2 = \frac{R}{2L}+ \sqrt{(\frac{R}{2L})^2+\frac{1}{LC}}    =  90000

so here

wc2 - wc1 will be

wc2 - wc1 = 90000 - 40000 = 50000

so

\frac{R}{L} =  50000

we consider here R is 500

so  L = \frac{500}{50000}  

L = 10 m H

and here total cut off frequency is

total cut off frequency = 40000 + 90000 = 130000

so capacitance will be

capacitance  C is =  \frac{1}{10\times 10^{-3}\times 130000^2}  

so C = 59.17 nF

quality factor Q will be

Q = \frac{130000}{50000}  

Q = 2.6

4 0
3 years ago
A horizontal opaque flat plate is well insulated on the edges and the lower surface. The top surface has an area of 8 m2 and it
Oliga [24]

Answer:

a = 0.8333

p = 0.1667

ε = 0.6243        

Explanation:

Given:

- Solar Irradiation G = 6000 W

- Irradiation absorbed G_abs = 5000 W

- Heat Loss by convection Q_convec = 750 W

- The temperature of surface remains constant @ 350 K

Find:

Determine the absorptivity, reflectivity, and emissivity of the plate

Solution:

- Absorptivity is the ratio of energy absorbed to the incident energy given as:

                                        a = G_abs / G

- Plug in values:              a = 5000 / 6000

                                        a = 0.8333

- Reflectivity is the ratio of energy not absorbed to the incident energy given as:

                                          p = (1 - G_abs) / G

- Plug in values:                p = 1000 / 6000

                                          p = 0.1667

- Emissivity is the of ratio amount energy that is radiated from the body to the black body:

- We will use energy balance on the plate surface:

                                          E_in - E_out = -k*A*dT / L

- We know that the plate temperature remains constant, dT = 0:

                                          E_in = E_out

                                          Q_abs = Q_convec + Q_rad

                                          Q_rad = Q_abs - Q_convec

- Plug in values:                Q_rad = 5000 -  750 = 4250 W

- The expression for surface radiation is given by:

                                           Q_rad = ε*A*б*T_b^4

- Re-arrange:                     ε = Q_rad / A*б*T_b^4  

- Plug values in:                ε = 4250 / 8*(5.67*10^-8)*(350)^4    

                                           ε = 0.6243        

4 0
3 years ago
Describe what you have been taught about the relationship between basic science research, and technological innovation before th
alexira [117]

Answer:

With the Breakthrough of Technology, the rate at which things are done are becoming much more easy. but without basic science, innovation towards technology cannot occur, so the both work hand in hand in the world of technology today.

Explanation:

Technological innovation and Basic science research plays a major role in the world of science and technology today, while we all want technology innovation the more, without basic science, innovation cannot come in place,

Just as we are going further in technology, breakthroughs and growth are been made which helps on the long run in science research which in turn has made things to be done much better and easily.

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3 years ago
What is an example of chemical electrical transformation
Stella [2.4K]

Answer:

stove

Explanation:

Chemical Energy is converted to Electrical Energy (stove)  this is because electricity is taking process and the chemical are going together to create heat.

4 0
3 years ago
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