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2 years ago

11

State five applications of thermochromic materials

1 answer:

rusak2 [61]2 years ago

8 0

Explanation:

The end-use industries of thermochromic materials include packaging, printing & coating, medical, textile, industrial, and others. Printing & coating is the fastest-growing end-use industry of thermochromic materials owing to a significant increase in the demand for thermal paper for POS systems. The use of thermochromic materials is gaining momentum for interactive packaging that encourages consumers to take a product off the shelf and use it.

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Consider a plane composite wall that is composed of two materials of thermal conductivities kA = 0.1 W/m*K and kB = 0.04 W/m*K a

nadya68 [22]

Answer:

q=39.15 W/m²

Explanation:

We know that

Thermal resistance due to conductivity given as

R=L/KA

Thermal resistance due to heat transfer coefficient given as

R=1/hA

Total thermal resistance

$R_{th}=\dfrac{L_A}{AK_A}+\dfrac{L_B}{AK_B}+\dfrac{1}{Ah_1}+\dfrac{1}{Ah_2}+\dfrac{1}{Ah_3}$

Now by putting the values

$R_{th}=\dfrac{0.01}{0.1A}+\dfrac{0.02}{0.04A}+\dfrac{1}{10A}+\dfrac{1}{20A}+\dfrac{1}{0.3A}$

$R_{th}=4.083/A\ K/W$

We know that

Q=ΔT/R

$Q=\dfrac{\Delta T}{R_{th}}$

$Q=A\times \dfrac{200-40}{4.086}$

So heat transfer per unit volume is 39.15 W/m²

q=39.15 W/m²

4 0

2 years ago

A manometer containing a fluid with a density of 60 lbm/ft3 is attached to a tank filled with air. If the gage pressure of the a

8090 [49]

Answer:

The fluid level difference in the manometer arm = 22.56 ft.

Explanation:

Assumption: The fluid in the manometer is incompressible, that is, its density is constant.

The fluid level difference between the two arms of the manometer gives the gage pressure of the air in the tank.

And P(gage) = ρgh

ρ = density of the manometer fluid = 60 lbm/ft³

g = acceleration due to gravity = 32.2 ft/s²

ρg = 60 × 32.2 = 1932 lbm/ft²s²

ρg = 1932 lbm/ft²s² × 1lbf.s²/32.2lbm.ft = 60 lbf/ft³

h = fluid level difference between the two arms of the manometer = ?

P(gage) = 9.4 psig = 9.4 × 144 = 1353.6 lbf/ft²

1353.6 = ρg × h = 60 lbf/ft³ × h

h = 1353.6/60 = 22.56 ft

A diagrammatic representation of this setup is presented in the attached image.

Hope this helps!

5 0

3 years ago

He is going ___ in the hot air ballon

Vladimir [108]

no artical shoul be used here

5 0

2 years ago

Which task best fits the role of a planning engineer?

Nonamiya [84]

Answer:

D

Explanation:

ensuring project end on time through carefully planning and organizing

8 0

2 years ago

A 4-stroke Diesel engine with a displacement of Vd = 2.5x10^-3m^3 produces a mean effective pressure of 6.4 bar at the speed of

yKpoI14uk [10]

Answer:

The power developed by engine is 167.55 KW

Explanation:

Given that

$V_d=2.5\times 10^{-3} m^3$

Mean effective pressure = 6.4 bar

Speed = 2000 rpm

We know that power is the work done per second.

So

$P=6.4\times 100\times 2.5\times 10^{-3}\times \dfrac{2\pi \times2000}{120}$

We have to notice one point that we divide by 120 instead of 60, because it is a 4 cylinder engine.

P=167.55 KW

So the power developed by engine is 167.55 KW

4 0

3 years ago