Estimate (a) the maximum, and (b) the minimum thermal conductivity values (in W/m-K) for a cermet that contains 76 vol% carbide particles in a metal matrix. Assume thermal conductivities of 30 and 67 W/m-K for the carbide and metal, respectively.

Answers

Answer 1

Answer:

The answer is below

Explanation:

Given that:

Volume of carbide ([tex]V_{C}[/tex]) = 76% = 0.76, Volume of Nickel ([tex]V_{M}[/tex]) = 100% - 76% = 24% = 0.24, thermal conductivities of carbide ([tex]E_{C}[/tex]) = 30 W/m-K and thermal conductivities of meta. ([tex]E_M[/tex]) = 67 W/m-K

a) The maximum thermal conductivity is given by:

Max = [tex]E_CV_c+E_mV_m=(0.76*30) + (0.24*67) = 38.88\ W/mK[/tex]

b) The minimum thermal conductivity is given by:

Min = [tex]\frac{E_ME_C}{E_MV_C+E_CV_M}=\frac{30*67}{(0.76*67)+(0.24*30)}=34.58\ W/mK[/tex]


Related Questions

Which of the following terms describes the path from an electrical source to a switch or plug?
transmitter
circuit breaker
raceway
breaker panel

Answers

Answer:

transmitter hope thus helped!

Explanation:

Raceway is the answer

"A raceway is an enclosed conduit that forms a physical pathway for electrical wiring."


Support with three reasons the decision to use a plastic material for the package in the following
scenario.
Situation: A client has hired Jose, a materials engineer, to develop a package for an item he has begun
to market. The object needs to be mailed to customers within three days of being ordered.

Answers

Answer:

its durable. it's cheap. its recyclable

Explanation:

Plastic is made of lots of recycled materials that make it very useful and cheap.

An astronomer of 65 kg of mass hikes from the beach to the observatory atop the mountain in Mauna Kea, Hawaii (altitude of 4205 m). By how much (in newtons) does her weight change when she goes from sea level to the observatory?

Answers

Answer:

[tex]0.845\ \text{N}[/tex]

Explanation:

g = Acceleration due to gravity at sea level = [tex]9.81\ \text{m/s}^2[/tex]

R = Radius of Earth = 6371000 m

h = Altitude of observatory = 4205 m

Change in acceleration due to gravity due to change in altitude is given by

[tex]g_h=g(1+\dfrac{h}{R})^{-2}\\\Rightarrow g_h=9.81\times(1+\dfrac{4205}{6371000})^{-2}\\\Rightarrow g_h=9.797\ \text{m/s}^2[/tex]

Weight at sea level

[tex]W=mg\\\Rightarrow W=65\times 9.81\\\Rightarrow W=637.65\ \text{N}[/tex]

Weight at the given height

[tex]W_h=mg_h\\\Rightarrow W_h=65\times 9.797\\\Rightarrow W_h=636.805\ \text{N}[/tex]

Change in weight [tex]W_h-W=636.805-637.65=-0.845\ \text{N}[/tex]

Her weight reduces by [tex]0.845\ \text{N}[/tex].

How many flip-flop values are complemented in an 8-bit binary ripple counter to reach the next count value after: 0110111 and 01010110?

Answers

Answer:

- Four (4) flip-flop values will complemented

- one (1) flip-flop value will complemented

Explanation:

To find how many flip flop number of bits complemented, we just need to figure out what the next count in the sequence is and find how many bits have changed.

taking a look at the a) 00110111

we need to just 1 to the value,

so

00110111 +  0000001  = 00111000        

So here, only the first four bits are complemented.

Therefore Four (4) flip-flop values will complemented

Next

b) 01010110

we also add 1 to the value

01010110  + 00000001  = 01010111

only the first bit is complemented.

Therefore one (1) flip-flop value will complemented

Determine the size of memory needed for CD recording of a piece of music, which lasts for 26 minutes, is done with a 20-bit Analog-to-Digital Converter (ADC) in stereo (2 channels), at the rate of 44.1 kSa/s, with the compression factor 6 (allow 10% error margin).

Answers

Answer: the size of memory needed for the CD recording is 28.7 MB

Explanation:

so in the case of stereo, the bitrate is;

⇒ 26 × 60 × 44.1 × 10³ × 2

=   137592 × 10³  

for 10 bit

⇒ 137592 × 10³ × 10

= 1375920 × 10³ bits

now divide by 8 (convert to bytes)

⇒ (1375920 × 10³) / 8

= 171,990,000 BYTE

divide by 1000 (convert to kilobytes)

= 171,990,000 / 1000

= 171,990 KILOBYTES

now Given that, the compression ratio is 6      

so  

171,990 / 6

= 28665 KB

we know that. 1 MB = 1000 KB

x MB = 28665 KB

x MB = 28665 / 1000

⇒ 28.665 MB ≈ 28.7 MB

Therefore the size of memory needed for the CD recording is 28.7 MB

In a compression test, a steel test specimen (modulus of elasticity 30 106 lb/in2 ) has a starting height 2.0 in and diameter 1.5 in. The metal yields (0.2% offset) at a load 140,000 lb. At a load of 260,000 lb, the height has been reduced to 1.6 in. Determine (a) yield strength and (b) fl ow curve parameters (strength coeffi cient and strain-hardening exponent). Assume that the cross-sectional area increases uniformly during the test.

Answers

Answer:

A) σ_y = 79096 lb/in² = 79.1 ksi

B) strain-hardening exponent = 0.102

(strength coefficient = 137838.78 lb/in²

Explanation:

A) Formula for volume is;

V = πd²h/4

We are given;

height 2.0 in and diameter 1.5 in

Thus;

V = (π × 1.5² × 2)/4

V = 3.53 in³

Area is;

A = πd²/4

A = (π × 1.5²)/4

A = 1.77 in²

Yield strength is gotten from the formula;

σ_y = Force/Area

We are given load = 140,000 lb

Thus;

σ_y = 140000/1.77

σ_y = 79096 lb/in²

B) We are given

modulus of elasticity: E = 30 × 10^(6) lb/in²

Formula for strain is;

ε = σ_y/E

ε = 79096/(30 × 10^(6))

ε = 0.00264

The metal yields (0.2% offset), thus;

strain offsets = 0.00264 + 0.002

strain offsets: ε1 = 0.00464

Thus;

(h_i - h_o)/h_o = 0.00464

(h_i/h_o) - 1 = 0.00464

(h_i/h_o) = 1.00464

h_i = h_o(1.00464)

h_o = 2 in

Thus; h_i = 2(1.00464) = 2.00928 in

Area = Volume/height = 3.53/2.00928 = 1.757 in²

True stress is;

σ = force/area = 140000/1.757

σ1 = 79681.27 lb/in²

At a load of 260,000 lb, the height has been reduced to 1.6 in. Thus;

Area = 3.53/1.6 = 2.206 in²

True stress is;

σ2 = 260000/2.206

σ2 = 117860.38 lb/in²

True strain;

ε2 = In(2/1.6)

ε2 = 0.223

From flow curve;

σ = kεⁿ

Thus;

σ1 = k(ε1)ⁿ

79681.27 = k(0.00464ⁿ) - - - (eq 1)

Also for σ2 = k(ε2)ⁿ;

117860.38 = k(0.223ⁿ) - - - - - (eq 2)

From eq 1,

k = 79681.27/0.00464ⁿ

Putting this for k in eq2 to get;

117860.38 = (0.223ⁿ) × 79681.27/0.00464ⁿ

117860.38/79681.27 = 0.223ⁿ/0.00464ⁿ

Solving for n, we have ≈ 0.102

Thus,K is;

k = 79681.27/0.00464^(0.102)

k = 137838.78 lb/in²

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