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Sergeeva-Olga [200]
2 years ago
8

Estimate the maximum expected thermal conductivity for a Cermet that contains 58 vol% titanium carbide (TiC) particles in a coba

lt matrix. Assume thermal conductivities of 27 and 69 W/m-K for TiC and Co respectively.
Engineering
1 answer:
Alik [6]2 years ago
8 0

Answer:

The right answer is "36.32 W/mk".

Explanation:

According to the question,

TiC = 76%

or,

      = 0.76

CO = 24%

or,

     = 0.24

Thermal conductivity of TiC,

= 26 W/mk

Thermal conductivity of CO,

= 69 W/mk

On applying the rule, we get

⇒  "K" \ of \ Cermet=(K)_{TiC} (V_f)_{TiC}+(K)_{CO} (V_m)_{CO}

On putting the given values, we get

⇒                            =(26)(0.76)+(69)(0.24)

⇒                            =19.76+16.56

⇒                            =36.32 \ W/mk

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An air-standard Diesel cycle engine operates as follows: The temperatures at the beginning and end of the compression stroke are
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This question is incomplete, the complete question is;

An air-standard Diesel cycle engine operates as follows: The temperatures at the beginning and end of the compression stroke are 30 °C and 700 °C, respectively. The net work per cycle is 590.1 kJ/kg, and the heat transfer input per cycle is 925 kJ/kg. Determine the a) compression ratio, b) maximum temperature of the cycle, and c) the cutoff ratio, v3/v2.

Use the cold air standard assumptions.

Answer:

a) The compression ratio is 18.48

b) The maximum temperature of the cycle is 1893.4 K

c) The cutoff ratio, v₃/v₂ is 1.946

Explanation:

Given the data in the question;

Temperature at the start of a compression T₁ = 30°C = (30 + 273) = 303 K

Temperature at the end of a compression T₂ = 700°C = (700 + 273) = 973 K

Net work per cycle W_{net = 590.1 kJ/kg

Heat transfer input per cycle Qs = 925 kJ/kg

a) compression ratio;

As illustrated in the diagram below, 1 - 2 is adiabatic compression;

so,

Tγ^{Y-1 = constant { For Air, γ = 1.4 }

hence;

⇒ V₁ / V₂ = ( T₂ / T₁ )^{\frac{1}{Y-1}

so we substitute

⇒ V₁ / V₂ = (  973 K / 303 K  )^{\frac{1}{1.4-1}

= (  3.21122  )^{\frac{1}{0.4}

= 18.4788 ≈ 18.48

Therefore, The compression ratio is 18.48

b) maximum temperature of the cycle

We know that for Air, Cp = 1.005 kJ/kgK

Now,

Heat transfer input per cycle Qs = Cp( T₃ - T₂ )

we substitute

925 = 1.005( T₃ - 700 )

( T₃ - 700 ) = 925 / 1.005

( T₃ - 700 ) = 920.398

T₃ = 920.398 + 700

T₃ = 1620.398 °C

T₃ = ( 1620.398 + 273 ) K

T₃ = 1893.396 K ≈ 1893.4 K

Therefore, The maximum temperature of the cycle is 1893.4 K

c)  the cutoff ratio, v₃/v₂;

Since pressure is constant, V ∝ T

So,

cutoff ratio S = v₃ / v₂  = T₃ / T₂

we substitute

cutoff ratio S = 1893.396 K / 973 K

cutoff ratio S = 1.9459 ≈ 1.946

Therefore, the cutoff ratio, v₃/v₂ is 1.946

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