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tatiyna
3 years ago
9

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.
Engineering
1 answer:
Georgia [21]3 years ago
8 0

Answer:

The answer is below

Explanation:

Given that:

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

a) The maximum thermal conductivity is given by:

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

b) The minimum thermal conductivity is given by:

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

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Explanation:

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7 0
4 years ago
A three-phase voltage source with a terminal voltage of 22kV is connected to a three-phase transformer rated 5MVA 22kV/220V. The
amid [387]

Answer:

A) See Attachment B) See Attachment C) 186.153

Explanation:

C) Per phase Voltage= 220/∛3

                                  = 6.037

S=V²/Z

  = (6.037)²/(0.01+i0.05)

  = 62051.282-i310256.41

Active power losses per phase= 62.051kW

total Active power losses= 62.051×3

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6 0
4 years ago
A subsurface exploration report shows that the average water content of a fine-grained soil in a proposed borrow area is 22% and
Morgarella [4.7K]

Answer:

shrinkage ratio  = 1.538

Explanation:

given data

water content = 22 %

dry density γ = 82 pcf

required dry density specified γ' = 96 pcf

required to produce = 50,000 yd³  = 50000 × 27 = 1350,000 ft³

solution

we get here first volume of borrow pit that is we know that

dry density ∝  \frac{1}{volume}  

so \frac{\gamma d}{\gamma 'd}  = \frac{v'}{v}

\frac{82}{96}  = \frac{1350000}{v}

v = 1580487.8 ft³

v = 58536.58 yd³

so here

shrinkage ratio will be as

shrinkage ratio = \frac{96}{62.4}  

shrinkage ratio  = 1.538

6 0
3 years ago
Air enters a 34 kW electrical heater at a rate of 0.8 kg/s with negligible velocity and a temperature of 60 °C. The air is disch
Flura [38]

Answer:

79 kW.

Explanation:

The equation for enthalpy is:

H2 = H1 + Q - L

Enthalpy is defined as:

H = G*(Cv*T + p*v)

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The gas state equation is:

p*v = R*T (with specific volume)

The specific gas constant for air is:

287 K/(kg*K)

Then:

T1 = 60 + 273 = 333 K

T2 = 200 + 273 = 473 K

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p2*v2 = 287 * 473 = 135.7 kJ/kg

The Cv for air is:

Cv = 720 J/(kg*K)

So the enthalpies are:

H1 = 0.8*(0.72 * 333 + 95.6) = 268 kW

H2 = 0.8*(0.72 * 473 + 135.7) = 381 kW

Ang the heat is:

Q = 34 kW

Then:

H2 = H1 + Q - L

381 = 268 + 34 - L

L = 268 + 34 - 381 = -79 kW

This is the work from the point of view of the air, that's why it is negative.

From the point of view of the machine it is positive.

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Montano1993 [528]

Answer:

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Explanation:

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