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3 years ago

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Estimate (a) the maximum, and (b) the minimum thermal conductivity values (in W/m-K) for a cermet that contains 85 vol% carbide

particles in a metal matrix. Assume thermal conductivities of 20 and 66 W/m-K for the carbide and metal, respectively.

1 answer:

velikii [3]3 years ago

5 0

Answer:

Explanation:

k_max = 26.9 w/mk

k_min = 22.33 w/mk

Explanation:

a) the maximum thermal conductivity is given as

K_MAX = k_m v_m + k_p v_p

where k_m is thermal conductvitiy of metal

k_p is thermal conductvitiy of carbide

v_m = proportion of metal in the cement = 0.15

v_p = proportion of carbide in the cement = 0.85

$K_MAX = k_m v_m + k_p v_p$

= 66*0.15 + 20*0.85

k_max = 26.9 w/mk

b) the minimum thermal conductivity is given as

$k_min = \frac{ k_{carbide} *k_{metal}}{k_{metal} v_{carbide} +k_{carbide} v_{metal}}$

= \frac{20*66}{20*0.15 +66*0.85}

k_min = 22.33 w/mk

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Two parallel pipelines spaced 0.5 m apart are buried in soil having a thermal conductivity of 0.5 W/m·K. The pipes have outer di

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Answer:

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3 years ago

A 75 ohm coaxial transmission line has a length of 2.0 cm and is terminated with a load impedance of 37.5 + j75 Ohm. If the diel

Hatshy [7]

Answer:

The load reflection coefficient, $\Gamma =0.62\angle 82.875^{\circ} \Omega$

Reflection coefficient at input, $\Gamma = 0.62\angle - 147.518^{\circ} \Omega$

SWR = 4.26

Given:

Characteristic impedance of the co-axial cable, $Z_{c} = 75 \Omega$

Length of the cable, L = 2.0 cm = 0.02 m

$Z_{Load} = 37.5 + j75 \Omega$

Dielectric constant, K = 2.56

frequency, f = 3.0 GHz = $3.0 \times 10^{9} Hz$

Explanation:

In order to calculate the reflection coefficient at load, we first calculate these:

The line input impedance $Z_{i}$ is given by:

$Z_{i} = Z_{c}\frac{Z_{Load} + jZ_{c} tan(\beta L)}{Z_{c} + jZ_{Load} tan (\beat L)}$ (1)

Now, we calculate the value of $\beta$ :

$\beta = \frac{2\pi}{\lambda'} = \farc{2\pi f\sqrt{K}}{c}$

(since, $\lambda' = \farc{c}{f\sqrt{K}}$ )

$\beta = \farc{2\pi f\sqrt{2.56}}{3\times 10^{8}} = 100.53$

Now, Substituting the value in eqn (1):

$Z_{i} = 75\frac{37.5 + j75 + j75 tan(100.53\times 0.02)}{75 + j(37.5 + j75) tan ( 100.53\times 0.02)} = 18.99 - j20.55 \Omega = 27.98\angle - 47.257^{\circ} \Omega$

Now, the load reflection coefficient is given by:

$\Gamma = \frac{Z_{Load} - Z_{c}}{Z_{c} + Z_{Load}}}$

Thus

$\Gamma = \frac{37.5 + j75 - 75}{75 + 37.5 + j75}} = 0.077 + j0.615 = 0.62\angle 82.875^{\circ} \Omega$

Similarly,

Reflection coefficient at input:

$\Gamma' = \frac{Z_{i} - Z_{c}}{Z_{c} + Z_{i}}}$

$\Gamma' = \frac{18.99 - j20.55 - 75}{75 + 18.99 - j20.55}} = - 0.523 - j0.334 = 0.62\angle - 147.518^{\circ} \Omega$

Now, the SWR is given by:

SWR, Standing Wave Ratio = $\frac{1 +|\Gamma|}{1 - |\Gamma|}$

SWR = $\frac{1 +|0.62|}{1 - |0.62|} = 4.26$

8 0

3 years ago

Consider a cubic workpiece of rigid perfect plastic material with side length lo. The cube is deformed plastically to the shape

Taya2010 [7]

Answer: ε₁+ε₂+ε₃ = 0

Explanation: Considering the initial and final volume to be constant which gives rise to the relation:-

l₀l₀l₀=l₁l₂l₃

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taking natural log on both sides

$ln(\frac{(lo*lo*lo)}{l1*l2*l3})=ln(1)$

Considering the logarithmic Laws of division and multiplication :

ln(AB) = ln(A)+ln(B)

ln(A/B) = ln(A)-ln(B)

$ln(\frac{(l1)}{lo})*ln(\frac{(l2)}{lo})*ln(\frac{(l3)}{lo}) = 0$

Use the image attached to see the definition of true strain defined as

ln(l1/1o)= ε₁

which then proves that ε₁+ε₂+ε₃ = 0

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3 years ago

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Neko [114]

Answer:

True

Explanation:

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3 years ago

Read 2 more answers

Write a single statement to print: user_word,user_number. Note that there is no space between the comma and user_number. Sample

kykrilka [37]

Answer:

cout<<"''<<user_word<<"' "<<user_number;

Explanation:

The above question was answered using C++ programming language.

The keyword cout represents print and it carries out print operation only.

It prints all variable in front of it.

Assume the values of user_word and user_number to be Charles and 20, respectively.

The output of the above instruction would be

'Charles' 20 just as it is in the sample output in the question.

In java programming language, it is

System.out.print("'"+user_word+"' "+user_number);

In Qbasic, it is

PRINT "'"+user_word+"' "+ user_number

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3 years ago