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Kaylis [27]
2 years ago
13

A freshly annealed glass containing flaws of maximum length of 0.1 microns breaks under a tensile stress of 120 MPa. If a sample

of this glass is now subjected to stress of 30 MPa, failure is found to occur after 10 days. Assuming that the specific surface energy (or fracture toughness) does not change, the average rate at which the crack has grown during the period of the test is
Engineering
1 answer:
almond37 [142]2 years ago
6 0

Answer:

0.16 micron per day

Explanation:

Given:

The initial crack length, a₁ = 0.1 micron = 0.1 × 10⁻⁶ m

Initial tensile stress, σ₁ = 120 MPa

Final stress = 30 MPa

now from Griffith's equation, we have

\sigma=[\frac{G_cE}{\pi\ a}]^\frac{1}{2}

where,

Gc and E are the material constants

now,

for the initial stage

120=[\frac{G_cE}{\pi\ (0.1\times10^{-6}}]^\frac{1}{2}  ........{1}

and for the final case

30=[\frac{G_cE}{\pi\ a_2}]^\frac{1}{2}   ............{2}

on dividing 1 by 2, we get

\frac{120}{30}=[\frac{a_2}{0.1\times10^{-6}}]^\frac{1}{2}

or

a₂ = 4² × 0.1 × 10⁻⁶ m

or

a₂ = 1.6 micron

Now,

the change from 0.1 micron to 1.6 micron took place in 10 days

therefore, the rate at which the crack is growing = \frac{1.6-0.1}{10}

or

average rate of change of crack = 0.16 micron per day

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Answer:

ΔQ = 4930.37 BTu

Explanation:

given data

height h = 8ft

Δt = 8  hours

length L = 24 feet

R value = 16.2 hr⋅°F⋅ft² /Btu

inside temperature t1 = 68°F

outside temperature t2 = 16°F

to find out

number of Btu conducted

solution

we get here number of Btu conducted by this expression that s

\frac{\Delta Q}{\Delta t} =\frac{-A}{R} (t2 -t1)     ......................1

here A is area that is = h × L = 8 × 24 = 1492 ft²

put here value we get

\frac{\Delta Q}{8} =\frac{-192}{16.2} (16-68)

solve it we get

ΔQ = 4930.37 BTu

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What are the factors that influence the power input to the compressor?
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2 years ago
Air is compressed by a 40-kW compressor from P1 to P2. The air temperature is maintained constant at 25°C during this process a
AlexFokin [52]

Answer:

the rate of entropy change of the air is -0.1342 kW/K

the assumptions made in solving this problem

- Air is an ideal gas.

- the process is isothermal ( internally reversible process ). the change in internal energy is 0.

- It is a steady flow process

- Potential and Kinetic energy changes are negligible.

Explanation:

Given the data in the question;

From the first law of thermodynamics;

dQ = dU + dW ------ let this be equation 1

where dQ is the heat transfer, dU is internal energy and dW is the work done.

from the question, the process is isothermal ( internally reversible process )

Thus, the change in internal energy is 0

dU = 0

given that; Air is compressed by a 40-kW compressor from P1 to P2

since it is compressed, dW = -40 kW

we substitute into equation 1

dQ = 0 + ( -40 kW )

dQ = -40 kW

Now, change in entropy of air is;

ΔS_{air = dQ / T

given that T = 25 °C = ( 25 + 273.15 ) K = 298.15 K

so we substitute

ΔS_{air =  -40 kW / 298.15 K

ΔS_{air =  -0.13416 ≈ -0.1342 kW/K

Therefore, the rate of entropy change of the air is -0.1342 kW/K

the assumptions made in solving this problem

- Air is an ideal gas.

- the process is isothermal ( internally reversible process ). the change in internal energy is 0.

- It is a steady flow process

- Potential and Kinetic energy changes are negligible.

7 0
2 years ago
Write equations used to calculate the diode reverse saturation current, the voltage at which diode goes into resistive behavior,
laiz [17]

Answer:

Diode equation for reverse saturation current

I_o = A\times e^{\frac{-Eg}{KT}} + B\times e^{\frac{-Eg}{2KT}}  

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Voltage at which high level injection occurs:Va=0.55 volt

Voltage at which avalanche multiplication occurs:V=5volts

Explanation:

we take here forward and reverse 0.7 volt and -5 volt  

As Diode current equation is express as

I_D = I_o \times (e^{\frac{V_D}{\eta V_T}} -1 )   ....................1

here I_D is total current through the diode and I_o is reverse saturated current and V_D is  voltage drop across diode and \eta is idealized factor and V_T is thermal voltage

so here we know that when Bios is forward than

V_D  = V_T     .................2

ans Bios is Reverse than  

V_D  = V_R      ..................3

so here

1.  diode reverse saturation current is express as

I_o = A\times e^{\frac{-Eg}{KT}} + B\times e^{\frac{-Eg}{2KT}}  

and

2. Voltage at which diode go into Reverse behavior will be

V_D  = V_R    = -5 volt

and

3. voltage at which high level injection occur that is

Va = 0.55 volt

and

4. voltage at which avalanche multiplication occurs is

Va = 5 volt

6 0
3 years ago
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