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4 years ago

Before turning in a test, it would be best to

Engineering

2 answers:

kumpel [21]4 years ago

5 0

Answer:

Explanation:

Doss [256]4 years ago

4 0

Answer:

check the grammar and double check the work.Explanation:

sorry if its wrong :P

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A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 MPa (50 ksi ) is exposed to a stress of 2023 MPa

Neko [114]

Answer:

Explanation:

The formula for critical stress is

$\sigma_c=\frac{K}{Y\sqrt{\pi a} }$

$\sigma_c =\texttt{critical stress}$

K is the plane strain fracture toughness

Y is dimensionless parameters

We are to Determine the Critical stress

Now replacing the critical stress with 54.8

a with 0.2mm = 0.2 x 10⁻³

Y with 1

$\sigma_c=\frac{54.8}{1\sqrt{\pi \times 0.2\times10^{-3}} } \\\\=\frac{54.8}{\sqrt{6.283\times10^{-4}} } \\\\=\frac{54.8}{0.025} \\\\=2186.20Mpa$

The fracture will not occur because this material can handle a stress of 2186.20Mpa before fracture. it is obvious that is greater than 2023Mpa

Therefore, the specimen does not failure for surface crack of 0.2mm

4 0

3 years ago

The constant A in Equation 17.2 is 12π4 R/5(θD) 3 where R is the gas constant and θD is the Debye temperature (K). Estimate θD f

kirill115 [55]

Answer:

The Debye temperature for aluminum is 375.2361 K

Explanation:

Molecular weight of aluminum=26.98 g/mol

T=15 K

The mathematical equation for the specific heat and the absolute temperature is:

$C_{v} =AT^{3}$

Substituting in the expression of the question:

$C_{v} =(\frac{12\pi ^{4}R }{5\theta _{D}^{3} } )T^{3}$

$\theta _{D} =(\frac{12\pi ^{4}RT^{3} }{5C_{v} } ) ^{1/3}$

Here

$C_{v} =4.6\frac{J}{kg-K} *\frac{1kg}{1000g} *\frac{26.98g}{1mol} =0.1241J/mol-K$

Replacing:

$\theta _{D} =(\frac{12\pi ^{4}*8.31*15^{3} }{5*0.1241} )^{1/3} =375.2361K$

3 0

3 years ago

Consider a system whose temperature is 18°C. Express this temperature in R, K, and °F.

zvonat [6]

Answer:

In Rankine 524.07°R

In kelvin 291 K

In Fahrenheit 64.4°F

Explanation:

We have given temperature 18°C

We have to convert this into Rankine R

From Celsius to Rankine we know that $T(R)=(T_{C}+273.15)\frac{9}{5}$

We have to convert 18°C

So $T(R)=(18+273.15)\frac{9}{5}=524.07^{\circ}R$

Conversion from Celsius to kelvin

$T(K)=(T_{C}+273)$

We have to convert 18°C

$T(K)=(18+273)=291K$

Conversion of Celsius to Fahrenheit

$T(F)=T_{C}\times \frac{9}{5}+32=64.4^{\circ}F$

7 0

4 years ago

A cold-rolled sheet metal that is 40 mm wide and has a thickness of 5.00 mm is going to be bent into a V shape with a 60° angle.

Artist 52 [7]

a= the force of gravity b= the amount of bicker to maple syrup ratio

5 0

3 years ago

Steam enters an adiabatic turbine at 10MPa and 500 C and leaves at 10 kPa with a quality of 90%. Neglecting the changes in kinet

Amanda [17]

Answer:

flow ( m ) = 4.852 kg/s

Explanation:

Given:

- Inlet of Turbine

P_1 = 10 MPa

T_1 = 500 C

- Outlet of Turbine

P_2 = 10 KPa

x = 0.9

- Power output of Turbine W_out = 5 MW

Find:

Determine the mass ow rate required

Solution:

- Use steam Table A.4 to determine specific enthalpy for inlet conditions:

P_1 = 10 MPa

T_1 = 500 C ---------- > h_1 = 3375.1 KJ/kg

- Use steam Table A.6 to determine specific enthalpy for outlet conditions:

P_2 = 10 KPa -------------> h_f = 191.81 KJ/kg

x = 0.9 -------------> h_fg = 2392.1 KJ/kg

h_2 = h_f + x*h_fg

h_2 = 191.81 + 0.9*2392.1 = 2344.7 KJ/kg

- The work produced by the turbine W_out is given by first Law of thermodynamics:

W_out = flow(m) * ( h_1 - h_2 )

flow ( m ) = W_out / ( h_1 - h_2 )

- Plug in values:

flow ( m ) = 5*10^3 / ( 3375.1 - 2344.7 )

flow ( m ) = 4.852 kg/s

3 0

3 years ago