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3 years ago

9

A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 MPa1m (50 ksi1in.) is exposed to a stress of 103

0 MPa (150,000 psi). Will this specimen experience fracture if the largest

1 answer:

topjm [15]3 years ago

3 0

Answer:

1382.67. MPa

Explanation:

Kindly check attachment for the step by step solution of the given problem.

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Tech A says you can find the typical angle of a V-block engine by dividing the number of cylinders by 720

Lady_Fox [76]

Answer:

Tech A is correct

Explanation:

Tech A is right as its V- angle is identified by splitting the No by 720 °. Of the piston at the edge of the piston.

Tech B is incorrect, as the V-Angle will be 720/10 = 72 for the V-10 motor, and he says 60 °.

8 0

2 years ago

Steam at 4 MPa and 350°C is expanded in an adiabatic turbine to 125kPa. What is the isentropic efficiency (percent) of this turb

guajiro [1.7K]

Answer:

$\eta_{turbine} = 0.603 = 60.3\%$

Explanation:

First, we will find actual properties at given inlet and outlet states by the use of steam tables:

AT INLET:

At 4MPa and 350°C, from the superheated table:

h₁ = 3093.3 KJ/kg

s₁ = 6.5843 KJ/kg.K

AT OUTLET:

At P₂ = 125 KPa and steam is saturated in vapor state:

h₂ = $h_{g\ at\ 125KPa}$ = 2684.9 KJ/kg

Now, for the isentropic enthalpy, we have:

P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K

Since s₂ is less than $s_g$ and greater than $s_f$ at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

$x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88$

Now, we will find $h_{2s}$ (enthalpy at the outlet for the isentropic process):

$h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg$

Now, the isentropic efficiency of the turbine can be given as follows:

$\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%$

3 0

3 years ago

Calculate the electroosmotic velocity of an aqueous solution through a glass capillary 5 cm long with a 0.5 mm internal diameter

natita [175]

Answer:

Electroosmotic velocity will be equal to $1.6\times 10^{-4}m/sec$

Explanation:

We have given applied voltage v = 100 volt

Length of capillary L = 5 mm = 0.005 m

Zeta potential of the capillary surface $\xi =80mV=0.08volt$

Dielectric constant of glass is between 5 to 10 here we are taking dielectric constant as $\epsilon =10$

Viscosity of glass is $\eta =10^8$

Electroosmotic velocity is given as $v_{eo}=\frac{\epsilon \xi }{\eta }\times \frac{v}{L}$

$v_{eo}=\frac{10\times 0.08 }{10^8 }\times \frac{100}{0.005}=1.6\times 10^{-4}m/sec$

So Electroosmotic velocity will be equal to $1.6\times 10^{-4}m/sec$

8 0

3 years ago

Wave flow of an incompressible fluid into a solid surface follows a sinusoidal pattern. Flow is two-dimensional with the x-axis

Artyom0805 [142]

Answer:

sorry , for my point

Explanation:

4 0

3 years ago

In using the drag coefficient care needs to be taken to use the correct area when determining the drag force. What is a typical

stealth61 [152]

Answer:

Explanation:

We know that Drag force $F_D$

$F_D=\dfrac{1}{2}C_D\rho AV^2$

Where

$C_D$ is the drag force constant.

A is the projected area.

V is the velocity.

ρ is the density of fluid.

Form the above expression of drag force we can say that drag force depends on the area .So We should need to take care of correct are before finding drag force on body.

Example:

When we place our hand out of the window in a moving car ,we feel a force in the opposite direction and feel like some one trying to pull our hand .This pulling force is nothing but it is drag force.

6 0

3 years ago