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tatyana61 [14]
3 years ago
7

The modulus of elasticity for a ceramic material having 6.0 vol% porosity is 303 GPa. (a) Calculate the modulus of elasticity (i

n GPa) for the nonporous material.
Engineering
1 answer:
Phantasy [73]3 years ago
8 0

Answer:

modulus of elasticity for the nonporous material is 340.74 GPa

Explanation:

given data

porosity = 303 GPa

modulus of elasticity = 6.0

solution

we get here  modulus of elasticity for the nonporous material Eo that is

E = Eo (1 - 1.9P + 0.9P²)    ...............1

put here value and we get Eo

303 = Eo ( 1 - 1.9(0.06) + 0.9(0.06)² )  

solve it we get

Eo = 340.74 GPa

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Steam at a pressure of 100 bar and temperature of 600 °C enters an adiabatic nozzle with a velocity of 35 m/s. It leaves the noz
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Answer:

Exit velocity V_2=1472.2 m/s.

Explanation:

Given:

At inlet:

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Properties of steam at 100 bar and 600°C

        h_1=3624.7\frac{KJ}{Kg}

At exit:Lets take exit velocity V_2

We know that if we know only one property inside the dome  then we will find the other property by using steam property table.

Given that dryness or quality of steam at the exit of nozzle  is 0.85 and pressure P=80 bar.So from steam table we can find the other properties.

Properties of saturated steam at 80 bar

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So the enthalpy of steam at the exit of turbine  

h_2=h_f+x(h_g-h_f)\frac{KJ}{Kg}

h_2=1316.61+0.85(2757.8-1316.61)\frac{KJ}{Kg}

 h_2=2541.62\frac{KJ}{Kg}

Now from first law for open system

h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w

In the case of adiabatic nozzle Q=0,W=0

3624.7+\dfrac{35^2}{2000}+0=2541.62+\dfrac{(V_2)^2}{2000}+0

V_2=1472.2 m/s

So Exit velocity V_2=1472.2 m/s.

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Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you
viktelen [127]

Answer:

\theta_1=15^o\\\theta_2=75^o

Explanation:

<u>Projectile Motion</u>

In projectile motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration (assuming no friction), and the acceleration in the vertical direction is always the acceleration of gravity. The basic formulas are shown below:

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x=V_{o}cos\theta t

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\theta_2=90^o-\theta_1

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\theta_1=15^o

And

\theta_2=90^o-15^o=75^o

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