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3 years ago

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Acoke can with inner diameter(di) of 75 mm, and wall thickness (t) of 0.1 mm, has internal pressure (pi) of 150 KPa and is suffe

red from a torque (T) of 100 Nmat one end (fixed at the other end). By neglecting the outside pressureand radial stresses, choose a proper 2D element and showcase the non-zero stresses on the element. Then, determine all three principal stresses and the maximum shear stress at that point.

1 answer:

NemiM [27]3 years ago

6 0

Answer:

All 3 principal stress

1. 56.301mpa

2. 28.07mpa

3. 0mpa

Maximum shear stress = 14.116mpa

Explanation:

di = 75 = 0.075

wall thickness = 0.1 = 0.0001

internal pressure pi = 150 kpa = 150 x 10³

torque t = 100 Nm

finding all values

∂1 = 150x10³x0.075/2x0,0001

= 0.5625 = 56.25mpa

∂2 = 150x10³x75/4x0.1

= 28.12mpa

T = 16x100/(πx75x10³)²

∂1,2 = 1/2[(56.25+28.12) ± √(56.25-28.12)² + 4(1.207)²]

= 1/2[84.37±√791.2969+5.827396]

= 1/2[84.37±28.33]

∂1 = 1/2[84.37+28.33]

= 56.301mpa

∂2 = 1/2[84.37-28.33]

= 28.07mpa

This is a 2 d diagram donut is analyzed in 2 direction.

So ∂3 = 0mpa

∂max = 56.301-28.07/2

= 14.116mpa

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a) the mass flow rate of the steam is $\mathbf{m_1 =6.92 \ kg/s}$

b) the exit velocity of the steam is $\mathbf{V_2 = 562.7 \ m/s}$

c) the exit area of the nozzle is $A_2$ = 0.0015435 m²

Explanation:

Given that:

A steam with 5 MPa and 400° C enters a nozzle steadily

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$P_1 =$ 5 MPa

$T_1$ = 400° C

Velocity V = 80 m/s

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$T_2$ = 300° C

From the properties of steam tables at $P_1 =$ 5 MPa and $T_1$ = 400° C we obtain the following properties for enthalpy h and the speed v

$h_1 = 3196.7 \ kJ/kg \\ \\ v_1 = 0.057838 \ m^3/kg$

From the properties of steam tables at $P_2 =$ 2 MPa and $T_1$ = 300° C we obtain the following properties for enthalpy h and the speed v

$h_2 = 3024.2 \ kJ/kg \\ \\ v_2= 0.12551 \ m^3/kg$

Inlet Area of the nozzle = 50 cm²

Heat lost Q = 120 kJ/s

We are to determine the following:

a) the mass flow rate of the steam.

From the system in a steady flow state;

$m_1=m_2=m_3$

Thus

$m_1 =\dfrac{V_1 \times A_1}{v_1}$

$m_1 =\dfrac{80 \ m/s \times 50 \times 10 ^{-4} \ m^2}{0.057838 \ m^3/kg}$

$m_1 =\dfrac{0.4 }{0.057838 }$

$\mathbf{m_1 =6.92 \ kg/s}$

b) the exit velocity of the steam.

Using Energy Balance equation:

$\Delta E _{system} = E_{in}-E_{out}$

In a steady flow process;

$\Delta E _{system} = 0$

$E_{in} = E_{out}$

$m(h_1 + \dfrac{V_1^2}{2})$ $= Q_{out} + m (h_2 + \dfrac{V_2^2}{2})$

$- Q_{out} = m (h_2 - h_1 + \dfrac{V_2^2-V^2_1}{2})$

$- 120 kJ/s = 6.92 \ kg/s (3024.2 -3196.7 + \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})$

$- 120 kJ/s = 6.92 \ kg/s (-172.5 + \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})$

$- 120 kJ/s = (-1193.7 \ kg/s + 6.92\ kg/s ( \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})$

$V_2^2 = 316631.29 \ m/s$

$V_2 = \sqrt{316631.29 \ m/s$

$\mathbf{V_2 = 562.7 \ m/s}$

c) the exit area of the nozzle.

The exit of the nozzle can be determined by using the expression:

$m = \dfrac{V_2A_2}{v_2}$

making $A_2$ the subject of the formula ; we have:

$A_2 = \dfrac{ m \times v_2}{V_2}$

$A_2 = \dfrac{ 6.92 \times 0.12551}{562.7}$

$A_2$ = 0.0015435 m²

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3 years ago