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3 years ago

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The one end of a hollow square bar whose side is (10+N/100) in with (1+N/100) in thickness is under a tensile stress 102,500 psi

and the other end is connected with a U bracket using a double-pin system. Find the minimum diameter of pin is used according to shear strength. Take the factor of safety as 1.5 and σ_all=243 ksi for pin material.
n=1

1 answer:

netineya [11]3 years ago

3 0

Answer:

The one end of a hollow square bar whose side is (10+N/100) in wit

Explanation:

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Searches related to Probability questions - A person frequents one of the two restaurants KARIM or NAZEER, choosing Chicken’s it

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c) 30%

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Explanation:

Person chooses Chicken's item : 70% = 0.7

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Visits in which he orders Afghani Chicken = 60% = 0.6

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Probability that he goes to KARIM and orders Afghani Chicken:

P = 0.7 * 0.6 = 0.42

b)

Two events are said to be independent when occurrence of one event does not affect the probability of the other event's occurrence. Here the person orders Afghani Chicken regardless of where he visits so the events are independent.

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P = 0.30 because he orders Afghani Chicken regardless of where he visits.

d)

Let A be the probability that he goes to KARIM:

P(A) = 0.7 * ( 1 - 0.6 ) = 0.28

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P(B) = 0.3 * 0.6 = 0.18

Let C be the probability that he goes to KARIM and orders Afghani chicken:

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P(A) + P(B) + P(C) = 0.28 + 0.18 + 0.42 = 0.88

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2 years ago

Steam at 5 MPa and 400 C enters a nozzle steadily with a velocity of 80 m/s, and it leavesat 2 MPa and 300 C. The inlet area of

Korolek [52]

Answer:

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

So;

Inlet:

$P_1 =$ 5 MPa

$T_1$ = 400° C

Velocity V = 80 m/s

Exit:

$P_2 =$ 2 MPa

$T_2$ = 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