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2 years ago

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Determine the amount of gamma and alpha phases in a 10-kg, 1060 steel casting as it is being cooled to the following temperature

s: (a) 900°C; (b) 728°C; and (c) 726°C.

1 answer:

Snezhnost [94]2 years ago

8 0

Answer:

Given that

Mass ,m=10 kg

a)

At 900°C

From iron carbon diagram

Percentage of alpha phase = 0 %

Percentage of gamma phase =100 %

b)

At 728°C

From iron carbon diagram

Percentage of alpha phase = 23 %

Percentage of gamma phase =77 %

c)

At 726°C

From iron carbon diagram

Percentage of alpha phase = 91 %

Percentage of gamma phase =9 %

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Air at 40°C flows over a long, 25-mm-diameter cylinder with an embedded electrical heater. In a series of tests, measurements we

Murljashka [212]

Answer:

a) See attachment

b) C = 21.626 W / m^2 .K , n = 0.593

Explanation:

a)

dT = 300 - 40 = 160 K

A = pi*D*l where D = 0.025 m

Power = P' * l = h * dT * A

P' l = h * pi * D * l* dT

Hence,

h = P' / (pi*D*dT)

We will use the above equation to compute for respective values of P'

Note: The results are tabulated and attached

b)

Assuming the dependence of the convection coefficient on the velocity to be of the form h=CV^n, determine the parameters C and n from the results of part (a).

Taking logarithm on both sides:

Ln (h) = Ln(C) + n*Ln(V)

n = (Ln(h2) - Ln(h1)) / (Ln(V2) - Ln(V1))

n = (Ln (32.223/22.037)) / Ln(2))

n = 0.593

Using regression we can compare:

C = 21.626 W/m^2 . K

7 0

2 years ago

Link BD consists of a single bar 36 mm wide and 18 mm thick. Knowing that each pin has a 12-mm diameter, determine the maximum v

MAXImum [283]

Answer:

hello the diagram attached to your question is missing attached below is the missing diagram

answer :

a) 48.11 MPa

b) - 55.55 MPa

Explanation:

First we consider the equilibrium moments about point A

∑ Ma = 0

( Fbd * 300cos30° ) + ( 24sin∅ * 450cos30° ) - ( 24cos∅ * 450sin30° ) = 0

therefore ;<em> Fbd = 36 ( cos ∅tan30° - sin∅ ) kN ----- ( 1 )</em>

A ) when ∅ = 0

Fbd = 20.7846 kN

link BD will be under tension when ∅ = 0, hence we will calculate the loading area using this equation

A = ( b - d ) t

b = 12 mm

d = 36 mm

t = 18

therefore loading area ( A ) = 432 mm^2

determine the maximum value of average normal stress in link BD using the relation below

бbd = $\frac{Fbd}{A}$ = 20.7846 kN / 432 mm^2 = 48.11 MPa

b) when ∅ = 90°

Fbd = -36 kN

the negativity indicate that the loading direction is in contrast to the assumed direction of loading

There is compression in link BD

next we have to calculate the loading area using this equation ;

A = b * t

b = 36mm

t = 18mm

hence loading area = 36 * 18 = 648 mm^2

determine the maximum value of average normal stress in link BD using the relation below

бbd = $\frac{Fbd}{A}$ = -36 kN / 648mm^2 = -55.55 MPa

4 0

1 year ago

) A given system has four sensors that can produce an output of 0 or 1. The system operates properly when exactly one of the sen

Rashid [163]

A given system has four sensors that can produce an output of 0 or 1. The system operates proper . An alarm must be raised when two or more sensors have the output of 1. Design the simplest circuit that can be used to raise the alarm ly when exactly one of the sensors has its output equal to Repeat problem #4 for a system that has 7 sensors. Hint: Before you slog through a truth table with 128 rows in it, think about whether SOP or POS might be a better approach.

7 0

2 years ago

A steam power plant operates on an ideal reheat- regenerative Rankine cycle and has a net power output of 80 MW. Steam enters th

trasher [3.6K]

Answer:

flow(m) = 54.45 kg/s

thermal efficiency u = 44.48%

Explanation:

Given:

- P_1 = P_8 = 10 KPa

- P_2 = P_3 = P_6 = P_7 = 800 KPa

- P_4 = P_5 = 10,000 KPa

- T_5 = 550 C

- T_7 = 500 C

- Power Output P = 80 MW

Find:

- The mass flow rate of steam through the boiler

- The thermal efficiency of the cycle.

Solution:

State 1:

P_1 = 10 KPa , saturated liquid

h_1 = 192 KJ/kg

v_1 = 0.00101 m^3 / kg

State 2:

P_2 = 800 KPa , constant volume process work done:

h_2 = h_1 + v_1 * ( P_2 - P_1)

h_2 = 192 + 0.00101*(790) = 192.80 KJ/kg

State 3:

P_3 = 800 KPa , saturated liquid

h_3 = 721 KJ/kg

v_3 = 0.00111 m^3 / kg

State 4:

P_4 = 10,000 KPa , constant volume process work done:

h_4 = h_3 + v_3 * ( P_4 - P_3)

h_4 = 721 + 0.00111*(9200) = 731.21 KJ/kg

State 5:

P_5 = 10,000 KPa , T_5 = 550 C

h_5 = 3500 KJ/kg

s_5 = 6.760 KJ/kgK

State 6:

P_6 = 800 KPa , s_5 = s_6 = 6.760 KJ/kgK

h_6 = 2810 KJ/kg

State 7:

P_7 = 800 KPa , T_7 = 500 C

h_7 = 3480 KJ/kg

s_7 = 7.870 KJ/kgK

State 8:

P_8 = 10 KPa , s_8 = s_7 = 7.870 KJ/kgK

h_8 = 2490 KJ/kg

- Fraction of steam y = flow(m_6 / m_3).

- Use energy balance of steam bleed and cold feed-water:

E_6 + E_2 = E_3

flow(m_6)*h_6 + flow(m_2)*h_3 = flow(m_3)*h_3

y*h_6 + (1-y)*h_3 = h_3

y*2810 + (1-y)*192.8 = 721

Compute y: y = 0.2018

- Heat produced by the boiler q_b:

q_b = h_5 - h_4 +(1-y)*(h_7 - h_8)

q_b = 3500 -731.21 + ( 1 - 0.2018)*(3480 - 2810)

Compute q_b: q_b = 3303.58 KJ/ kg

-Heat dissipated by the condenser q_c:

q_c = (1-y)*(h_8 - h_1)

q_c= ( 1 + 0.2018)*(2810 - 192)

Compute q_c: q_c = 1834.26 KJ/ kg

- Net power output w_net:

w_net = q_b - q_c

w_net = 3303.58 - 1834.26

w_net = 1469.32 KJ/kg

- Given out put P = 80,000 KW

flow(m) = P / w_net

compute flow(m) flow(m) = 80,000 /1469.32 = 54.45 kg/s

- Thermal efficiency u:

u = 1 - (q_c / q_b)

u = 1 - (1834.26/3303.58)

u = 44.48 %

5 0

2 years ago

The small washer is sliding down the cord OA. When it is at the midpoint, its speed is 28 m/s and its acceleration is 7 m/s 2 .

Neporo4naja [7]

Answer:

Velocity components

$V_r = -16.28 m/s$

$V_z = -22.8 m/s$

$V_q = 0 m/s$

For Acceleration components;

$a_r = -4.07m/s^2$

$a_z = -5.70m/s^2$

$a_q = 0m/s^2$

Explanation:

We are given:

$Speed v_o = 28 m/s$

$Acceleration a_o= 7 m/s^2$

We first need to find the radial position r of washer in x-y plane.

Therefore

$r = \sqrt{300^2 + 400^2}$

r = 500 mm

To find length along direction OA we have:

$L = \sqrt{500^2 + 700^2}L = 860 mm$

Therefore, the radial and vertical components of velocity will be given as:

$V_r = V_o*cos(Q)$

$V_z = V_o*sin(Q)$

Where Q is the angle between OA and vector r.

Therefore,

$V_r = 28 * \frac{r}{L} = > 28 * \frac{500}{860}$

$V_r = -16.28 m/s$

• $V_z = 28 * \frac{700}{860} = -22.8$

• $V_q = 0 m/s$

The radial and vertical components of acceleration will be:

$a_r = a_o*cos(Q)$

$a_z = a_o*sin(Q)$

Therefore we have:

• $a_r = 7* \frac{500}{860} = -4.07m/s^2$

• $a_z = 7 * \frac{700}{860} = -5.70 m/s^2$

• $a_q = 0 m/s^2$

Note : image is missing, so I attached it

3 0

2 years ago