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3 years ago

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A cylindrical rod of 1040 Steel originally 10.5 mm in diameter is to be cold worked by drawing to a final diameter of 8.5 mm. Th

e circular cross section will be maintained during deformation. A cold-worked tensile strength in excess of 750 MPa and a ductility of at least 12 %EL are desired. Explain, with details and calculations, how this may be accomplished.

1 answer:

zavuch27 [327]3 years ago

4 0

Answer:

Check the explanation

Explanation:

The desired Cold working <em><u>(which is any process or procedure of metalworking in which the metal body is formed at a lower temperature beyond its recrystallization, typically at the ambient temperature.)</u></em> tensile strength in excess of 750 MPa and a ductility of at least 12 %EL can be seen in the attached image below.

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Which battery produces more volts per cell, maintenance type or maintenance free ?

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Double aa battery. It’s is a very high bolted battery that can power up a lot of things

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3 years ago

An air-standard dual cycle has a compression ratio of 9.1 and displacement of Vd = 2.2 L. At the beginning of compression, p1 =

jok3333 [9.3K]

Answer:

a) T₂ is 701.479 K

T₃ is 1226.05 K

T₄ is 2350.34 K

T₅ is 1260.56 K

b) The net work of the cycle in kJ is 2.28 kJ

c) The power developed is 114.2 kW

d) The thermal efficiency, $\eta _{dual}$ is 53.78%

e) The mean effective pressure is 1038.25 kPa

Explanation:

a) Here we have;

$\frac{T_{2}}{T_{1}}=\left (\frac{v_{1}}{v_{2}} \right )^{\gamma -1} = \left (r \right )^{\gamma -1} = \left (\frac{p_{2}}{p_{1}} \right )^{\frac{\gamma -1}{\gamma }}$

Where:

p₁ = Initial pressure = 95 kPa

p₂ = Final pressure =

T₁ = Initial temperature = 290 K

T₂ = Final temperature

v₁ = Initial volume

v₂ = Final volume

$v_d$ = Displacement volume =

γ = Ratio of specific heats at constant pressure and constant volume cp/cv = 1.4 for air

r = Compression ratio = 9.1

Total heat added = 4.25 kJ

1/4 × Total heat added = $c_v \times (T_3 - T_2)$

3/4 × Total heat added = $c_p \times (T_4 - T_3)$

$c_v$ = Specific heat at constant volume = 0.718×2.821× 10⁻³

$c_p$ = Specific heat at constant pressure = 1.005×2.821× 10⁻³

v₁ - v₂ = 2.2 L

$\left \frac{v_{1}}{v_{2}} \right =r \right = 9.1$

v₁ = v₂·9.1

∴ 9.1·v₂ - v₂ = 2.2 L = 2.2 × 10⁻³ m³

8.1·v₂ = 2.2 × 10⁻³ m³

v₂ = 2.2 × 10⁻³ m³ ÷ 8.1 = 2.72 × 10⁻⁴ m³

v₁ = v₂×9.1 = 2.72 × 10⁻⁴ m³ × 9.1 = 2.47 × 10⁻³ m³

Plugging in the values, we have;

${T_{2}}= T_{1} \times \left (r \right )^{\gamma -1} = 290 \times 9.1^{1.4 - 1} = 701.479 \, K$

From;

$\left (\frac{p_{2}}{p_{1}} \right )^{\frac{\gamma -1}{\gamma }}= \left (r \right )^{\gamma -1}$ we have;

$p_{2} = p_{1}} \times \left (r \right )^{\gamma } = 95 \times \left (9.1 \right )^{1.4} = 2091.13 \ kPa$

1/4×4.25 = $0.718 \times 2.821 \times 10^{-3}\times (T_3 - 701.479)$

∴ T₃ = 1226.05 K

Also;

3/4 × Total heat added = $c_p \times (T_4 - T_3)$ gives;

3/4 × 4.25 = $1.005 \times 2.821 \times 10^{-3} \times (T_4 - 1226.05)$ gives;

T₄ = 2350.34 K

$\frac{T_{4}}{T_{5}}=\left (\frac{v_{5}}{v_{4}} \right )^{\gamma -1} = \left (\frac{r}{\rho } \right )^{\gamma -1}$

$\rho = \frac{T_4}{T_3} = \frac{2350.34}{1226.04} = 1.92$

$T_{5} = \frac{T_{4}}{\left (\frac{r}{\rho } \right )^{\gamma -1}}= \frac{2350.34 }{\left (\frac{9.1}{1.92 } \right )^{1.4-1}} =1260.56 \ K$

b) Heat rejected = $c_v \times (T_5 - T_1)$

$Therefore \ heat \ rejected = 0.718 \times 2.821 \times 10^{-3}\times (1260.56 - 290) = 1.966 kJ$

The net work done = Heat added - Heat rejected

∴ The net work done = 4.25 - 1.966 = 2.28 kJ

The net work of the cycle in kJ = 2.28 kJ

c) Power = Work done per each cycle × Number of cycles completed each second

Where we have 3000 cycles per minute, we have 3000/60 = 50 cycles per second

Hence, the power developed = 2.28 kJ/cycle × 50 cycle/second = 114.2 kW

d)

$Thermal \ efficiency, \, \eta _{dual} = \frac{Work \ done}{Heat \ supplied} = \frac{2.28}{4.25} \times 100 = 53.74 \%$

The thermal efficiency, $\eta _{dual}$ = 53.78%

e) The mean effective pressure, $p_m$ , is found as follows;

$p_m = \frac{W}{v_1 - v_2} =\frac{2.28}{2.2 \times 10^{-3}} = 1038.25 \ kPa$

The mean effective pressure = 1038.25 kPa.

3 0

4 years ago

A 0.06-m3 rigid tank initially contains refrigerant- 134a at 0.8 MPa and 100 percent quality. The tank is connected by a valve t

lesantik [10]

Answer:

a) 0.50613

b) 22.639 kJ

Explanation:

From table A-11 , we will make use of the properties of Refrigerant R-13a at 24°C

first step : calculate the volume of R-13a ( values gotten from table A-11 )

V = m1 * v1 = 5 * 0.0008261 = 0.00413 m^3

next : calculate final specific volume ( v2 )

v2 = V / m2 = 0.00413 / 0.25 ≈ 0.01652 m^3/kg

<u>a) Calculate the mass of refrigerant that entered the tank </u>

v2 = Vf + x2 * Vfg

v2 = Vf + [ x2 * ( Vg - Vf ) ] ----- ( 1 )

where: Vf = 0.0008261 m^3/kg, V2 = 0.01652 m^3/kg , Vg = 0.031834 m^3/kg ( insert values into equation 1 above )

x2 = ( 0.01652 - 0.0008261 ) / 0.031834

= 0.50613 ( mass of refrigerant that entered tank )

<u>b) Calculate the amount of heat transfer </u>

Final specific internal energy = u2 = Uf + ( x2 + Ufg ) ----- ( 2 )

uf = 84.44 kj/kg , x2 = 0.50613 , Ufg = 158.65 Kj/kg

therefore U2 = 164.737 Kj/kg

The mass balance ( me ) = m1 - m2 --- ( 3 )

energy balance( Qin ) = ( m2 * u2 ) - ( m1 * u1 ) + ( m1 - m2 ) * he

therefore Qin = 41.184 - 422.2 + 403.655 = 22.639 kJ

3 0

3 years ago

What is meant by the dead state. Discuss its significance.

Jlenok [28]

Answer and Explanation:

Dead state can be defined as the state in which both the are ambient properties and the system properties are coincident to each other.

Since the surroundings are similar and these surroundings are fixed also, there will be no kinetic energy and also the gravitational potential energy for both the systems is also equal.

Practically, in thermodynamic studies, the Gravitational potential energy of the system is negligible. This state holds significance because it leads to the termination of all the processes that are spontaneous in nature.

4 0

4 years ago

Vehicles begin arriving at an amusement park 1 hour before the park opens, at a rale of four vehicles per minute. The gale to th

Alina [70]

Answer:

a) ≈ 30 mins

b) 8 vpm

Explanation:

<u>a) Determine how long after the first vehicle arrival will the queue dissipate</u>

The time after the arrival of the first vehicle for the queue to dissipate

= 29.9 mins ≈ 30 mins

<u>b) Determine the average service rate at the parking lot gate </u>

U = A / t

where : A = 240 vehicles , t = 30

U = 240 / 30 = 8 Vpm

attached below is a detailed solution of the given problems above

3 0

3 years ago