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3 years ago

What historical event allowed both aerospace fields to make enormous strides forward? *

Engineering

1 answer:

galina1969 [7]3 years ago

6 0

Answer:

The world wars. Most notably world War II

Explanation:

The demand for aircrafts during these events led to extensive research into the design of aircrafts. Aircraft advanced within these years from a simple design to a more complex design; capable of carrying fire power and even became bomb equipped. Also, the material of choice of production moved from wood to metal and the engine was improved on to gain more speed and maneuverability.

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What is manufacturing machining ?

zzz [600]

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6. Which type of document would Alex, a computer programmer, sign in order to protect his intellectual property rights prior to

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Answer:

Explanation:

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3 years ago

A peripheral milling operation is performed on the top surface of a rectangular workpart which is 400 mm long and 60 mm wide. Th

oee [108]

Answer:

N = 278.5 rpm

F = 348.15 mm/min

machine time = 10.34 seconds

material removal rate = 33420000 mm³/min

Explanation:

given data

length L = 400 mm

width W = 60 mm

diameter D = 80 mm

no of cutting n = 5

velocity V = 70 m/min = 70000 mm/min

chip load P = 0.25 mm/tooth

depth = 5 mm

solution

we know velocity that is express as

velocity V = $\pi$ D N ........1

N = $\frac{70000}{\pi \times 80}$

N = 278.5 rpm

and

now we get feed rate in milling operation is

F n P N

F = 5 × 0.25 × 278.5

F = 348.15 mm/min

and

now we get actual machine time to make 1 pass across surface of worl is

machine time = $\frac{W}{F}$

machine time = $\frac{60}{348.15}$

machine time = 0.1723 min = 10.34 seconds

and

max material removal rate during these cut is

material removal rate = A × d × N

material removal rate = 400 × 60 × 5 × 278.5

material removal rate = 33420000 mm³/min

7 0

3 years ago

A piston-cylinder assembly contains 5kg of water that undergoes a series of processes to form a thermodynamic cycle. Process 1à2

Nikitich [7]

Answer:

The net work done is 272.38 kJ

Explanation:

The parameters given are;

Mass of water = 5 kg

p₁ = 20 bar

T₁ = 360°C

v₁ = 0.141147 m³/kg

Process 1 to 2 = Constant pressure process

p₂ = 20 bar

Process 2 to 3 = Constant volume process

p₃ = 5 bar

Process 3 to 4 = Constant pressure process

Process 4 to 1 = Polytropic process pv = Constant

For Stage 1 to 2, we have;

p₂ = 20 bar

From the steam tables for superheated steam, we have;

T₂ = 212.385°C

v₂ = 0.0995805 m³/kg

Work done = p₂×(v₂ - v₁) = 2×10⁶ × (0.0995805 - 0.141147 ) = -83133 J/kg

For the 5 kg, we have;

$W_{1-2}$ = -83133 J/kg × 5 = -415,665 J

Stage 2 to 3: Constant volume cooling

v₂ = v₃ = 0.0995805 m³/kg

p₃ = 5 bar

T₃ = 151.836°C

(0.0995805 - 0.00109256)/(0.374804 - 0.00109256) = 0.2635 liquid vapor mixture

Work done, $W_{2-3}$ = 0

Stage 3 to 4: Constant pressure heating

p₃ = p₄ = 5 bar

v₄/T₄ = v₃/T₃

v₄ = 0.374804 m³/kg

T₄ = v₄×T₃/v₃ = 0.374804*(273.15 + 151.836)/0.0995805 = 1599.6 K = 1326.4 °C

Work done = p₄×(v₄ - v₃) = 5×10⁵ × (0.374804 - 0.0995805 ) = 137611.75 J/kg

For the 5 kg, we have;

$W_{3-4}$ = 137,611.75 J/kg × 5 = 688,058.75 J

Stage 4 to 1: Polytropic process

$\dfrac{p_{4}}{p_{1}} = \left (\dfrac{V_{1}}{V_{4}} \right )^{n} = \left (\dfrac{T_{4}}{T_{1}} \right )^{\dfrac{n}{n-1}}$

Which gives;

$\dfrac{5}{20} = \left (\dfrac{0.141147 }{0.374804} \right )^{n}$

n = log(5/20) ÷log(0.141147/0.374804) = 1.42

Work done, $W_{pdv}$ , is given as follows;

$W_{pdv} = \dfrac{p_4 \times v_4 -p_4 \times v_4 }{n-1}$

Which gives;

$W_{pdv} = \dfrac{5\times 0.374804 -20\times 0.141147 }{1.42-1} = -2.259 \, J$

For the 5 kg, we have;

$W_{4-1}$ = -2.259 J/kg × 5 = -11.2967 J

The net work done, $W_{Net}$ , is therefore;

$W_{Net}$ = $W_{1-2}$ + $W_{3-4}$ + $W_{4-1}$

-415,665 + 688,058.75 -11.2967 = 272,382.45 J = 272.38 kJ.

5 0

3 years ago