ENGINEERING COLLEGE

Assume that in orthogonal cutting the rake angle is 15° and the coefficient of friction is 0.15. a. Determine the percentage change in chip thickness when the coefficient of friction is doubled. Justify your answer. b. Determine the percentage change in chip thickness when the rake angle is increased to 25o . Justify your answer.

Answers

Answer 1

Answer:

Answer:

Δr=20.45 %

Explanation:

Given that

Rake angle α = 15°

coefficient of friction ,μ = 0.15

The friction angle β

tanβ = μ

tanβ = 0.15

β=8.83°

2φ + β - α = 90°

φ=Shear angle

2φ + 8.833° - 15° = 90°

φ = 48.08°

Chip thickness r given as

$r=(tan\phi)/(cos\alpha +sin\alpha\ tan\phi)$

$r=(tan48.08^(\circ))/(cos15^(\circ) +sin15^(\circ)\ tan48.08^(\circ))$

r=0.88

New coefficient of friction ,μ' = 0.3

tanβ' = μ'

tanβ' = 0.3

β'=16.69°

2φ' + β' - α = 90°

φ'=Shear angle

2φ' + 16.69° - 25° = 90°

φ' = 49.15°

Chip thickness r' given as

$r'=(tan\phi')/(cos\alpha +sin\alpha\ tan\phi')$

$r'=(tan44.15^(\circ))/(cos49.15^(\circ) +sin49.15^(\circ)\ tan44.15^(\circ))$

r'=0.70

Percentage change

$\Delta r=(r-r')/(r)* 100$

$\Delta r=(0.88-0.70)/(0.88)* 100$

Δr=20.45 %

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Air expands adiabatically through a nozzle from a negligible initial velocity to a final velocity of 300 m/s, what is the temperature drop of the air, if air is assumed to be an ideal gas for which CP = (7/2)R?

Answers

The temperature drop of air if air is assumed to be an ideal gas for which C_p = ⁷/₂R is; Δt = 1546 K

We are given;

Final velocity; v₂ = 300 m/s

C_p = ⁷/₂R

At constant pressure, the change in enthalpy is;

Δh = C_p × Δt

Now, from first law of thermodynamics;

h₂ + (v₂²/2) = h₁ + (v₁²/2)

We are told initial velocity is negligible and as such v₁ = 0 m/s

Thus;

h₂ + (v₂²/2) = h₁ + 0

(h₁ - h₂) = (v₂²/2)

Thus; Δh = v₂²/2

Finally;

C_p × Δt = v₂²/2

Δt = v₂²/2/(C_p)

Δt = (300²/2)/(⁷/₂R)

where R is ideal gas constant = 8.314 Kj/kg.mol

Thus;

Δt = (300²/2)/(⁷/₂ × 8.314)

Δt = 1546 K

Read more at; brainly.com/question/24188841

Answer:

ΔH+U²/2=0

and

ΔH= $C{p$ ×ΔT

∴to get the temperature drop of air, you make ΔT subject of the formula

ΔT=-U²/2Cp

=-300²/2× $(7)/(2)$ ×8.314

∴ΔT=-1546K

Explanation:

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Answers

Answer:

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Explanation:

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Answers

Answer:

The problem is that the pumps would consume more energy than the generators would produce.

Explanation:

Water has a potential energy associated with the height it is at. The higher it is, the higher the potential energy. When water flows down into the turbines that energy is converted to kinetic energy and then into electricity.

A pump uses electricity to add energy to the water to send it to a higher potential energy state.

Ideally no net energy woul be hgenerate or lost, because the generators would release the potential energy and pumps would store it again in the water. However the systems are not ideal, everything has an efficiency and losses. The losses would accumulate and the generator would be generating less energy than the pumps consume, so that system wastes energy.

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Answers

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Given the latent heat of fusion (melting) and the latent heat of vaporisation for water are Δhs = 333.2 kJ/kg and Δhv = 2257 kJ/kg, respectively. Use these values to estimate the total energy required to melt 100 kg of ice at 0 °C and boil off 40 kg of water at 100 °C. a) 239,028 kJ b) 95,250 kJ c) 185,500 kJ d) 362,628 kJ e) 123,600 kJ

Answers

Answer:

C)185,500 KJ

Explanation:

Given that

Latent heat fusion = 333.23 KJ/kg

Latent heat vaporisation = 333.23 KJ/kg

Mass of ice = 100 kg

Mass of water = 40 kg

Mass of vapor=60 kg

Ice at 0°C ,first it will take latent heat of vaporisation and remain at constant temperature 0°C and it will convert in to water.After this water which at 0°C will take sensible heat and gets heat up to 100°C.After that at 100°C vapor will take heat as heat of vaporisation .

Sensible heat for water Q

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