In order to break even, your minimum selling price must be _________

In an orthogonal cutting operation, the tool has a rake angle = 12°. The chip thickness before the cut = 0.32 mm and the cut yie

Snezhnost [94]

Answer:

The shear plane angle and shear strain are 28.21° and 2.155 respectively.

Explanation:

(a)

Orthogonal cutting is the cutting process in which cutting direction or cutting velocity is perpendicular to the cutting edge of the part surface.

Given:

Rake angle is 12°.

Chip thickness before cut is 0.32 mm.

Chip thickness is 0.65 mm.

Calculation:

Step1

Chip reduction ratio is calculated as follows:

$r=\frac{t}{t_{c}}$

$r=\frac{0.32}{0.65}$

r = 0.4923

Step2

Shear angle is calculated as follows:

$tan\phi=\frac{rcos\alpha}{1-rsin\alpha}$

Here, $\phi$ is shear plane angle, r is chip reduction ratio and $\alpha$ is rake angle.

Substitute all the values in the above equation as follows:

$tan\phi=\frac{rcos\alpha}{1-rsin\alpha}$

$tan\phi=\frac{0.4923cos12^{\circ}}{1-0.4923sin12^{\circ}}$

$tan\phi=\frac{0.48155}{0.8976}$

$\phi=28.21^{\circ}$

Thus, the shear plane angle is 28.21°.

(b)

Step3

Shears train is calculated as follows:

$\gamma=cot\phi+tan(\phi-\alpha)$

$\gamma=cot28.21^{\circ}+tan(28.21^{\circ}-12^{\circ})$ $\gamma = 2.155$ .

Thus, the shear strain rate is 2.155.

6 0

3 years ago

Develop a chase plan that matches the forecast and compute the total cost of your plan. (Negative amounts should be indicated by

vladimir2022 [97]

Answer:

The answer for the question :

"Develop a chase plan that matches the forecast and compute the total cost of your plan. (Negative amounts should be indicated by a minus sign. Leave no cells blank - be certain to enter "0" wherever required. Omit the "$" sign in your response.)"

is explained in the attachment.

Explanation:

4 0

3 years ago

Assume you have four fins, each with a mass of 8.0 grams. What is the total weight of these four fins? (Hint: watch your units!)

soldier1979 [14.2K]

The answer is 32.0 grams

7 0

3 years ago

Ô tô có khối lượng m (kg) đặt tại trung tâm h . Khoảng cách từ h tới 2 bánh xe hai bên của a (m) và b (m) , khoảng cách vết bánh

Nutka1998 [239]

Answer:

wiwhwnwhwwbbwbwiwuwhwhehehewhehehheheheehehehehhehehwh

Explanation:

jwhwhwhwhwhwwhhahwhahahwh

6 0

3 years ago

An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures befo

SVEN [57.7K]

This question is incomplete, the complete question is;

An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures before and after the isothermal compression are 150 and 300 kPa, respectively. If the net work output per cycle is 0.5 kJ, determine (a) the maximum pressure in the cycle, (b) the heat transfer to air, and (c) the mass of air. Assume variable specific heats for air.

Answer:

a) the maximum pressure in the cycle is 30.01 Mpa

b) the heat transfer to air is 0.7058 KJ

c) mass of Air is 0.002957 kg

Explanation:

Given the data in the question;

We find the relative pressure of air at 1200 K (T1) and 350 K ( T4)

so from the "ideal gas properties of air table"

Pr1 = 238

Pr4 = 2.379

we know that Pressure P1 is only maximum at the beginning of the expansion process,

so

now we express the relative pressure and pressure relation for the process 4-1

P1 = (Pr2/Pr4)P4

so we substitute

P1 = (238/2.379)300 kPa

P1 = 30012.6 kPa = 30.01 Mpa

Therefore the maximum pressure in the cycle is 30.01 Mpa

b)

the Thermal heat efficiency of the Carnot cycle is expressed as;

ηth = 1 - (TL/TH)

we substitute

ηth = 1 - (350K/1200K)

ηth = 1 - 0.2916

ηth = 0.7084

now we find the heat transferred

Qin = W_net.out / ηth

given that the net work output per cycle is 0.5 kJ

we substitute

Qin = 0.5 / 0.7084

Qin = 0.7058 KJ

Therefore, the heat transfer to air is 0.7058 KJ

c)

first lets express the change in entropy for process 3 - 4

S4 - S3 = (S°4 - S°3) - R.In(P4/P3)

S4 - S3 = - (0.287 kJ/Kg.K) In(300/150)kPa

= -0.1989 Kj/Kg.K = S1 - S2

so that; S2 - S1 = 0.1989 Kj/Kg.K

Next we find the net work output per unit mass for the Carnot cycle

W"_netout = (S2 - S1)(TH - TL)

we substitute

W"_netout = ( 0.1989 Kj/Kg.K )( 1200 - 350)K

= 169.065 kJ/kg

Finally we find the mass

mass m = W_ net.out / W"_netout

we substitute

m = 0.5 / 169.065

m = 0.002957 kg

Therefore, mass of Air is 0.002957 kg

5 0

3 years ago

In order to break even, your minimum selling price must be __________ your variable costs.