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zloy xaker [14]
3 years ago
14

In an orthogonal cutting operation, if the friction coefficient between the tool and the chip is decreased while the material an

d the rake angle are kept constant, then the shear angle will increase.
True
False
Engineering
2 answers:
levacccp [35]3 years ago
5 0

Answer:

The Correct Answer is False.              

Explanation:

It is logical that the cutting force increases as the depth of cut increases and rake angle decreases. Deeper cuts remove more material, thus requiring a higher cutting force. As the rake angle, α, decreases, the shear angle, φ , decreases and hence shear energy dissipation and cutting forces increase.

Alexus [3.1K]3 years ago
3 0

Answer:

True

Explanation:

- The shear plane angle is given by the following formula (Merchant's Equation):

                             ∅ = 45 + α/2 - β/2

Where,                  α: Rake Angle

                             β: Friction angle

- It is seen from the relation above that shear plane angle depends on friction angle (β) and rake angle (α).

- The friction angle (β) is related by the friction coefficient (u) with the following relation:

                            u = tan ( β )

                            β = arctan (u)

- The expression becomes:

                            ∅ = 45 + α/2 - arctan (u)/2

- Its given that rake angle (α) remains constant, while friction coefficient (u) between the tool and the chip is decreased.

- For the equation to hold true then shear plane angle (∅) must increase.                          

                             

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Question 1 An ice making machine operates on the ideal vapour-compression cycle using refrigerant R134a. The refrigerant enters
-BARSIC- [3]

Answer:

check the attached image for the correct solution

Explanation:

In an ideal vapor-compression refrigeration cycle, the refrigerant enters the compressor as a saturated vapor and is cooled to the saturated liquid state in the condenser. It is then throttled to the evaporator pressure and vaporizes as it absorbs heat from the refrigerated space.

the correct workings/calculation is in the attached image

4 0
4 years ago
An engine indicator is used to determine the following: (a) speed (d) m.e.p, and IHP (e) BHP (b) temperature (c) volume of cylin
Greeley [361]

Answer:

Option (d) MEP and IHP

Explanation:

MEP stands for Mean Effective Pressure and IHP stands for Indicated Horse Power

In engines (Internal Combustion), engine indicator is generally to indicate the indicate the changes in pressure inside the cylinder of an Internal Combustion Engine or IC engines. Once, Mean Effective Pressure of the engine is calculated it further helps to calculate the Horse power and both these quantities, i.e., MEP and IHP are displayed on the engine indicator.

8 0
3 years ago
A gas is compressed from an initial volume of 0.42 m3 to a final volume of 0.12 m3. During the quasi-equilibrium process, the pr
Georgia [21]

Answer:

W=-52 800\ \text{J}=-52.8\ \text{kJ}

Explanation:

First I sketched the compression of the gas with the help of the given pressure change process relation. That is your pressure change due to change in volume.

To find the area underneath the curve (the same as saying to find the work done) you should integrate the given relation for pressure change:

W=\int_{0.42}^{0.12}-1200V+500dV=-52.8\ \text{kJ}

6 0
3 years ago
A hole of diameter D = 0.25 m is drilled through the center of a solid block of square cross section with w = 1 m on a side. The
attashe74 [19]

Answer:

q=4.013\:\:kW\\\\T_1=278.91\:\:^{\circ}C\\\\T_2=275.82\:\:^{\circ}C

Explanation:

R_{conv,1}=(h_1\pi D_1L)^{-1}=(50*0.25*2)^{-1}=0.01273\:\:K/W\\\\R_{conv,2}=(h^2*4wL)^{-1}=(4*4*1)^{-1}=0.0625\:\:K/W\\\\R_{cond(2D)}=(Sk)^{-1}=(8.59*150)^{-1}=0.00078\:\:K/W

So, heat rate can be calculated as follows:

q=\frac{T_{\infty,1}-T_{\infty,2}}{R_{conv,1}+R_{conv,2}+R_{cond(2D)}} =\frac{330-25}{0.076} =4.013\:\:kW

Surface temperatures can be calculated as follows:

T_1=T_{\infty,1}-qR_{conv,1}=330-51.09=278.91\:\:^{\circ}C\\\\T_2=T_{\infty,2}+qR_{conv,2}=25+250.82=275.82\:\:^{\circ}C

6 0
4 years ago
Which of the following ranges depicts the 2% tolerance range to the full 9 digits provided?
Lyrx [107]

Answer:

the only one that meets the requirements is option C .

Explanation:

The tolerance of a quantity is the maximum limit of variation allowed for that quantity.

To find it we must have the value of the magnitude, its closest value is the average value, this value can be given or if it is not known it is calculated with the formula

         x_average = ∑ x_{i} / n

The tolerance or error is the current value over the mean value per 100

         Δx₁ = x₁ / x_average

         tolerance = | 100 -Δx₁  100 |

bars indicate absolute value

let's look for these values ​​for each case

a)

    x_average = (2.1700000+ 2.258571429) / 2

    x_average = 2.2142857145

fluctuation for x₁

        Δx₁ = 2.17000 / 2.2142857145

        Tolerance = 100 - 97.999999991

        Tolerance = 2.000000001%

fluctuation x₂

        Δx₂ = 2.258571429 / 2.2142857145

        Δx2 = 1.02

        tolerance = 100 - 102.000000009

        tolerance 2.000000001%

b)

    x_average = (2.2 + 2.29) / 2

    x_average = 2,245

fluctuation x₁

         Δx₁ = 2.2 / 2.245

         Δx₁ = 0.9799554

         tolerance = 100 - 97,999

         Tolerance = 2.00446%

fluctuation x₂

          Δx₂ = 2.29 / 2.245

          Δx₂ = 1.0200445

          Tolerance = 2.00445%

c)

   x_average = (2.211445 +2.3) / 2

   x_average = 2.2557225

       Δx₁ = 2.211445 / 2.2557225 = 0.9803710

       tolerance = 100 - 98.0371

       tolerance = 1.96%

       Δx₂ = 2.3 / 2.2557225 = 1.024624

       tolerance = 100 -101.962896

       tolerance = 1.96%

d)

   x_average = (2.20144927 + 2.29130435) / 2

   x_average = 2.24637681

       Δx₁ = 2.20144927 / 2.24637681 = 0.98000043

       tolerance = 100 - 98.000043

       tolerance = 2.000002%

       Δx₂ = 2.29130435 / 2.24637681 = 1.0200000017

       tolerance = 2.0000002%

e)

   x_average = (2 +2,3) / 2

   x_average = 2.15

   Δx₁ = 2 / 2.15 = 0.93023

   tolerance = 100 -93.023

   tolerance = 6.98%

   Δx₂ = 2.3 / 2.15 = 1.0698

   tolerance = 6.97%

Let's analyze these results, the result E is clearly not in the requested tolerance range, the other values ​​may be within the desired tolerance range depending on the required precision, for the high precision of this exercise the only one that meets the requirements is option C .

4 0
4 years ago
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