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2 years ago

3

In a turning operation on stainless steel with hardness = 200 HB, the cutting speed = 200 m/min, feed = 0.25 mm/rev, and depth o

f cut = 3.5 mm. How much power will the lathe draw in performing this operation if its mechanical efficiency = 90%?

1 answer:

Sveta_85 [38]2 years ago

8 0

Answer:

Gross power perform this operation is 9.07 kW

Explanation:

Given :

Cutting speed $v = 200 \times 10^{3}$ $\frac{mm}{min}$

Feed $f = 0.25$ mm

Depth of cut $d = 3.5$ mm

Mechanical efficiency $\eta = 0.90$

First find the material removal rate,

$R = v \times f \times d$

$R = 200 \times 10^{3} \times 0.25 \times 3.5$

$R = 175000$ $\frac{mm^{2} }{min}$

$R = \frac{175000}{60} = 2916.67$ $\frac{mm^{2} }{sec}$

From the standard table,

Specific cutting energy $U = 2.8 \frac{J}{mm^{2} }$

Cutting power,

$P = 2916.67 \times 2.8$

$P = 8166.67$ W

We know that gross power is given by,

$P_{gross} = \frac{P}{\eta}$

$P_{gross} = \frac{8166.67}{0.90}$

$P_{gross} = 9074.07$ W

$P_{gross} = 9.07$ kW

Therefore, gross power perform this operation is 9.07 kW

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Answer:

Temperature at center of apples = 11.2⁰C

Temperature at surface of apples = 2.7⁰C

Amount of Heat transferred = 17.2kJ

Explanation:

The properties of apple are given as:

k = 0.418 W/m.°C

ρ = 840 kg/m³

Cр = 3.81 kJ/kg.°C

α = 1.3*10 ⁻⁷ m²/s

h = 8 W/m².°C

d = 0.09m

r = 0.045m

t = 1 hour = 3600s

<h2>Solution</h2>

Biot number is given as:

$Bi = \frac{hr}{k}= \frac{8\cdot0.045}{0.418}=0.861$

The constants λ₁ and A₁ corresponding to Biot number (from the table) are:

λ₁ = 1.476

A₁ = 1.239

Fourier Number is:

$T = \frac{a\cdot{t}}{r^2} = \frac{(1.3\cdot10^{-7})(3600)}{0.045^2}= 0.231> 0.2$

As Fourier Number > 0.2 , one term approximates solutions are applicable

The temperature at the center of apples, The temperature at surface of apples and Amount of heat transfer is found in the ATTACHMENT.

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2 years ago

Water (cp = 4180 J/kg·°C) enters the 2.5 cm internal diameter tube of a double-pipe counter-flow heat exchanger at 17°C at a rat

aksik [14]

Answer:

Length = 129.55m, 129.55m

Explanation:

Given:

cp of water = 4180 J/kg·°C

Diameter, D = 2.5 cm

Temperature of water in = 17°C

Temperature of water out = 80°C

mass rate of water =1.8 kg/s.

Steam condensing at 120°C

Temperature at saturation = 120°C

hfg of steam at 120°C = 2203 kJ/kg

overall heat transfer coefficient of the heat exchanger = 700 W/m2 ·°C

U = 700 W/m2 ·°C

Since Temperature of steam is at saturation,

temperature of steam going in = temperature of steam out = 120°C

Energy balance:

Heat gained by water = Heat loss by steam

Let specific capacity of steam = 2010kJ/Kg .°C

Find attached the full solution to the question.

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2 years ago

Water is flowing in a metal pipe. The pipe OD (outside diameter) is 61 cm. The pipe length is 120 m. The pipe wall thickness is

Yuki888 [10]

Answer:

1113kN

Explanation:

The ouside diameter OD of the pipe is 61cm and the thickness T is 0.9cm, so the inside diameter ID will be:

Inside Diameter = Outside Diameter - Thickness

Inside Diameter = 61cm - 0.9cm = 60.1cm

Converting this diameter to meters, we have:

$60.1cm*\frac{1m}{100cm}=0.601m$

This inside diameter is useful to calculate the volume V of water inside the pipe, that is the volume of a cylinder:

$V_{water}=\pi r^{2}h$

$V_{water}=\pi (\frac{0.601m}{2})^{2}*120m$

$V_{water}=113.28m^{3}$

The problem gives you the water density d as 1.0kg/L, but we need to convert it to proper units, so:

$d_{water}=1.0\frac{Kg}{L}*\frac{1L}{1000cm^{3}}*(\frac{100cm}{1m})^{3}$

$d_{water}=1000\frac{Kg}{m^{3}}$

Now, water density is given by the equation $d=\frac{m}{V}$ , where m is the water mass and V is the water volume. Solving the equation for water mass and replacing the values we have:

$m_{water}=d_{water}.V_{water}$

$m_{water}=1000\frac{Kg}{mx^{3}}*113.28m^{3}$

$m_{water}=113280Kg$

With the water mass we can find the weight of water:

$w_{water}=m_{water} *g$

$w_{water}=113280kg*9.8\frac{m}{s^{2}}$

$w_{water}=1110144N$

Finally we find the total weight add up the weight of the water and the weight of the pipe,so:

$w_{total}=w_{water}+w_{pipe}$

$w_{total}=1110144N+2500N$

$w_{total}=1112644N$

Converting this total weight to kN, we have:

$1112644N*\frac{0.001kN}{1N}=1113kN$

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2 years ago

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Answer:

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Explanation:

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2 years ago

A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the

bagirrra123 [75]

The magnitude of applied stress in the direction of 101 is 12.25 MPA and in the direction of 011, it is not defined.

Explanation:

Given:

tensile stress is applied parallel to the [100] direction

Shear stress is 0.5 MPA.

To calculate:

The magnitude of applied stress in the direction of [101] and [011].

Formula:

zcr=σ cosФ cosλ

Solution:

For in the direction of 101

cosλ = (1)(1)+(0)(0)+(0)(1)/√(1)(2)

cos λ = 1/√2

The magnitude of stress in the direction of 101 is 12.25 MPA

In the direction of 011

We have an angle between 100 and 011

cosλ = (1)(0)+(0)(-1)+(0)(1)/√(1)(2)

cosλ = 0

Therefore the magnitude of stress to cause a slip in the direction of 011 is not defined.

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2 years ago