I want to solve the question

Tool life testing on a lathe under dry cutting conditions gauge 'n' and 'C' of Taylor tool life equation as 0.12 and 130 m/min.

Tom [10]

Answer:

So % increment in tool life is equal to 4640 %.

Explanation:

Initially n=0.12 ,V=130 m/min

Finally C increased by 10% , V=90 m/min

Let's take the tool life initial condition is $T_1$ and when C is increased it become $T_2$ .

As we know that tool life equation for tool

$VT^n=C$

At initial condition $130\times (T_1)^{0.12}=C$ ------(1)

At final condition $90\times (T_2)^{0.12}=1.1C$ -----(2)

From above equation

$\dfrac{130\times (T_1)^{0.12}}{90\times (T_2)^{0.12}}=\dfrac{1}{1.1}$

$T_2=47.4T_1$

So increment in tool life = $\dfrac{T_2-T_1}{T_1}$

= $\dfrac{47.4T_1-T_1}{T_1}$

So % increment in tool life is equal to 4640 %.

7 0

3 years ago

A 200-L tank (see Fig. P4.107) initially contains water at 100 kPa and a quality of 1%. Heat is transferred to the water, thereb

Tomtit [17]

Answer:

25.46 MJ

Explanation:

continuity : m3 - m1 = -me

Energy Equation: m3u3 - m1u1 = -meue + 1Q3

See the image attached (Well typed out format)

5 0

4 years ago

A car is moving at 68 miles per hour. The kinetic energy of that car is 5 × 10 5 J.How much energy does the same car have when i

Blababa [14]

Answer:

The car has an energy of 1.017 × 10⁵ J when it moves at 97 miles per hour

Explanation:

Given the data in the question;

Initial velocity v₁ = 68 miles per hour = 30.398 meter per seconds

let mass of the car be m

kinetic energy of that car is 5 × 10⁵ J

so

E₁ = $\frac{1}{2}$ mv²

we substitute

5 × 10⁵ = $\frac{1}{2}$ × m × ( 30.398 )²

5 × 10⁵ = m × 462.019

m = 5 × 10⁵ / 462.019

m = 1082.2065 kg

Now, Also given that; v₂ = 97 miles per hour = 43.362 meter per seconds

E₂ = $\frac{1}{2}$ mv₂²

we substitute

E₂ = $\frac{1}{2}$ × 1082.2065 × ( 43.362 )²

E₂ = $\frac{1}{2}$ × 1082.2065 × 1880.263

E₂ = 1.017 × 10⁵ J

Therefore, The car has an energy of 1.017 × 10⁵ J when it moves at 97 miles per hour

6 0

3 years ago

How much power is needed to operate a Carnot heat pump if the pump receives heat 10°C and delivers 50 kW of heat at 40°C? at A)

mariarad [96]

Answer:

Power needed to pump=4.79 KW.

Explanation:

Given that: $T_{1}=283K,T_{2}=313K,Q_{H}=50KW$

We know that coefficient of performance of heat pump

COP= $\dfrac{T_{H}}{T_{H}-T_{L}}$

So COP= $\dfrac{313}{313-283}$

COP=10.43

COP= $\frac{Q_{H}}{W_{in}}$

10.43 = $\frac{50}{W_{in}}$

$W_{in}$ =4.79 KW

So power needed to pump=4.79 KW.

3 0

3 years ago

The ???? − i relationship for an electromagnetic system is given by ???? = 1.2i1/2 g where g is the air-gap length. For current

Artemon [7]

Answer:

a) The mechanical force is -226.2 N

b) Using the coenergy the mechanical force is -226.2 N

Explanation:

a) Energy of the system:

$\lambda =\frac{1.2*i^{1/2} }{g} \\i=(\frac{\lambda g}{1.2} )^{2}$

$\frac{\delta w_{f} }{\delta g} =\frac{g^{2}\lambda ^{3} }{3*1.2^{2} }$

$f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{g^{2}\lambda ^{3} }{3*1.2^{2} }$

If i = 2A and g = 10 cm

$\lambda =\frac{1.2*i^{1/2} }{g} =\frac{1.2*2^{1/2} }{10x10^{-2} } =16.97$

$f_{m}=-\frac{g^{2}\lambda ^{3} }{3*1.2^{2} }=-\frac{16.97^{3}*2*0.1 }{3*1.2^{2} } =-226.2N$

b) Using the coenergy of the system:

$f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{1.2*2*i^{3/2} }{3*g^{2} }=-\frac{1.2*2*2^{3/2} }{3*0.1^{2} } =-226.2N$

8 0

3 years ago