What is not required for current to flow through a conductor

A plant has ten machines and currently operates two 8-hr shift per day, 5 days per week, 50 weeks per year. the ten machines pro

Zarrin [17]

Answer:

83.6%

Explanation:

On its current schedule, the plant can theoretically produce ...

(30 pc/h/machine)(10 machine)(8 h/shift)(2 shift/day)(5 day/wk)(50 wk/yr)

= (30)(10)(8)(2)(5)(50) pc/yr = 1,200,000 pc/yr

__

On the proposed schedule, this production becomes ...

(30 pc/mach)(10 mach)(8 h/sh)(3 sh/da)(6 da/wk)(51 wk/yr)

= (30)(10)(8)(3)(6)(51) pc/yr = 2,203,200 pc/yr

The increase in capacity is ...

(2,203,200/1,200,000 -1) × 100% = 83.6% . . . capacity increase

_____

<em>Additional comment</em>

The number of parts per shift did not change. Only the number of shifts per year changed. It went up by a factor of (3/2)(6/5)(51/50) = 1.836. Hence the 83.6% increase in capacity.

We have to assume that maintenance and repair are done as effectvely as before in the reduced down time that each machine has.

3 0

2 years ago

Describe with an example how corroded structures can lead to environment pollution?

raketka [301]

According to EonCoat, corrosion is the process of decay on a material caused by a chemical reaction with its environment. Corrosion of metal occurs when an exposed surface comes in contact with a gas or liquid, and the process is accelerated by exposure to warm temperature, acids, and salts.” (1)
Although the word ‘corrosion’ is used to describe the decay of metals, all natural and man-made materials are subject to decay, and the level of pollutants in the air can speed up this process.

5 0

3 years ago

Steam enters a turbine at 120 bar, 508oC. At the exit, the pressure and quality are 50 kPa and 0.912, respectively. Determine th

Archy [21]

Answer:

The turbine produces <u>955.53 KW</u> power.

Explanation:

Taking the turbine as a system. Applying Law of Conservation of Energy:

m(h₁ - h₂) - Heat Loss = P

where,

m = mass flow rate of steam = 1.31 kg/s

h₁ = enthalpy at state 1 (120 bar and 508°C)

h₂ = enthalpy at state 2 (50 KPa and x = 0.912)

Heat Loss = 225 KW

P = Power generated by turbine = ?

First, we find h₁ from super steam tables:

At,

T = 508°C

P = 120 bar = 12000 KPa = 12 MPa

we find that steam is in super-heated state with enthalpy:

Due to unavailibility of values in chart we approximate the state to 500° C and 12.5 MPa:

h₁ = 3343.6 KJ/kg

Now, for state 2, we have:

P = 50 KPa and x = 0.912

From saturated steam table:

h₂ = hf₂ + x(hfg₂) = 340.54 KJ/kg + (0.912)(2304.7 KJ/kg)

h₂ = 2442.4 KJ/kg

Now, using values in the conservation equation:

(1.31 kg/s)(3343.6 KJ/kg - 2442.4 KJ/kg) - 225 KW = P

5 0

3 years ago

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

A 4-L pressure cooker has an operating pressure of 175 kPa. Initially, one-half of the volume is filled with liquid and the othe

vodomira [7]

Answer:

the highest rate of heat transfer allowed is 0.9306 kW

Explanation:

Given the data in the question;

Volume = 4L = 0.004 m³

V $_f$ = V $_g$ = 0.002 m³

Using Table ( saturated water - pressure table);

at pressure p = 175 kPa;

v $_f$ = 0.001057 m³/kg

v $_g$ = 1.0037 m³/kg

u $_f$ = 486.82 kJ/kg

u $_g$ 2524.5 kJ/kg

h $_g$ = 2700.2 kJ/kg

So the initial mass of the water;

m₁ = V $_f$ /v $_f$ + V $_g$ /v $_g$

we substitute

m₁ = 0.002/0.001057 + 0.002/1.0037

m₁ = 1.89414 kg

Now, the final mass will be;

m₂ = V/v $_g$

m₂ = 0.004 / 1.0037

m₂ = 0.003985 kg

Now, mass leaving the pressure cooker is;

m $_{out$ = m₁ - m₂

m $_{out$ = 1.89414 - 0.003985

m $_{out$ = 1.890155 kg

so, Initial internal energy will be;

U₁ = m $_f$ u $_f$ + m $_g$ u $_g$

U₁ = (V $_f$ /v $_f$ )u $_f$ + (V $_g$ /v $_g$ )u $_g$

we substitute

U₁ = (0.002/0.001057)(486.82) + (0.002/1.0037)(2524.5)

U₁ = 921.135288 + 5.030387

U₁ = 926.165675 kJ

Now, using Energy balance;

E $_{in$ - E $_{out$ = ΔE $_{sys$

QΔt - m $_{out$ h $_{out$ = m₂u₂ - U₁

QΔt - m $_{out$ h $_g$ = m₂u $_g$ - U₁

given that time = 75 min = 75 × 60s = 4500 sec

so we substitute

Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675

Q(4500) - 5103.7965 = 10.06013 - 926.165675

Q(4500) = 10.06013 - 926.165675 + 5103.7965

Q(4500) = 4187.690955

Q = 4187.690955 / 4500

Q = 0.9306 kW

Therefore, the highest rate of heat transfer allowed is 0.9306 kW

5 0

3 years ago