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Lunna [17]
3 years ago
9

In a surface grinding operation, the wheel diameter = 8.0 in, wheel width = 1.0 in, wheel speed = 6000 ft/min, work speed = 40 f

t/min, infeed = 0.002 in, and crossfeed = 0.20 in. The number of active grits per square inch of wheel surface = 400. Determine
(a) the average length per chip,
(b) the metal removal rate
(c) the number of chips formed per unit and the average volume per chip?
(d) If the tangential cutting force on the workpiece is 50 lb, what is the specific energy calculated for this job?
Engineering
1 answer:
Katen [24]3 years ago
7 0

Answer: the answer will be d because it is the right one to be

Explanation:

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Conditions of special concern: i. Suggest two reasons each why distillation columns are run a.) above or b.) below ambient press
lutik1710 [3]

Solution :

Methods for selling pressure of a distillation column :

a). Set, \text{based on the pressure required to condensed} the overhead stream using cooling water.

  (minimum of approximate 45°C condenser temperature)

b). Set, \text{based on highest temperature} of bottom product that avoids decomposition or reaction.

c). Set, \text{based on available highest } not utility for reboiler.

Running the distillation column above the ambient pressure because :

The components to be distilled have very high vapor pressures and the temperature at which they can be condensed at or below the ambient pressure.

Run the reactor at an evaluated temperature because :

a). The rate of reaction is taster. This results in a small reactor or high phase conversion.

b). The reaction is endothermic and equilibrium limited increasing the temperature shifts the equilibrium to the right.

Run the reaction at an evaluated pressure because :

The reaction is gas phase and the concentration and hence the rate is increased as the pressure is increased. This results in a smaller reactor and /or higher reactor conversion.

The reaction is equilibrium limited and there are few products moles than react moles. As increase in pressure shifts the equilibrium to the right.

7 0
3 years ago
In the fully developed region of flow in a circular pipe, does the velocity profile change in the flow direction?
taurus [48]

Answer:

<em>No, the velocity profile does not change in the flow direction.</em>

Explanation:

In a fluid flow in a circular pipe, the boundary layer thickness increases in the direction of flow, until it reaches the center of the pipe, and fill the whole pipe. If the density, and other properties of the fluid does not change either by heating or cooling of the pipe, <em>then the velocity profile downstream becomes fully developed, and constant, and does not change in the direction of flow.</em>

3 0
3 years ago
A 60-cm-high, 40-cm-diameter cylindrical water tank is being transported on a level road. The highest acceleration anticipated i
dlinn [17]

Answer:

h_{max} = 51.8 cm

Explanation:

given data:

height of tank = 60cm

diameter of tank =40cm

accelration = 4 m/s2

suppose x- axis - direction of motion

z -axis - vertical direction

\theta = water surface angle with horizontal surface

a_x =accelration in x direction

a_z =accelration in z direction

slope in xz plane is

tan\theta = \frac{a_x}{g +a_z}

tan\theta = \frac{4}{9.81+0}

tan\theta =0.4077

the maximum height of water surface at mid of inclination is

\Delta h = \frac{d}{2} tan\theta

            =\frac{0.4}{2}0.4077

\Delta h  0.082 cm

the maximu height of wwater to avoid spilling is

h_{max} = h_{tank} -\Delta h

            = 60 - 8.2

h_{max} = 51.8 cm

the height requird if no spill water is h_{max} = 51.8 cm

3 0
4 years ago
A well-insulated, rigid tank has a volume of 1 m3and is initially evacuated. A valve is opened,and the surrounding air enters at
DiKsa [7]

Answer:

0.5 kW

Explanation:

The given parameters are;

Volume of tank = 1 m³

Pressure of air entering tank = 1 bar

Temperature of air = 27°C = 300.15 K

Temperature after heating  = 477 °C = 750.15 K

V₂ = 1 m³

P₁V₁/T₁ = P₂V₂/T₂

P₁ = P₂

V₁ = T₁×V₂/T₂ = 300.15 * 1 /750.15 = 0.4 m³

dQ = m \times c_p \times (T_2 -T_1)

For ideal gas, c_p = 5/2×R = 5/2*0.287 = 0.7175 kJ

PV = NKT

N = PV/(KT) = 100000×1/(750.15×1.38×10⁻²³)

N = 9.66×10²⁴

Number of moles of air = 9.66×10²⁴/(6.02×10²³) = 16.05 moles

The average mass of one mole of air = 28.8 g

Therefore, the total mass = 28.8*16.05 = 462.135 g = 0.46 kg

∴ dQ = 0.46*0.7175*(750.15 - 300.15) = 149.211 kJ

The power input required = The rate of heat transfer = 149.211/(60*5)

The power input required = 0.49737 kW ≈ 0.5 kW.

3 0
3 years ago
PLEASE QUICK!!! what phrase describes an ad hominem fallacy?
Igoryamba

Answer:

personal attack

Explanation:

it is personal attack

5 0
3 years ago
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