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3 years ago

What is the difference between a single-model production line and a mixed-model production line?

1 answer:

6 0

The unique model production line is responsible for producing identical pieces. For this purpose the balancing of the assembly line is only responsible for assembling a model throughout the line.

This is a considerable difference compared to the mixed model assembly line where many models are assembled during the same production line, that is, it produces parts or products that have slight changes accommodated in them, with slight variations in their model or products of soft variety

The choice of the type of production depends on the type of company and its own demand, always prioritizing the efficiency in the operation. Generally, the mixed model tends to be chosen when demand is very large and customer demand is required to be met. In others it is considered a plant model in which half of the line is mixed and the other one is the only model in order to keep the efficiency balanced.

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Write torsion equation and explain the importance of each components.<br>

Elanso [62]

The equations are based on the following assumptions

1) The bar is straight and of uniform section
2) The material of the bar is has uniform properties.
3) The only loading is the applied torque which is applied normal to the axis of the bar.
4) The bar is stressed within its elastic limit.

Nomenclature

T = torque (Nm)
l = length of bar (m)
J = Polar moment of inertia.(Circular Sections) ( m^4)
J' = Polar moment of inertia.(Non circluar sections) ( m^4 )
K = Factor replacing J for non-circular sections.( m^4)
r = radial distance of point from center of section (m)
ro = radius of section OD (m)
τ = shear stress (N/m^2)
G Modulus of rigidity (N/m^2)
θ = angle of twist (radians)

4 0

3 years ago

3. A 4-m × 5-m × 7-m room is heated by the radiator of a steam-heating system. The steam radiator transfers heat at a rate of 10

Natali [406]

Answer:

14.52 minutes

14 minutes and 31 seconds

Explanation:

Let's first start by mentioning the specific heat of air at constant volume. We consider constant volume and NOT constant pressure because the volume of the room remains constant while pressure may vary.

Specific heat at constant volume at 27°C = 0.718 kJ/kg*K

Initial temperature of room (in kelvin) = 283.15 K

Final temperature (required) of room = 293.15 K

Mass of air in room= volume * density= (4 * 5 * 7) * (1.204 kg/m3) = 168.56kg

Heat required at constant volume: 0.718 * (change in temp) * (mass of air)

Heat required = 0.718 * (293.15 - 283.15) * (168.56) = 1,210.26 kJ

Time taken for temperature rise: heat required / (rate of heat change)

Where rate of heat change = 10000 - 5000 = 5000 kJ/hr

Time taken = 1210.26 / 5000 = 0.24205 hours

Converted to minutes = 0.24205 * 60 = 14.52 minutes

4 0

4 years ago

The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine-generator. The elevat

ankoles [38]

Answer:

a) 75%

b) 82%

Explanation:

Assumptions:

$\text{The mechanical energy for water at turbine exit is negligible.} \\ \\ \text{The elevation of the lake remains constant.}$

Properties: The density of water $\delta = 1000 kg/m^3$

Conversions:

$165 \ ft \ to \ meters = 50 m \\ \\7000 \ lbm/s \ to \ kilogram/sec = 3175 kg/s \\ \\1564 \ hp \ to \ kilowatt = 1166 kw \\ \\$

Analysis:

Note that the bottom of the lake is the reference level. The potential energy of water at the surface becomes gh. Consider that kinetic energy of water at the lake surface & the turbine exit is negligible and the pressure at both locations is the atmospheric pressure and change in the mechanical energy of water between lake surface & turbine exit are:

$e_{mech_{in}} - e_{mech_{out}} = gh - 0$

Then;

$gh = (9.8 m/s^2) (50 m) \times \dfrac{1 \ kJ/kg}{1000 m^2/s^2}$

gh = 0.491 kJ/kg

$\Delta E_{mech \ fluid} = m(e_{mech_{in}} - e_{mech_{out}} ) \\ \\ = 3175 kg/s \times 0.491 kJ/kg$

= 1559 kW

Therefore; the overall efficiency is:

$\eta _{overall} = \eta_{turbine- generator} = \dfrac{W_{elect\ out}}{\Delta E_{mech \fluid}}$

$= \dfrac{1166 \ kW}{1559 \ kW}$

= 0.75

= 75%

b) mechanical efficiency of the turbine:

$\eta_{turbine- generator} = \eta_{turbine}\times \eta_{generator}$

thus;

$\eta_{turbine} = \dfrac{\eta_{[turbine- generator]} }{\eta_{generator}} \\ \\ \eta_{turbine} = \dfrac{0.75}{0.92} \\ \\ \eta_{turbine} = 0.82 \\ \\ \eta_{turbine} = 82\%$

6 0

3 years ago

Assuming Stokes behavior, calculate the terminal settling velocity in standard air () for the following particles: (a) diameter

inna [77]

Answer:

Detailed solution is given below :

4 0

4 years ago

What type of power source is used for gmaw

nataly862011 [7]

Answer:

The GMAW process commonly uses a constant voltage power source (GMAW-CV) that allows for a relatively constant welding voltage output over a range of welding currents. For GMAW-CV the welder selects the wire feed speed (WFS) on the wire feeder unit and an appropriate voltage on the welding power supply.

4 0

3 years ago

Read 2 more answers