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Digiron [165]
3 years ago
9

Which of the following best describes maintenance?

Engineering
1 answer:
Mama L [17]3 years ago
6 0

Answer:

Equipment Reliability and Maintenance

SUBSTANTIAL CAPITAL investments, in the form of facilities and equipment, are required for manufacturing almost all goods of economic significance. The productivity of these investments is a fundamental element of competition among companies and nations. Events that slow or interrupt the manufacturing process or degrade the product impair the competitiveness of a manufacturing enterprise.

The term equipment reliability and maintenance (ERM) encompasses not only equipment, such as machines, tools, and fixtures, but also the technical, operational, and management activities, ranging from equipment specifications to daily operation and maintenance, required to sustain the performance of manufacturing equipment throughout its useful life. This chapter addresses all causes of diminished or degraded output. The panel considers ERM to be a significant factor in the competitiveness of manufacturing firms, an assessment supported by the case studies in the section on present practice (pp. 57-63).

Historically, the evolution of ERM can be traced from breakdown maintenance and repair to preventive maintenance to predictive maintenance. Breakdown maintenance and repair is the after-the-fact restoration of failed equipment. Preventive maintenance is the systematic servicing of equipment to reduce the possibility of failure. Predictive maintenance, in use in U.S. industry for only four or five years, is usually understood to involve the use of computer software to detect conditions that might eventually lead

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A polymeric extruder is turned on and immediately begins producing a product at a rate of 10 kg/min. An operator realizes 20 min
hodyreva [135]

Answer:

The plot of the function production rate m(t) (in kg/min) against time t (in min) is attached to this answer.

The production rate function M(t) is:

m(t)=[H(t)\cdot10+H(t-20)\cdot5-H(t-80)\cdot14+H(t-81)\cdot9]kg/min (1)

The Laplace transform of this function is:

\displaystyle m(s)=[\frac{10+5e^{-20s}-14e^{-80s}+9e^{-81s}}{s}]kg/min    (2)

Explanation:

The function of the production rate can be considered as constant functions by parts in the domain of time. To make it a continuous function, we can use the function Heaviside (as seen in equation (1)). To join all the constant functions, we consider at which time the step for each one of them appears and sum each function multiply by the function Heaviside.

For the Laplace transform we use the following rules:

\mathcal{L}[f(x)+g(x)]=\mathcal{L}[f(x)]+\mathcal{L}[g(x)]=F(s)+G(s)    (3)

\mathcal{L}[aH(x-b)]=\displaystyle\frac{ae^{-bs}}{s}    (4)

4 0
4 years ago
At the grocery store you place a pumpkin with a mass of 12.5 lb on the produce spring scale. The spring in the scale operates su
UkoKoshka [18]

Answer:

x=2.19in

Explanation:

This is the equation that relates the force and displacement of a spring

F=Kx

m=mass=12.5lbx1slug/32.14lb=0.39slug

F=mg=0.39*32.2=12.52Lbf

then we calculate the spring count in lbf / ft

K=F/x

K=5.7lbf/1in=5.7lbf/in=68.4lbf/ft

Finally we calculate the displacement with the initial equation

X=F/k

x=12.52/68.4=0.18ft=2.19in

8 0
4 years ago
Add the following vector given in rectangular form and illustrated the process graphically A = 16+j12, B= 6+j10.4
MariettaO [177]

Answer:

A=16+j12…'B=6+j10.4

Explanation:

add the following vector given in

3 0
2 years ago
Limestone (calcium carbonate) particles
Aleksandr-060686 [28]

Answer:

606506 026216n ytj dtyjjtj t

Explanation:

tyjtyjtyjtyj ty

3 0
3 years ago
An uninsulated, thin-walled pipe of 100-mm diameter is used to transport water to equipment that operates outdoors and uses the
Viefleur [7K]

Answer:

4.6 mm

Explanation:

Given data includes:

thin-walled pipe diameter = 100-mm =0.1 m

Temperature of pipe T_p = -15° C = (-15 +273)K =258 K

Temperature of water T_w = 3° C = (3 + 273)K = 276 K

Temperature of ice T_i = 0° C = (0 +273)K =273 K

Thermal conductivity (k) from the ice table = 1.94 W/m.K  ;  R = 0.05

convection coefficient Lh_l =2000 W/m².K

The energy balance can be expressed as:

q_{conduction} =q_{convention}

where;

q_{conduction} = \frac{2\pi LK(T_i-T_p)}{In(R/r)}       -------------   equation (1)

q_{convention} = \pi DLh_l(T_w-T_i)  ------------ equation(2)

Equating both equation (1) and (2); we have;

\frac{2\pi LK(T_i-T_p)}{In(R/r)} = \pi DLh_l(T_w-T_i)

Replacing the given data; we have:

\frac{2\pi (1)(1.94)(273-258)}{In(0.05/r)} = \pi (0.1)*2000(276-273)

\frac{182.84}{In(\frac{0.05}{r}) } = 1884.96

In(\frac{0.05}{r})*1884.96 = 182.84

In(\frac{0.05}{r}) = \frac{182.84}{1884.96}

In(\frac{0.05}{r}) =0.0970

\frac {0.05}{r} =e^{0.0970}

\frac {0.05}{r} =1.102

r=\frac{0.05}{1.102}

r = 0.0454

The thickness (t) of the ice layer can now be calculated as:

t = (R - r)

t = (0.05 - 0.0454)

t = 0.0046 m

t = 4.6 mm

6 0
4 years ago
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