Answer:
Tradicionalmente, se han distinguido 5 tipos de mantenimiento, que se diferencian entre sí por el carácter de las tareas que incluyen:
Explanation:
Mantenimiento Correctivo: Es el conjunto de tareas destinadas a corregir los defectos que se van presentando en los distintos equipos y que son comunicados al departamento de mantenimiento por los usuarios de los mismos.
Mantenimiento Preventivo: Es el mantenimiento que tiene por misión mantener un nivel de servicio determinado en los equipos, programando las intervencions de sus puntos vulnerables en el momento más oportuno. Suele tener un carácter sistemático, es decir, se interviene aunque el equipo no haya dado ningún síntoma de tener un problema.
Mantenimiento Predictivo: Es el que persigue conocer e informar permanentemente del estado y operatividad de las instalaciones mediante el conocimiento de los valores de determinadas variables, representativas de tal estado y operatividad. Para aplicar este mantenimiento, es necesario identificar variables físicas (temperatura, vibración, consumo de energía, etc.) cuya variación sea indicativa de problemas que puedan estar apareciendo en el equipo. Es el tipo de mantenimiento más tecnológico, pues requiere de medios técnicos avanzados, y en ocasiones, de fuertes conocimientos matemáticos, físicos y/o técnicos.
Mantenimiento Cero Horas (Overhaul): Es el conjunto de tareas cuyo objetivo es revisar los equipos a intervalos programados bien antes de que aparezca ningún fallo, bien cuando la fiabilidad del equipo ha disminuido apreciablemente de manera que resulta arriesgado hacer previsiones sobre su capacidad productiva. Dicha revisión consiste en dejar el equipo a Cero horas de funcionamiento, es decir, como si el equipo fuera nuevo. En estas revisiones se sustituyen o se reparan todos los elementos sometidos a desgaste. Se pretende asegurar, con gran probabilidad un tiempo de buen funcionamiento fijado de antemano.
Mantenimiento En Uso: es el mantenimiento básico de un equipo realizado por los usuarios del mismo. Consiste en una serie de tareas elementales (tomas de datos, inspecciones visuales, limpieza, lubricación, reapriete de tornillos) para las que no es necesario una gran formación, sino tal solo un entrenamiento breve. Este tipo de mantenimiento es la base del TPM (Total Productive Maintenance, Mantenimiento Productivo Total).
A recessed luminaire bears no marking indicating that it is ""Identified for Through- Wiring."" Is it permitted to run branch-circuit conductors other than the conductors that supply the luminaire through the integral junction box on the luminaire?
Answer:
No it is not permitted
Explanation:
It is not permitted because as per NEC 410.21 policy no other conductor is allowed to be passed through integral junction box luminaries unless such conductor supply recessed luminaries.
The marking will show that the Luminaries is of the right construction or right installation to ensure that the the conductors ( in the outer boxes ) will not be exposed to temperatures greater than the conductor rating, hence the lack of marking makes it not to be permitted.
The load on a bolt consists of an axial pull of 10 KN together with a transverse shear force of 5 KN. Find the diameter of bolt required according to I. Maximum principal stress theory; 2. Maximum shear stress theory; 3. Maximum principal strain theory, 4. Maximum strain energy theory, and 5 Maximum distortion energy theory. Take permissible tensile stress at elastic limit = 100 MPa and poisson's ratio = 0.3
Answer:
hey. its a big question. solved from *c hegg
Explanation:
A steel component with ultimate tensile strength of 800 MPa and plane strain fracture toughness of 20 MPam is known to contain a tunnel (internal) crack of length 1.4 mm. This alloy is being considered for use in a cyclic loading application for which the design stresses vary from 0 to 410 MPa. Would you recommend this alloy for this application
Complete question:
A steel component with a tensile strength of 800 MPa and fracture toughness Kic=20 MPa Nm is known to contain internal cracks (also called tunnel cracks) with the maximum length of 1.4 mm. This steel is being considered for use in a cyclic loading application for which the designed stresses vary from 0 to 420 MPa. Would you recommend using this steel in this application?
a. Not sure. Because cyclic loading is applied. Fatigue test is needed in order to make the recommendation.
b. Yes, this because the tensile strength of steel is much higher than the applied highest stress of 420 MPa.
c. Yes, this because the calculated critical stress to fracture for the cracks is higher than the highest applied stress of 420 MPa and the steel can withstand the stress of 420 MPa.
d. No. Although the calculated critical stress to fracture for the cracks is slightly higher than the highest applied stress of 420 MPa and the steel may withstand the static stress of 420 MPa, the cyclic loading may cause rapid fatigue fracture.
Answer:
A. Not sure. Because cyclic loading is applied. Fatigue test is needed in order to make the recommendation.
Explanation:
we are not sure if to recommend this alloy for this application given that this material has already been left to experience fatigue degradation. the cyclic load application brings about a growth in the crack. We know that cyclic loading is continuous loading that is useful for the testing of fatigue. Therefore the answer to this question is option a. We cannot make recommendations except fatigue testing has been carried out.
thank you!
A structure is designed using 4 circular columns. Due to a quirky design, the four columns will all carry different loads of 1800 N, 2100 N, 2275 N, and 2200 N. A factor of safety of 5 is used to design the columns. The diameter of each of the columns is supposed to be 50 cm, at most. Determine the maximum height of the structure (i.e. the column height) so that the structure will not fail. Assume that all columns may be modeled as Euler columns for your analysis. Assume a pinned-pinned boundary condition for your analysis, and assume the elastic modulus of the column material is 10 MPa.
Answer:
5.16 M
Explanation:
Loads ; 1800N, 2100N, 2275N, 2200N
safety factor = 5
diameter of each column = 50 cm = 0.5 m
Elastic modulus = 10 MPa
Calculate the max height of structure
moment of inertia for a circular section ( I ) = πd^4 / 64
lets represent the required maximum height of the column as L
Applying Euler column theory
The bucker load of the column = ( attached below )
attached below is the remaining solution
what is geo technical
phân tích phương pháp gia công plasma
Answer:
can you plz write in English language so we can give you answer
Your organization recently purchased 20 Android tablets for use by the organization's management team. To increase the security of these devices, you want to ensure that only specific apps can be installed. Which of the following would you implement?
A. Credential Manager.
B. App whitelisting.
C. App blacklisting.
D. Application Control.
An Otto cycle with air as the working fluid has a compression ratio of 8.2. The ambient temperature is 298K. The maximum temperature of the Otto cycle is 2000K. Under cold air standard conditions, the thermal efficiency of this cycle is ____
Answer:
The correct answer is "57%".
Explanation:
Given:
Temperature:
[tex]T_a = 298 \ K[/tex]
[tex]T_3 = 2000 \ K[/tex]
Compression ratio (r),
= [tex]8.2[/tex]
For Otto cycle, the thermal efficiency will be:
⇒ [tex]\eta =1-\frac{1}{(r)^{\nu-1}}[/tex]
By substituting the values, we get
[tex]=1-\frac{1}{(8.2)^{1.4-1}}[/tex]
[tex]=1-\frac{1}{(8.2)^{0.4}}[/tex]
[tex]=57[/tex] (%)
Design a ductile iron pumping main carrying a discharge of 0.35 m3/s over a distance of 4 km. The elevation of the pumping station is 140 m and that of the exit point is 150 m. The required terminal head is 10 m. Estimate the pipe diameter and pumping head using the explicit design procedure g
Answer:
[tex]D=0.41m[/tex]
Explanation:
From the question we are told that:
Discharge rate [tex]V_r=0.35 m3/s[/tex]
Distance [tex]d=4km[/tex]
Elevation of the pumping station [tex]h_p= 140 m[/tex]
Elevation of the Exit point [tex]h_e= 150 m[/tex]
Generally the Steady Flow Energy Equation SFEE is mathematically given by
[tex]h_p=h_e+h[/tex]
With
[tex]P_1-P_2[/tex]
And
[tex]V_1=V-2[/tex]
Therefore
[tex]h=140-150[/tex]
[tex]h=10[/tex]
Generally h is give as
[tex]h=\frac{0.5LV^2}{2gD}[/tex]
[tex]h=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]
Therefore
[tex]10=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]
[tex]D=^5\frac{8*(0.35)^2*0.003*4000}{3.142^2*9.81*10}[/tex]
[tex]D=0.41m[/tex]
A steam turbine receives steam at 1.5MPa and 220oC, and exhausts at 50kPa, 0.75 dry. Neglecting heat losses and changes in kinetic and potential energy, estimate the work output per kg steam.
If when allowance is made for friction, radiation and leakages losses, the actual work of that estimated in (a), calculate the power output of the turbine when consuming 600kg of steam per minute.
Answer:
Can you make friend with me ?
A 40-mm-diameter solid steel shaft, used as a torque transmitter, is replaced with a hollow shaft having a 40-mm outer diameter and a 36-mm inner diameter. If both materials have the same strength, what is the percentage reduction in torque transmission
Answer:
65.61%
Explanation:
we have the following information to answer this question
diameter of the solid steel shaft = 40 mm
outer diametr of the hollow shaft = 40mm
inner diametr pf the hollow shaft = 36mm
[tex]percentage reduction in torque transmission = \frac{Tsolid-Thollow}{Tsolid} *100[/tex]
= (40³ - (40⁴-36⁴)/40)/40³ * 100
= (40³ - 22009.6)/40³ * 100
= 41990.4/64000 * 100
= 0.6561 x 100
= 65.61%
percentage reduction in torque transmission = 65.61%
determine the number of flipflops required to build a binary counter that count from 0 to 2043
Answer:
10 flip -flops are required to build a binary counter circuit to count to from 0 to 1023 .
Explanation:
______ is not a type of digital signaling technique
Answer:
Data Rate Signaling
what is tracer lathe machine
Answer: The tracer lathe is a roughing operation for the output shaft on rear wheel drive transmissions.
Explanation:
how to solve circuit theory using mesh analysis
Explanation:
Find a minimal set of cycles that covers all vertices and edges of the circuit graph. For each cycle, define a "mesh" current, and write the Kirchhoff's Voltage Law (KVL) equation with respect to each of the edges in the cycle. Where an edge is part of more than one cycle, all current(s) defined for the edge will contribute to the voltage there.
This will give as many equations as there are mesh currents. Solve the resulting system of equations. The (signed) sum of the mesh currents through any edge is the current in that circuit branch.
__
Example
Consider the attached circuit. It shows mesh currents I1, I2, and I3 in graph cycles with those numbers. The KVL equations are ...
mesh 1: I1(R3 +R2 +R1) -I2·R1 -I3·R2 = Vi (the voltage across the current source)
mesh 2: -I1·R1 +I2(r1 +1/(sC)) -I3(1/(sC)) = Vs
mesh 3: -I1·R2 -I2(1/(sC)) +I3(R2 +sL +1/(sC)) = 0
You will note that the matrix of equation coefficients is symmetric.
__
In this example, you will end with I1 as a function of Vi. If I1 is a given source value, that relation can be used to find Vi.
the heat treatment process depends on 3 steps according to the time-temperature cycle. Name and Briefly describe each one
Explanation:
there are three stages of heat treatment
1.hit the metal slowly to ensure that the metal maintains a uniform temperature
2.soak or hold the metal at a specific temperature for a alloted period of time
3.cool the metal to room temperature
Consider the need to insulate a flat wall using three different insulating materials, each having the same thickness. The materials have constant thermal conductivities, k, of 1, 2, and 3 Wm-1 K-1 . You are asked to advise workers on the proper installation order of the three materials so that the final composite will be the most effective at reducing heat loss from the wall. Which statement is true
Answer: The options related to your question is missing attached below is the missing option
a.) The insulation must be arranged in order of increasing k starting with the lowest k material at the wall surface.
b.) The insulation must be arranged in order of decreasing k starting with the highest k material at the wall surface.
c.) The insulation can be arranged in any order.
d.) It is impossible to decide because the temperatures are not specified and the order will depend on the ∆T.
e) The insulation with the lowest k should be nearest the wall, and the order of the other two is irrelevant.
answer:
The insulation must be arranged in order of increasing k starting with the lowest k material at the wall surface. ( A )
Explanation:
The correct statement is ; The insulation must be arranged in order of increasing k starting with the lowest k material at the wall surface.
Because the insulation has to be arranged following the order of it's increasing thermal conductivity ( K-value )
You are given a pot of hot tea (350 mL) at a temperature of 85°C and a quantity of ice at -12°C. Determine the absolute minimum amount of ice you can add to the hot tea so that at equilibrium you have iced tea (ie. a very, very small amount of ice and some water). You can assume tea has the same density (1.00 g/mL) and specific heat (4190 J/kgK) as liquid water. The heat of fusion of H2O is 3.33x10^5 J/kg. The specific heat of ice is 2090 J/kgK.
Answer:
348 g
Explanation:
The heat gained by the ice equals the heat lost by the hot tea.
Also, since we require a very, very small amount of ice and some water, to still have some ice, the temperature of the hot tea has to decrease to the melting point of ice which is 0°C. So, the final temperature of the mixture is 0 °C.
So, mc(T - T') + mL = -MC(T - T") where m = mass of ice, c = specific heat of ice = 2090 J/kgK., T = final temperature of mixture = 0 °C, T' = initial temperature of ice = -12 °C, L = heat of fusion of H2O = 3.33 × 10⁵ J/kg, M = mass of hot tea = ρV where ρ = density of tea = 1.00 g/mL and V = volume of hot tea = 350 mL. So, M = ρV = 1.00 g/mL × 350 mL = 350 g = 0.350 kg, C = specific heat of tea = 4190 J/kgK and T" = initial temperature of tea = 85 °C.
Making m subject of the formula, we have
mc(T - T') + mL = -MC(T - T")
m[c(T - T') + L] = -MC(T - T")
m = -MC(T - T")/[c(T - T') + L]
substituting the values of the variables into the equation, we have
m = -MC(T - T")/[c(T - T') + L]
m = -0.350 kg × 4190 J/kgK(0 °C - 85 °C.)/[2090 J/kgK(0 °C. - (-12 °C)) + 3.33 × 10⁵ J/kg]
m = -1466.5 J/kK(- 85 °C)/[2090 J/kgK(0 °C + 12 °C)) + 3.33 × 10⁵ J/kg]
m = 124652.5 J/[2090 J/kgK(12 °C) + 3.33 × 10⁵ J/kg]
m = 124652.5 J/[25080 J/kg + 3.33 × 10⁵ J/kg]
m = 124652.5 J/358080 J/kg
m = 0.3481 kg
m = 348.1 g
m ≅ 348 g
So, the minimum amount of ice to be added is 348 g.
Carbon dioxide at a temperature of 0oC and a pressure of 600 kPa (abs) flows through a horizontal 40-mm- diameter pipe with an average velocity of 2 m/s. Determine the friction factor if the pressure drop is 235 N/m2 per 10-m length of pipe.
Answer:
f = 0.04042
Explanation:
temperature = 0°C = 273k
p = 600 Kpa
d = 40 millemeter
e = 10 m
change in P = 235 N/m²
μ = 2m/s
R = 188.9 Nm/kgk
we solve this using this formula;
P = ρcos*R*T
we put in the values into this equation
600x10³ = ρcos * 188.9 * 273
600000 = ρcos51569.7
ρcos = 600000/51569.7
=11.63
from here we find the head loss due to friction
Δp/pg = feμ²/2D
235/11.63 = f*10*4/2*40x10⁻³
20.21 = 40f/0.08
20.21*0.08 = 40f
1.6168 = 40f
divide through by 40
f = 0.04042
Microsoft Project là phần mềm có sẵn trong bộ Office 365, đúng (True) hay sai (False)?
Describe the main component in an electronic control unit (ECU) used in an automobile. Indicate the function of each ingredient.
8.50 nC of charge is uniformly distributed along a thin rod of length L = 9.20 cm, which is then bent into a semicircle. What is the magnitude of the electric field at the center of the circle
what is an OTG USB? how is it useful
Answer:
An OTG or On The Go adapter (sometimes called an OTG cable, or OTG connector) allows you to connect a full sized USB flash drive or USB A cable to your phone or tablet through the Micro USB or USB-C charging port
Explanation:
pls mark brainliest
answer this qustion plz
Answer:
click here and I'm so happy I have been doing this weekend I can see the same way you could come back in a bit and then we are all of you to know you do for a bit of time with me or
thì nghiệm nén xác định cường độ của bê tông trên ba mẫu thí nghiệm hình trụ HxD=300x150(mm). kết quả thu được lực phá hoại P1=45200daN, P2=46800daN, P3=46000daN. hãy xác định cường độ chịu nén của bê tông theo TCNV 3118:1993
spanish
Explanation:
the above question is written in spanish
The base of an aluminum block, which is fixed in place, measures 90 cm by 90 cm, and the height of the block is 60 cm. A force, applied to the upper face and parallel to it, produces a shear strain of 0.0060. The shear modulus of aluminum is . What is the displacement of the upper face in the direction of the applied force
The question is incomplete. The complete question is :
The base of an aluminum block, which is fixed in place, measures 90 cm by 90 cm, and the height of the block is 60 cm. A force, applied to the upper face and parallel to it, produces a shear strain of 0.0060. The shear modulus of aluminum is [tex]3.0 \times 10^{10} \ Pa[/tex] . What is the displacement of the upper face in the direction of the applied force?
Solution :
The relation between shear modulus, shear stress and strain,
[tex]$\text{Shear modulus, S =} \frac{\text{Shear stress}}{\text{shear strain}}$[/tex]
Shear stress = shear modulus (S) x shear strain
[tex]$=3 \times 10^{10} \times 0.0060$[/tex]
[tex]$=1.80 \times 10^8$[/tex] Pa
[tex]$=180 \times 10^6$[/tex] Pa
[tex]$=180 \ MPa$[/tex]
The length represents the distance between the fixed in place portion and where the force is being applied.
Therefore,
[tex]$\text{Displacement} = \text{shear strain} \times \text{length}$[/tex]
= 0.006 x 60 cm
= 0.360 cm
= 3.6 mm
Thus, the displacement of the upper face is 3.6 mm in the direction of the applied force.
True or false: You can create a network with two computers.
Answer
True
Question 1: Determine the maximum load P the steel bracket can withstand if the steel bracket has a circular cross section with a diameter of 1.2 in, and has an allowable normal stress of allow
Complete Question
Complete Question is attached below
Answer:
[tex]P=1124.2ibf[/tex]
Explanation:
From the question we are told that:
Diameter [tex]d=1.2in[/tex]
Allowable Normal stress [tex]\sigma=27.5ksi=27.5 * 10^3 psi[/tex]
Generally the equation for Bending Stress is mathematically given by
[tex]\phi= \frac{32M}{ \pi d^3}[/tex]
[tex]\phi= {32 * 4 P}{\pi * 1.2^3}[/tex]
[tex]\phi=23.58 psi[/tex]
Generally the equation for Direct Normal Stress is mathematically given by
[tex]\sigma'=\frac{4P}{ \phi * 1.2^2}[/tex]
[tex]\sigma'= 0.88P psi[/tex]
Therefore
Total Normal stress
[tex]\sigma_T=23.58 + 0.88[/tex]
[tex]\sigma_T=24.46P[/tex]
Generally the equation for Allowable Stress is mathematically given by
[tex]\sigma=\sigma_T P[/tex]
[tex]P=\frac{\sigma}{\sigma_T}[/tex]
[tex]P=\frac{27.5 * 10^3}{24.46P}[/tex]
[tex]P=1124.2ibf[/tex]
Question 1. If a fiber weight 3.0 g and composite specimen weighing 4.g. The composite specimen weighs 2.0 g in water. If the specific gravity of the fiber and matrix is 2.4 and 1.3, respectively, find the 1. Theoretical density of composite 2. Experimental density 3. Void fraction
Answer:
Explanation:
From the given information:
weight of fiber [tex]w_f[/tex] = 3.0 g
weight of composite specimen [tex]w_c[/tex] = 4.0 g
specimen composite weight in water [tex]C_{wm}[/tex] = 2.0 g
specific gravity of fiber [tex]S_f[/tex] = 2.4
specific gravity of matrix [tex]S_m[/tex] = 1.3
The weight of the matrix = weight of the composite - the weight of fiber
⇒ (4.0 - 3.0) g
= 1.0 g
The theoretical density of the composite [tex]\rho_{ct}[/tex] can be determined by using the formula:
[tex]\dfrac{1}{\rho_{ct}} = \dfrac{w_f}{w_cS_f}+ \dfrac{w_m}{w_cS_m}[/tex]
[tex]\dfrac{1}{\rho_{ct}} = \dfrac{3.0}{(4.0 \times 2.4)}+ \dfrac{1.0}{(4.0\times 1.3)}[/tex]
[tex]\dfrac{1}{\rho_{ct}} = \dfrac{3.0}{9.6}+ \dfrac{1.0}{5.2}[/tex]
[tex]\dfrac{1}{\rho_{ct}} =0.505\\[/tex]
[tex]\rho_{ct} =\dfrac{1}{0.505}[/tex]
[tex]\mathbf{\rho_{ct} = 1.980 \ g/cm^3}[/tex]
The experimental density [tex]\rho _{ce}[/tex] is determined by using the equation:
[tex]\rho _{ce} = \dfrac{w_f + w_c}{\dfrac{w_f }{S_f} + \dfrac{w_c }{S_m} }[/tex]
[tex]\rho _{ce} = \dfrac{3.0 + 4.0}{\dfrac{3.0 }{2.4} + \dfrac{4.0 }{1.3} }[/tex]
[tex]\rho _{ce} = \dfrac{3.0 + 4.0}{1.250 +3.077 }[/tex]
[tex]\mathbf{\rho _{ce} = 1.620 \ g/cm^3}[/tex]
The void fraction is: [tex]= \dfrac{\rho_{ct}-\rho_{ce}}{\rho_{ct}}[/tex]
[tex]= \dfrac{1.980-1.620}{1.980}[/tex]
= 0.1818
Consider CO at 500 K and 1000 kPa at an initial state that expands to a final pressure of 200 kPa in an isentropic manner. Report the final temperature in units of K and using three significant digits.
Answer:
[tex]T_2=315.69k[/tex]
Explanation:
Initial Temperature [tex]T_1=500K[/tex]
Initial Pressure [tex]P_1=1000kPa[/tex]
Final Pressure [tex]P_2=200kPa[/tex]
Generally the gas equation is mathematically given by
[tex]\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{n-1}{n}}[/tex]
Where
n for [tex]CO=1.4[/tex]
Therefore
[tex]\frac{T_2}{500}=\frac{200}{1000}^{\frac{1.4-1}{1.4}}[/tex]
[tex]T_2=315.69k[/tex]
