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Scilla [17]
3 years ago
13

What is the function maintenance? List some important steps for vibration monitoring based maintenance.

Engineering
1 answer:
Mrrafil [7]3 years ago
4 0

The maintenance is in charge of controlling that all the machines of a company are constantly running in order to avoid damages that cause loss of money when the machines fail.

The maintenance based on vibration monitoring allows to predict failures in some rotating machines such as:

1. worn bearings

2.alignment

3.balance

4. affected gears

5. bent shafts

6. rocks

7.gags

8. eccentricity

9. failures of electrical origin

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A car is about to start but it blows up. what is the problem with the car<br> ?
ratelena [41]

Answer:

because there is a bomb

6 0
3 years ago
Read 2 more answers
The Ethernet (CSMA/CD) alternates between contention intervals and successful transmissions. Assume a 100 Mbps Ethernet over 1 k
Vesnalui [34]
<h3><u>CSMA/CD Protocol: </u></h3>

Carrier sensing can transmit the data at anytime only the condition is before sending the data sense carrier if the carrier is free then send the data.

But the problem is the standing at one end of channel, we can’t send the entire carrier. Because of this 2 stations can transmit the data (use the channel) at the same time resulting in collisions.

There are no acknowledgement to detect collisions, It's stations responsibility to detect whether its data is falling into collisions or not.

<u>Example: </u>

T_{P}=1 H r, at time t = 10.00 AM, A starts, 10:59:59 AM B starts at time 11:00 AM collision starts.

12:00 AM A will see collisions

Pocket Size to detect the collision.

\begin{aligned}&T_{t} \geq 2 T_{P}\\&\frac{L}{B} \geq 2 T_{P}\\&L \geq 2 \times T_{P} \times B\end{aligned}

CSMA/CD is widely used in Ethernet.

<u>Efficiency of CSMA/CD:</u>

  • In the previous example we have seen that in worst case 2 T_{P} time require to detect a collision.
  • There could be many collisions may happen before a successful completion of transmission of a packet.

We are given number of collisions (contentions slots)=4.

\text { Propagation day }=\frac{\text {distance}}{\text {speed}}

Distance = 1km = 1000m

\begin{aligned}&\text { Speed }=2 \times 10^{8} \mathrm{m} / \mathrm{sec}\\ &T_{P}=\frac{1000}{2 \times 10^{8}}=(0.5) \times 10^{-5}=5 \times 10^{-6}\\ &T_{t}=5 \mu \mathrm{sec}\end{aligned}

7 0
3 years ago
. Air at 200 C blows over a 50 cm x 75 cm plain carbon steel (AISI 1010) hot plate with a constant surface temperature of 2500 C
MrRissso [65]

Answer:

The inside temperature, T_{in} is approximately 248 °C.

Explanation:

The parameters given are;

Temperature of the air = 20°C

Carbon steel surface temperature 250°C

Area of surface = 50 cm × 75 cm = 0.5 × 0.75 = 0.375 m²

Convection heat transfer coefficient = 25 W/(m²·K)

Heat lost by radiation = 300 W

Assumption,

Air temperature = 20 °C

Hot plate temperature = 250 °C

Thermal conductivity K = 65.2 W/(m·K)

Steady state heat transfer process

One dimensional heat conduction

We have;

Newton's law of cooling;

q = h×A×(T_s - T_{\infty) + Heat loss by radiation

= 25×0.325×(250 - 20) + 300

= 2456.25 W

The rate of energy transfer per second is given by the following relation;

P = \dfrac{K \times A \times \Delta T}{L}

Thermal conductivity K = 65.2 W/(m·K)

Therefore;

2456.25  = \dfrac{65.2 \times 0.375 \times (250 - T_{in})}{0.02}

T_{in} = 250 - \dfrac{2456.25  \times 0.02}{65.2 \times 0.375} = 247.99 ^{\circ}C

The inside temperature, T_{in} = 247.99 °C  ≈ 248 °C.

3 0
4 years ago
LC3 Programming ProblemUse .BLKW to set up an array of 10 values, starting at memory location x4000, as in lab 4.Now programmati
irga5000 [103]

Answer:

Check the explanation

Explanation:

Code

.ORIG x4000

;load index

LD R1, IND

;increment R1

ADD R1, R1, #1

;store it in ind

ST R1, IND

;Loop to fill the remaining array

TEST LD R1, IND

;load 10

LD R2, NUM

;find tw0\'s complement

NOT R2, R2

ADD R2, R2, #1

;(IND-NUM)

ADD R1, R1, R2

;check (IND-NUM)>=0

BRzp GETELEM

;Get array base

LEA R0, ARRAY

;load index

LD R1, IND

;increment index

ADD R0, R0, R1

;store value in array

STR R1, R0,#0

;increment part

INCR

;Increment index

ADD R1, R1, #1

;store it in index

ST R1, IND

;go to test

BR TEST

;get the 6 in R2

;load base address

GETELEM LEA R0, ARRAY

;Set R1=0

AND R1, R1,#0

;Add R1 with 6

ADD R1, R1, #6

;Get the address

ADD R0, R0, R1

;Load the 6th element into R2

LDR R2, R0,#0

;Display array contents

PRINT

;set R1 = 0

AND R1, R1, #0

;Loop

;Get index

TOP ST R1, IND

;Load num

LD R3,NUM

;Find 2\'s complement

NOT R3, R3

ADD R3, R3,#1

;Find (IND-NUM)

ADD R1, R1,R3

;repeat until (IND-NUM)>=0

BRzp DONE

;load array address

LEA R0, ARRAY

;load index

LD R1, IND

;find address

ADD R3, R0, R1

;load value

LDR R1, R3,#0

;load 0x0030

LD R3, HEX

;convert value to hexadecimal

ADD R0, R1, R3

;display number

OUT

;GEt index

LD R1, IND

;increment index

ADD R1, R1, #1

;go to top

BR TOP

;stop

DONE HALT

;declaring variables

;set limit

NUM .FILL 10

;create array

ARRAY .BLKW 10 #0

;variable for index

IND .FILL 0

;hexadecimal value

HEX .FILL x0030

;stop

.END

7 0
3 years ago
Your family has asked you to estimate the operating costs of your clothes dryer for the year. The clothes dryer in your home has
trasher [3.6K]

Answer:

The costs to run the dryer for one year are $ 9.03.

Explanation:

Given that the clothes dryer in my home has a power rating of 2250 Watts, and to dry one typical load of clothes the dryer will run for approximately 45 minutes, and in Ontario, the cost of electricity is $ 0.11 / kWh, to calculate the costs to run the dryer for one year the following calculation must be performed:

1 watt = 0.001 kilowatt

2250/45 = 50 watts per minute

45 x 365 = 16,425 / 60 = 273.75 hours of consumption

50 x 60 = 300 watt = 0.3 kw / h

0.3 x 273.75 = 82.125

82.125 x 0.11 = 9.03

Therefore, the costs to run the dryer for one year are $ 9.03.

8 0
3 years ago
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