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larisa [96]
3 years ago
11

An automated transfer line is to be designed. Based on previous experience, the average downtime per occurrence = 5.0 min, and t

he probability of a station failure that leads to a downtime occurrence is 0.01. The total work content time = 39.2 min and is to be divided evenly among the workstations, so the ideal cycle time for each station = 39.2/n, where n is the number of workstations.
Determine; (a) the optimum number of stations on the line that will maximize production rate and b) the production rate Rp and proportion uptime for answer to part (a).
Engineering
1 answer:
IRINA_888 [86]3 years ago
3 0

Answer:

a) 28 stations

b) Rp = 21.43

E = 0.5

Explanation:

Given:

Average downtime per occurrence = 5.0 min

Probability that leads to downtime, d= 0.01

Total work time, Tc = 39.2 min

a) For the optimum number of stations on the line that will maximize production rate.

Maximizing Rp =minimizing Tp

Tp = Tc + Ftd

=  \frac{39.2}{n} + (n * 0.01 * 5.0)

= \frac{39.2}{n} + (n * 0.05)

At minimum pt. = 0, we have:

dTp/dn = 0

= \frac{-39.2}{n^2} + 0.05 = 0

Solving for n²:

n^2 = \frac{39.2}{0.05} = 784

n = \sqrt{784} = 28

The optimum number of stations on the line that will maximize production rate is 28 stations.

b) Tp = \frac{39.2}{28} + (28 * 0.01 * 5)

Tp = 1.4 +1.4 = 2.8

The production rate, Rp =

\frac{60min}{2.8} = 21.43

The proportion uptime,

E = \frac{1.4}{2.8} = 0.5

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Explanation:

As we know that density of water is 1000 kilograms per cubic meter of water hence we infer that 1 cubic meter of water will have a weight of 1000 kilograms of 1 metric tonnes which is beyond the lifting capability of strongest man on earth let alone a normal human being who can just lift a weight of 100 kilograms thus we conclude that we cannot lift 1 cubic meter of liquid water.

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A refrigerator has a cooling load of 50 kW. It has a COP of 2. It is run by a heat engine which consumes 50 kW of heat to supply
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<u>Given information</u>

Cooling load=50 kW

COP=2

Consumption=50 kW

<u>Calculations</u>

Revised input is given by cooling load/COP=50/2=25 kW

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When hermetic refrigerant motor-compressors are designed to operate continuously at currents greater than 156 percent of the rat
ANEK [815]

The nameplate of a hermetic refrigerant motor-compressor that is designed to operate continuously at currents greater than 156% of the rated-load current is marked with branch-circuit selection current.

<h3>What is a hermetic refrigerant motor-compressor?</h3>

A hermetic refrigerant motor-compressor can be defined as a mechanical device that is designed and developed by combining a compressor and motor in a single outer-welded steel shell.

Basically, a hermetic refrigerant motor-compressor is used in the following areas:

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According to HSE, the nameplate of a hermetic refrigerant motor-compressor that is designed to operate continuously at currents greater than 156% of the rated-load current is marked with branch-circuit selection current, so as to ensure safety for end users and technicians.

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Flow rate= 0.79128*10^-3 Ns

For Air

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Explanation:

For the flow rate of water in pipe.

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Flow rate =1.2717*10^-3 Ns

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