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dexar [7]
3 years ago
9

A batch of parts is produced on a semi-automated production machine in a sequential batch production operation. Batch quantity i

s 300 units. Setup takes 55 min. A worker loads and unloads the machine each cycle, which takes 0.75 min. Machine processing time is 3.46 min/cycle, and tool handling time is negligible. One part is produced each cycle. Determine:
a. Average cycle time
b. Time to complete the batch
c. Average production rate.
Engineering
1 answer:
vredina [299]3 years ago
4 0

Answer:

a) 4.21 min

b) 21.9666 hrs

c) 1.3657 Pc/hr

Explanation:

Given that;

Batch quantity = 300 units

time setup = 55min

unload/loading time = 0.75

processing time = 3.46

a) Average cycle time;

Average Cycle Time TC = loading time + processing time

TC = 0.75 + 3.46

Tc = 4.21 min

b) Time to complete the batch

Time to complete the batch Tb = setup time + process time + non operation time

Tb = (55min * 1hr/60min) + (300 * 3.46 * 1hr/60min) + (300 * 0.75 * 1hr/60min)

Tb = 0.9166 + 17.3 + 3.75

Tb = 21.9666 hrs

c) Average production rate

Average production rate Rp = 1 / ( Tb / batch size)

we substitute

Rp = 1 / ( 21.9666 / 300 )

Rp = 1 / 0.7322

Rp = 1.3657 Pc/hr

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Consider a steady isentropic airflow through a convergent channel with an inletto-exit cross section area ratio of 2. If the inl
Nina [5.8K]

Answer:

for m1=0.2

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for m1=2.5

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Explanation:

Hello!

To solve this problem follow the steps below!

1.As it is an isentropic system, it means that the temperature of the fluid remains constant, so the volumetric flow is the same, remember that this is calculated by calculating the product between the velocity and the area.

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We can divide both sides of the equation by the speed of sound(V)

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2. The mach number is defined as the ratio between the speed of a fluid and the speed of sound,

we rearrange the equation

M1A1=M2A2

\frac{M1A1}{A2} =M2

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M1(2)=M2

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0.2(2)=M2=0.4

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8 0
3 years ago
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masha68 [24]

Answer:

(a). max possible efficiency = 55.62%

(b). max power output = = 133.5 MW

Explanation:

From the question we were given the Maximum temperature in the system as

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Minimum temperature in the system Tmin = 70°C

the Heat supplied to the boiler Qb = 240000 KJ/s

we use the temperature conversion factor from °C to K

given T(K) = T (°C) + 273

⇒ Tmax = 500 + 273 = 773 K

⇒ Tmin = 70 + 273 = 343 K

(a). we are to determine the maximum possible thermal efficiency;

(Πth)max = 1 - Tmin/Tmax

(Πth)max = 1 - 343/773  = 0.5562

(Πth)max = 55.62%

(b). to determine the maximum possible power output for the plant we have;

(Πth)max = Wmax/Qb

where Wmax rep the maximum power output

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3 0
3 years ago
A rotating shaft is subjected to a steady torsional stress of 13 ksi and an alternating bending stress of 22 ksi.
mixas84 [53]

Answer:

A) б1 = 28 ksi and  б2 = -6.02 ksi

B) 1.25

Explanation:

Given data :

Torsional stress = 13 ksi

Alternating bending stress = 22ksi

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б1,2 = \frac{22}{2} ± √(22/2)² + 13²

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therefore б1 = 28 ksi  hence б2 = -6.02 ksi

B) determine the fatigue factor of safety  

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( б1 - б2 )²  + ( б2 - б3 )² + ( б3 - б1 )²  ≤  2 ( Sy / FOS ) ²

( 28 + 6.02 ) ² + ( 6.02 - 0 )² + ( 0 - 28 )² ≤  2 ( 60 / FOS ) ²

solving for FOS = 1.9

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3 0
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