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Fynjy0 [20]
3 years ago
15

A consultant has proposed that a pulse-jet baghouse with bags that are 15 cm in diameter and 5 m in length. Estimate the net num

ber of bags required if the manufacturer’s recommended air-to-cloth ratio for aggregate plants in 0.05 m/s, and a requirement that 1/8 of the bags are off-line for cleaning
Engineering
1 answer:
Dmitry_Shevchenko [17]3 years ago
5 0

Answer:

0,5 bags

Total number of bags = 0.5 +  4 = 4.5 bags or 5 bags

Explanation:

Convert 15cm to meters  = 15/100 = 0.15m

Flow rate Q = Area x velocity

= π d x v

= π (0.15) x 1

= 0.4712

A = Q/v = 0.4712/ 0.05

=<u> 9.435 m^2</u>

<u></u>

The net number of bags =  A/π dh

= 9.425/π (0.15)(5)

= <u>4 bags </u>

<u></u>

Requirement that 1/8 of the bags are off-line for cleaning

= Net number of bags / 8

= 4/8

= <u>0.5 bag or 1bag</u>

<u></u>

<u>Total number of bags </u>

<u>= 4bags + 1bag </u>

<u>= 5bags </u>

<u></u>

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A 5000-lb truck is being used to lift a 1000-lb boulder B that is on a 200-lb pallet A. Knowing that the truck starts from rest
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Explanation:

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Read 2 more answers
A pitfall cited in Section 1.10 is expecting to improve the overall performance of a computer by improving only one aspect of th
Oxana [17]

Answer:

a) For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time

(1-0.2)*70 s =56s

The reduction on this case is 70-56 s=14s

And since the new total time would be given by 250-14=236 s

b) For this case the total time is reduced 20%  so that means that the new total time would be (1-0.2)=0.8 times the original total time (1-0.2) *250s =200 s

The original time for INT operations is calculated as:

250 = 70+85+40 +t_{INT}

t_{INT}=55s

For this part the only time that was changed is assumed the INT operations so then:

200 = 70+85+40 \Delta t_{INT}

And then: \Delta t_{INT}= 200-70-85-40=5 s

c) A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.

Explanation:

From the info given we know that a computer running a program that requires 250 s, with 70 s spent executing FP instructions, 85 s executed L/S instructions and 40 s spent executing branch instructions.

Part 1

For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time

(1-0.2)*70 s =56s

The reduction on this case is 70-56 s=14s

And since the new total time would be given by 250-14=236 s

Part 2

For this case the total time is reduced 20%  so that means that the new total time would be (1-0.2)=0.8 times the original total time (1-0.2) *250s =200 s

The original time for INT operations is calculated as:

250 = 70+85+40 +t_{INT}

t_{INT}=55s

For this part the only time that was changed is assumed the INT operations so then:

200 = 70+85+40 \Delta t_{INT}

And then: \Delta t_{INT}= 200-70-85-40=5 s

And we can quantify the decrease using the relative change:

\% Change = \frac{5s}{55 s} *100 = 9.09\% of reduction

Part 3

A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.

8 0
3 years ago
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