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Fynjy0 [20]
2 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]2 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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When determining risk, it is necessary to estimate all routes of exposure in order to determine a total dose (or CDI). Recognizi
Allushta [10]

Answer:

The following are the solution to this question:

Explanation:

The Formula for calculating CDI:

\bold{CDI = \frac{C \times CR \times EF \times ED}{BW \times AT}}

_{where} \\ CDI = \text{Chronic daily Intake rate}  (\frac{mg}{kg-day})} \\\\\text{C = concentration of Toluene}\\\\\text{CR = contact rate} \frac{L}{day}\\\\\text{EF = Exposure frequency} \frac{days}{year}\\\\\text{ED = Exposure duration (in years)} = 10 \ \ years\\\\\text{BW = Body weight (kg) = 70 kg for adult}\\\\ \text{AT = average period of exposure (days) }

calculating the value of AT:

=  365 \frac{days}{year}  \times  70 \ year  \\\\ = 25550 \ days

 calculating the value of Intake based drinking:

C = 1 \ \frac{mg}{L}

CR = 2 \frac{L}{day} Considering that adult females eat 2 L of water a day,

EF = 350 \frac{days}{year} for drink

calculating the CDI value:

\to CDI = \frac{(1 \times 2 \times 350 \times 10)}{(70 \times  25550)}\\\\

             = \frac{(2 \times 3500)}{(70 \times  25550)}\\\\ = \frac{(7000)}{(70 \times  25550)}\\\\ = \frac{(100)}{(25550)}\\\\=0.00391 \frac{mg}{ kg-day}

Centered on inhalation, intake:

C = \frac{1 \mu g} { m^3} \ \ \  or \ \ \ \ 0.001  \ \ \frac{mg}{m^3}\\\\CR = 20  \frac{m^3}{day}\\\\EF = 15 \frac{min}{day}  \ \ or\ \  5475 \frac{min}{yr} \ \ \  or \ \  3.80 \frac{days}{year}\\

calculating the value of CDI:

\to CDI = \frac{(0.001 \times 20 \times 3.80 \times 10)}{(70 \times 25550)}

             = \frac{(0.76)}{(1788500)}\\\\= 4.24 \times 10^{-7} \ \ \frac{mg}{kg-day}

7 0
3 years ago
Calculate the number of vacancies per cubic meter at 1000∘C for a metal that has an energy for vacancy formation of 1.22 eV/atom
Makovka662 [10]

Answer:

C

Explanation:

N = Na.P/A------(1)

Na = avogadro's number = 6.02210²³

P = density

A = atomic weight of metal

When we substitute into equation 1 above we get

1.0x10²⁹atoms/m³

From here we calculate the number of vacancies

T = 1000⁰c = 1273K

The formula to use is

Nv= Nexo(Qt/K.T) -----(2)

Qt = 1.22eV

K = Boltzmann's constant = 8.6210x10^-5

When we substitute values into equation 2

We get Nv = 1.49 x 10²⁴m-3

Therefore option c is correct

Check attachment for a more detailed calculation of this question

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3 years ago
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4 0
2 years ago
Read 2 more answers
Calculate the equivalent capacitance of the three series capacitors in Figure 12-1
GrogVix [38]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

Calculate the equivalent capacitance of the three series capacitors in Figure 12-1

a) 0.01 μF

b) 0.58 μF

c) 0.060 μF

d) 0.8 μF

Answer:

The equivalent capacitance of the three series capacitors in Figure 12-1 is 0.060 μF

Therefore, the correct option is (c)

Explanation:

Please refer to the attached Figure 12-1 where three capacitors are connected in series.

We are asked to find out the equivalent capacitance of this circuit.

Recall that the equivalent capacitance in series is given by

$ \frac{1}{C_{eq}} =  \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}} $

Where C₁, C₂, and C₃ are the individual capacitance connected in series.

C₁ = 0.1 μF

C₂ = 0.22 μF

C₃ = 0.47 μF

So the equivalent capacitance is

$ \frac{1}{C_{eq}} =  \frac{1}{0.1} + \frac{1}{0.22} + \frac{1}{0.47} $

$ \frac{1}{C_{eq}} =  \frac{8620}{517}  $

$ C_{eq} =  \frac{517}{8620}  $

$ C_{eq} =  0.0599  $

Rounding off yields

$ C_{eq} =  0.060 \: \mu F $

The equivalent capacitance of the three series capacitors in Figure 12-1 is 0.060 μF

Therefore, the correct option is (c)

5 0
3 years ago
You are a designer of a new processor. You have to choose between two possible implementations (called M1 and M2) of the same ar
Kaylis [27]

Answer:

A ) CPI : M1 = 2.4 , M2 = 2.65

B ) MIPS : M1 = 1083, M2 = 1056

C ) The machine that has a better performance based on MIPS is M1 and this is  by 27 million number of instructions per sec

Explanation:

A) The CPI for each machine

CPI = ( Total number of execution cycles ) / ( instruction counter executed )

For Machine 1 ( M1 )

we have to make some assumptions : number of instructions = 10

number of times A was executed = 4 , Number of times B was executed = 2.5 , number of times C was executed = 2.5, Number of times D was executed = 1. and this was based on the frequency given above

hence CPI for M1 =[ ( 1 * 4 ) + ( 3 * 2.5 ) + ( 3 * 2.5 ) + ( 5 * 1 ) ] / 10

       CPI  for M1 = 2.4

For Machine 2 ( M2 )

we have to make some assumptions : number of instructions = 10

number of times A was executed = 4 , Number of times B was executed = 2.  number of times C was executed = 1.5, Number of times D was executed = 2.5 times. and this was based on the frequency given above

Hence CPI  for M1 = [ ( 2 * 4 ) + ( 2 * 2 ) + ( 3 * 1.5 ) + ( 4 * 2.5 ) ] / 10

            CPI for M2 = 2.65

B ) Calculate the native MIPS  ratings for M1 and M2

MIPS = ( instruction counts ) / ( Execution time * 10^6 )

For M1

Assumptions : number of instructions executed = 10

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first we calculate the total execution time which is equal to :

= [ ( 1 * 4 ) + ( 3 * 2.5 ) + ( 3 * 2.5 ) + ( 5 * 1 ) ] * 0.3846 * 10 ^-9

= 9.2304 * 10 ^-9 secs

therefore the MIPS for M1

= 10 / ( 9.2304 * 10^-9 ) * 10^6  = 1083

                                         

For M2

Assumptions : number of instructions executed = 10

                        each clock cycle = 0.3846 * 10^-9.      frequency = 2.8 Ghz

first we calculate the total execution time which is equal to :

= [ (2*4) + (2*2) + (3 * 1.5 ) + ( 4 * 2.5 ) ] * 0.3846 * 10^-9 = 9.4631 * 10^-9 secs

therefor the MIPS for M2

= 10 / ( 9.4631*10^-9) * 10^6 = 1056

C ) The machine that has a better performance based on MIPS is M1 and this is by 27 million number of instructions per sec

8 0
2 years ago
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