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docker41 [41]
3 years ago
7

A pulse-jet baghouse is desired for a finished cement plant. Calculate the number of bags required to filter 500 m3/min of air w

ith a dust loading of 3.0 g/m3. Each bag is 3.0 m long with a 0.13 m diameter. If the average pressure drop is 1.0 kPa and the main fan is 60% efficient, calculate the fan power in kW. If the pulse air volumetric flow rate is 0.5% of the filter airflow rate and the pulse air pressure is 6.0 atm, calculate the power drawn by a 50% efficient compressor (in kW).
Engineering
1 answer:
omeli [17]3 years ago
3 0

Answer:

1) <em>4.41 * 10^-4 kw </em>

<em>2) </em>2.20 * 10^-4 kw

Explanation:

Given data:

Filter = 500 m^3/min

dust velocity = 3g/m^3

bag ; length = 3 m , diameter = 0.13 m

change in pressure = 1 kPa

efficiency = 60%

<u>1) Calculate the Fan power </u>

First :

Calculate the total dust loading = 3 * 500 = 1500 g

To determine the Fan power we will apply the relation

n_{o}  = \frac{\frac{p}{eg*Q*h} }{1000}    = \frac{\frac{p}{(3*10^{-3})* 981*( 500/60) *3  } }{1000}

<em>fan power ( </em>n_{0} )<em> = 4.41 * 10^-4 kw </em>

<u>2) calculate power drawn </u>

change in P = 6 atm = 6 * 10^5 pa

efficiency compressor = 50%

hence power drawn = 4.41 * 10^-4 kw  * 50% = 2.20 * 10^-4 kw

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According to OSHA standards, the air in the building that John works in is unsafe. The type of regulation that OSHA engages in i
ioda

Answer:

social regulation.

Explanation:

Social regulation are rules set aside to protect the environment or restrain activities that poses threat to public health and safety, examples includes environment pollution which includes lands, air, water etc, unhealthy work environment, etc. This rules identify activities that are allowed or under sanction for individuals, firms and government, breaking this rules most times comes with heavy fines or sanctions.

Social regulation help to see to the safety and well being of our environment, it serves as a guide for human activities.

7 0
3 years ago
Read 2 more answers
A labor-intensive process to manufacture a product has a fixed cost of $338,000 and a variable cost of $143 per unit. An automat
ozzi

Answer:

no of unit is 17941

Explanation:

given data

fixed cost = $338,000

variable cost = $143 per unit

fixed cost = $1,244,000  

variable cost = $92.50 per unit

solution

we consider here no of unit is = n

so here total cost of labor will be sum of fix and variable cost i.e

total cost of labor = $33800 + $143 n  ..........1

and

total cost of capital intensive  = $1,244,000 + $92.5 n   ..........2

so here in both we prefer cost of capital if cost of capital intensive less than cost of labor

$1,244,000 + $92.5 n  <  $33800 + $143 n

solve we get

n > \frac{906000}{50.5}

n > 17941

and

cost of producing less than selling cost so here

$1,244,000 + $92.5 n < 197 n

solve it we get

n > \frac{1244000}{104.5}  

n > 11904

so in both we get greatest no is 17941

so no of unit is 17941

3 0
3 years ago
A sheet of steel 3-mm thick has nitrogen atomospheres on both sides at 900 C and is permitted to achieve a steady-state di usion
kati45 [8]

Answer:

X_B = 1.8 \times 10^{-3} m = 1.8 mm

Explanation:

Given data:

Diffusion constant for nitrogen is = 1.85\times 10^{-10} m^2/s

Diffusion flux = 1.0\times 10^{-7} kg/m^2-s

concentration of nitrogen at high presuure = 2 kg/m^3

location on which nitrogen  concentration is 0.5 kg/m^3   ......?

from fick's first law

J = D \frac{C_A C_B}{X_A X_B}

Take C_A as point  on which nitrogen concentration is 2 kg/m^3

x_B = X_A + D\frac{C_A -C_B}{J}

Assume X_A is zero at the surface

X_B = 0 + ( 12\times 10^{-11} ) \frac{2-0.5}{1\times 10^{-7}}

X_B = 1.8 \times 10^{-3} m = 1.8 mm

4 0
3 years ago
Two dogbone specimens of identical geometry but made of two different materials: steel and aluminum are tested under tension at
makkiz [27]

Answer:

\dot L_{steel} = 3.448\times 10^{-4}\,\frac{in}{min}

Explanation:

The Young's module is:

E = \frac{\sigma}{\frac{\Delta L}{L_{o}} }

E = \frac{\sigma\cdot L_{o}}{\dot L \cdot \Delta t}

Let assume that both specimens have the same geometry and load rate. Then:

E_{aluminium} \cdot \dot L_{aluminium} = E_{steel} \cdot \dot L_{steel}

The displacement rate for steel is:

\dot L_{steel} = \frac{E_{aluminium}}{E_{steel}}\cdot \dot L_{aluminium}

\dot L_{steel} = \left(\frac{10000\,ksi}{29000\,ksi}\right)\cdot (0.001\,\frac{in}{min} )

\dot L_{steel} = 3.448\times 10^{-4}\,\frac{in}{min}

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3 years ago
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What is the uppermost part of the tree referred to as?
tangare [24]

Answer:

canopy I believe it is called

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