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stellarik [79]
3 years ago
7

A company has a stack that emits a hazardous air pollutant. The ground mass concentration directly downwind of the plume sometim

es exceeds the safe limit (call it cj,safe) and the Department of Environmental Protection (DEP) is requiring a correction to this situation. The company decides to look into building a taller stack, rather than investing in a more sophisticated and very expensive APCS. They hire you as a consultant to calculate how tall the new stack would have to be in order to satisfy DEP.
Required:
a. Generate an equation to determine the stack height hs required such that, for the reflecting ground case, the ground level concentration on the ground directly downwind (y = 0 and z = 0) does not exceed the safe level at any location along the ground.
b. Repeat for the absorbing ground case.
c. Discuss. Is this a good solution for the company’s problem? For the environment?
Engineering
2 answers:
fomenos3 years ago
4 0
Nobody about to do all of that lol
fredd [130]3 years ago
3 0

Answer:

do the wam wam

Explanation:

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An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500
Rashid [163]

Answer:

Exit temperature = 32 °C

Explanation:

We are given;

Initial Pressure;P1 = 100 KPa

Cp =1000 J/kg.K = 1 KJ/kg.k

R = 500 J/kg.K = 0.5 Kj/Kg.k

Initial temperature;T1 = 27°C = 273 + 27K = 300 K

volume flow rate;V' = 15 m³/s

W = 130 Kw

Q = 80 Kw

Using ideal gas equation,

PV' = m'RT

Where m' is mass flow rate.

Thus;making m' the subject, we have;

m' = PV'/RT

So at inlet,

m' = P1•V1'/(R•T1)

m' = (100 × 15)/(0.5 × 300)

m' = 10 kg/s

From steady flow energy equation, we know that;

m'•h1 + Q = m'h2 + W

Dividing through by m', we have;

h1 + Q/m' = h2 + W/m'

h = Cp•T

Thus,

Cp•T1 + Q/m' = Cp•T2 + W/m'

Plugging in the relevant values, we have;

(1*300) - (80/10) = (1*T2) - (130/10)

Q and M negative because heat is being lost.

300 - 8 + 13 = T2

T2 = 305 K = 305 - 273 °C = 32 °C

13000 + 300 - 8000 = T2

6 0
3 years ago
A reservoir is 1 km wide and 10 km long and has an average depth of 100m. Every hour, 0.1% of the reservoir's volume drops throu
Ksju [112]

Answer:

250.7mw

Explanation:

Volume of the reservoir = lwh

Length of reservoir = 10km

Width of reservoir = 1km

Height = 100m

Volume = 10x10³x10³x100

= 10⁹m³

Next we find the volume flow rate

= 0.1/100x10⁹x1/3600

= 277.78m³/s

To get the electrical power output developed by the turbine with 92 percent efficiency

= 0.92x1000x9.81x277.78x100

= 250.7MW

7 0
3 years ago
The atmosphere within a room is at 70 °F dry-bulb temperature, 50 percent degree of saturation, and 14.696 psia pressure. The in
Gre4nikov [31]

Answer:

Given that the temperature of the window is below the dew point it will condensate.

Explanation:

A psychrometric chart (like the one attached) will give you the information needed. This chart is for 14.696 psia.

On the bottom horizontal axes you have the dry-bulb temperature, in this case 70°F, going up from this point you can reach the 50% relative humidity curve (red point on chart), going horizontally from this point to the 100% relative humidity you get the dew point temperature (the point at which moisture will condensate) (blue point on chart). In this case the dew point is 50°C. Given that the temperature of the window is below the dew point it will condensate.

6 0
3 years ago
Which option distinguishes the type of software the team should use in the following scenario?
Vladimir79 [104]
C geographic mapping software
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3 years ago
If it is desired to lay off a distance of 10,000' with a total error of no more than ± 0.30 ft. If a 100' tape is used and the
Ira Lisetskai [31]

Answer:

± 0.003 ft

Explanation:

Since our distance is 10,000 ft and we need to use a full tape measure of 100 ft. We find that 10,000 = 100 × 100.

Let L' = our distance and L = our tape measure

So, L' = 100L

Now by error determination ΔL' = 100ΔL

Now ΔL' = ± 0.30 ft

ΔL = ΔL'/100

= ± 0.30 ft/100

= ± 0.003 ft

So, the maxim error per tape is ± 0.003 ft

3 0
3 years ago
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