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Katarina [22]
3 years ago
9

A smelter emits SO2 at a rate of 10,000 kg/day. The stack is 200 m tall and the plume rise is 100 m. The windspeed is 2 m/s at a

height of 10 m and it is an overcast day. Estimate the ground-level SO2 concentration in units of µg/m3. a. 6000 m directly downwind b. 6000 m downwind and 300 m off the centerline
Engineering
1 answer:
Olegator [25]3 years ago
5 0

Answer:

lol

Explanation:

lol

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If the bending moment (M) is 4,176 ft-lb and the beam is an 1 beam, calculate the bending stress (psi) developed at a point with
SpyIntel [72]

Answer:

Bending stress at point 3.96 is \sigma_b = 1.37 psi

Explanation:

Given data:

Bending Moment M is 4.176 ft-lb = 50.12 in- lb

moment of inertia I = 144 inc^4

y = 3.96 in

\sigma_b = \frac{M}{I} \times y

putting all value to get bending stress

\sigma_b = \frac{50.112}{144} \times 3.96  

\sigma_b =  1.37 psi

Bending stress at point 3.96 is \sigma_b = 1.37 psi

3 0
3 years ago
Why is it important to know where your online information comes from?
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It is very important to know where online information comes from in order to validate, authenticate and be sure it's the right information

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2 years ago
Which one of the following questions about population growth is the only TRUE statement?A) The size of a population can never ex
tatyana61 [14]

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6 0
3 years ago
what would my engine exhaust sound like if i made a custom muffler that was just smooth on the inside, would it have an echoing
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3 years ago
A freezer is maintained at 20°F by removing heat from it at a rate of 75 Btu/min. The power input to the freezer is 0.70 hp, and
Igoryamba

Answer:

Explanation:

Cop of reversible refrigerator = TL / ( TH - TL)

TL = low temperature of freezer = 20 °F

TH = temperature of air around = 75 °F

Heat removal rate QL = 75 Btu/min

W actual, power input = 0.7 hp

conversion on F to kelvin = (T (°F) + 460 ) × 5 / 9

COP ( coefficient of performance) reversible = (20 + 460) × 5/9 / (5/9 ( ( 75 +460) - (20 + 460) ))

COP reversible = 480 / 55 = 8.73

irreversibility expression, I = W actual - W rev

COP r = QL / Wrev

W rev = QL /  COP r  where 75 Btu/min = 1.76856651 hp  where W actual = 0.70 hp

a) W rev =  1.76856651 hp  /  8.73  = 0.20258 hp is reversible power

I = W actual - W rev

b) I = 0.7 hp - 0.20258 hp = 0.4974 hp

c) the second-law efficiency of this freezer = W rev / W actual =  0.20258 hp / 0.7 hp = 0.2894 × 100 = 28.94 %

8 0
3 years ago
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