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3 years ago

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Engineering

1 answer:

san4es73 [151]3 years ago

7 0

Answer:

I dont know. I have the same question. Can you help me out?

Explanation:

idk

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The water behind Hoover Dam is 206m higher than the Colorado river below it. At what rate must water pass through the hydraulic

Leokris [45]

Answer:

m' = 4948.38 kg/s

Explanation:

For a case of 100% efficiency, the power produced must be equal to the rate of potential energy conversion

GIVEN THAT

Power = 100 MW

rate of Potential energy = (m')*g*h

100*10^6 = (m')*9.81*206

m' = 4948.38 kg/s

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3 years ago

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How do information systems support the activities in a supply chain?

Degger [83]

Answer:

the no. of activities supply in a cahin like in the figuration wise they supply the chain

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2 years ago

At a certain location, wind is blowing steadily at 7 m/s. Determine the mechanical energy of air per unit mass and the power gen

Kaylis [27]

Answer:

Explanation:

From the information given;

The velocity of the wind blow V = 7 m/s

The diameter of the blades (d) = 80 m

Percentage of the overall efficiency $\eta_{overall} = 30\%$

The density of air $\rho = 1.25 kg/km^3$

Then, we can use the concept of the kinetic energy of the wind blowing to estimate the mechanic energy of air per unit mass by using the formula:

$e_{mechanic} = \dfrac{mV^2}{2}$

here;

m = $\rho AV$

= $1.25 \times \dfrac{\pi}{4}(80)^2 \times 7$

= 43982.29 kg/s

∴

$W = e_{mechanic} = \dfrac{mV^2}{2}$

$= \dfrac{43982.29 \times 7^2}{2}$

$= 1077566.105 \ W$

$\mathbf{ =1077.566 \ kW}$

The actual electric power is:

$W_{electric} = \eta_{overall} \times W$

$W_{electric} = 0.3 \times 1077.566$

$\mathbf{W_{electric} =323.26 \ kW}$

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3 years ago

What are the 5 major forest types?

Nataly [62]

Answer:

1. Equatorial Evergreen or Rainforest

2. Tropical forest

3. Mediterranean forest

4. Temperate broad-leaved forest

5. Warm temperate forest

Explanation:

4 0

4 years ago

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An induced-draft cooling tower cools 90,000 gallons per minute of water from 84 to 68oF. Air at 14.61 psia, 70oF dry bulb and 60

belka [17]

Answer:

a. V = 109.64 × 10⁵ ft/min

b. Mw = 654519.54 kg/hr

Explanation:

Given Parameters

mass flow rate of water, Mw = 90000g/min = 6607.33 kg/s

inlet temperature of water, T1 = 84 F = 28.89 C

outlet temperature of water, T2 = 68 F = 20 C

specific heat capacity of water, c = 4.18kJ/kgK

rate of heat remover from water, Qw is given by

Qw = 6607.33[28.89 - 20] * 4.18

Qw = 245529.545kw

For air, inlet condition

DBT = 70 F hi = 43.43 kJ/kg

WBT = 60 F wi = 0.00874 kJ/kg

u1 = 0.8445 m/kg

oulet condition,

DBT = 70 F RH = 100.1

h1 = 83.504kJ/kg

Wo = 0.222kJ/kg

check the attached file for complete solution

3 0

3 years ago