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PolarNik [594]
3 years ago
11

8. When supplying heated air for a building, one often chooses to mix in some fresh outside air with air that has been heated fr

om the building as it passes through the furnace. An insulated mixing chamber is used to combine two streams of air to be used in a building. One stream of air, brought from the outside, enters at 2 kg/s, at a pressure of 120 kPa, and a temperature of 5oC. The second stream of air, coming from the building’s furnace, has a mass flow rate of 8 kg/s, a pressure of 120 kPa, and a temperature of 35oC. The combined stream is then delivered to the warm space at 120 kPa. Determine the rate of entropy generation for this mixing chamber.

Engineering
1 answer:
Afina-wow [57]3 years ago
5 0

Answer:

Check the explanation

Explanation:

Kindly check the attached images below to see the step by step explanation to the question above.

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Answer:

We would need background context,

Explanation:

Then I would be happy to help!

4 0
3 years ago
Pumped-storage hydroelectricity is a type of hydroelectric energy storage used by electric power systems for load balancing. The
NikAS [45]

Answer:

A) energy loss E = pgQtH

Where p = density in kg/m3

g = gravity acceleration in m/s2

Q = flow rate in m3/s

t = time taken for flow in sec

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B) power required to run pump;

P = pgQH

Explanation:

Detailed explanation and calculation is shown in the image below

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Liquid benzene and liquid n-hexane are blended to form a stream flowing at a rate of 1700 lbm/h. An on-line densitometer (an ins
Taya2010 [7]
Let me think of that
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3 years ago
1.
Levart [38]

Answer:

This is a for loop.

Explanation:

6 0
3 years ago
A gas stream flowing at 1000 cfm with a particulate loading of 400 gr/ft3 discharges from a certain industrial plant through an
Makovka662 [10]

<u>Solution and Explanation:</u>

Volume of gas stream = 1000 cfm (Cubic Feet per Minute)

Particulate loading = 400 gr/ft3 (Grain/cubic feet)

1 gr/ft3 = 0.00220462 lb/ft3

Total weight of particulate matter = 1000 \mathrm{cfm} \times 400 \mathrm{gr} / \mathrm{tt} 3 \times .000142857 \mathrm{lb} / \mathrm{ft} 3 \times 60=3428.568 \mathrm{lb} / \mathrm{hr}

Cyclone is to 80 % efficient

So particulate remaining = 0.20 \times 3428.568 \mathrm{lb} / \mathrm{hr}=685.7136

emissions from this stack be limited to = 10.0 lb/hr

Particles to be remaining after wet scrubber = 10.0 lb/hr

So particles to be removed = 685.7136- 10 = 675.7136

Efficiency = output multiply with 100/input = 98.542 %

4 0
3 years ago
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