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valkas [14]
3 years ago
9

Describe a me best answer gets brainliest

Engineering
2 answers:
SCORPION-xisa [38]3 years ago
7 0

Answer:

Human: an idiot who does idiot stuff everyday

Explanation:

Soloha48 [4]3 years ago
6 0
A person hope this helps 6382(272
You might be interested in
Five kg of nitrogen gas (N2) in a rigid, insulated container fitted with a paddle wheel is initially at 300 K, 150 kPa. The N2 g
andrew-mc [135]

Answer:

A) attached below

B) 743 KJ

C) 1.8983 KJ/K

Explanation:

A) Diagram of system schematic and set up states

attached below

<u>B) Calculate the amount of work received from the paddle wheel </u>

assuming ideal gas situation

v1 = v2 ( for a constant volume process )

work generated by paddle wheel = system internal energy

dw = mCv dT .     where ; Cv = 0.743 KJ/kgk

     = 5 * 0.743 * ( 500 - 300 )

     = 3.715 * 200 = 743 KJ

<u>C) calculate the amount of entropy generated  ( KJ/K )</u>

S2 - S1 = 1.8983 KJ/K

attached below is the detailed solution

4 0
3 years ago
A gas stream contains 18.0 mole% hexane and the remainder nitrogen. The stream flows to a condenser, where its temperature is re
Anna [14]

Answer:

A. 72.34mol/min

B. 76.0%

Explanation:

A.

We start by converting to molar flow rate. Using density and molecular weight of hexane

= 1.59L/min x 0.659g/cm³ x 1000cm³/L x 1/86.17

= 988.5/86.17

= 11.47mol/min

n1 = n2+n3

n1 = n2 + 11.47mol/min

We have a balance on hexane

n1y1C6H14 = n2y2C6H14 + n3y3C6H14

n1(0.18) = n2(0.05) + 11.47(1.00)

To get n2

(n2+11.47mol/min)0.18 = n2(0.05) + 11.47mol/min(1.00)

0.18n2 + 2.0646 = 0.05n2 + 11.47mol/min

0.18n2-0.05n2 = 11.47-2.0646

= 0.13n2 = 9.4054

n2 = 9.4054/0.13

n2 = 72.34 mol/min

This value is the flow rate of gas that is leaving the system.

B.

n1 = n2 + 11.47mol/min

72.34mol/min + 11.47mol/min

= 83.81 mol/min

Amount of hexane entering condenser

0.18(83.81)

= 15.1 mol/min

Then the percentage condensed =

11.47/15.1

= 7.59

~7.6

7.6x100

= 76.0%

Therefore the answers are a.) 72.34mol/min b.) 76.0%

Please refer to the attachment .

4 0
3 years ago
The symmetrical load below is connected to a three-phase network. A line current of 25A has been measured. The load resistors ha
boyakko [2]

Answer:

The line voltage of the three phase network is 346.41 V

Explanation:

Star Connected Load

Resistance, R₁ = R₂ = R₃ = 18 Ω

For a star connected load, the line current = the phase current, that is we have

I_L = 25 \, A =  I_{Ph}

Whereby the the voltage across each resistance = V_R is given by the relation;

V_R = I_{Ph} × R

Hence;

V_{Ph} = V_R = I_{Ph} × R  = 25 × 8 = 200 V

Therefore we have;

The line voltage, V_{L} = √3 × V_{Ph} = √3 × 200 = 346.41 V.

Hence, the line voltage of the three phase network = 346.41 V.

3 0
4 years ago
150 lb of force is applied at the end of a shaft that is 2 1/2 ft long. It is applied
Alona [7]

Answer:

Torque: 500

Power: 1700

Explanation:

Since the force is 150 lbs and if the shaft is rotating at 350 rpms, then you will need 1700 for the power in order to keep it running.

5 0
3 years ago
Which ranks careers in the Architecture and Construction career cluster based on the typical number of years for each degree, go
timama [110]

Answer:

hmm tdyufrjhfjjbdjjkljgjjddtu

8 0
3 years ago
Read 2 more answers
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