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umka21 [38]
3 years ago
8

Heat is transferred at a rate of 2 kW from a hot reservoir at 800 K to a cold reservoir at 300 K. Calculate the rate at which th

e entropy of the two reservoirs changes. (Round the final answer to six decimal places.) The rate at which the entropy of the two reservoirs changes is kW/K.

Engineering
1 answer:
andre [41]3 years ago
6 0

Answer:

4.17x10^-3 kW/K

Explanation:

Detailed explanation and calculation is shown in the image below

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A steam reformer operating at 650C and 1 atm uses propane as fuel for hydrogen production. At the given operating conditions, th
dusya [7]

Answer:

Explanation:

a) for shifting reactions,

Kps =  ph2 pco2/pcoph20

=[h2] [co2]/[co] [h2o]

h2 + co2 + h2O + co + c3H8 = 1

it implies that

H2 + 0.09 + H2O + 0.08 + 0.05 = 1

solving the system of equation yields

H2 = 0.5308,

H2O = 0.2942

B)  according to Le chatelain's principle for a slightly exothermic reaction, an increase in temperature favors the reverse reaction producing less hydrogen. As a result, concentration of hydrogen in the reformation decreases with an increasing temperature.

c) to calculate the maximum hydrogen yield , both reaction must be complete

C3H8 + 3H2O ⇒ 3CO + 7H2( REFORMING)

CO + H2O ⇒ CO2 + H2 ( SHIFTING)

C3H8 + 6H2O ⇒ 3CO2 + 10 H2 ( OVER ALL)

SO,

Maximum hydrogen yield

= 10mol h2/3 molco2 + 10molh2

= 0.77

⇒ 77%

3 0
3 years ago
In a wind-turbine, the generator in the nacelle is rated at 690 V and 2.3 MW. It operates at a power factor of 0.85 (lagging) at
Juli2301 [7.4K]

To solve this problem we will apply the concepts related to real power in 3 phases, which is defined as the product between the phase voltage, the phase current and the power factor (Specifically given by the cosine of the phase angle). First we will find the phase voltage from the given voltage and proceed to find the current by clearing it from the previously mentioned formula. Our values are

V = 690V

P_{real} = 2.3MW

Real power in 3 phase

P_{real} = 3V_{ph}I_{ph} Cos\theta

Now the Phase Voltage is,

V_{ph} = \frac{V}{\sqrt{3}}

V_{ph} = \frac{690}{\sqrt{3}}

V_{ph} = 398.37V

The current phase would be,

P_{real} = 3V_{ph}I_{ph} Cos\theta

Rearranging,

I_{ph}=\frac{P_{real}}{3V_{ph}Cos\theta}

Replacing,

I_{ph}=\frac{2.3MW}{3( 398.37V)(0.85)}

I_{ph}= 2.26kA/phase

Therefore the current per phase is 2.26kA

6 0
3 years ago
Which of the following is NOT an ASE certification? Select one:
stiks02 [169]

The option that is not an ASE certification is . A/C and Refrigerants handling certification (609).

<h3>What is ASE certification?</h3>

The term ASE is known to be a body that tends to promotes excellence in regards to vehicle repair, service as well as parts distribution.

Note that in the world today more than a quarter of  million of people are known to possess ASE certifications.

Since ASE Certified professionals work in in all areas of the transportation industry. one can say that The option that is not an ASE certification is. A/C and Refrigerants handling certification (609).

Learn more about ASE certification from

brainly.com/question/5533417

#SPJ1

8 0
2 years ago
Historical thinking skills can be divded into there main processes
morpeh [17]

Answer:

asking questions, drawing conclusions, and gathering information.

Explanation:

hope this helps!

7 0
4 years ago
Lithium at 20°C is BCC and has a lattice constant of 0.35092 nm. Calculate a value for the atomic radius of a lithium atom in na
Nutka1998 [239]

Answer:

the atomic radius of a lithium atom is 0.152 nm

Explanation:

Given data in question

structure = BCC

lattice constant  (a) = 0.35092 nm

to find out

atomic radius of a lithium atom

solution

we know structure is BCC

for BCC radius formula is \sqrt{3} /4 × a

here we have known a value so we put a in radius formula

radius =  \sqrt{3} /4 × a

radius =  \sqrt{3} /4 × 0.35092

radius = 0.152 nm

so the atomic radius of a lithium atom is 0.152 nm

5 0
3 years ago
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