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stepan [7]
3 years ago
12

A coal-burning power plant generates electrical power at a rate of 650 megawatts (MW), or 6.50 × 108 J/s. The plant has an overa

ll efficiency of 38% for the conversion of heat to electricity. Given this efficiency, how much heat energy in joules is produced in one year of operation??
Engineering
1 answer:
Vinvika [58]3 years ago
7 0

Answer:

Energy produce in one year =20.49 x 10¹⁶ J/year

Explanation:

Given that

Plant produce 6.50 × 10⁸ J/s of energy.

It produce  6.50 × 10⁸ J in 1 s.

We know that

1 year = 365 days

1 days = 24 hr

1 hr = 3600 s

1 year = 365 x 24 x 3600 s

1 year = 31536000 s

So energy produce in 1 year = 31536000 x  6.50 × 10⁸ J/year

          Energy produce in one year = 204984 x 10¹² J/year

          Energy produce in one year =20.49 x 10¹⁶ J/year

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Wet steam at 15 bar is throttled adiabatically in a steady-flow process to 2 bar. The resulting stream has a temperature of 130°
cricket20 [7]

Answer:

\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}

Explanation:

The adiabatic throttling process is modelled after the First Law of Thermodynamics:

m\cdot (h_{in} - h_{out}) = 0

h_{in} = h_{out}

Properties of water at inlet and outlet are obtained from steam tables:

State 1 - Inlet (Liquid-Vapor Mixture)

P = 1500\,kPa

T = 198.29\,^{\textdegree}C

h = 2726.9\,\frac{kJ}{kg}

s = 6.3068\,\frac{kJ}{kg\cdot K}

x = 0.967

State 2 - Outlet (Superheated Vapor)

P = 200\,kPa

T = 130\,^{\textdegree}C

h = 2726.9\,\frac{kJ}{kg}

s = 7.1776\,\frac{kJ}{kg\cdot K}

The change of entropy of the steam is derived of the Second Law of Thermodynamics:

\Delta s = 7.1776\,\frac{kJ}{kg\cdot K} - 6.3068\, \frac{kJ}{kg\cdot K}

\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}

6 0
3 years ago
Guys, can you rate this toolbox, please. It is my DT project, made for car trips, what do you think of the idea/design/usability
Maurinko [17]

Answer:

Honestly overall i think it looks fantastic

Explanation:

It looks like some really nice clean craftsmanship and i love the use of some different colors for some drawers to make it pop. the only con that i can possibly think of is that with it being wood and you moving it from place to place, some rubber feet or something that would prevent it from scratching/damaging anything else if it doesn't already (cant really see under it). other then that one thing i think it looks really good. well done.

3 0
3 years ago
What happens when force is placed on a square/rectangle?
mart [117]

im not sure i need to see a photo and also is this science

7 0
3 years ago
Read 2 more answers
Air enters the combustor of a jet engine at p1=10 atm, T1=1000°R, and M1=0.2. Fuel is injected and burned, with a fuel/air mass
snow_lady [41]

Answer:

M2 = 0.06404

P2 = 2.273

T2 = 5806.45°R

Explanation:

Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.

Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,

To1 = (1.008)*(1000) = 1008 ºR

R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)

F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga

For the air q = cp(To2– To1)

(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2

Table A.3 of steam table gives P/P* = 2.273,

T/T* = 0.2066,

To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =

F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit

5 0
3 years ago
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