Steam enters the turbine of a cogeneration plant at 7 mpa and 500°c. one-fourth of the steam is extracted from the turbine

10 answers
Question:

Steam enters the turbine of a cogeneration plant at 7 mpa and 500°c. one-fourth of the steam is extracted from the turbine at 600-kpa pressure for process heating. the remaining steam continues to expand to 10 kpa. the extracted steam is then condensed and mixed with feedwater at constant pressure and the mixture is pumped to the boiler pressure of 7 mpa the mass flow rate of steam through the boiler is 30 kg/s. disregarding any pressure drops and heat losses in the piping, and assuming the turbine and the pump to be isentropic, determine the net power produced and the utilization factor of the plant.

Answers

3 square feet for edgenuity

3 square feet per minute

30x + 30y = 450 20x + 25y = 345

Here 345 represents the total area (in sq ft) shoveled; x repres. Melinda's rate and y represents Paul's rate.  We need to solve for x and y and then subtract the smaller from the larger.

30x + 30y = 450   Multiply this eqn by 2:      60x + 60y = 900
20x + 25y = 345   Mult this eqn by -3:          -60x -75y  = -1035
                                                                   
                                                                              -15y  = -135
                                                                                    y = 9

This comes to 9 square feet per minute.
Now, let's find x:    30x + 30y = 450; 30x + 270 = 450; 30x = 180; x = 6

Melinda can shovel 6 sq ft in 1 min, whereas Paul can shovel 9.

The difference is (9 - 6) sq ft / min,     or     3 sq ft / min   (answer)

A

( 3 square feet per minute)

your welcome

Hello mate!

we have the system of equations

30x + 30y = 450

20x + 25y = 345

where x is the number of square feet of snow that Melinda shovels in one minute, and y is the number of square feet of snow that Paul shovels in one minute.

Let's solve the system of equations, first, isolate one of the therms in the first equation, for example, x.

30x+30y = 450

x + y = 450/30 = 15

x = 15 - y

now we can replace it in the second equation and solve it for y.

20x + 25y = 345

20(15 - y) + 25y = 345

300 - 20y + 25y = 345

5y = 45

y = 45/5 = 9

Paul can shovel 9 square feets of snow in one minute, now replace it in the first equation and solve it for x.

x + y = 450/30 = 15

x + 9 = 15

x = 15 - 9 = 6

so melinda sholves 6 square feet of snow in one minute.

The question is:

How much more can Paul shovel in 1 minute than Melinda?

We need to see the difference y - x = 9 - 6 = 3

this means that paul shovels 3 more square feet per minute than Melinda.

3 more square feet.

Step-by-step explanation:

x = The amount Melinda can shovel in 1 minute

y = The amount Paula can shovel in 1 minute

So make two equations:

30(x + y) = 450       x + y = 15       x = 15 - y

20x + 25y = 345     4x + 5y = 69

Plug in:

4(15 - y) + 5y = 69

60 - 4y + 5y = 69

y = 9

x = 6

9 - 6 = 3

30x + 30y = 450

20x + 25y = 345

3 square ft per minute

30x+30y=450

20x+25y=345

3 square ft per minute

a

Step-by-step explanation:

Answer is A

Step-by-step explanation:

Took it on Edg.

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