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dlinn [17]
3 years ago
15

A power plant operates on a regenerative vapor power cycle with one open feedwater heater. Steam enters the first turbine stage

at 12 MPa, 5608C and expands to 1 MPa, where some of the steam is extracted and diverted to the open feedwater heater operating at 1 MPa. The remaining steam expands through the second turbine stage to the condenser pressure of 6 kPa. Saturated liquid exits the open feedwater heater at 1 MPa. The net power output for the cycle is 330 MW. For isentropic processes in the turbines and pumps.
Determine:
a. the cycle thermal efficiency.
b. the mass flow rate into the first turbine stage, in kg/s.
c. the rate of entropy production in the open feedwater heater, in kW/K.

Engineering
1 answer:
faltersainse [42]3 years ago
4 0

Answer:

a) 0.489

b) 54.42 kg/s

c) 247.36 kW/s

Explanation:

Note that all the initial enthalpy and entropy values were gotten from the tables.

See the attachment for calculations

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An open vat in a food processing plant contains 500 L of water at 20°C and atmospheric pressure. If the water is heated to 80°C,
tester [92]

Answer:

percentage change in volume is 2.60%

water level rise is 4.138 mm

Explanation:

given data

volume of water V = 500 L

temperature T1 = 20°C

temperature T2 = 80°C

vat diameter = 2 m

to find out

percentage change in volume and how much water level rise

solution

we will apply here bulk modulus equation that is ratio of change in pressure   to rate of change of volume to change of pressure

and we know that is also in term of change in density also

so

E = -\frac{dp}{dV/V}  ................1

And -\frac{dV}{V} = \frac{d\rho}{\rho}   ............2

here ρ is density

and we know ρ  for 20°C = 998 kg/m³

and ρ  for 80°C = 972 kg/m³

so from equation 2 put all value

-\frac{dV}{V} = \frac{d\rho}{\rho}

-\frac{dV}{500*10^{-3} } = \frac{972-998}{998}

dV = 0.0130 m³

so now  % change in volume will be

dV % = -\frac{dV}{V}  × 100

dV % = -\frac{0.0130}{500*10^{-3} }  × 100

dV % = 2.60 %

so percentage change in volume is 2.60%

and

initial volume v1 = \frac{\pi }{4} *d^2*l(i)    ................3

final volume v2 = \frac{\pi }{4} *d^2*l(f)    ................4

now from equation 3 and 4 , subtract v1 by v2

v2 - v1 =  \frac{\pi }{4} *d^2*(l(f)-l(i))

dV = \frac{\pi }{4} *d^2*dl

put here all value

0.0130 = \frac{\pi }{4} *2^2*dl

dl = 0.004138 m

so water level rise is 4.138 mm

8 0
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Explanation:

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A program contains the following function definition: int cube(int number) { return number * number * number; } Write a stateme
Nonamiya [84]

Answer:

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We can then use the "=" operator to assign the return output of the function to a variable (e.g. <em>int result = cube(4)</em>)

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If you stretch a rubber hose and pluck it, you can observe a pulse traveling up and down the hose. What happens to the speed of
stich3 [128]

Answer:

Explanation:if you stretch the hose more tightly the speed of the pulse will reduce..

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