Answer:
Explanation:
The detailed steps and appropriate calculation with analysis is as shown in the attachment.
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.
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.
1 answer:
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Answer:
<u><em>To answer this question we assumed that the area units and the thickness units are given in inches.</em></u>
The number of atoms of lead required is 1.73x10²³.
Explanation:
To find the number of atoms of lead we need to find first the volume of the plate:
<u>Where</u>:
A: is the surface area = 160
t: is the thickness = 0.002
<u><em>Assuming that the units given above are in inches we proceed to calculate the volume: </em></u>
Now, using the density we can find the mass:
Finally, with the Avogadros number () and with the atomic mass (A) we can find the number of atoms (N):
Hence, the number of atoms of lead required is 1.73x10²³.
I hope it helps you!
Answer:
18 kJ
Explanation:
Given:
Initial volume of air = 0.05 m³
Initial pressure = 60 kPa
Final volume = 0.2 m³
Final pressure = 180 kPa
Now,
the Work done by air will be calculated as:
Work Done = Average pressure × Change in volume
thus,
Average pressure = = 120 kPa
and,
Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³
Therefore,
the work done = 120 × 0.15 = 18 kJ