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cricket20 [7]
3 years ago
12

A heat exchanger takes compressed liquid water and converts it into steam with a temperature and pressure of 20 Mpa and 480 C; r

espectively. Steam is used to generate work �! in a turbine with an isentropic efficiency of 78%. Steam that exits the turbine at 14 Mpa is used in a second identical turbine (same efficiency) after reheating it to 500 C (with the same heating exchanger). Once the steam exits the second turbine it passes through a condenser (working at a pressure of 6 Kpa) before being pumped (with an internal reversible and adiabatic pump) back to the heat exchanger.
A. Work done by the first turbine (KJ/Kg).
B. Work done by the second turbine (KJ/Kg).
C. Quality of steam leaving the second turbine (5pts).
D. Cycle thermal efficiency (5pts).

Engineering
1 answer:
belka [17]3 years ago
6 0

Answer:

See explaination

Explanation:

Lets first consider the term Isentropic efficiency. The isentropic efficiency of a compressor or pump is defined as the ratio of the work input to an isentropic process, to the work input to the actual process between the same inlet and exit pressures. IN practice, compressors are intentionally cooled to minimize the work input.

Please kindly check attachment for the step by step solution of the given problem.

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I’m doing a project on renewable energy. There are 6 energy sources. Solar, wind, geothermal, hydroelectric, tidal, and biomass.
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Fluid power is a. The technology that deals with the generation, control, and transmission of power-using pressurized fluids b.
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The purpose of pasteurizing milk is to A. Kill pathogens B. Break down milk fat C. Add vitamins and minerals D. Prevent spoilage
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3 years ago
A closed system consists of 0.3 kmol of octane occupying a volume of 5 m³. Determine (a) the weight of the system, in N, and (b)
Leni [432]

Answer:

a) m=336.18N

b) Vn=16.67m/kmol

Vm=0.1459m^3/kg

Explanation:

To calculate the mass of the octane(m):

Number of mole of octane (n) =0.3kmol(given)

Molarmass of octane (M) =114.23kg/kmol

m=n*M

m=(0.3kmol)*(114.23kg/kmol)

m=34.269kg

To calculate for the weight of octane(W):

W=g*m

W=(9.81m/s^2)*(34.269kg)

W=336.18N

b) For specific volumes of Vn and Vm:

Given volume of octane (V) =5m^3

Vm=V/m

Vm=5m^3/34.269kg

Vm=0.1459m^3/kg

And Vn will be :

Vn=V/m=5m^3/0.3kmol

Vn=16.67m/Kmol

Therefore, the answers are:

a) m=336.18N

b) Vn=16.67m/kmol

Vm=0.1459m^3/kg

7 0
3 years ago
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