A simple Rankine cycle uses water as the working fluid. The boiler operates at 6000 kPa and the condenser at 50 kPa. At the entrance of the turbine the temperature is 450 deg C. The isentropic efficiency of the turbine is 94 percent, pressure and pump losses are negligible, and the water in the condenser is subcooled by 6.3 degC. The boiler is sized for a mass flow rate of 20 kg/s. Determine the rate at which heat is added in the boiler, the power required to operate the pumps, the net power produced by the cycle, and the thermal efficiency.

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Answer 1

Answer:

the rate at which heat is added in the boiler = 59597.4 kW

the power required to operate the pumps = 122.57 kW

The net power produced by the cycle = 17925 kW.

The thermal efficiency = 30%.

Explanation:

The specific enthalpy of saturated liquid is equal to the enthalpy of the first point which is equal to 314 kJ/ kg.

The second enthalpy is calculated from the pump work. Therefore, the second enthalpy = first enthalpy point + specific volume of water [ the pressure of the boiler - the pressure of the condenser].

The second enthalpy = 314  + 0.00103 [ 6000 - 50 ] = 320.13 kJ/kg.

The specific enthalpy for the third point = 3300 kJ/kg.

Therefore, the rate at which heat is added in the boiler = 20 × [3300 - 320.13] = 59597.4 kW.

The rate at which heat is added in the boiler = 59597.4 kW.

Also, the power required to operate the pumps = 20 × 0.00103 [6000 - 50] = 122.57 kW.

The power produced by the turbine = 20 [ 300 - ( the fourth enthalpy value)].

The fourth enthalpy value = 3300 - 0.94 [ 3300 - 2340] = 2397.6 kJ/kg

Thus, the power produced by the turbine = 20 [ 300 - 2397.6] = 18048 kW.

The power produced by the turbine  =  18048 kW.

The net power produced =  18048  + 122.57 = 17925 kW.

The thermal efficiency = [net power produced] / [the rate at which heat is added in the boiler].

The thermal efficiency = 17925/ 59597.4 = 30%.


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Answer:

fart

Explanation:

difference between theory and practice?​

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Answer:

There is a huge difference between theory vs. practice. Theory assumes an outcome, while practice allows you to test the theory and see if it is accurate.

Theory and Practice Explained

Practice is the observation of disparate concepts (or a phenomenon) that needs explanation. A theory is a proposed explanation of the relationship between two or more concepts, or an explanation for how/why a phenomenon occurs.

(a) Calculate the heat flux through a sheet of steel that is 10 mm thick when the temperatures oneither side of the sheet are held constant at 300oC and 100oC, respectively.(b) Determine the heat loss per hour if the cross-sectional area of the sheet is 0.25 m2.(c) What will be the heat loss per hour if a sheet of soda-lime glass is used instead

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Answer:

do the wam wam

Explanation:

The heat flux is =1038kW/m² , the heat lost per hour is =259.5 kW, the heat lost per hour using a sheet of soda- lime glass.

Calculation of heat flux

The thickness of steel( t) = 10mm = 10× 10^-³m

The temperature difference on both sides = 300-100

∆T = 200°C

But the formula for heat flux = q = k∆T/t

Where K = thermal conductivity for steel = 51.9W/mK.

Substitute the variables into the formula for heat flux;

q = 51.9 × 200/10 × 10-³

q = 10380 × 10³/10

q = 10380000/10

q = 1038000 W/m² = 1038kW/m²

To calculate the heat lost per hour if the cross sectional area is = 0.25 m2 use the formula q × A

= 1038kW/m² × 0.25 m2

= 259.5 kW.

Learn more about heat flux here:

https://brainly.com/question/15217121

Solved this question??????????????????

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I do not know!!!!!!!!

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