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bija089 [108]
3 years ago
11

Tubes through which water flows as it is brought from 0.8 MPa, 150C to 240C at essentially constant pressure in the boiler of a

power plant. The total mass flow rate of the water is 100 kg/s. Combustion gases passing over the tubes cool from 1067 to 547C at essentially constant pressure. The combustion gases can be modeled as air as an ideal gas. There is no significant heat transfer from the boiler to its surroundings. Surrounding (dead state) temperature and pressure are given as 25C and 1 atm, respectively. Determine i) the exergetic efficiency of the boiler ii) rate of exergy destruction as kW iii) mass flow rate of the combustion gases as kg/s
Chemistry
1 answer:
motikmotik3 years ago
4 0

Answer:

The correct answer is i) 50.2 % ii) 13440.906 kW and iii) 71.986 kg/s.

Explanation:

In order to find the mass flow rate of the combustion of gases, there is a need to use the energy balance equation:  

Mass of water × specific heat of water (T2 -T1)w = mass of gas × specific heat of gas (T2-T1)g

100 × 4.18 × [(240 + 273) - (150 + 273)] = mass of gas × 1.005 × [(1067+273) - (547+273)]

Mass of gas = 71.986 kg/s

The entropy generation of water can be determined by using the formula,  

(ΔS)w = mass of water × specific heat of water ln(T2/T1)w

= 100 × 4.18 ln(513/423)

= 80.6337 kW/K

Similarly the entropy generation of water will be,

(ΔS)g = mass of gas × specific heat of gas ln(T2/T1)g

= 71.986 × 1.005 ln (820/1340)

= -35.53 kW/K

The rate of energy destruction will be,  

Rate of energy destruction = To (ΔS)gen

= T₀ [(ΔS)w + (ΔS)g]

= (25+273) [80.6337-53.53)

Rate of energy destruction = 13440.906 kW

The availability of water will be calculated as,  

= mass of water (specific heat of water) [(T₁-T₂) -T₀ ln T₁/T₂]

= 100 × 4.8 [(513-423) - 298 ln 513/423]

= 13591.1477 kW

The availability of gas will be calculated as,  

= mass of gas (specific heat of gas) [(T₁-T₂) - T₀ ln T₁/T₂]

= 71.986 × 1.005 × [(1340-820) - 298 ln 1340/820]

= 27031.7728 kW

The exergetic efficiency can be calculated as,  

= Gain of availability / loss of availability  

= 13591.1477/27031.7728

= 0.502

The exergetic efficiency is 50.2%.  

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Answer : The correct option is, (A) 1.8\times 10^{-10}M, basic.

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pOH=-\log (5.5\times 10^{-5})

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Now we have to calculate the pH.

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pH=-\log [H_3O^+]

9.74=-\log [H_3O^+]

[H_3O^+]=1.8\times 10^{-10}M

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From the pH value, 9.74 we conclude that the solution is basic in nature because the value of pH is greater than 7.

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