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Svetllana [295]
2 years ago
5

Carbon dioxide enters an adiabatic compressor at 100 kPa and 300 K at a rate of 0.5 kg/s and leaves at 600 kPa and 450 K. Neglec

ting kinetic energy changes, determine (a) the volume flow rate of the carbon dioxide at the compressor inlet and (b) the power input to the compressor. Amswers: (a) 0.283 m^3/s, (b) 68.8 kW
Chemistry
1 answer:
Sergeeva-Olga [200]2 years ago
5 0

Answer:

(a) Q_1=0.283m^3/s

(b) W=68.8kW

Explanation:

Hello,

(a) Based on the inlet conditions, we see that such carbon dioxide has an ideal behavior, so the inlet specific volume is:

v_1=\frac{RT}{PM}=\frac{8.314\frac{Pa*m^3}{mol*K}*300K}{100000Pa*44g/mol} =0.000567m^3/kg

Thus, the volumetric flow rate turns out:

Q_1=0.000567m^3/g*0.5kg/s*1000=0.283m^3/s

(b) Now, by writing and energy balance we have:

mh_1+W=mh_2

Thus, since the ideal gas enthalpy does not depend on the pressure, in Cengel's A-20 table, the corresponding enthalpies at 300K and 450 K are 9431kJ/kmol and 15483kJ/kmol whose values in kJ/kg are 214.34 kJ/kg and 351.89 kJ/kg respectively, thus:

W=0.5kg/s*(351.89-214.34)kJ/kg=68.8kW

Best regards.

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a quantity of gas has a volume of 400.0 mL when confined under a pressure of 600.0mm Hg. what will be the new volume of the gas
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1200 mL

Explanation:

Given data

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  • Initial volume (V₁): 400.0 mL
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  • Final volume (V₂): ?

For a gaseous sample, there is an inverse relationship between the pressure and the volume. If we consider the gas as an ideal gas, we can find the final volume using Boyle's law.

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3 0
2 years ago
A single-effect evaporator is concentrating a feed of 9072 kg/h of a 10 wt % solution of NaOH in water at temperature of 288.8 K
Aloiza [94]

Answer:

a) steam used = 8440 kg/hr

b)  Because as per calculation done in a), Steam is sufficient to evaporate 7257.6 kg/hr water from feed. Since feed is less than this amount, therefore whole water will be evaporated and 100% solid product will be obtained in bottom collection

Explanation:

Saturated steam pressure = 42KPag=1.42 bar

From steam table, Steam temperature = 110 C

Latent heat of this steam = 2230 KJ/kg

Process side pressure = 20 Kpaa = .2 bar

Water latent heat at this Pressue = 2360 KJ/kg

Water boiling Point at this Pressure =60 C

Feed Inlet temperature = 15.6 C

Total Heat required = Heat required to rise feed temperature from 15.6 to 60 degree C + Evaporation of water to concentrate the feed

Total Water evaporation required = 9072*(100-10)/100-9072*.1/.5*.5=7257.6 Kg/hr

Specific heat of feed assumed = 4.2 KJ/kg/K

=> Total heat required = 9072*4.2*(60-15.6)+7257.6*2360=18819682 KJ/hr

LMTD = ((110-60)-(110-15.6))/LN((110-60)/(110-15.6))=69.8 C

U, Overall Heat transfer coefficient = 1988 W/m2/K

Total heat required = U*A*LMTD

=> Area required of evaporator, A=18819682 *1000/3600/1988/59.8=44 m2

Steam used = 18819682/2230=8440 kg/hr

Energy required to condense vaporised feed = 7257.6*2360=17127936 KJ/hr

=> Steam efficiency = (18819682-17127936)/18819682=9%

b) Because as per calculation done in a), Steam is sufficient to evaporate 7257.6 kg/hr water from feed. Since feed is less than this amount, therefore whole water will be evaporated and 100% solid product will be obtained in bottom collection

6 0
3 years ago
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