A gaseous stream containing 19.95 kmol h^-1 of oxygen, 75.05 kmol h^-1 of nitrog

Question

A gaseous stream containing 19.95 kmol h^-1 of oxygen, 75.05 kmol h^-1 of nitrogen and 20 kmol h^-1 water vapour at 150 degree C and 1.2 bar is cooled to 50 degree C and 1.15 bar in a partial condenser resulting in the condensation of part of the water vapour. The gaseous-liquid mixture leaving the condenser is in equilibrium the partial pressure of the water vapour in the gaseous stream saturated pressure of water at 50 degree C. Assume that both oxygen and nitrogen are essentially insoluble in water. Using the data given below calculate a) The component flow rates in both the liquid and gaseous streams leaving the condenser b) The heat load on the condenser Data Saturated vapour pressure water at 50 degree C See steam tables Mean specific heat capacity liquid water 50 degree C to 150 degree C. 4.251 kJ kg^-1 K^-1 Mean specific heat capacity water vapour 50 degree C to 150 degree C 1.892 kJ kg^-1 K^-1 Latent heat of water at 50 degree C 2382.1 kJ kg^-1 Mean specific heat capacity of nitrogen gas 50 degree C to 150 degree C 1.044 kJ kg^-1 K^-1 Mean specific heat capacity of oxygen gas 50 degree C to 150 degree C 0.986 kJ kg^-1 K^-1 Relative molar mass water 18 Relative molar mass nitrogen 28 Relative molar mass oxygen 32 Answer a) Liquid 8.591 kmol h^-1 water Gaseous 19.95 kmol h^-1 oxygen, 75.05 kmol h^-1 nitrogen and 11.409 kmol h^-1 water b) 0.1997 MW

Explanation / Answer

First, one has to determine during the process of cooling, how much water is condensed.

Since after partial condensation, only water vapor is partly condensed, flow rate of other gases like nitrogen and water vapor remain same at 19.95 kmol/hr and 75.07 kmoles/hr. This is water vapor free gas = 75.07+19.95 kmole/hr=95.02 kmoles/hr

Since at 50 deg.c, the gas is saturated with water vapor,

Saturation vapor pressure, logPsat(mm Hg)= 8.07131-1730.63/(t(deg.c)+233.426

At 50 deg.c, vapor pressure of water = 8.07131- 1730.63/(50+233.426)= 92.3 mm Hg

At saturation condition, partial pressure of water vapor = vapor presure of liquid

Partial pressure of water vapor= 92.3 mm Hg

Total pressure at the exit of condenser= 1.15 bar = 1.15*0.9869*760 mm Hg =863 mm Hg ( 1 bar=0.9869 atm and 1 atm= 760mm Hg)

Hence moles of water vapor/ moles of water vapor free gas= partial pressure of water vapor/ partial pressure of water vapor free gas

Moles of water vapor/ 95.02= 92.3/ (863-92.3)

Moles of water vapor =11.3 kmoles/hr

This many moles of water vapor is there in the gas phase.

Moles of water vapor condensed= 20-11.3= 8.7 kmoles/hr

From 150 deg.c to 50 deg.c, water vapor undergoes one sensinle heat loss of 20 kmole/hr of water vapor. The sensinble heat loss= moles of water vapor* specific heat of water vapor* temperature difference

Since specific heat as well as latent heat is given in Kj/kg.deg.c, moles need to be converted to mass by multiplying with respective molar masses of the substances.

(20*1000 gmoles/hr*18 gm/mole /1000) Kg*1.892 KJ/kg.deg.c*(150-50)= 68112KJ/hr.

At 50 deg.c, 8.7kmoles/ hr of water vapor has to be condensed by removing latent heat of vaporization and this heat= (8.7*1000*18/1000)Kg*2382.1Kj/kg =373037 Kj/hr

While oxygen and nitrogen loose only sensible heat

Sensible heat loss by nitrogen = (75.05 kmoles/hr*1000*28/1000) Kg/hr*1.044*(150-50) Kj/hr=219386 Kj/hr

Sensilbe heat loss by oxygen= (19.95*1000*32/1000) kg/hr* 0.986*(150-50)= 62946 Kj/hr

Total heat to be removed = sensible heat of water + latent heat of water + sensible heat of nitrogen+ sensible heat of oxygen= 68112+373037+219386+62946=723481Kj/hr= 723481/3600 Kj/sec= 200.967 KW = 0.200 Mw ( 1MW= 1000 KW)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.