Steam at 12 bar and 310 degree C is expanded through, a nozzle to saturated stea
ID: 1053121 • Letter: S
Question
Steam at 12 bar and 310 degree C is expanded through, a nozzle to saturated steam at 125 degree C. Negligible heat is transferred from the nozzle to its surroundings The approach velocity of the steam is negligible. Calculate the pressure drop across the nozzle in pounds per square inch Calculate the exit steam velocity in miles per hour (mph) A turbine discharges 500 ob_m/h of saturated steam at 12 bar. It is desired to generate steam at 300 degree C and 12 bar by mixing the turbine as charge with a second stream of superheated steam at 350 degree C and 12 bar. Calculate the heat required to produce 750 lb_m/h of the product steam exiting the mixer is adiabatic. Calculate the now rate (lb_m/h) of product steam exiting the mixer if the mixer is adiabatic Air at 45 degree C and 87% relative humidity is to be cooled to 15 degree C and fed into a plant area at a rate of 600 m^3/min. Calculate the rate at which water condenses (in kg/h) Calculate the cooling requirement in tons (1 ton of cooling = 12000 Btu/h), assuming that the enthalpy of water vapor is that of saturated steam at the steam at the same temperature. The enthalpy of dry air is given by the expression: A(kJ/mol) = 0.0291*(T-25) where T is in degree CExplanation / Answer
Saturation presure of steam at 125 deg.c =2.32 bar ( from steam tables)
Inlet pressrue of steam =12 bar
Pressure drop across the nozzle = 12-2.32= 9.68 bar
1 bar = 14.38 psi
9.68 bar = 9.68*14.38 psi = 139.1984 psi
Since the loss in enthalpy of steam results in velcoity at the exit of nozzle
deltaH + delU2/2= 0
deltaH (H2-H1)= -deltaU2/2 ,since the inlet velocity = 0
H1-H2 = U2/2
enthalpy of super heated steam at inlet pressure of 12 bar and 310 deg.c , H1= 3250 Kj/Kg
enthalpy of saturated vapor at 125 deg.c,H2 = 2713 Kj/Kg
Enthalpy change = 3250-2713= 537 Kj/Kg
537= u2/2
Hence u2= 1074 and u= 32.77 m/sec=32.77*60*60 m/hr=117972 m/hr
1m= 0.00062 mile/hr, hence velocity = 117972*0.00062 mile/hr=73.14 miles/hr
2.
. Flow rate of steam at 300 deg.c and 12 bar =750-500= 250 lb/hr
Enthalpy of saturated steam at 12 bar = 2682.2 J/g = 2682.2*0.4299btu/lb =1153 btu/lb
It is supplied at 500 lb/hr. So total enthalpy= 500*1153 btu/hr =576500 btu/hr
Enthalpy of stram at 350 deg.c and 12 bar, = 3759.5 j/g= 3759.5*0.4299 btu/lb= 1616.21 btu/lb
Enthalpy of this stream = 250*1616.21=404052.5 btu/hr
Enthalpy of mixed stream =576500+404052.5= 980552.5 btu/hr
Enthalpy of steam at 300 deg.c and 12 bar = 3649.5 btu/lb = 3649.5*0.4299 btu/lb = 1569 btu/lb
Its enthalpy =750*1569 btu/hr =1176750 btu/hr
Enthalpy chanage during mixing = 1176750-980552.5= 196197.5 btu/hr
This is the heat that need to be supplied
Let x= produt flow rate, then flow rate of steam at 350 deg.c and 12 bar= x-500
Under adiabatic conditions
Inlet enthalpy = exit enthalpy
576500+(x-500)*1616.21= x*1569
576500 +1616.21x- 1616.21*500 =x*1569
Hence x= 4905.846 lb/hr
3.
Basis : 1 mole of propane and butane miture
Molar masses : propane = 44 and n- butane =58
Moles : propane = 0.3 and butane =0.7
Mass of 1 mole = 0.3*44+0.7*58= 53.8 gm
Moles in 250 kg/hr= 250* 1000/53.8= 4647 moles/hr
For stream-2 , mass of 1 mole =0.6*44+0.4*58= 49.6 gm
Moles in 150 kg/hr= 150*1000/49.6= 3024 moles/hr
Total moles of mixture = 3024+4647 =7671 moles/hr
Flow rate from V= nRT/P= 7671* 0.0821L.atm/mole.K* (227+273.15)/1.1=286354 lit/hr
1lit/hr= 0.000588 ft3/min
286354 lit/hr= 286354*0.000588 ft3/min=168.4 ft3/min
2. Enthalpy of 250 kg/hr of propane butane mixture = 4647*0.3*(20685-1772)+ 4647*0.7*(27442-2394) =107845253 j/hr
Enthalpy of 150 kg/hr of propane butane mixture = 3024*0.6*(20685-1772)+ 3024*0.4*(27442-2394) =64613808 j/hr
Total enthalpy ( heat requirement) =172459061 j/hr= 4.14*109 j/day=4.14*109*0.000947=3919650 btu/day
4.
Relative humidity = 100*( partial pressure of vapor/ vapor pressure of liquid at 15 deg.c)
0.87= partial pressure of water vapor/ vapor pressure
0.87 =partial pressure of water vapor/ 71.9
Partial pressure of water vapor = 71.9*0.87=62.553 mm Hg
Total moles in 600 m3/min =600*1000 L/min, T= 45 deg.c =45+273.15= 318.15K , R =0.0821 L.atm/mole.K and n= PV/RT= 1*600*1000/(0.0821*318.15)=22971 moles/min
Moles of water vapor/ total moles = partial pressure of water vapor/ total pressure
Moles of water vapor/22971= 62.553/(760-62.553)
Moles of water vapor = 2060.235 moles/min
At 15 deg.c, it is assumed to be saturated and hence partial pressure = vapor pressure = 12.8 mm Hg
Moles of dry air = 22971-2060.235=20910.77 moles/min
During cooling, only the water moles change and moles of dry air will not change
Moles of water vapor at 15 deg.c/ moles of dry air =partial pressure of water vapor/partial pressure of dry air
Moles of water vapor/20910.77= 12.8/(760-12.8)
Moles of water vapor = 358.2145 moles/min
Moles of water vapor condensed = 2060.235 ( initial moles)- 358.2145 ( final moles)=1702.021 moles/min
Enthalpy of Air during change from 45 to 15 deg.c = 0.0291*20910.77{(45-25) – (15-25)}
=0.0291*20910.77* 30 joules=18255.1 joules/min
Enthalpy change of water when it got condensed = 1702.021 moles/min* 18 g/mole*2600 J/gm=79663428 joules/min
Total heat to be removed = 18255.1+79663428 = 79681683 joules/min=79681683*0.056869 btu/hr =4531418 btu/hr= 4531418/12000 ton/hr= 3776 tons/hr