give all the answers(a,b,c,d,e) accurate ooo AT&T; LTE 12:06 AM 61% Open book, o

Question

give all the answers(a,b,c,d,e) accurate

ooo AT&T; LTE 12:06 AM 61% Open book, open notes. If enough information is not supplied in the problem statement, use your engineering judgment to make an assumption (state the assumption). (10 points) 100 ms of air at 100 and 5 bar expands through 1 valve and is cooled in heat exchanger, The exit pressure is 2 bar and the exit temperature is 40 °C. Standard conditions (STP) are 1 am and 0 Answer the following questions related to this process. a. How many moles of air pass through the process? b. What is air density at each set of cooditions (inlet, outlet, STP? .What is the outlet air volume at exit conditions? d How many standard cubic meters of air pass through the system? eWhat is the mass density at standard conditices?

Explanation / Answer

flow rate of air, V =100 m3, 1m3=1000L, 100m3= 100*1000L= 105 L, pressure, P= 5 bar ( 1 bar=0.9869 atm), 5 bar= 5*0.9869 atm =4.9345 atm, T= 100deg.c= 100+273= 373K

from gas law, n= PV/RT, R=0.0821 L.atm/mole.K

n= 4.9345*105/(0.0821*373)=16113.55 moles

density = mass/ volume, mass of air =molar mass* no of moles

Air contains 21% O2 and 79% N2, molar mass of air =0.21*32+0.79*28 = 29 g/mole

mass of air = 16113.55* 29 gm=467293 gm=467293/1000Kg =467.3 kg

density of air = 467.3/ 100 kg/m3 = 4.673 kg/m3

2, at the, mass remain the same and only volume changes which can be calculated from gas law

P1V1/T1= P2V2/T2, P1,V1, T1 are inlet conditions, while P2,V2 and T2 are outlet conditions

P1= 5 bar, V1= 100m3, T1= 373K, P2= 2bar, T2= 40 deg.c= 40+273= 313K

V2= P1V1*T2/T1P2 = 5*100*313/(373*2)= 210 m3, volume of air at the exit conditions

Density = 467.3/210 kg/m3 =2.225 kg/m3

2. 1 kg mole of any gas contains 22.4 m3 at STP

16113.55/1000 kg moles of gas contains 16.11355 kg moles of gas occupies 22.4*16.11355 m3 =360.94 m3

3. mass density at STP = mass/volume =   467.3/360.94 kg/m3 =1.29 kg/m3

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