A reagent for an atmospheric pressure industrial reaction mu

A reagent for an atmospheric pressure industrial reaction must be vaporized at elevated temperature when being used.   The reagent is stored as a liquid in an insulated tank of capacity 9.2 m³ with an internal heating element that can deliver 57 kW of power at its maximum setting.   When not in use, the reagent is kept at room temperature (25^oC).

When the reagent is used, the reagent must first be heated up to its vaporization temperature. When that temperature is reached, vapor is generated for the reaction.

a)      On a particular day, the tank is initially 60% full. Given the data below, what is the shortest length of time required before vapor begins to be generated from the tank?

b)      What is the power requirement to generate 300 kg/hr of vapor for the reaction once the vessel is heated?

DATA:            Heat capacity of liquid reagent:   C_p,L = 176 J/mol/^oC

                        Heat capacity of vapor reagent:   C_p,V = 126 J/mol/^oC

                        Heat of vaporization of reagent: H_v = 34.9 kJ/mol

                        Boiling point of reagent:                 T_b = 133^oC

                        Molecular weight of reagent:     MW = 110

                        Liquid density of reagent:              _L = 840 kg/m³

Solution

solution:

1)here given tank vome=V=9.2 m3

2)on that tank volume=V1=.6*V=5.52 m3

3)mass of reactant=density*V1=840*5.52=4636.8 kg

molecular weight=m=110 gm

4)here number of mole of reagent is

n=m/M=4636.8*10^3/110=42152.72 mole

5)for shorter time if all mole are contact with heatre to maximumcapcity Pmax=57000 j/s

hence power is given by for just vapor generation if all mole are in contact with heater

P=n\'Cp(Tb-Ts)+n\'Hv

hence time is n\'=n/t

hence minimum time is give nas

t=nCp(Tb-Ts)+nHv/P=42152.7*176*(133-25)+42152.7*34900/57000

t=39866.10 sec

and if single mole is in contact with heater the time would

t=1*176*(133-25)+1*34900/57000=0.945754 sec

3)for power generation to be m\'=300 kg/hr,power requirement for all moles is

P=n\'Cp(Tb-Ts)+n\'Hv

here n\'=m\'/M=300/110=2727.27 mol/hr

P=2727.27*176(133-25)+2727.27*34900=147021.818 kj/hr=40.8393 kj/s=40.8393 kw