Che320 Tutorial Kit

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COVENANT UNIVERSITY  NIGERIA

TUTORIAL KIT OMEGA SEMESTER

PROGRAMME: CHEMICAL ENGINEERING COURSE: CHE 320

DISCLAIMER The contents of this document are intended for practice and leaning purposes at the undergraduate level. The materials are from different sources including the internet and the contributors do not in any way claim authorship or ownership of them. The materials are also not to be used for any commercial purpose.

CHE320: CHEMICAL ENGINEERING PROCESS ANALYSIS I Contributor: Dr! O"#$#%&'#n! En(r! )Mr* O+&,u-i M! E . Mr O"#%i-&+i T&-it#/o

Q1.

Distinguish amongst the three types of a system.

Q2. A house is 45 ft! and has 12 ft ceilings. "or comfort# the home owner specifies .$ changes of air per hour. The outside air temperature is %&" dry bulb and '$.5& wet  bulb. The air indoors is '5&" dry bulb 5( relative humidity. )hat is the amount of cooling re*uired to provide the fresh air+ Q3(i) State three reasons for studying energy balances of industrial processes. (ii)

Distinguish succinctly between extensive and intensive properties, giving two exaples in each property.

Q4.

,iven the ambient temperature is '&" measured by a dry bulb thermometer and -&" measured by a wet bulb thermometer# what is the relative humidity+

Q!.

Distinguish succinctly between path and point functions with illustration, and give two exaples of each function.

Q-.

)hat is the heat duty for a mier that mies %.2 moles of / 20 with 1 mole of .2 mole fraction /204 if the inlet and outlet streams are all to be at 25 o+

Q".

#alculate the heat re$uired in %& to raise ' %g of nitrogen () oxide, * '+, fro 3! o# C v o

to '! # in a constantvolue vessel if the heat capacity, this teperature range is given by: C v  R

=

−$

5

4.$23 + 1.214 × 1 T  − .%23 × 1 T 

, of *'+ at constant volue in

−2

C v

 -/ diensionless,

 -/ &0(ol 1) and T  -/

1  (ii) *eon gas is to be heated in an insulated plasa deposition chaber with a volue of 2.! l by an electric resistance heater. nitially, the gas, which can be treated as an ideal gas, is at 3.! a and 3"! 1. 4he 5!oh heater draws current at 6!7 for 5! inutes. 8hat are the 9nal gas teperature and pressure at e$uilibriu 4he ass of the heater is '! g and its heat capacity is .3! &0(g 1). ;ssue that the heat transfer to the chaber fro the gas at this low pressure and in the short tie is negligible. 4he olal heat

Q<. #alculate the adiabatic =ae teperature of li$uid butane burned with 3> excess air. ?oth the air and li$uid butane enter at '!o#.

Q@. f an ideal gas at 3! 1 and '! %a is enclosed in a cylinder by a frictionless piston, and the gas slowly forces the piston up so that the volue of gas expands fro .' to .2 3, calculate the wor% done by the gas on the piston if two diAerent paths are used to go fro the initial state to the 9nal state - R<.352 &0(ol 1)/: ath : expansion occurs at constant pressure ( p  '! %a). ath : expansion occurs at constant teperature ( T   3! 1). S%etch the pV  diagra for both paths. (ii) ;ir is being copressed fro 5 %a and '!! 1 (where it has an enthalpy of 2<@ %&0%g) to 5 %a and '"< 1 (where it has an enthalpy of !@ %&0%g). 4he exit velocity of  the air fro the copressor is "! 0s. 8hat is the power re$uired (in %8) for the copressor if the load is 5! %g0h of air 4he olecular weight of air is '<.@.

Q1. ethane is burned with 5( ecess air in a furnace. "ig below shows the stream compositions and those variables whose values are specified. The process occurs with each stream at 1atm. Determine if the number of degrees of freedom for the process is ero.  A table is needed to ma6e the count of the variables and e*uilibrium. The 7nergy balance is assumed to reduce to Q 8 9/: replaced as a variable with p and T  ;umber of variables in the process < pecies in

"1

1

"2

2

"$

5

Total no of species is 3

tream =ressure

$

Q

1

7tent of reaction >2 reactions?

2

Total

2

C  p

Q55. 4he olal heat capacity, as: C  p  R

=

−$

, of nbutane at constant pressure is expressed

−-

1.%$5 + $-.%15 × 1 T  − 11.42 × 1 T 

2

C  p

 -/ diensionless,

 -/ &0(g ol 1) and

T  -/ 1 

#onvert this e$uation into a for so that the speci9c heat capacity can be expressed over the entire teperature range in &0(%g o#).

Q5'. 4he conversion of solid wastes to innocuous gases can be accoplished in incinerators in an environental acceptable fashion. Bowever, hot exhaust gases often ust be cooled or diluted with air. ;n econoic feasibility study indicates that solid unicipal waste can be burned to a gas of the following coposition (on a dry basis): #+': @.'>C #+: 5.!>C + ': ".3> and *': <'.>. 8hat is the enthalpy diAerence for this gas per g ol between the botto and the top of the stac% if the teperature at the botto of the stac% is 3 o# and the teperature at the top is @! o# gnore the water vapour in the gas. *eglect any energy eAects resulting fro the ixing of the gaseous coponents. ;ssue ideal gas ixture.  4he heat capacity e$uations are given in the table below. a

b × 1 2

c

#+'

36.55

2.'33

'.<<"

".262

#+

'<.@!

.255

.3!2<

'.''

+'

'@.5

5.5!<

.6"6

5.355

*'

'@.

.'5@@

.!"'3

'.<"5

#oponent

C  p = a + bT  + cT 

2

+ dT 

$

× 1

5

b × 1 %

saturation ixing ratio (3 g0%g). (?) 4hen you ight be able to 9gure out the ixing ratio in your head. ;ir that is 9lled to !> of its capacity could hold up to 3 g0%g. Balf of 3 is 5!, that is the ixing ratio. +r you can substitute into the relative huidity forula and solve for the ixing ratio (the details are shown above). Hinally, to deterine the dew point, we iagine cooling the air. 4he saturation ixing ratio decreases, the ixing ratio stays constant, and the relative huidity increases. n this exaple the KB reachs 5> when the air has cooled to " H.  4hat is the dew point teperature.

Q5<.

 Noo% at the horiGontal axis to 9nd "6H. Oove a pencil up this line to eet the intersection with the exponential line for 2> relative huidity. Oove the pencil to the right to read the dew point. 4his is !H. 4he evaporator coil is warer than the dew point so it will not condense water fro the air. Q'.  4he total volue of the house is 2! x 5'  !2, ftM. 8e need to change !2, x .3  56,' ftM0hour (which e$uates to '" ftM0inute or cf). Hro the psychroetric chart, the enthalpy of the incoing air is 3". ?4I0lb and the speci9c volue is 52.' ftM0lb. 4herefore the energy of the incoing air is 56,' x 3" 0 52.'  2','55 ?4I0hour. Siilarly, the enthalpy of the air indoors is 56,' x '<.5 0 53."  33,''< ?4I0hour. 4he heat diAerence is <,@<2 ?4I0h, or about ."! tons.

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