Non reactive systems Chapter 8 PDF

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Spring 2016
Dr. William Epling
Chapter 8 Notes...

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Balances on Nonreactive Systems (Ch 8) Here we will combine energy and material balances to evaluate processes. In doing so, we need to know how to evaluate internal energy and enthalpy changes for substances other than water (steam tables)!

Let’s start with learning how to estimate or calculate the internal energy and enthalpy changes. First U and H vary with:

Changes in pressure with constant temperature For solids and liquids

So for a solid or liquid a pressure change at constant temperature can lead to the assumption that ∆ U´ 0

And since

´ ´ +∆ P V´ ∆ H=∆ U

For an ideal gas



∆ H´ V´ ∆ P

Changes in temperature One way to change the temperature is through heat transfer. For a closed system with no work and negligible kinetic and potential energy changes, Q = ∆U. For an open system, with no work and negligible ´ H´ . kinetic and potential energy changes, Q=∆ Internal energy and enthalpy changes are strong functions of temperature. Heat capacities are used to calculate changes in internal energy or enthalpy as a function of T. C

U = U(V,T)

( ∂∂VU ) dV + ( ∂∂ UT ) dT

dU =

T

V

( )

CV =

∂U´ ∂T

V

But what if there is a change in volume?

A ( T 1 , V´ 1 ) → A ( T 2 , V´ 2 )

can be broken into steps!

And state function calculations can be calculated through such a process (because they are only state dependent!). Therefore

∆ U´

would be

For ideal gases T2

∆ U´ =∫ CV ( T ) dT

´ ≠ f ( V´ ) U

∆ U´ 1 +∆U´ 2

and therefore

for any change in temperature (even if

T1

V´

changes)

Similarly, ∆ U´ is a weak function of specific volume for solids and liquids, so this equation is relatively accurate for those phases. For a real (non-ideal) gas, it is only valid if the specific volume is constant.

The constant pressure heat capacity, CP, is

C P ( T) =

´

(∂∂TH )

P

H = H(P,T)

( ∂∂HP ) dP+ ( ∂∂ HT ) dT

dH=

T

P

So at constant P, ´ =C P dT dH

Or T2

´ ∫ C P ( T ) dT ∆ H= T1

If we go through a similar thought process as we did earlier, but with pressure change included: A(T1,P1)  A(T2,P2) and we want to evaluate the enthalpy (a state function) change

For an ideal gas, A(T1,P1)  A(T1,P2)

For a solid or liquid

∆ H´ ≅ V´ ∆ P

For the second step (T change) T2

∆´H=∫C P dT T1

Therefore, for an ideal gas T2

´ ∫ C P dT ∆ H=

or for a real gas at constant

T1

pressure For a solid or liquid T2

´ V ´ ∆ P+∫C P dT ∆ H= T1

So now we have relationships to find the specific enthalpy and internal energy changes. But they are all functions of heat capacity (CP). Heat capacities can be tabulated or estimated through a common polynomial relationship.

CP = a + bT + cT2 + dT3 (a, b, c and d are tabulated in B.2)

Also, for liquids and solids, CP ~ CV For ideal gases CP = CV + R What is the heat capacity of liquid benzene at 50°C?

I want to calculate the amount of heat required to heat a flowing stream of chlorine gas from 100 to 200°C. The flow rate is a steady 5.0 kmol/s and the pressure is constant. KE and PE impacts are negligible.

Estimate the specific enthalpy of steam at 350°C and 100 bar relative to steam at 100°C and 1 bar using the steam tables and then through the calculation using CP described above.

Note: another method to estimate heat capacities for a liquid and solid is Kopp’s rule (the CP is a weighted sum of contributions from the elements in the compound).

The above dealt with pressure and temperature effects for single phase systems. Let’s now evaluate phase changes.

Think about the internal energy of liquid versus vapor water at 100°C and 1 atm. Which has more internal energy per mole?

The heat required to change from one phase to the other is called the latent heat. Heat of fusion Heat of vaporization

We can estimate latent heats Trouton’s rule estimates boiling point ∆ H´ V ≈ 0.088 T b (K ) ∆ H´ V ≈ 0.109 T b (K)

Chen’s equation

∆ H´ V

at the normal

for polar liquids water, low MW alcohols

[ ()

T b 0.0331 ∆ H´ V =

Tb −0.0327+0.0297 log10 P c Tc

( )

1.07−

Tb Tc

]

Tb – normal boiling point (K) Tc – critical temperature (K) Pc – critical pressure (atm)

If you know ∆ H´ at one temperature you can estimate it at another using Watson’s correlation: V

∆ H´ V (T 2) T c −T 2 = ∆ H´ V (T 1) T c −T 1

(

)

0.38

Then of course there is also the Clausius Clapeyron equation: ¿ ´V d (ln p ) −∆ H = R 1 d T

( )

And of course you can also find tabulated values for heats of vaporization and fusion – usually at the normal boiling point

Benzene vapor enters a continuous condenser at 580°C and exits as a liquid at 25°C. The condensate is drained into 1.75 m3 drums, each of which takes 2.0 minutes to fill. Calculate the rate at which heat is transferred from the benzene in the condenser.

A fuel gas containing 95 mol% CH4 and 5.0 mol% ethane is combusted with 25% excess air. The stack gas leaves the furnace at 900°C and is cooled to 450°C in a waste heat boiler. A waste heat boiler is a heat exchanger in which heat lost by the cooling gases is used to produce steam from liquid water. A.

Using a 100 mol basis of fuel gas fed, calculate the amount of heat that must be transferred from the gas to go from 900°C to 450°C. B. How much saturated steam at 50 bar can be produced from boiler feed water at 40°C? C. At what rate must fuel gas be burned to produce 1250 kg steam per hour? What is the volumetric flow rate of the gas leaving the boiler?

We have designed two adiabatic evaporators in series (a double effect evaporator) to produce fresh water from seawater, which contains 3.5 wt% dissolved salts. Seawater enters the first effect (evaporator) at 300K at a rate of 5000 kg/h, and saturated steam at 4.00 bar (abs) is fed into the heat exchange tube bundle in this first evaporator. This steam condenses at 4.00 bar, and the heat given up to do this is used to evaporate some of the seawater. The pressure in the first effect is 0.60 bar. The exiting brine contains 5.50 wt% salt and is fed to the second effect. The evaporated steam from the first effect is the feed to the heat exchange bundle of this evaporator. This condenses for the heat exchange. The second effect operates at 0.20 bar. Assume the brine solutions have the same properties as pure water. At what rate must steam be supplied to the first effect? What is the production rate of fresh water? What is the salt concentration of the final brine solution?

Heat of mixing or solution When you mix two solutions, you can change the intermolecular interactions.

Energy balance of mixing (no pressure change, negligible kinetic and potential energy changes) Q = ∆H = H2 – H1 = Hmix – (Hcomp1 + Hcomp2)

An ideal mixture has

∆ H´ solution =0

Otherwise we need to find data to help us calculate this value. One source is table B.11. For this, the heat of solution is the enthalpy change when 1 mol of the solute is dissolved in r moles of a liquid solvent, at constant temperature. r is tabulated as mol H2O/mol solute in this table. We also have charts (8.5-1) as an example.

We want to produce 1 L of 8 molar HCl aqueous solution (SG = 1.12, CP = 2.76 J/g/°C) by absorbing gaseous HCl into water. The H2O and HCl enters at 25°C and 790 Torr (abs). If I want the solution to be at 40°C, how much heat must be transferred? What is the volume of HCl fed to the absorber?...