Worked Examples on Exergy. Paras 32 to 37 of Notes.

Background Information

Assumed properties and constants for air.

Assumed temperature and pressure of surroundings

Take a datum at these conditions

A worksheet function to work out specific entropy of air

PARA 32.

Air expands from 4 to 3.5 bar in a pipe. Find exergy losses.

Applying SFEE, there is no heat loss and no shaft work. So we can state

Also

On a basis of one kg of fluid

Heat exchange: No loss of enthalpy, and hence no problem

Work transfer: 11 kJ/kg NOT available for work transfer.

PARA 33.

Water at 2.0 bar enters a steady flow as saturated liquid and leaves as saturated vapour. Find the change in the Gibbs function. Explain the findings.

Take the following data from tables

per kilogram

Very little change. Discharge fluid to surroundings at Ts. Hence Ts = Tf = Tg = To. Then b = g, and we see that Db = 0 and hence no work can be done.

PARA 34.

Carbon monoxide is burnt stoichiometrically with oxygen, in steady-flow. The start and end points of the process are both at 300K, equal to the temperature of the surroundings.

Taking a basis of 1kmol of carbon monoxide, find the change in enthalpy of the mixture and the maximum possible available work. (Assume that the surroundings are also at 300K).

The temperatures of the surroundings, start and end points are identical, so again b = g throughout. Molar values of s and h are tabulated in steam tables

(Note that a datum of 0K applies here).

Entropies (see steam tables).

Enthalpies

Enthalpies are relative to 0°C, whereas the datum for chemical reaction is 25°C. Close enough to ignore the above. Then the enthalpy change equals the heat of

reaction which is.

The maximum available work is

per kmol

PARA 35.

The properties of a waste gas can be approximated by those of air. The gas is supplied at 5 bar, 900K, and expanded to 2 bar in a turbine. The isentropic efficiency is 85%. Find (i) the useful work done by the gas (ii) the useful work for an isentropic process (iii) Wmax for isentropic efficiencies of 85 and 100% (iv) The effectiveness - W / Wmax

Properties of air and inlet condition.

Kelvin

Outlet Condition, and temperature at outlet for an isentropic process

The useful work for an isentropic process. Must be on the basis of kJ/kg

The useful work for quoted isentropic efficiency.

Note also

Assumed Temperature of the Surroundings.

Maximum Work Non-Isentropic Process.

|Wmax| = b1 - bo

= (h1 - To.s1) -(ho -Toso)

=420 KJ per Kg.

Effectiveness is defined as actual work divided by thermodynamic maximum, or

USING the above values for work, e = 42% for the non isentropic process and 50% for the isentropic process.

PARA 36.

Waste gas at 2 bar , 600K exchanges heat to a waste heat boiler, and so leaves at 2 bar and 400K. Find the loss of enthalpy of the gas, and the exergy loss.

Para 37)

Waste gas (mostly air) leaves a burner at 900K. Evaluate the following proposals.

The pressure of the gas will not change. Mechanical power can be sold at 12p per MJ, whereas thermal energy can be sold at 4 p per MJ.

(i) The gas is to be used to drive a heat engine with and effectiveness of 0.4. The exhaust from the heat engine is at 500K. It then exchanges heat with a central heating sytem, and is exhausted to ambient at 380K.

(ii) The gas enters the heat exchanger directly, at 900K.

(i)

Similarly

bo = 289 KJ per Kg.

Saving = (b1 - b0) . e . 0.012 (pence per KJ) = 1.4 pence per kilogram

Thermal part

Total saving = 1.9 pence per kilogram of gas.

(ii)

 per kg