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THERMO Spoken Here! ~ J. Pohl © ~ 2017 www.ThermoSpokenHere.com (C4280-149)
Ambient Pressure1 atm
Tank Cross-section Area2 m2
Shaft Rate60 rpm
Shaft Torque100 Nm
Mixing Time60 min
Ingredients Depth
Initial
1.6m
Ingredients Depth
Final
1.8m

Batch Mix Event

The apparatus mixes the ingredients of a polymer that is used to seal and fire-proof the interiors of aircraft fuel tanks. Ingredients are poured in from the top which remains open to the atmosphere. The stirring propeller rotates at 60 rpm, driven by the motor below. The table contains information regarding the event.

Apply the energy equation. What does it tell us?
♦  Although this is a "batch" or increment type event, we will begin with the general, rate-form of the energy equation.

(1)The table contains data of the event.

As our system, we take the ingredients of one batch. The event, as depicted, involves two manners of work. These are shaft work and the work of expansion as the ingredients force back the ambient atmosphere at their upper surface. We expand the work summation.

(2)The work at the upper surface is a bit involved.

We take two steps as Equations (3).

(3)3

The above is our usual first-order differential equation. The first step toward solution is to separate variables. Equation (4) shows that task completed.

(4)When variables are separated, the resulting form is called differential.

Next apply the integration operator then write the limits and integrate. The system energy (E) integrates readily. We write it as its increment. The sum of heat rates over the time is expressed generally.

(5)It is a bit more writing but we prefer
to keep the energy equation intact rather that to calculate the works apart from it.

We see that the work of the shaft rotation is positive meaning "energy to the ingredients." The work interaction at the surface of the batch occurs as the batch expands and "backs-off" (pushes away) the atmosphere. The vector force at the boundary (integral of pressure over area) acts opposite to its displacement ~ the work should be negative in sign; it is! This problem is fictitious; there is no value in running the numbers any further.

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