Basic Thermodynamics ~ J. Pohl © www.THERMOspokenhere.com (D230) -  (D232)

Pressure Cooker (other)

Problems concerning pressure cookers can be difficult because they are transient. At the time the chef places the cooker on the stove it contains water and air. In the Cengel and Boles Example the authors avoid difficulties by "starting the problem" at a special time (after the chef placed the cooker on the stove and turned on the heat. That special time is once the cooker pressure had attained 175kPa and was devoid of air. (The example statement does not say "no air" but the initial sketch Fig. 4-55 does. Below is an alternate solution of this example).

Thermodynamic analysis is complicated in that there is no specific, easiest way of solution. This is to say there is no specific "place" to begin and "specific" things do not have to be done in any order. Analysis involves mass, state (initial and final), energy, property equations, heat, work and solution techniques. There are many paths to the solution of any thermo problem. Below is demonstrated one way.

The event of the pressure cooker (water within the interior volume as system) involves changes of mass and energy. The mass equation is easier than the energy equation. Therefore we develope the mass equation first. The generic mass equation (suited for all systems and all events) is:

1 (1)General Form of the Mass Equation.

This is a first order differential equation with time as the independent variable. The event perspective of this cooker is a "start-stop" event. Starts at time, t=0+ (with conditions stated) and ends after 30 minutes. So we integrate over those limits.

2 (2)2

Left of equality integrates! The integral right of equality, cannot be integrated without knowing the time dependence of "mass out." All Engineers are obliged to get an (at least one) answer. An answer MUST be obtained - What to do? Apply the Mean Value Therorem of calculus. Replace the time dependent integrand, [m-dotout(t)] with its "average," [m-dotout,avg], then integrate. So what we did is replace a time dependent term we did not know with another term we don't know. That is what we did. Read on to see why?

3 (3)3

In the above equation the product "m-dotout,avg times Δt" is an amount of mass. What is left of the equality is an increment of mass, Δm = m2- m1. But physically "m-dotout,avg" times "Δt" is not an increment. Thus "m-dotout,avg times Δt" is written as a lower case " δ " type of increment.

These steps to arrive at the above mass equation are used for many problems. For this Pressure Cooker, the authors have made it clear that the initial state of the water is two-phase. It is not known whether the second (final state after 30 minutes of heat) will be two phase or all vapor. Consequently,

ASSUME: The second state is two phase, then extend the mass equation:
         [mvapor,2 + mliquid,2] - [mvapor,1 + mliquid,1] = δmass out

In this case the initial state is known to be two phase water at 175kPa with a volume of 0.006 m3 and a mass of 1.0 kilograms. Immediately we have:
        mvapor,1 + mliquid,1 = 1.0 kg

Also Properties of Water - Saturation Tables at p = 175kPa provide:
        Tsat = 116°C,
        vf = 0.001 m3/kg,   vg = 1.00 m3/kg,
        uf = 487 kJ3/kg,   ug = 2038 kJ3/kg,
        hf = 487 kJ3/kg,   hg =2701 kJ3/kg,

For the initial state:
      mvapor,1vvapor + mliquid,1vg = 0.006 m 3 or
      mvapor,1 (0.001 m3/kg) + mliquid,1 (1.00 m3/kg) = 0.006 m 3
So having done all we can with the mass equation we see we have three equations with five unknowns.

Initial mass of the system:
      (1)  mvapor,1 + mliquid,1 = 1.0 kg

Volume of the system - State (1):
      (2)  mvapor,1 (0.001 m3/kg) + mliquid,1 (1.00 m3/kg) = 0.006 m 3

Mass Equation for the Event:
      (3)  [mvapor,2 + mliquid,2] - [mvapor,1 + mliquid,1] = δmass out

The unknown terms are: mvapor,1, mliquid,1, mvapor,2, mliquid,2 and δmass out.

Next we address the energy equation.

(4)4

If we were experts in "pressure cooking" we would not have to write the equation. We're not; write the equation.

Since the only substance is water; drop the first,second and third summation. There is no entering mass; drop the "in" terms. The cooker is constant volume; the differential of work "dV" is zero. Kinetic and potential energy changes will be negligible. Assume no heat fron the cooked to space. With these facts and simplifications the energy equation becomes:

(5)5

Next integrate the above equation:

(6)6

Our result contains no new unknown entities. Hence, with the above, we have 4 equations and 5 unknowns. These equations are:
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Initial mass of the system:
      (1)  mvapor,1 + mliquid,1 = 1.0 kg

Volume of the system:
      (2)  mvapor,1 (0.001 m3/kg) + mliquid,1 (1.00 m3/kg) = 0.006 m 3

Mass Equation for the Event:
      (3)  [mvapor,2 + mliquid,2] - [mvapor,1 + mliquid,1] = δmass out

Energy Equation for the Event:
      (4)   [mvapor,2uvapor,2 + mliquid,2uliquid,2] - [mvapor,1uvapor,1 + mliquid,1uliquid,1] = δmass outhmass out

The unknown terms are: mvapor,1, mliquid,1, mvapor,2, mliquid,2 and δmass out.

So we have 4 equations and 5 unknowns. This cannot be solved. We need as many independent equations as there are unknowns. What is the new equation we need? Answer: the second volume is also 0.006 m3.

Volume of the system in the second state:
      (5)  mvapor,2 (0.001 m3/kg) + mliquid,2 (1.00 m3/kg) = 0.006 m 3

The numbers are not important. There are sufficient equations to calculate the unknowns. We leave it here!

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