A 4-L pressure cooker has an operating pressure of 175 kPa. Initially, one-half of the volume is

A 4-L pressure cooker has an operating pressure of 175 kPa....

A $4-\mathrm{L}$ pressure cooker has an operating pressure of $175 \mathrm{kPa}$. Initially, one-half of the volume is filled with liquid water and the other half by water vapor. The cooker is now placed on top of a $750-\mathrm{W}$ electrical heating unit that is kept on for 20 min. Assuming the surroundings to be at $25^{\circ} \mathrm{C}$ and $100 \mathrm{kPa},$ determine $(a)$ the amount of water that remained in the cooker and $(b)$ the exergy destruction associated with the entire process. Answers: (a) $1.51 \mathrm{~kg},$ (b) $689 \mathrm{~kJ}$.

Key Concepts

Phase Equilibrium in Two?Phase Systems

When a system contains both a liquid and its vapor, it is in a state of phase equilibrium, meaning that the properties of the liquid and vapor phases are related by saturation conditions at a given pressure and temperature. In engineering applications, determining the mixture quality, specific volumes, and other thermodynamic properties under these conditions is essential for predicting system behavior during heating or cooling processes.

Energy Balance in a Closed System

The energy balance for a closed system is a fundamental concept in thermodynamics which accounts for all forms of energy entering and leaving the system, including heat, work, and changes in internal energy. In processes where a system is subjected to a known heat input (such as electrical heating), energy conservation is used to determine changes in state, phase shifts, and the eventual final condition of the system.

Thermodynamic Properties of Saturated Water

The properties of water in both the liquid and vapor phases at saturation conditions are critical in analyzing processes involving water. These properties, including specific volume and latent heat, are used to establish the initial state of the system and to track changes during heating, as well as to calculate the amount of water remaining in a different phase after the process.

Exergy (Availability) Analysis

Exergy analysis evaluates the maximum useful work obtainable from a system as it exchanges energy with its environment. Unlike energy, exergy accounts for the quality of energy and is dependent on the surrounding (reference) conditions. It provides a measure of how efficiently energy inputs are utilized, thereby highlighting the potential for work extraction in processes.

Irreversibility and Exergy Destruction

Exergy destruction is a measure of the irreversibilities present in a process, which stem from factors like friction, heat transfer across finite temperature differences, and mixing. This concept quantifies the loss of potential work due to entropy generation, allowing engineers to assess and optimize the performance of systems by identifying where energy quality is degraded.

Transcript

00:02 Hi friends, the initial mass in the cooker am but mass of liquid plus mass of paper that is b by two specific volume of liquid plus specific volume of paper 4 into 10 to the power minus 3 upon 2 into 1 upon 0 .001057 plus 1 .0037.

00:50 So it is to be 1 .89 414 kg.

00:59 The internal energy and entropy, internal energy, initial internal energy, you will get m into u of liquid plus m into u of.

01:13 Vapor b by 2 into u upon alpha liquid plus u upon alpha vapor putting the value 4 into 10 to the power minus 3 upon 286 .82 upon 0 .00157 plus 25 to 24 .55 upon 1 .0037 plus 25 to 4 .4 .5 upon 1 .003037 so it is to be 926 .2 kilojoude and entropy m into s of liquid plus m into s for vapor.

02:14 And it could be written as b by 2 s upon alpha of liquid plus s upon alpha for vapor.

02:23 Putting the value 4 into 10 to the power minus 3 upon 2 1 .485 upon 0 .0017 plus 7 .17.

02:39 Divided by 1 .0037.

02:43 So it is to be 2 .8241 kilojoules per kg.

02:59 Now, final quality and final mass.

03:04 Quality is defined as alpha for liquid into power into delta t minus m1 into h out plus u1 plus v into h out minus u liquid divided by b into u vapor minus alpha vapor alpha vapor alpha liquid into rate power into delta t minus m1 h out plus u one after substituting the value of all parameter given quality of fluid you will get 0 .001626.

04:07 So final mass in the cooker would be v.

04:14 Alpha 2, that is v upon alpha liquid, plus q2, alpha vapor minus alpha liquid...

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