Two springs with the same spring constant are installed in a...

Two springs with the same spring constant are installed in a massless piston/cylinder arrangement with the outside air at $100 \mathrm{kPa}$. If the piston is at the bottom, both springs are relaxed, and the second spring comes in contact with the piston at $V=2 \mathrm{m}^{3} .$ The cylinder (Fig. P4.11 5 ) contains ammonia initially at $-2^{\circ} \mathrm{C}, x=0.13$ $V=1 \mathrm{m}^{3},$ which is then heated until the pressure finally reaches $1200 \mathrm{kPa}$. At what pressure will the piston touch the second spring? Find the final temperature and the total work done by the ammonia.

Key Concepts

Work in Processes with Variable Loads

When a system, such as a piston-cylinder device with springs, operates under conditions where the load (or opposing force) changes with displacement, calculating the work done requires integrating the variable pressure (or force) over the change in volume or distance. This concept is central to thermodynamic analyses that involve work interactions beyond simple constant-pressure or constant-volume conditions.

Phase Change and Quality in Mixtures

Phase change refers to the transformation between different states of matter when energy is added or removed, while quality is a measure of the proportion of vapor in a saturated mixture. Understanding these concepts is fundamental when analyzing processes involving heating of a fluid mixture, as the properties of the mixture change significantly during phase transitions.

Thermodynamic Properties of Fluids

This key concept covers the properties of fluids, including the relationships between pressure, temperature, specific volume, and phase composition. In thermodynamic processes, particularly those involving mixtures and phase changes, a detailed knowledge of these properties enables accurate predictions of state changes as energy is added or removed.

Spring Mechanics and Hooke’s Law

Spring mechanics describes the relationship between the force exerted by a spring and its displacement from equilibrium, summarized by Hooke’s law. In systems where springs interact with pistons, the spring force increases linearly with compression, affecting the overall pressure within the cylinder. Recognizing this behavior is crucial for understanding how the spring influences the system's dynamics.

Force Equilibrium in Piston Systems

This concept involves understanding how the forces on a piston are balanced at mechanical equilibrium. In many piston problems, the internal pressure force is counteracted by external loads, such as atmospheric pressure and spring forces. Establishing equilibrium is essential for determining when additional mechanical components, like a second spring, become active in the system.

Transcript

00:01 All right guys, we are given the following data for ammonia, that is d1, is equal to minus 2 degrees centigrade.

00:08 We've got p nod is equal to 100 kilopascal.

00:13 We've got p3 is equal to 1 ,200 kilopascal.

00:21 And then we've got v1 is equal to 1 cubic meters.

00:26 And v2 is equal to 2 cubic meters.

00:30 And we've got x is equal to 0 .13.

00:34 So from saturated ammonia table b2 .1 corresponding to temperature that is minus 5 degrees centigrade, we can obtain pressure specific volume and internal energy.

00:47 So we have we've got pressure is equal to 354 .9 kilo -pascal.

00:59 Then we've got vf that is 0 .001550 which is cubic meters per kg.

01:11 Then we've got vfg that is equal to 0 .34493 cubic meters per kg and uf that is equal to 156 .76 0 .76 kilojoules per kg then we've got ufg is equal to 1157 kilojoules per kg.

01:47 So from saturated ammonia table b2 .1 .1.

01:51 Corresponding to temperature 0 degrees centigrade minus 0 degrees centigrade we can obtain specific we can obtain pressure specific volume and internal energy that is p is equal to 429 .6 kilo pascal then we've got vf is equal to 0 .001566 cubic meters per kg then we've got vf is equal to 0 .001566 cubic meters per kg then we've got vf is equal to 0.

02:31 We've got vfg that is equal to 0 .28763 cubic meters per kg.

02:45 Then we've got uf is equal to 179 .69 .69 kilojoules per kg.

02:55 And we've got ufg is equal to 1138 .3 kilojoules per kg.

03:08 So using the method of interpolation, we can calculate initial pressure.

03:12 Saturated and evaporated specific volume and specific internal energy so we've got p1 that is equal to 354 .9 plus minus 2 minus 5 multiplied by 24 .5 sorry 429 .5 minus 354 .9 and that is equal to 400 kilo -pascal then we've got vf that is equal to 0 .001550 plus minus 2 minus 5 divided by 0 minus 5 multiplied by 0 .001566 minus 0 .00156 minus 0 and that is equal to 0 .00156 cubic meters per kg then we've got vfg is equal to 0 .34493 plus minus 2 minus 5 divided by 0 minus 5 multiplied by 0 .28763 minus 0 .3 .3.

04:59 And that is equal to 0 .3 .3.

05:01 And that is equal to 0 .3 .3 106 cubic meters per kg so moving further guys we've got the value of uf is equal to 156 .76 plus minus 2 minus 5 0 minus minus minus 5 multiplied by 179 0 .69 0 .76 is equal to 170 .5 to kilojoules per kg.

05:58 Then we've got ufg is equal to 1157 plus minus two into minus five, minus two minus minus five.

06:17 So zero we have minus five in the bracket multiplied by one one three eight point three minus one...

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