00:01
In this problem, we have a cylinder which is initially filled with saturated liquid water at a specified pressure.
00:07
The water is heated electrically as it is stirred by a paddle wheel at constant pressure.
00:11
We need to calculate the voltage of the current source and show the process on a pv diagram.
00:18
So here we have our pv diagram and we can see that our pressure remains constant while our volume increases as the temperature increases.
00:31
So we assume that the cylinder is stationary and thus the kinetic and potential energy changes are zero.
00:39
The cylinder is well insulated and there is no heat transfer.
00:42
The thermal energy store in the cylinder itself is negligible and the compression expansion process is quasi -equilibrium.
00:49
Now looking at this closed system, since no mass enters or leaves, we have the energy balance equation, e -in minus e -out is equal to delta e.
00:59
The energy in is due to the work done by the current so that's the work done by the current plus the work done by the pedal wheel minus the boundary work out of the system and this is equal to a change in internal energy doubt to you remember q kinetic energy and the potential energy are all zero so we can write this as the work done by the electricity current, the electrical current, plus the work done by the paddle wheel is equal to m into a change in enthalpy h2 minus h1.
01:58
Since we know that delta u plus the boundary work, delta u plus boundary work, is equal to delta h change in enthalpy.
02:06
The electrical work done is due to vi, delta t.
02:15
So a power, times a time delta t is the work done by the electricity plus the work done by the paddle wheel is equal to this change in enthalpy m h2 minus h1 so now we have an equation for our energy balance and we can write the properties in each of the states for water so in state one we have p1 the initial pressure is 175 kilo -pascals and this is for a saturated liquid this corresponds to enthalpy h1 which is hf at 175 kilopascals and this is 487 .01 kilojoules the kg so this this we get from our tables specific volume v1 is able to the saturated liquid volume at the same pressure 175 kioskals again from our tables this is 0 .001057 cubic meters per kg.
04:10
P2 is the same the pressure of the second state is 175 kospes since the pressure is constant the quality x2 is 50 percent or 0 .5 to enthalpy of the second state h2, being hf plus x2 times hg minus hf, which is hfg.
04:44
And this is 487 .01 plus 0 .5 times 221223 .1...