Update 1_Properties.ipynb

Author: pareal

notebooks_en/1_Properties.ipynb

@@ -75,7 +75,7 @@
     "\n",
     "Examples of familiar properties are Temperature $T$, Pressure $P$, mass $m$, volume $V$, and density $\\rho$.  Properties that scale in proportion to mass are called extensive.  Those that do not change if mass is changed are called **intensive**.\n",
     "\n",
-    "One way to visualize the difference between extensive and intensive properties is to imaging a block of matter with uniform properties (every spatial point in the block looks the same).  If the block is cut in half, which properties are also cut in half, and which ones remain the same?  The following figure shows some examples:\n",
+    "One way to visualize the difference between extensive and intensive properties is to imagine a block of matter with uniform properties (every spatial point in the block looks the same).  If the block is cut in half, which properties are also cut in half, and which ones remain the same?  The following figure shows some examples:\n",
     "\n",
     "<img src=\"../images/extensive vs intensive.png\" width=\"500\">\n",
     "\n",
@@ -229,14 +229,14 @@
    "source": [
     "m = ct.Methane()\n",
     "\n",
-    "# set the temperature of water to 99 C, 1 atmosphere\n",
+    "# set the temperature of methane to -162 C, 1 atmosphere\n",
     "T = -162 + 273.15\n",
     "m.TP = T, P_atm\n",
     "print('Temperature =',m.T,'K')\n",
     "print('Specific volume =',m.v,'m^3/kg')\n",
     "print('\\n')    # print a blank line\n",
     "\n",
-    "# set the temperature of water to 101 C, 1 atmosphere\n",
+    "# set the temperature of methane to -161 C, 1 atmosphere\n",
     "T = -161 + 273.15\n",
     "m.TP = T, P_atm\n",
     "print('Temperature =',m.T,'K')\n",
@@ -366,12 +366,17 @@
     "\n",
     "w.PX = P_atm, 0    # set the quality x = 0 to find the point when boiling starts (at sea level)\n",
     "vf = w.v           # the saturated liquid specific volume\n",
+    "print('vf=',vf,'m3/kg')\n",
     "\n",
     "w.PX = P_atm, 0.5  # set the quality x = 0.5 to find the point when boiling is halfway done\n",
     "vhalf = w.v        # specific volume of the liquid-vapor mixture\n",
+    "print('vhalf=',vhalf,'m3/kg')\n",
     "\n",
     "w.PX = P_atm, 1    # set the quality x = 1 to find the point when boiling ends\n",
     "vg = w.v           # the saturated vapor specific volume\n",
+    "print('vg=',vg,'m3/kg')\n",
+    "\n",
+    "print('vhalf/vg=',vhalf/vg)\n",
     "\n",
     "T_array = np.linspace(15,150) + 273.15    # temperature range: 15-150 deg. C, converted to K\n",
     "v_array = np.zeros_like(T_array)          # create an array of 0s to store the specific volumes at each temperature\n",
@@ -393,7 +398,18 @@
     "\n",
     "The star on the right represents the completion of boiling at this pressure.  This is called a **saturated vapor** state.  Thermodynamic properties in this state are usually denoted with a subscript 'g' for gas.  Hence, the specific volume at this state is denoted $v_g$.\n",
     "\n",
-    "Note that the middle star does not occur halfway between the other two stars.  Remember that this middle star represents the point at which half the mass is liquid and half is vapor.  The overall density of the liquid+vapor at this point is closer to the density of the pure vapor because the liquid does not take up nearly as much volume, so overall the density is very low.  Hence, the specific volume, which is just the inverse of density, is close to its pure vapor value."
+    "Note that the middle star does not occur halfway between the other two stars. Remember that this middle star represents the point at which half the mass is liquid and half is vapor. Even thoug the specific volume vhalf is very close to half of vg (saturated vapour specific volume), the use of a logarythmic axis in a range that goes from 0.001 to 1.6 means 0.8 is plotted much closer to 1.6. If a linear axis is used, the point is indeed in the middle."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "plt.plot(v_array,T_array-273.15,'b',[vf, vhalf, vg],[Tsat-273.15, Tsat-273.15, Tsat-273.15],'r*')\n",
+    "plt.ylabel('Temperature [$^\\circ C$]')\n",
+    "plt.xlabel('Specific volume [$m^3/s$]');"
    ]
   },
   {
@@ -534,11 +550,11 @@
     "\n",
     "Let's look at some examples of thermodynamic processes (converting a substance from one state to another), and how we can plot the path of a process on a T-v diagram.\n",
     "\n",
-    "**Example:** Water is stored inside a rigid tank with an initial temperature $T_1 = 500 ^\\circ C$ and pressure $P_1 = 2.0 MPa$.  The water is cooled until $T_2 = 150 ^\\circ C$.  Find the final pressure $P_2$ and plot this process on a T-v diagram.\n",
+    "**Example:** Water is stored inside a closed rigid tank with an initial temperature $T_1 = 500 ^\\circ C$ and pressure $P_1 = 2.0 MPa$.  The water is cooled until $T_2 = 150 ^\\circ C$.  Find the final pressure $P_2$ and plot this process on a T-v diagram.\n",
     "\n",
     "<img src=\"../images/ex water cooling.png\" width=\"200\">\n",
     "\n",
-    "Because the process takes place in a rigid tank, the mass and volume of water remain constant throughout the process.  Therefore, the specific volume remains constant too.  We can write this as $v_1 = v_2$.  Let's summarize:\n",
+    "The tank is closed, therefore the mass of water remains constant. Also, because the process takes place in a rigid tank, the volume of water remain constant throughout the process.  Therefore, the specific volume remains constant too.  We can write this as $v_1 = v_2$.  Let's summarize:\n",
     "\n",
     "**State 1:**  ***Known:*** $T_1,P_1$, ***Unknown:*** $v_1$\n",
     "\n",
@@ -669,7 +685,7 @@
    "source": [
     "<a id='phasestate'></a>\n",
     "\n",
-    "### Determining Phase from Therodynamic State\n",
+    "### Determining Phase from Thermodynamic State\n",
     "\n",
     "From plots like the T-v diagram shown above, it is easy to determine the phase of the substance (in this case water) at the beginning and end of the process.  The way we do this is to look at where the points lie relative to the saturation curves.\n",
     "\n",
@@ -917,6 +933,13 @@
     "css_file = '../style/custom.css'\n",
     "HTML(open(css_file, \"r\").read())"
    ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": []
   }
  ],
  "metadata": {
@@ -935,7 +958,7 @@
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    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.6.5"
+   "version": "3.7.3"
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