more nuce 2100 final

NUCE_2100/final_exam/.ipynb_checkpoints/Final Exam-checkpoint.ipynb

@@ -432,6 +432,48 @@
     "print(f'h_9 = {h_9:.3f} BTU/lbm')"
    ]
   },
+  {
+   "cell_type": "markdown",
+   "id": "06f7a7be-2537-46fc-91cc-8279bdfd96e7",
+   "metadata": {},
+   "source": [
+    "Now we look at the closed feedwater heater, which conserves energy:\n",
+    "$$\\dot m_8 h_8 + \\dot m_2 h_2 = \\dot m_3 h_3 + \\dot m_{10} h_{10}$$\n",
+    "$$\\dot m_7 y h_8 + \\dot m_7 (1-y) h_2 = \\dot m_7 y h_3 + \\dot m_7 (1-y) h_{10}$$\n",
+    "$$y h_8 + (1-y) h_2 = y h_3 + (1-y) h_{10}$$\n",
+    "$$y h_8 + h_2 - y h_2 = y h_3 + h_{10} - y h_{10}$$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "63541c28-394a-4611-be12-5f32e8ea131f",
+   "metadata": {},
+   "source": [
+    "$$y(h_8 - h_2 - h_3 + h_{10}) = h_{10}-h_2 $$\n",
+    "$$y = \\frac{h_{10}-h_2}{h_8 - h_2 - h_3 + h_{10}} $$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "id": "5bed5dc9-34a9-4aa0-b0c7-c6637ab28774",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "#Find h_2\n",
+    "h_2 = givens.get(2000)['hf']\n",
+    "print(f'h_2 = {h_2:.3f} BTU/lbm')\n",
+    "\n",
+    "#Find h_10\n",
+    "h_10 = givens.get(2000)['hf']\n",
+    "print(f'h_2 = {h_2:.3f} BTU/lbm')\n",
+    "\n",
+    "#Find h_3\n",
+    "h_10 = givens.get(155)['hf']\n",
+    "print(f'h_2 = {h_2:.3f} BTU/lbm')\n",
+    "\n"
+   ]
+  },
   {
    "cell_type": "markdown",
    "id": "df89be07-8172-44bf-900a-46e2108cdd9a",
@@ -439,9 +481,31 @@
    "source": [
     "Now we also don't know $W_{LPT}$. We need a second equation to find $y$. We get it from the closed water heater, which conserves energy between the two streams:\n",
     "\n",
-    "$$ \\dot m_3 h_3 + \\dot m_10 h_10 = \\dot m_2 h_2 + \\dot m_8 + h_8 $$\n",
+    "$$ \\dot m_3 h_3 + \\dot m_{10} h_{10} = \\dot m_2 h_2 + \\dot m_8 h_8 $$\n",
     "\n",
-    "But THIS equation has us going further down the rabbit hole. Must find pump works, then HPT work, then finally we can take the LPT work from that. THEN! We can find y.\n"
+    "But THIS equation has us going further down the rabbit hole. Must find pump works, then HPT work, then finally we can take the LPT work from that. THEN! We can find y. Things we know though:\n",
+    "$$\\dot m_3 = \\dot m_7 = \\dot m_{11}$$\n",
+    "$$\\dot m_{10} = \\dot m_2 = \\dot m_1 = m_9$$\n",
+    "$$\\dot m_4 = \\dot m_5 = \\dot m_6 = m_7$$\n",
+    "and also when looking at pumps, we can find the work done by each pump, and work done by the high pressure turbine. We find all of these, and then we can figure out what that LPT work is.\n",
+    "\n",
+    "**P1 Work**:\n",
+    "$$W_{P1} = \\dot m_7 (1-y) (h_2 - h_1)$$\n",
+    "**P2 Work**:\n",
+    "$$W_{P2} = \\dot m_7 y (h_{11} - h_3)$$\n",
+    "\n",
+    "**Boiler Heat**:\n",
+    "$$Q_{Boiler} = \\dot m_7 (h_{4} - h_5)$$\n",
+    "\n",
+    "**Condenser Heat**:\n",
+    "$$Q_{Condenser} = \\dot m_7 (1-y) (h_9 - h_1)$$\n",
+    "\n",
+    "**CFWH Balance**:\n",
+    "$$\\dot m_7 y (h_3 - h_7) = \\dot m_7 (1-y) (h_{10} - h_2)$$\n",
+    "$$y (h_3 - h_7) = (1-y) (h_{10} - h_2)$$\n",
+    "\n",
+    "**HPT Work**:\n",
+    "$$W_{HPT} = \\dot m_7 (h_6 - h_5)$$\n"
    ]
   },
   {

NUCE_2100/final_exam/Final Exam.ipynb

@@ -432,6 +432,48 @@
     "print(f'h_9 = {h_9:.3f} BTU/lbm')"
    ]
   },
+  {
+   "cell_type": "markdown",
+   "id": "06f7a7be-2537-46fc-91cc-8279bdfd96e7",
+   "metadata": {},
+   "source": [
+    "Now we look at the closed feedwater heater, which conserves energy:\n",
+    "$$\\dot m_8 h_8 + \\dot m_2 h_2 = \\dot m_3 h_3 + \\dot m_{10} h_{10}$$\n",
+    "$$\\dot m_7 y h_8 + \\dot m_7 (1-y) h_2 = \\dot m_7 y h_3 + \\dot m_7 (1-y) h_{10}$$\n",
+    "$$y h_8 + (1-y) h_2 = y h_3 + (1-y) h_{10}$$\n",
+    "$$y h_8 + h_2 - y h_2 = y h_3 + h_{10} - y h_{10}$$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "63541c28-394a-4611-be12-5f32e8ea131f",
+   "metadata": {},
+   "source": [
+    "$$y(h_8 - h_2 - h_3 + h_{10}) = h_{10}-h_2 $$\n",
+    "$$y = \\frac{h_{10}-h_2}{h_8 - h_2 - h_3 + h_{10}} $$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "id": "5bed5dc9-34a9-4aa0-b0c7-c6637ab28774",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "#Find h_2\n",
+    "h_2 = givens.get(2000)['hf']\n",
+    "print(f'h_2 = {h_2:.3f} BTU/lbm')\n",
+    "\n",
+    "#Find h_10\n",
+    "h_10 = givens.get(2000)['hf']\n",
+    "print(f'h_2 = {h_2:.3f} BTU/lbm')\n",
+    "\n",
+    "#Find h_3\n",
+    "h_10 = givens.get(155)['hf']\n",
+    "print(f'h_2 = {h_2:.3f} BTU/lbm')\n",
+    "\n"
+   ]
+  },
   {
    "cell_type": "markdown",
    "id": "df89be07-8172-44bf-900a-46e2108cdd9a",
@@ -439,9 +481,31 @@
    "source": [
     "Now we also don't know $W_{LPT}$. We need a second equation to find $y$. We get it from the closed water heater, which conserves energy between the two streams:\n",
     "\n",
-    "$$ \\dot m_3 h_3 + \\dot m_10 h_10 = \\dot m_2 h_2 + \\dot m_8 + h_8 $$\n",
+    "$$ \\dot m_3 h_3 + \\dot m_{10} h_{10} = \\dot m_2 h_2 + \\dot m_8 h_8 $$\n",
     "\n",
-    "But THIS equation has us going further down the rabbit hole. Must find pump works, then HPT work, then finally we can take the LPT work from that. THEN! We can find y.\n"
+    "But THIS equation has us going further down the rabbit hole. Must find pump works, then HPT work, then finally we can take the LPT work from that. THEN! We can find y. Things we know though:\n",
+    "$$\\dot m_3 = \\dot m_7 = \\dot m_{11}$$\n",
+    "$$\\dot m_{10} = \\dot m_2 = \\dot m_1 = m_9$$\n",
+    "$$\\dot m_4 = \\dot m_5 = \\dot m_6 = m_7$$\n",
+    "and also when looking at pumps, we can find the work done by each pump, and work done by the high pressure turbine. We find all of these, and then we can figure out what that LPT work is.\n",
+    "\n",
+    "**P1 Work**:\n",
+    "$$W_{P1} = \\dot m_7 (1-y) (h_2 - h_1)$$\n",
+    "**P2 Work**:\n",
+    "$$W_{P2} = \\dot m_7 y (h_{11} - h_3)$$\n",
+    "\n",
+    "**Boiler Heat**:\n",
+    "$$Q_{Boiler} = \\dot m_7 (h_{4} - h_5)$$\n",
+    "\n",
+    "**Condenser Heat**:\n",
+    "$$Q_{Condenser} = \\dot m_7 (1-y) (h_9 - h_1)$$\n",
+    "\n",
+    "**CFWH Balance**:\n",
+    "$$\\dot m_7 y (h_3 - h_7) = \\dot m_7 (1-y) (h_{10} - h_2)$$\n",
+    "$$y (h_3 - h_7) = (1-y) (h_{10} - h_2)$$\n",
+    "\n",
+    "**HPT Work**:\n",
+    "$$W_{HPT} = \\dot m_7 (h_6 - h_5)$$\n"
    ]
   },
   {