"
]
},
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"source": [
"Regen = VaporDome(substance, (\"s\", \"T\"))\n",
"\n",
"Regen.add_process(st_1, st_2, \"isentropic\")\n",
"Regen.add_process(st_2, st_3, \"isobaric\")\n",
"Regen.add_process(st_3, st_4, \"isentropic\")\n",
"Regen.add_process(st_4, st_5, \"isentropic\")\n",
"Regen.add_process(st_5, st_6, \"isobaric\")\n",
"Regen.add_process(st_6, st_7, \"isentropic\")\n",
"Regen.add_process(st_7, st_8, \"isobaric\")\n",
"Regen.add_process(st_8, st_9, \"isentropic\")\n",
"Regen.add_process(st_9, st_10, \"isobaric\")\n",
"Regen.add_process(st_10, st_1, \"isobaric\")\n",
"\n",
"Regen.add_process(st_11, st_12, \"isoenthalpic\")\n",
"Regen.add_process(st_12, st_7, \"isobaric\")\n",
"Regen.add_process(st_2, st_11, \"isobaric\")\n",
"Regen.add_process(st_4, st_7, \"isobaric\")"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {},
"outputs": [],
"source": [
"y_p = (h_10 - h_9) / (h_2 - h_11)\n",
"Qdot_in = (h_1 - h_10 + (1 - y_p) * (h_3 - h_2)).to(\"kJ/kg\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"\n",
"\n",
"**Answer:** The heat transfer input is $\\dot{Q}_{in}/\\dot{m} =$ 2808.15 kJ/kg\n",
"\n",
"
"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### 2. the thermal efficiency"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"To find the thermal efficiency, we need to calculate the net work output of the power plant. This involves all three turbine stages and both of the pump stages. First, we need to calculate the mass fractions of the flow, $y'$ and $y''$. To find these values, we need to draw control volumes around the feedwater heaters, moving from high pressure to low pressure. $y'$ has already been found, so we need to find $y''$\n",
"\n",
"\\begin{align*}\n",
"0 &= \\dot{m}_7 h_7 + \\dot{m}_{12} h_{12} + \\dot{m}_{4} h_4 - \\dot{m}_8 h_8\n",
"\\end{align*}\n",
"\n",
"Dividing through by $\\dot{m}_1$ and noting that $\\dot{m}_8 = \\dot{m}_1$, $\\dot{m}_{12} = \\dot{m}_2$, and $\\dot{m}_7 = \\dot{m}_5$,\n",
"\n",
"\\begin{align*}\n",
"0 &= (1 - y' - y'')h_7 + y' h_{12} + y'' h_4 - h_8 \\\\\n",
"y'' &= \\frac{h_8 - h_7 + y'(h_7 - h_{12})}{h_4 - h_7}\n",
"\\end{align*}"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {},
"outputs": [],
"source": [
"y_pp = (h_8 - h_7 + y_p * (h_7 - h_12)) / (h_4 - h_7)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Now, for the turbine stages,\n",
"\n",
"\\begin{align*}\n",
"\\dot{W}_t &= \\dot{m}_1(h_1 - h_2) + \\dot{m}_3 (h_3 - h_4) + \\dot{m}_5 (h_4 - h_5) \\\\\n",
"\\frac{\\dot{W}_t}{\\dot{m}_1} &= h_1 - h_2 + (1 - y')(h_3 - h_4) + (1 - y' - y'')(h_4 - h_5)\n",
"\\end{align*}"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {},
"outputs": [],
"source": [
"Wdot_t = h_1 - h_2 + (1 - y_p) * (h_3 - h_4) + (1 - y_p - y_pp) * (h_4 - h_5)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"and for the pump stages\n",
"\n",
"\\begin{align*}\n",
"\\dot{W}_p &= \\dot{m}_6(h_6 - h_7) + \\dot{m}_8(h_8 - h_9) \\\\\n",
"\\frac{\\dot{W}_p}{\\dot{m}_1} &= (1 - y' - y'')(h_6 - h_7) + h_8 - h_9\n",
"\\end{align*}"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {},
"outputs": [],
"source": [
"Wdot_p = (1 - y_p - y_pp) * (h_6 - h_7) + h_8 - h_9\n",
"eta = ((Wdot_t + Wdot_p) / Qdot_in).to(\"dimensionless\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"\n",
"\n",
"**Answer:** The thermal efficiency of the cycle is $\\eta =$ 0.4617 = 46.17%\n",
"\n",
"
"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### 3. the rate of heat transfer out of the condenser into the cooling water"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Drawing a control volume around the condenser, we can find the rate of heat transfer as\n",
"\n",
"\\begin{align*}\n",
"\\dot{Q}_{out} &= \\dot{m}_5(h_6 - h_5) \\\\\n",
"\\frac{\\dot{Q}_{out}}{\\dot{m}_1} &= (1 - y' - y'') (h_6 - h_5)\n",
"\\end{align*}"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {},
"outputs": [],
"source": [
"Qdot_out = (1 - y_p - y_pp) * (h_6 - h_5).to(\"kJ/kg\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"\n",
"\n",
"**Answer:** The rate of heat transfer is $\\dot{Q}_{out}/\\dot{m}_1 =$ -1511.72 kJ/kg\n",
"\n",
"
"
]
}
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