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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 9 Condensation and Boiling Heat Transfer"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exa 9.1"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Value of reynolds no. is 49.7 so the laminar assumption was correct\n",
"The heat transfer is 2368.0 w\n",
"Total mass flow condensate is 3.78 kg/h\n"
]
}
],
"source": [
"#Example Number 9.1\n",
"# condensation on vertical plate\n",
"\n",
"#Variable declaration\n",
"\n",
"\t# we have to check the reynolds no. to that film is laminar or turbulent\n",
"Tf = (100+98)/2 \t\t\t# [degree celsius]\n",
"Tw = 98 \t\t\t\t# [degree celsius]\n",
"RHOf=960 \t\t\t\t# [kg/cubic meter] \n",
"MUf=2.82*10**(-4) \t\t\t# [kg/m s]\n",
"Kf=0.68 \t\t\t\t# [W/m degree celsius]\n",
"g=9.81 \t\t\t\t\t# [m/s**(2)]\n",
"L=0.3 \t\t\t\t\t# [m]\n",
"\t# RHOf(RHOf-RHOv)~RHOf**(2)\n",
"\t# let us assume laminar film condensate \n",
"Tsat=100 \t\t\t\t# [degree celsius]\n",
"Tg=100 \t\t\t\t\t# [degree celsius]\n",
"Hfg=2255*10**(3) \t\t\t# [J/kg]\n",
"\n",
"#Calculation\n",
"\n",
"hbar=0.943*((RHOf**(2)*g*Hfg*Kf**(3)/(L*MUf*(Tg-Tw)))**(0.25)) \t# [W/sq m deg celsius]\n",
"h=hbar \t\t\t\t\t# [W/square meter degree celsius]\n",
"\t# checking reynolds no. with equation(9-17)\n",
"Ref=4*h*L*(Tsat-Tw)/(Hfg*MUf) \n",
"\n",
"print \"Value of reynolds no. is\",round(Ref,1),\"so the laminar assumption was correct\" \n",
"\t# the heat transfer is now calculated from \n",
"A=0.3*0.3 \t\t\t\t# [square meter]\n",
"q=hbar*A*(Tsat-Tw) \t\t\t# [W]\n",
"mdot=q/Hfg \t\t\t\t# [kg/h]\n",
"\n",
"#Result\n",
"\n",
"print \"The heat transfer is\",round(q),\"w\" \n",
"mdot=mdot*3600 \t\t\t\t# [kg/h]\n",
"print \"Total mass flow condensate is\",round(mdot,2),\"kg/h\" "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exa 9.2"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Total surface area is 3.99 square meter/m\n",
"Heat transfer is 100.1 kW/m\n",
"Total mass flow of condensate is 159.8 kg/h\n"
]
}
],
"source": [
"#Example Number 9.2\n",
"# condensation on tube tank\n",
"\n",
"# Variable declaration\n",
"\n",
"\t# the condensate properties are obtained from previous example\n",
"\t# replacing L by n*d\n",
"Tw=98 \t\t\t\t# [degree celsius]\n",
"RHOf=960 \t\t\t# [kg/cubic meter] \n",
"MUf=2.82*10**(-4) \t\t# [kg/m s]\n",
"Kf=0.68 \t\t\t# [W/m degree celsius]\n",
"g=9.81 \t\t\t\t# [m/s^(2)]\n",
"Tsat=100 \t\t\t# [degree celsius]\n",
"Tg=100 \t\t\t\t# [degree celsius]\n",
"Hfg=2255*10**(3) \t\t# [J/kg]\n",
"d=0.0127 \t\t\t# [m]\n",
"n=10 \n",
"\n",
"#Calculation\n",
"\n",
"hbar=0.725*((RHOf**(2)*g*Hfg*Kf**(3)/(n*d*MUf*(Tg-Tw)))**(0.25)) # [W/sq m deg C]\n",
"\t# total surface area is \n",
"n=100 \n",
"Al=n*22*d/7 \t\t\t# [square meter]\n",
"print \"Total surface area is\",round(Al,2),\"square meter/m\" \n",
"\t# so the heat transfer is \n",
"Ql=hbar*Al*(Tg-Tw)\t\t # [W]\n",
"\n",
"#Result \n",
"\n",
"print \"Heat transfer is\",round(Ql/1000,2),\"kW/m\" \n",
"\t# total mass flow of condensate is then \n",
"mdotl=Ql/Hfg \t\t\t# [kg/h]\n",
"mdotl=mdotl*3600\t\t# [kg/h]\n",
"print \"Total mass flow of condensate is\",round(mdotl,1),\"kg/h\" "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exa 9.4"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Heat transfer in a 1.0 m length of tube is 2810.0 W/m\n"
]
}
],
"source": [
"#Example Number 9.4\n",
"# Flow boiling\n",
"# Variable declaration\n",
"\n",
"p =0.5066\t\t\t# [MPa] pressure of water \n",
"d = 0.0254 \t\t\t# [m] diameter of tube \n",
"Tw = 10.0 \t\t\t# [degree celsius]\n",
"\t\t\t\t# for calculation we use equation (9-45), noting that \n",
"dT = 10.0 \t\t\t# [degree celsius]\n",
"\t\t\t\t# the heat transfer coefficient is calculated as \n",
"\n",
"import math\n",
"h = 2.54*Tw**(3)*math.exp(p/1.551) \t# [W/square meter degree celsius]\n",
"\n",
"\t\t# the surface area for a 1-m length of tube is \n",
"L = 1 \t\t\t\t# [m]\n",
"import math\n",
"A = math.pi*d*L \t\t# [square meter]\n",
"\n",
"\t\t# so the heat transfer is \n",
"q = h*A*dT \t\t\t# [W/m]\n",
"print \"Heat transfer in a 1.0 m length of tube is\",round(q),\"W/m\" "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exa 9.5"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Heat flux obtained is 22.8 kW/square meter\n",
"If the pan operates as a pressure cooker at 0.17 MPa the increase in heat flux is 23.0 percent\n"
]
}
],
"source": [
"#Example Number 9.5\n",
"# water boiling in a pan \n",
"\n",
"# Variable declaration\n",
"\n",
"p = 0.101\t\t\t\t# [MPa] pressure of water \n",
"dT_x = 8 \t\t\t\t# [degree celsius]\n",
"p1 = 0.17 \t\t\t\t# [MPa] given operating pressure\n",
"\t# we will use the simplified relation of table 9-13(page no.-506) for the \t\testimates.we do not know the value of q_by_A and so must choose one of the two \trelation for a horizontal surface from the table\n",
"\t# we anticipate nucleate boiling, so choose\n",
"h = 5.56*dT_x**(3) \t\t\t# [W/square meter degree celsius]\n",
"\t\t\t\t\t# and the heat flux is \n",
"q_by_A = h*dT_x \t\t\t# [W/square meter]\n",
"\t# for operation as a pressure cooker we obtain the value of h from \t\tequation(9-44)\n",
"\n",
"#Calculation\n",
"\n",
"hp = h*(p1/p)**(0.4) \t\t\t# [W/square meter degree celsius]\n",
"\t# the corresponding heat flux is \n",
"q_by_A1 = hp*dT_x \t\t\t# [W/square meter]\n",
"\n",
"#Result\n",
"\n",
"print \"Heat flux obtained is\",round(q_by_A/1000,1),\"kW/square meter\" \n",
"per_inc = 100*(q_by_A1-q_by_A)/q_by_A \n",
"\n",
"print \"If the pan operates as a pressure cooker at 0.17 MPa the increase in heat flux is\",round(per_inc),\"percent\" \n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exa 9.6"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Thus the heat transfers 14.0 times the heat of a pure copper rod with a substantial temperature gradient\n"
]
}
],
"source": [
"#Example Number 9.6\n",
"# heat-flux comparisons \n",
"\n",
"# Variable declaration\n",
"\n",
"Tw = 200.0 \t\t\t\t# [degree celsius] water temperature \n",
"L = 0.08 \t\t\t\t# [m] length of solid copper bar\n",
"dT = 100.0 \t\t\t\t# [degree C] temp differential in copper bar\n",
"\t#using the data of table 9-4(page no.-508)\n",
"\t# the heat flux per unit area is expressed as q_by_A = -k*del_T/dx\n",
"\t# from table A-2(page no.-) the thermal conductivity of copper is \n",
"k = 374.0 \t\t\t\t# [W/m degree celsius]\n",
"\n",
"#Calculation\n",
"\n",
"q_by_A = -k*(-dT)/L \t\t\t# [W/square meter]\n",
"\n",
"\t# from table 9-4(page no.-508) the typical axial heat flux for a water heat \t\tflux for a water heat pipe is \n",
"q_by_A_axial = 0.67 \t\t\t# [kW/csquare meter]\n",
"q_by_A = q_by_A/(1000*10**(4.0)) \t# [kW/csquare meter]\n",
"time=q_by_A_axial/q_by_A\n",
"\n",
"#Result\n",
"\n",
"print \"Thus the heat transfers \",round(time),\"times the heat of a pure copper rod with a substantial temperature gradient\" "
]
}
],
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