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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 05:First law applied to Flow Processes"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex5.1:pg-97"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" Example 5.1\n",
"\n",
" The rate of work input is 116.0 kW\n",
"\n",
" The ratio of the inlet pipe diameter and outet pipe diameter is 0.0 \n"
]
}
],
"source": [
"# Part(a)\n",
"import math\n",
"V1 = 0.95 # Inlet volume flow rate in m**3/kg\n",
"\n",
"P1 = 100 # Pressure at inlet in kPa\n",
"\n",
"v1 = 7 # velocity of flow at inlet in m/s\n",
"\n",
"V2 = 0.19 # Exit volume flow rate in m**3/kg\n",
"\n",
"P2 = 700 # Pressure at exit in kPa \n",
"\n",
"v2 = 5 # velocity of flow at exit in m/s\n",
"\n",
"w = 0.5 # mass flow rate in kg/s\n",
"\n",
"u21 = 90 # change in internal energy in kJ/kg\n",
"\n",
"Q = -58 # Heat transfer in kW\n",
"\n",
"W = - w*( u21 + (P2*V2-P1*V1) + ((v2**2-v1**2)/2) ) + Q # W = dW/dt \n",
"\n",
"print \"\\n Example 5.1\"\n",
"\n",
"print \"\\n The rate of work input is \",abs(W) ,\" kW\"\n",
"\n",
"#The answers given in textbook is wrong\n",
"\n",
"# Part (b)\n",
"\n",
"A = (v2/v1)*(V1/V2) # A = A1/A2\n",
"\n",
"d_ratio = math.sqrt(A) # d = d1/d2\n",
"\n",
"print \"\\n The ratio of the inlet pipe diameter and outet pipe diameter is \",d_ratio ,\" \"\n",
"\n",
"#The answers vary due to round off error\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex5.2:pg-98"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" Example 5.2\n",
"\n",
" The internal energy decreases by 20.0 kJ\n"
]
}
],
"source": [
"import math\n",
"V1 = 0.37 # volume flow rate at inlet in m**3/kg\n",
"\n",
"P1 = 600# Inlet pressure in kPa\n",
"\n",
"v1 = 16 # Inlet velocity of flow in m/s\n",
"\n",
"V2 = 0.62 # volume flow rate at exit in m**3/kg \n",
"\n",
"P2 = 100# Exit pressure in kPa\n",
"\n",
"v2 = 270 # Exit velocity of flow in m/s\n",
"\n",
"Z1 = 32 # Height of inlet port from datum in m\n",
"\n",
"Z2 = 0 #Height of exit port from datum in m\n",
"\n",
"g = 9.81 # Acceleration due to gravity\n",
"\n",
"Q = -9 # Heat transfer in kJ/kg\n",
"\n",
"W = 135 # Work transfer in kJ/kg\n",
"\n",
"U12 = (P2*V2-P1*V1) + ((v2**2-v1**2)/2000) + (Z2-Z1)*g*1e-3 + W - Q # Change in internal energy in kJ\n",
"\n",
"\n",
"\n",
"print \"\\n Example 5.2\"\n",
"\n",
"print \"\\n The internal energy decreases by \",round(U12) ,\" kJ\"\n",
"\n",
"#The answers vary due to round off error\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex5.3:pg-99"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" Example 5.3\n",
"\n",
" The steam flow rate is 53.5854836932 Kg/s\n"
]
}
],
"source": [
"import math\n",
"\n",
"P1 = 4 # Boiler pressure in MPa\n",
"\n",
"t1 = 400 # Exit temperature at boiler in degree Celsius\n",
"\n",
"h1 = 3213 # Enthalpy at boiler exit in kJ/kg\n",
"\n",
"V1 = 0.073 # specific volume at boiler exit in m**3/kg\n",
"\n",
"P2 = 3.5 # Pressure at turbine end in MPa\n",
"\n",
"t2 = 392 # Turbine exit temperature in degree Celsius\n",
"\n",
"h2 = 3202 # Enthalpy at turbine exit in kJ/kg\n",
"\n",
"V2 = 0.084 # specific volume at turbine exit in m**3/kg\n",
"\n",
"Q = -8.5 # Heat loss from pipeline in kJ/kg\n",
"\n",
"v1 = math.sqrt((2*(h1-h2+Q)*1e3)/(1.15**2-1)) # velocity of flow in m/s\n",
"\n",
"A1 = (math.pi/4)*0.2**2 # Area of pipe in m**2\n",
"\n",
"w = (A1*v1)/V1 # steam flow rate in Kg/s\n",
"\n",
"\n",
"\n",
"print \"\\n Example 5.3\"\n",
"\n",
"print \"\\n The steam flow rate is \",w ,\" Kg/s\"\n",
"\n",
"#The answers vary due to round off error\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex5.4:pg-100"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" Example 5.4\n",
"\n",
" The amount of heat that should be supplied is 703.880549402 Kg/h\n"
]
}
],
"source": [
"import math\n",
"h1 = 313.93 # Enthalpy of water at heater inlet in kJ/kg\n",
"\n",
"h2 = 2676 # Enthalpy of hot water at temperature 100.2 degree Celsius\n",
"\n",
"h3 = 419 #Enthalpy of water at heater inlet in kJ/kg\n",
"\n",
"w1 = 4.2 # mass flow rate in kg/s\n",
"\n",
"\n",
"\n",
"print \"\\n Example 5.4\"\n",
"\n",
"w2 = w1*(h3-h1)/(h2-h3)# Steam rate \n",
"\n",
"print \"\\n The amount of heat that should be supplied is \",w2*3600 ,\" Kg/h\"\n",
"\n",
"\n",
"\n",
"#The answers vary due to round off error\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex5.5:pg-100"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" Example 5.5\n",
"\n",
" The rate of heat transfer to the air in the heat exchanger is 1577.85 kJ/s\n",
"\n",
" The power output from the turbine assuming no heat loss is 298 kW\n",
"\n",
" The velocity at the exit of the nozzle is 552.358579186 m/s\n"
]
}
],
"source": [
"import math\n",
"t1 = 15 # Heat exchanger inlet temperature in degree Celsius\n",
"\n",
"t2 = 800 # Heat exchanger exit temperature in degree Celsius\n",
"\n",
"t3 = 650 # Turbine exit temperature in degree Celsius\n",
"\n",
"t4 = 500 # Nozzle exit temperature in degree Celsius\n",
"\n",
"v1 = 30 # Velocity of steam at heat exchanger inlet in m/s\n",
"\n",
"v2 = 30# Velocity of steam at turbine inlet in m/s\n",
"\n",
"v3 = 60 # Velocity of steam at nozzle inlet in m/s\n",
"\n",
"w = 2 # mass flow rate in kg/s\n",
"\n",
"cp = 1005 # Specific heat capacity of air in kJ/kgK\n",
"\n",
"\n",
"\n",
"print \"\\n Example 5.5\"\n",
"\n",
"Q1_2 = w*cp*(t2-t1) # rate of heat transfer\n",
"\n",
"print \"\\n The rate of heat transfer to the air in the heat exchanger is \",Q1_2/1e3 ,\" kJ/s\"\n",
"\n",
"\n",
"\n",
"W_T = w*( ((v2**2-v3**2)/2) + cp*(t2-t3)) # power output from the turbine\n",
"\n",
"print \"\\n The power output from the turbine assuming no heat loss is \",W_T/1000 ,\" kW\"\n",
"\n",
"v4 = math.sqrt( (v3**2) + (2*cp*(t3-t4)) ) # velocity at the exit of the nozzle\n",
"\n",
"print \"\\n The velocity at the exit of the nozzle is \",v4 ,\" m/s\"\n",
"\n",
"#The answers vary due to round off error\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex5.6:pg-102"
]
},
{
"cell_type": "code",
"execution_count": 16,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" Example 5.6\n",
"\n",
" Velocity of exhaust gas is 541.409855832 m/s\n"
]
}
],
"source": [
"import math\n",
"\n",
"ha = 260 # Enthalpy of air in kJ/kg\n",
"\n",
"hg = 912 # Enthalpy of gas in kJ/kg\n",
"\n",
"Va = 270 # Velocity of air in m/s\n",
"\n",
"wf = 0.0190 # mass of fuel in Kg\n",
"\n",
"wa = 1 # mass of air in Kg\n",
"\n",
"Ef = 44500 # Chemical energy of fuel in kJ/kg\n",
"\n",
"Q = 21 # Heat loss from the engine in kJ/kg\n",
"\n",
"\n",
"\n",
"print \"\\n Example 5.6\"\n",
"\n",
"Eg = 0.05*wf*Ef/(1+wf) # As 5% of chemical energy is not released in reaction\n",
"\n",
"wg = wa+wf # mass of flue gas\n",
"\n",
"Vg = math.sqrt(2000*(((ha+(Va**2*0.001)/2+(wf*Ef)-Q)/(1+wf))-hg-Eg)) \n",
"\n",
"\n",
"\n",
"print \"\\n Velocity of exhaust gas is \",Vg ,\" m/s\"\n",
"\n",
"#Answer given in textbook is wrong\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex5.8:pg-103"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" Example 5.8\n",
"\n",
" The rate at which air flows out of the tank is 0.85 kg/h\n"
]
}
],
"source": [
"import math\n",
"# Given that\n",
"\n",
"V = 0.12 # Volume of tank in m**3\n",
"\n",
"p = 1 # Pressure in MPa\n",
"\n",
"T = 150 # Temperature in degree centigrade\n",
"\n",
"P = 0.1 # Power to peddle wheel in kW\n",
"\n",
"print \"\\n Example 5.8\"\n",
"\n",
"u0 = 0.718*273 # Internal energy at 0 degree Celsius\n",
"\n",
"# Function for internal energy of gas\n",
"\n",
"def f1(t):\n",
" u = u0+(0.718*t)\n",
" pv = 0.287*(273+t)\n",
" return (u,pv)\n",
" \n",
"U,PV=f1(T)\n",
" \n",
" \n",
"hp = U+PV # At 150 degree centigrade\n",
"m_a = P/hp\n",
" \n",
"print \"\\n The rate at which air flows out of the tank is \",round(m_a*3600,2) ,\" kg/h\"\n",
"\n",
"#The answers vary due to round off error\n",
"\n",
"\n",
"\n"
]
}
],
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|