Q: At present day, the rate of formation of fossil fuels (coal, petroleum, natural gas) is estimated to be about $4\text{ billion kW}$ ($4 \times 10^9\text{ kW}$) and the world's consumption of energy is of the order of $10^{17}\text{ Btu/year}$. Show that the rate of consumption of fossil fuel is of the same order of magnitude as the rate of their formation and these energy sources are therefore non-renewable.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $1\text{ Btu} = 1055\text{ J}$, $1\text{ year} = 365 \times 24 \times 3600 = 3.15 \times 10^7\text{ s}$
The rate of formation of fossil fuel $= 4\text{ billion kW} = 4 \times 10^9 \times 10^3\text{ W} = \mathbf{4 \times 10^{12}\text{ J s}^{-1}}$
The world's consumption of energy $= 10^{17}\text{ Btu/year} = \frac{10^{17} \times 1055\text{ J}}{1 \times 3.15 \times 10^7\text{ s}} = \mathbf{3.35 \times 10^{12}\text{ J s}^{-1}}$
Hence, the rate of consumption of fossil fuel is of the same order of magnitude as the rate of their formation and these energy sources are therefore non-renewable.
Q: A home requires $62\text{ kWh}$ of heat on winter day to maintain a constant indoor temperature of $20^\circ\text{C}$. Assume that the temperature of hot water outlet is $60^\circ\text{C}$. (i) How much collector surface area does it need for a solar heating system that has a $20\%$ efficiency? (ii) How large does the storage tank have to be to provide this much energy? (The average solar radiation per day in winter is about $6.5\text{ kWh m}^{-2}$)
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $\Delta Q_{\text{req}} = 62\text{ kWh}$, $\Delta Q_{\text{input}} = 6.5\text{ kWh m}^{-2}$, $\eta = 20\% = 0.2$
(i) $\eta = \frac{\Delta Q_{\text{output}}}{\Delta Q_{\text{input}}} \implies \Delta Q_{\text{output}} = 6.5 \times 0.2 = 1.3\text{ kWh m}^{-2}$
Collector surface area $= \frac{\Delta Q_{\text{req}}}{\Delta Q_{\text{output}}} = \frac{62}{1.3} = \mathbf{47.69 \approx 48\text{ m}^2}$
(ii) $1\text{ kWh} = 3.6 \times 10^6\text{ J}$, $\Delta T = 60 - 20 = 40^\circ\text{C}$, $c = 4184\text{ J kg}^{-1}\text{ K}^{-1}$
$\Delta Q_{\text{req}} = m c \Delta T \implies m = \frac{\Delta Q_{\text{req}}}{c \Delta T} = \frac{62 \times 3.6 \times 10^6}{4184 \times 40} = 1333.6\text{ kg}$
Volume of storage tank should be $\mathbf{1334\text{ L}}$
Q: Determine the input power of the wind if the wind speed is $20\text{ m s}^{-1}$ and blade length is $20\text{ m}$. Density of air is $1.3\text{ kg m}^{-3}$.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $v = 20\text{ m s}^{-1}$, $\rho = 1.3\text{ kg m}^{-3}$, $r = 20\text{ m}$, $P = ?$
$A = \pi r^2 = 3.142 \times (20)^2 = 1256.8\text{ m}^2$
$P = \frac{1}{2}\rho A v^3 = \frac{1}{2} \times 1.3 \times 1256.8 \times (20)^3 = \mathbf{6.535 \times 10^6\text{ W} = 6.535\text{ MW}}$
Q: At a hydroelectric power plant, the water pressure head is at a height of $300\text{ m}$ and the water flow available is $100\text{ m}^3\text{ s}^{-1}$. If the turbine generator efficiency is $60\%$, estimate the electric power available from the plant. ($g = 9.8\text{ m s}^{-2}$, $\rho = 1000\text{ kg m}^{-3}$)
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $h = 300\text{ m}$, $\dot{V} = 100\text{ m}^3\text{ s}^{-1}$, $\eta = 60\% = 0.6$, $g = 9.8\text{ m s}^{-2}$
$P = \eta \rho \dot{V} g h = 0.6 \times 1000 \times 100 \times 9.8 \times 300 = \mathbf{176.4 \times 10^6\text{ W} = 176.4\text{ MW}}$
Q: The band gap of GaAs is $1.4\text{ eV}$. Calculate the minimum wavelength of light for photovoltaic generation in a GaAs solar cell.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $E = 1.4\text{ eV} = 1.4 \times 1.6 \times 10^{-19}\text{ J}$, $h = 6.626 \times 10^{-34}\text{ J s}$, $c = 3 \times 10^8\text{ m s}^{-1}$
$\lambda = \frac{h c}{E} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{1.4 \times 1.6 \times 10^{-19}} = \mathbf{8.87 \times 10^{-7}\text{ m} = 887\text{ nm}}$
Q: A $100\text{ ft} \times 50\text{ ft}$ building has a flat roof. What is the average solar energy received by the roof in a month? (The solar radiation per day in winter is about $1600\text{ Btu ft}^{-2}$)
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $A = 100 \times 50 = 5000\text{ ft}^2$, $t = 30\text{ days}$, Radiation $= 1600\text{ Btu ft}^{-2}\text{ day}^{-1}$
Total Energy $= 1600 \times 5000 \times 30 = \mathbf{2.4 \times 10^8\text{ Btu}}$
Q: Calculate the solar energy received by a standard hot water collector of dimensions $1\text{ m} \times 2\text{ m}$, over one hour at around noon, if the irradiance stays fairly constant at about $800\text{ W m}^{-2}$.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $A = 1 \times 2 = 2\text{ m}^2$, $t = 1\text{ h} = 3600\text{ s}$, Irradiance $= 800\text{ W m}^{-2}$
Energy $= \text{Irradiance} \times A \times t = 800 \times 2 \times 3600 = \mathbf{5.76 \times 10^6\text{ J}}$
Q: An important hydroelectric plant has a head of $100\text{ m}$ and water volumetric flow rate of $10\,000\text{ m}^3\text{ s}^{-1}$. The turbine generator efficiency is $60\%$. What is the maximum power that the plant can produce? (Density of water = $1000\text{ kg m}^{-3}$)
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $h = 100\text{ m}$, $\dot{V} = 10\,000\text{ m}^3\text{ s}^{-1}$, $\eta = 60\% = 0.6$, $\rho = 1000\text{ kg m}^{-3}$
$P = \eta \rho \dot{V} g h = 0.6 \times 1000 \times 10\,000 \times 9.8 \times 100 = \mathbf{5880 \times 10^6\text{ W} = 5880\text{ MW}}$
Q: A dam which maximum head of $200\text{ m}$ produces $2000\text{ MW}$ electrical power. What is the rate of falling water on the turbines? The turbine generator efficiency is $50\%$. (Density of water = $1000\text{ kg m}^{-3}$)
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $h = 200\text{ m}$, $P = 2000\text{ MW} = 2000 \times 10^6\text{ W}$, $\eta = 50\% = 0.5$
$\dot{V} = \frac{P}{\eta \rho g h} = \frac{2000 \times 10^6}{0.5 \times 1000 \times 9.8 \times 200} = \mathbf{2040\text{ m}^3\text{ s}^{-1}}$
Q: The wind is blowing at $10\text{ m s}^{-1}$, how much total power produced by the wind turbine if the blades are $45\text{ m}$ long? Density of air is $1.3\text{ kg m}^{-3}$.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $v = 10\text{ m s}^{-1}$, $\rho = 1.3\text{ kg m}^{-3}$, $r = 45\text{ m}$
$A = \pi r^2 = 3.142 \times (45)^2\text{ m}^2$
$P = \frac{1}{2}\rho A v^3 = \frac{1}{2} \times 1.3 \times 3.142 \times (45)^2 \times (10)^3 = \mathbf{4.136 \times 10^6\text{ W} = 4.136\text{ MW}}$
Q: The hydro turbine receives water from a reservoir at an elevation of $100\text{ m}$ above it. What is the minimum water flow in $\text{kg s}^{-1}$ to produce a steady turbine output of $50\text{ MW}$ if the turbine generator efficiency is $75\%$?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $h = 100\text{ m}$, $\eta = 75\% = 0.75$, $P = 50\text{ MW} = 50 \times 10^6\text{ W}$
$P = \eta \left(\frac{m}{t}\right) g h \implies \frac{m}{t} = \frac{P}{\eta g h}$
$\frac{m}{t} = \frac{50 \times 10^6}{0.75 \times 9.8 \times 100} = \mathbf{68027\text{ kg s}^{-1}}$