QoD

L2 L3

In $\text{pu}1m3$ of town gas, there is $28.2\%-30.7\%$ of methane, taking the average we get $29.35\%$

$$\begin{array}{rl}pV& =nRT\\ (101325)(1\times 0.2945)& =n(8.31)(25+273)\\ n& \approx 12.0\text{pu}mol/m^3\end{array}$$

By the chemical reaction of burning methane, we can get the energy released per mole of methane

$$\text{ce}C{H}_4+2O_2->2H_{2}O+C{O}_2+Energy$$ $$\begin{array}{rl}\text{Net Energy Release}& =(4\times 460+2\times 732)-(4\times 410+2\times 498)\\ & =668\text{ kJ/mol}\end{array}$$

The total energy released per $\text{pu}m3$ of methane is as follows

$$\begin{array}{rl}\text{Total Energy Release}& =12\text{pu}mol/m^3\times 668\text{ kJ/mol}\\ & =8016\text{pu}kJ/m^3\\ & =8.016\text{pu}MJ/m^3\end{array}$$

The result is higher than the Calorific Value claimed by Towngas (17.27 * 0.2935 = 5.07 MJ/m^3), this is likely due to the energy lost to the environment in the process

Oil can also be burnt in the Towngas composition

L4

a

Work done on the gas corresponds to the area under the curve

$$\begin{array}{rl}{W}_{on}& =-\int pdV\\ & =-\left((1\times 1.013\times 10^5)(1-3)+\frac{(2-1)(1.013\times 10^5)(1-3)}{2}\right)\\ & =303900\text{ J}\end{array}$$

b

Consider the gas at state $a$

$$\begin{array}{rl}pV& =nRT\\ (2\times 1.013\times 10^5)(1)& =n(8.31)(27+273)\\ n& \approx 81.3\text{ mol}\end{array}$$

Consider the gas at state $c$

$$\begin{array}{rl}pV& =nRT\\ (1\times 1.013\times 10^5)(3)& =(81.3)(8.31)T\\ T& \approx 450\text{ K}\end{array}$$

Consider the process $c\to a$

$$\begin{array}{rl}\Delta U& =\frac{j}{2}nR\Delta T\\ & =\frac{3}{2}(81.3)(8.31)((27+273)-450)\\ & \approx -152000\text{ J}\end{array}$$

c

$$\begin{array}{rl}Q& =W_{on}-\Delta U\\ & =303900-(-152000)\\ & =455900\text{ J}\end{array}$$

L5

The ratio of heat capacity of $\text{ce}C{O}_2\approx 1.31$ at around $0^{\circ }\text{C}$

https://en.wikipedia.org/wiki/Carbon_dioxide_(data_page)

The pressure of $\text{ce}C{O}_2$ inside a soda can at $4^{\circ }\text{C}$ is around $117\text{ kPa}$

https://hypertextbook.com/facts/2000/SeemaMeraj.shtml

$$\begin{array}{rl}pV& =nRT\\ V& =\frac{nRT}{p}\end{array}$$ $$\begin{array}{rl}{p}_1{V}_1^{\gamma }& =p_2{V}_2^{\gamma }\\ p_1{\left(\frac{nR{T}_1}{p_1}\right)}^{\gamma }& =p_2{\left(\frac{nR{T}_2}{p_2}\right)}^{\gamma }\\ p_1^{1-\gamma }{T}_1^{\gamma }& =p_2^{1-\gamma }{T}_2^{\gamma }\end{array}$$ $$\begin{array}{rl}{p}_1^{1-\gamma }{T}_1^{\gamma }& =p_2^{1-\gamma }{T}_2^{\gamma }\\ (117000{)}^{1-1.31}(4+273{)}^{1.31}& =(101325{)}^{1-1.31}{T}_2^{1.31}\\ T_2& \approx 268\text{ K}\end{array}$$

L6

a

$$\begin{array}{rl}\Delta S& =\Delta Q\left(\frac{1}{T_H}-\frac{1}{T_C}\right)\\ & =1200\left(\frac{1}{650}-\frac{1}{350}\right)\\ & \approx -1.58\end{array}$$

$\Delta S<0$, therefore the process is impossible

b

For a process to be reversible, $\Delta S=0$

$$\begin{array}{rl}1200\left(\frac{1}{650}-\frac{1}{T_c}\right)& =0\\ T_c& =650\text{ K}\end{array}$$

L7

a

Assume the AC runs in Carnot Cycle and electricity power convert 100% to power in the compressor of the AC

i

$$\begin{array}{rl}{\text{COP}}_{19^{\circ }\text{C}}& =\frac{T_C}{T_H-T_C}\\ & =\frac{19+273}{(33+273)-(19+273)}\\ & \approx 20.9\end{array}$$

ii

$$\begin{array}{rl}{\text{COP}}_{25^{\circ }\text{C}}& =\frac{T_C}{T_H-T_C}\\ & =\frac{25+273}{(33+273)-(25+273)}\\ & \approx 37.3\end{array}$$

b

$$\begin{array}{rl}{W}_{19^{\circ }\text{C}}& =\frac{Q_c}{{\text{COP}}_{19^{\circ }\text{C}}}\\ & =\frac{100}{20.9}\\ & \approx 4.78\end{array}$$ $$\begin{array}{rl}{W}_{25^{\circ }\text{C}}& =\frac{Q_c}{{\text{COP}}_{25^{\circ }\text{C}}}\\ & =\frac{100}{37.3}\\ & \approx 2.68\end{array}$$ $$\begin{array}{rl}\text{Energy saved in \%}& =\frac{4.78-2.68}{4.78}\\ & =43.9\%\end{array}$$

L8

a

Airplane model: Airbus A320neo
Engine model: CFM International LEAP

https://en.wikipedia.org/wiki/Airbus_A320neo_family

Cruising altitude: 12100m
https://www.ana.co.jp/en/jp/guide/prepare/seatmap/international/a320_neo/

Temperature: -55${}^{\circ }\text{C}$
https://www.omnicalculator.com/physics/altitude-temperature

Turbine intake temperature: about 1000${}^{\circ }\text{C}$

https://www.easa.europa.eu/en/downloads/20810/en

b

Assume that we are calculating the maximum theoretical efficiency

$$\begin{array}{rl}e& =1-\frac{T_c}{T_h}\\ & =1-\frac{-55+273}{1000+273}\\ & \approx 82.9\%\end{array}$$

c

The efficiency of jet engine is higher than Otto and Diesel engines, since the temperature inside the engine is very high while the environment’s temperature is very low, making $\frac{T_c}{T_h}$ a smaller number, therefore its efficiency would be higher.

L9

a

By Lenz’s law, if the upward magnetic field is decreasing, an opposing upward magnetic field will be induced. Therefore an anti-clockwise current would be induced.

b

$$\begin{array}{rl}ϵ& =NA\frac{\Delta B}{\Delta t}\\ & =(10000)(\pi \times 0.1^2)\left(\frac{0.05}{1}\right)\\ & \approx 15.7\text{ V}\\ & \\ P& =\frac{V^2}{R}\\ & =\frac{15.7^2}{1}\\ & \approx 246\text{ W}\end{array}$$

L10

a

$$\frac{7477\times 10^6}{11}\approx 6.80\times 10^8$$

b

$$\begin{array}{rl}I& =\frac{P}{V}\\ & =\frac{6.80\times 10^8}{400\times 10^3}\\ & =1700\text{ A}\\ & \\ \text{Power loss}& =I^{2}R\\ & =1700^2\left(\frac{500}{11}\times 0.03\right)\\ & \approx 3.94\times 10^6\text{ W}\\ & \\ \text{Power loss \%}& =\frac{3.94\times 10^6}{6.80\times 10^8}\\ & \approx 0.579\%\end{array}$$

c

$$\begin{array}{rl}I& =\frac{P}{V}\\ & =\frac{6.80\times 10^8}{220}\\ & \approx 3.09\times 10^6\text{ A}\\ & \\ \text{Power loss}& =I^{2}R\\ & =(3.09\times 10^6{)}^2\left(\frac{500}{11}\times 0.03\right)\\ & \approx 1.30\times 10^{13}\text{ W}\\ & \\ \text{Power loss \%}& =\frac{1.30\times 10^{13}}{6.80\times 10^8}\\ & \approx 1910000\%\end{array}$$

L13

a

$$\begin{array}{rl}{P}_{\text{solar}}& =148.7\times (1.5\times 12)\times 0.25\\ & =669.15\text{ W}\\ & \\ E_{\text{solar}}& =669.15\times 60\times 60\times 24\\ & \approx 5.78\times 10^7\text{ J}\end{array}$$

A 40-inch LCD monitor consumes $120\text{ W}$ according to the following figure
https://www.researchgate.net/figure/Power-Comparison-between-CRT-LCD-Plasma-and-LED-Televisions-In-watts_fig2_336778177

$$\begin{array}{rl}{P}_{\text{consumption}}& =3\times 30+10+120\\ & =220\text{ W}\\ & \\ E_{\text{consumption}}& =220\times 60\times 60\times 24\\ & \approx 1.9\times 10^7\text{ J}\end{array}$$

Since $E_{\text{consumption}}<E_{\text{solar}}$, it is enough for the solar panel to power those devices for 24 hours

b

On average, a person uses $6.2\times 10^9\text{J}$ per year from page 36 of this report
https://www.emsd.gov.hk/filemanager/en/content_762/HKEEUD2022.pdf

$$\begin{array}{rl}\text{Required percentage}& =\frac{148.7\times 500\times 0.25\times 60\times 60\times 24}{\frac{6.2\times 10^9}{365}}\times 100\%\\ & \approx 9454\%\end{array}$$ $$\begin{array}{rl}\text{Maximum number of inhabitants}& =\frac{148.7\times 500\times 60\times 60\times 24}{\frac{6.2\times 10^9}{365}}\\ & \approx 378<1200\end{array}$$

Therefore it is not enough to power the entire building

L14

$$\begin{array}{rl}\text{Energy required per year}& =3204\times 8\times 10^9\times 10^3\times 3600\\ & \approx 9.23\times 10^{19}\text{ J}\end{array}$$ $$\begin{array}{rl}\text{Uranium required per year}& =\frac{9.23\times 10^{19}}{1\times 10^9\times 60\times 60\times 24\times 365}\times 162\\ & \approx 4.74\times 10^5\text{ tons}\end{array}$$ $$\begin{array}{rl}\text{Time the supply can last}& =\frac{28\times 10^6}{4.74\times 10^5}\\ & \approx 59.1\text{ years}<1000\text{ years}\end{array}$$

Therefore it is not sustainable in its status

L15

a

$$\begin{array}{rl}\Delta m& =2.01410+3.01605-4.00260-1.00866\\ & =0.01889\text{ amu}\\ & \\ E& =\Delta m{c}^2\\ & =(0.01889\times 1.66\times 10^{-27})(3\times 10^8{)}^2\text{ J}\\ & \approx 2.82\times 10^{-12}\text{ J}\\ & =2.82\times 10^{-12}\times \frac{1}{1.6\times 10^{-19}}\times 10^{-6}\text{ MeV}\\ & \approx 17.6\text{ MeV}\end{array}$$

For deuterium-tritium fusion, the energy release per total parent nuclei mass is

$$\frac{17.6}{2.01410+3.01605}\approx 3.50\frac{\text{MeV}}{\text{amu}}$$

U-235 fission releases about 200 MeV of energy, the energy release per total parent nuclei mass is

$$\frac{200}{235.04393}\approx 0.85\frac{\text{MeV}}{\text{amu}}$$

Therefore deuterium-tritium fusion has a higher energy release per total parent nuclei mass

b

1. Lack of natural tritium supply
2. Produces large quantities of radioactive waste
3. Difficult to obtain net energy gain

L16

Denote $p$ as Boeing-747 (plane)
Denote $c$ as sedan (car)

Fuel energy density of Boeing-747 is around 9.6 kWh/L
https://jetpackaviation.com/the-state-of-battery-technology/

Fuel consumption of Boeing-747 is around 12 L/km
https://lovethemaldives.com/faq/how-much-fuel-does-a-plane-use-per-km#:~:text=The%20747%20burns%20approximately%205,liters%20of%20fuel%20per%20kilometer

Passenger capacity of Boeing-747 is around 400 people
https://en.wikipedia.org/wiki/Boeing_747

The transportation efficiency of Boeing-747 is

$$e_p=9.6\times 12\times \frac{1}{400}=0.288\frac{\text{kWh}}{\text{p-km}}$$

Fuel energy density of sedan is around 34.2 MJ/L = 9.5 kWh/L
https://en.wikipedia.org/wiki/Energy_density

The transportation efficiency of sedan is

$$e_c=9.5\times \frac{4.1}{100}\times \frac{1}{4}=0.097375\frac{\text{kWh}}{\text{p-km}}$$

Since $e_c<e_p$, the sedan is more efficient

$$\begin{array}{rl}9.6\times 12\times \frac{1}{n_p}& =9.5\times \frac{4.1}{100}\times \frac{1}{n_c}\\ \frac{n_p}{n_c}& \approx 295.76\end{array}$$

Their efficiency would be the same if the number of passengers on Boeing-747 is about 295.76 times the one on the sedan

Therefore one scenario could be Boeing-747 has 296 passengers and the sedan has 1 passenger

L17

1. 10 minutes hot shower (40 °C)
2. 10 minutes coffee maker
3. 10 km minibus
4. 5 hours laptop (30 W)
5. 0.6 L boiling water
6. 10 km minibus
7. 10 W led lamp 5 hours
8. 5 hours laptop (30 W)
9. 0.6 L boiling water
10. 10 minutes hot shower (40 °C)

Assume we heat up the water for the shower from 15 °C to 40 °C, and the water flow is 10 L/min
From L17 p.7, the power consumption of a coffee maker is 500 W
From L17 p.6, the energy consumption of a minibus is 0.25 kWh/person-km = 900000 J/person-km
Assume we heat up the water for cup noodles from 20 °C to 100 °C

Take the specific heat capacity of water to be 4184 J kg^-1 K

$$\begin{array}{rlrl}E& =2\times (10\times 10)(4184)(40-15)& & \text{Shower}\\ & +500\times 60\times 10& & \text{Coffee}\\ & +2\times 900000\times 10& & \text{Minibus}\\ & +2\times 30\times 60\times 60\times 5& & \text{Laptop}\\ & +2\times 0.6\times 4184\times (100-20)& & \text{Cup noodles}\\ & +10\times 60\times 60\times 5& & \text{Lamp}\\ & \approx 4.10\times 10^7\text{ J}& & \end{array}$$

L18

From L18 p.10, a typical diet has an embodied energy of around 5 kWh per kWh eaten

If we walk at a speed of 5.63 km/hour, we consume 267 Calories/hour

https://www.brianmac.co.uk/energyexp.htm

$$\begin{array}{rl}{E}_{\text{100km walk}}& =\frac{100}{5.63}\times 267\times \frac{4184}{1000\times 3600}\times 5\\ & \approx 27.56\text{ kWh}\end{array}$$

To produce 6 liters of petrol, 42 kWh of embodied energy is required

https://en.wikipedia.org/wiki/Embodied_energy

The fuel consumption of a passenger car is around 5 L/100km

https://ourworldindata.org/grapher/fuel-efficiency-new-vehicles?tab=table

$$E_{\text{100km drive}}=\frac{5}{6}\times 42=35\text{ kWh}$$

Since $E_{\text{100km walk}}<E_{\text{100km drive}}$, walking is more energy efficient

The fuel energy density is around 9.5 kWh/L

https://ourworldindata.org/grapher/fuel-efficiency-new-vehicles?tab=table

$$E_{\text{100km drive}}=5\times 9.5=47.5\text{ kWh}$$

Since $E_{\text{100km walk}}<E_{\text{100km drive}}$, walking is more energy efficient

L19

a

36.97 TWh was generated in 2021

https://en.wikipedia.org/wiki/Electricity_sector_in_Hong_Kong

Fuel mix is as follows

Natural gas: 48%
Coal: 24%
Nuclear and renewable energy: 28%

https://www.emsd.gov.hk/energyland/en/energy/energy_use/energy_scene.html

Death per energy generated by different sources is as follows

Natural gas: 2.82 deaths/TWh
Coal: 24.62 deaths/TWh
Nuclear and renewable energy: 1.39 deaths/TWh

https://ourworldindata.org/grapher/death-rates-from-energy-production-per-twh

$$\begin{array}{rl}\text{Deaths}& =36.97\times 0.48\times 2.82\\ & +36.97\times 0.24\times 24.62\\ & +36.97\times 0.28\times 1.39\\ & \approx 283\end{array}$$

b

Solar: 0.02 deaths/TWh
Wind: 0.04 deaths/TWh

https://ourworldindata.org/grapher/death-rates-from-energy-production-per-twh

$$\begin{array}{rl}\text{Deaths}& =36.97\times 0.5\times 0.02\\ & +36.97\times 0.5\times 0.04\\ & \approx 1.11\end{array}$$

L20

a

1. Corrosion of buildings’ surface, e.g. limestone and marble surfaces
2. Water contamination in areas like harbor or river

b

1. Taking public transportation to reduce the emission of pollutants
2. Reduce energy consumption, such that less fossil fuels will be consumed for energy generation