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Problem 1 (10 pts) Assume that coal has a sulfur content of 3% by weight. If all the sulfur is converted into SO2 during the combustion process, (a) how much SO₂ is produced per tonne of coal? (b) How much is produced per megajoule of energy produced? The molecular mass of sulfur is 32 gram mole. Sulfur content in 1 to
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1. (25 pts) A solar panel is mounted facing due south in a location at 40° N latitude, with a tilt angle of 38° (south facing). If the sun shines on the panel for the entire day on the spring equinox with clear skies, calculate the total energy incident per square meter on the panel during the course of the day in kWh/m². (Use the direct normal values shown in the table below). Hour 1 2 3 4 5 6 7 8 9 10 11 12 13 15 16 17 18 19 20 21 22 23 24 Normal Irradiance [w/m2] 0 0 0 0 0 130.00 480.00 756.00 875.00 892.00 908.00 915.00 908.00 892.00 875.00 756.00 480.00 130.00 0 0 0 0 0 0 2. (10 pts) A house at 48° N latitude has a roof that faces due south, and is elevated to an angle of 28°. A solar panel is mounted to the roof. What is the angle of incidence between the sun and the array. 1 p.m. on June 1?

Problem 1 (10 pts) Assume that coal has a sulfur content of 3% by weight. If all the sulfur is converted into SO2 during the combustion process, (a) how much SO₂ is produced per tonne of coal? (b) How much is produced per megajoule of energy produced? The molecular mass of sulfur is 32 gram mole. Sulfur content in 1 tonne of coal is given (or may be found from what's given); you need to find out the content of SO2 when I tonne of coal is burned. So, find out the ratio (of molar mass) of S in SO2; from here you can determine the SO₂ content./nProblem 2 (10 pts) A typical gasoline-powered vehicle requires about 3.7 MJ of primary energy (i.e., energy content of the gasoline) to travel 1 km. How many kilograms of CO2 are produced annually by a vehicle that is driven 40,000 km per year? From the energy content of gasoline (octane), you may find out the amount (kg) of gasoline required to drive the vehicle annually. The chemical reaction for burning octane is given by: 2C8H18 + 250216CO2 + 18H20 From here, you can find out how many grams of octane yields what amount of CO2, which may then be used to find the total CO2 produced.

Course work: Design a heat pump to heat the James Watt Building South This course work assignment is part of the final assessment of this course, and it accounts for 25 marks (25% of the total assessment). Objective: The peak heating demand of the James Watt Building South is assumed to be 400 kW. Its heating network system delivers heat by pumping hot water at 80 *C through radiators. The measured river water temperature for most of the heating season is in the range 6-10 °C. After extracting heat, the cold water should be returned to the river with a temperature no less than 4 °C. You are expected to apply the learned thermodynamics knowledge and skills to conduct a conceptual design of a two-stage water source heat pump to extract heat from the water of river Kelvin to heat the James Watt Building South.

1 The temperatures of the hot and cold reservoirs are: = 1050°F = 565°C = 839 K T=554°F = 295°C = 563 K This gives an ideal Carnot efficiency of 100 ×(1-563K/839K) = 33 % Thus the total solar insolation must be 1 GW/0.33 = 3 GW. The installation in Figure 9.4 is in California and from Figure 8.2 the average insolation can be estimated to be 225 W/m. The area needed to produce the necessary power is 1.3 x 10 m. For a circular array this corresponds to an area about 4 km in diameter. This is too large of an area. Why?

3. Problem 10.2 (10 pts) Area per turbine is (3.14) x (5 m) = 78.5 m² per turbine. The power per turbine will be P = 4.08 × 10³ W The average spacing of the wind turbines downwind will be 10 times the rotor diameter or 100 m and the crosswind spacing will be 3 times the rotor diameter or 30 m. Thus each turbine will occupy a land area of (30 m) x (100 m) = 3000 m². Thus, 1 km² square of land area will accommodate (10 m/km)/(3000 m²) = 333 turbines with a total output of 1.36 MW (b) From equation (9.3), P = 30.2 MW. This is more than 20 times the wind power output.