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Words 950

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An introduction to the absorption refrigeration cycle

Absorption cycle versus vapour-compression cycle

There are two fundamental differences between the absorption refrigeration cycle and the vapour-compression refrigeration cycle with which we are more familiar. The first difference is that the compressor is replaced by an absorber, a pump and a generator. The second is that, in addition to the refrigerant, the absorption refrigeration cycle uses a secondary fluid, called the absorbent. The condenser, expansion device, and evaporator sections, however, are the same.

Refrigerant solutions and the absorbent, are mixed inside the chiller in various concentrations. The term dilute solution refers to a mixture that has a relatively high refrigerant content and low absorbent content. A concentrated solution has a relatively low refrigerant content and high absorbent content. An intermediate solution is a mixture of dilute and concentrated solutions.

The absorption refrigeration cycle

The four basic components of the absorption refrigeration cycle are the generator and condenser on the high-pressure side, and the evaporator and absorber on the low-pressure side. The pressure on the high-pressure side of the system is approximately ten times greater than that on the low-pressure side.

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Starting on the high-pressure side of the cycle, the purpose of the generator is to deliver the refrigerant vapour to the rest of the system. It accomplishes this by separating the water (refrigerant) from the lithium bromide-and-water solution.

In the generator, a high-temperature energy source, typically steam or hot water, flows through tubes that are immersed in a dilute solution of refrigerant and...

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...Chemistry Lab Chemistry Lab Aim: To investigate the specific heat capacity of water, copper and aluminium by using a caloriometer. Results: 1. Ohms Law: I=v/r, equation 3 = E=vlt hence you get E= v2t/r when ohms law is applied to equation 3. 2. De ionized water is used as it is the purest form of water, as it doesn’t contain ions from the soil like normal water which allow the conduction of electricity as normal water allows electrons to flow, thus preventing any extraneous variables to affect the experiment. 3. The value for Tf = 100o as water has a boiling point of 100o 4. SH = 1.00cal.g-1.c-1 m= density x volume (Tf – Ti) = 100 – 19.9 = 1 x 1,000ml = 80.1 = 1,000g EH2O = SH.m.(Tf – Ti) = 1 x 1,000 x 80.1 = 80,100cal Converting to Joules (1cal = 4.184j) 1000*4.184* 80.1= 335,138.4 joules g-1 deg-1 EKettle= V2.t / R V= 240 t= 219s R= 34.7 2402 * 219/34.7 =363,527.38 joules = E of kettle 5. The amount calories required to raise 1 gram of water by 1 degree Celsius is specific heat. Hence 80,100 calories are required to heat up the water, this shows the association between question 4. 6. In question 4, it can be observed that the energy produced by the kettle was 363,527.38 joules, however only = 335,138.4 joules g-1 deg-1 of energy was required to raise the water 80.1 degrees Celsius.......

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...Data/Results Specific Heat Capacity of a Metal (Copper) | Trial 1 | Trial 2 | Mass of metal (g) | 11.6813 g | 12.1214 g | Mass of empty calorimeter (g) | 4.7507g | 4.7507 g | Mass of calorimeter & water (g) | 24.2425 g | 24.2422 g | Mass of water (g) | 24.2425-4.7507=19.4918 g | 24.2422-4.7507=19.4915 g | Volume added to calorimeter (mL) | 20.0 mL | 20.0 mL | 1 | Time (s) | 0 | 30 | 60 | 90 | 120 | 150 | 180 | 210 | 240 | 270 | 300 | 330 | 360 | 390 | 420 | 1 | Temp(°C) | 26.1 | 26.2 | 26.3 | 26.3 | 26.2 | 26.2 | 26.2 | 26.9 | 27.1 | 27.4 | 27.2 | 27.2 | 27.1 | 27.1 | 27.1 | | | | | | | | | | | | | | | | | | 2 | Time (s) | 0 | 30 | 60 | 90 | 120 | 150 | 180 | 210 | 240 | 270 | 300 | 330 | 360 | 390 | 420 | 2 | Temp(°C) | 24.0 | 24.1 | 24.2 | 24.2 | 24.2 | 24.2 | 24.2 | 28.0 | 27.9 | 27.8 | 27.7 | 27.7 | 27.7 | 27.6 | 27.6 | Change in temperature of the water (°C) | 1.0°C | 3.6°C | Energy gained by the water (J) | 81.552J | 293.59J | Change in temperature of the metal (°C) | -72.9°C | -72.4°C | Calculated specific heat capacity (J/g°C) | 0.0958 J/g°C | 0.335 J/g°C | Molar mass approximation (g/mol) | 261.02g/mol | 74.73g/mol | Percent error of specific heat capacity (%) | 75.1% | 13.0% | Percent error of molar mass (%) | -310.76% | -17.60% | Enthalpy of Neutralization (Nitric Acid) | Trial 1 | Trial 2 | Volume of acid (mL) | 50.0 mL | 50.0 mL | Concentration of acid (M) | 1.1 M | 1.1 M | Volume of......

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