Retrieved from Ĭalculate the amount of primary coolant, which is required to evaporate 1 kg of feedwater in a typical steam generator. Thermodynamic Properties of Water, NISTIR 5078. The properties are also tabulated for single-phase states ( compressed water or superheated steam) on a grid of temperatures and pressures extending to 2000 ✬ and 1000 MPa.įurther comprehensive authoritative data can be found at the NIST Webbook page on thermophysical properties of fluids. In these tables the basic and key properties, such as pressure, temperature, enthalpy, density and specific heat, are tabulated along the vapor-liquid saturation curve as a function of both temperature and pressure. Their properties are tabulated in so called “ Steam Tables”. Water and steam are a common medium because their properties are very well known. Steam Tables – common parameters in energy systems
In almost all thermal power stations (coal, gas, nuclear), water is used as the working fluid (used in a closed loop between boiler, steam turbine and condenser), and the coolant (used to exchange the waste heat to a water body or carry it away by evaporation in a cooling tower). Water and steam also reacts with metals commonly found in industries such as steel and copper that are oxidized faster by untreated water and steam. It is especially effective to transport heat through vaporization and condensation of water because of its very large latent heat of vaporization.Ī disadvantage is that water moderated reactors have to use high pressure primary circuit in order to keep water in liquid state and in order to achieve sufficient thermodynamic efficiency. It used due to its availability and high heat capacity, both for cooling and heating. Assume the ideal gas law.Water and steam are a common fluid used for heat exchange in the primary circuit (from surface of fuel rods to the coolant flow) and in the secondary circuit. HFC134a (MW=102) is compressed in an adiabatic compressor from 244K and saturated vapor to 316K and 0.9856MPa. HFC134a is to be considered as the working fluid in a prospective refrigeration system. (e) Estimate the compressor's efficiency.Ģ. (d) Estimate the reversible work required for this compressor. (c) Write the entropy balance required to estimate the efficiency of the compressor. (b) Estimate the actual work required for this compressor.
(a) Write the energy balance for the compressor. Hint: what temperature is implied by the pressure of 0.148MPa for the "evaporator." (cf. The adiabatic compressor's actual exit temperature is 120☏. The evaporator coils operate at 0.148MPa. The process fluid is to be propane with a compression/expansion ratio (ie. An ordinary vapor compression cycle (OVC) is to be considered for cryogenic cooling. Re-watching this video after you have solved the problem using the chart will help you to understand a lot about the influences of molecular interactions and their significance in accounting for the work that goes into designing a chemical engineering process.ġ. We solve the same problem later using more accurate property estimates. In this video, we use the ideal gas assumption. This is enough information to compute the efficiency of the compressor. Compressor efficiency using an ideal gas assumption (, 13min) Propane is compressed from -100F and 1 bar to 180F and 10 bar.
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