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Coal equivalent for chilled water and heat transfer oil

2024-05-31View Original

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For a certain project under feasibility study, where the chilled water is supplied at -15°C and the heat transfer oil at 300°C, how should the equivalent amount of standard coal be calculated?
Reply #22024-05-31
To calculate the equivalent standard coal amount for chilled water and heat transfer oil, it is necessary to know their energy consumption coefficients. In China, it is generally calculated in accordance with ** or industry standards. Generally, the equivalent standard coal coefficient for chilled water and the equivalent standard coal coefficient for heat transfer oil can be found in relevant energy consumption standard documents. During calculation, the equivalent amount of standard coal is obtained by multiplying the actual energy consumption by the corresponding standard coal conversion factor. If specific coefficients are not available, it is necessary to refer to standard coefficients for similar energy sources or consult a professional energy assessment agency. .
Reply #32024-06-03
The coefficient for converting chilled water into standard coal can be found in the \"GBT 50441-2016 Standards for Calculating Energy Consumption in Petrochemical Design\", whereas no such coefficient for converting heat transfer oil into standard coal exists in the relevant energy consumption standards.
Reply #42024-06-03
In such cases, one can refer to the equivalent standard coal coefficient for similar energy sources, or consult a professional energy assessment agency to obtain the equivalent standard coal coefficient for heat transfer oil. For chilled water, the equivalent amount of standard coal can be calculated directly using the coefficient for converting to standard coal provided in the \"GBT 50441-2016 Standards for Calculating Energy Consumption in Petrochemical Design\". For heat transfer oil, if exact coefficients cannot be obtained immediately, it may be necessary to use approximate values or make estimates based on specific usage conditions; it is also recommended to continue seeking more accurate data or consult professional organizations directly. .
Reply #52024-06-03
It is quite difficult to determine the unit for chilled water, which is MJ of phase-change cooling capacity, according to the \"GBT 50441-2016 Standards for Energy Consumption Calculation in Petrochemical Design\". Currently, brine or ethylene glycol is generally used for chilled water. There is a lack of data for comparison.
Reply #62024-06-03
Calculating the energy consumption of chilled water is indeed quite complex, especially when media such as brine or ethylene glycol are used. Since the unit specified in \"GBT 50441-2016 Standard for Calculating Energy Consumption in Petrochemical Design\" is MJ of phase-change cooling capacity, it is necessary to convert the actual energy consumption into equivalent energy units for comparison and calculation. Typically, it can be handled through the following steps: 1. Determine the energy efficiency parameters of the chilled water system, including the type of coolant, temperature, flow rate, etc. 2. Estimate the actual energy consumption of the system, namely the electrical power or other forms of energy input, based on the system’s operating parameters. 3. Convert energy consumption to energy units (for example, from kWh to MJ). 4. Use the standard coal equivalent coefficient provided in “GBT 50441-2016” to convert energy into an amount of standard coal equivalent. If the data is missing or not accurate enough, it may be necessary to use specialized software for simulation calculations, or to turn to professional agencies for a more accurate assessment. At the same time, it is also possible to consider comparative analysis with cases of similar projects in order to find relevant application data within the industry. When an accurate calculation is not possible, it is recommended to use conservative estimates or range values to express energy consumption, in order to prevent errors from having a significant impact on project evaluation. .
Reply #72024-06-04
To calculate the equivalent standard coal quantity for chilled water at -15°C and heat transfer oil at 300°C, we need to know the calorific value or energy conversion factor of these two fluids, as well as their respective flow rates or consumption amounts. However, since the question does not provide specific calorific value or flow rate data, we can only offer a general calculation method. 1. Chilled water (-15°C): Chilled water is commonly used in cooling systems, and its “equivalent amount of standard coal” depends on the heat it removes. To calculate this amount of heat, we need to know the flow rate of the chilled water (usually measured in m³/h or L/s) and the temperature difference (i.e., the increase in temperature of the chilled water from -15°C to a certain temperature; a common return water temperature is 7°C). The formula for calculating heat (Q) is: Q = m × c × ΔT, where m is the mass flow rate of the chilled water (kg/s), which needs to be obtained by converting the volume flow rate and density. c is the specific heat capacity of water, approximately 4.18 kJ/(kg·℃). ΔT is the temperature difference (°C). After obtaining the heat quantity Q, we need to know the amount of standard coal corresponding to each unit of heat (this is usually an empirical value or a conversion factor specified by energy policies). For example, if 1 kWh (i.e., 3.6 MJ) of heat is equivalent to 0.1229 kg of standard coal, then we can convert Q to kWh and multiply it by this conversion factor to obtain the amount in terms of standard coal. 2. Heat transfer oil (300°C): Heat transfer oil is commonly used in high-temperature heat transfer systems. Similar to chilled water, its “equivalent standard coal amount” also depends on the heat it transfers. To calculate this amount of heat, we need to know the flow rate of the heat transfer oil, its inlet and outlet temperatures, as well as its specific heat capacity (which may vary depending on the type of heat transfer oil). The formula for calculating heat (Q) is the same as that for chilled water: Q = m × c × ΔT. Here, m represents the mass flow rate of the heat transfer oil, c is its specific heat capacity, and ΔT is the temperature difference between the inlet and outlet. Similarly, after obtaining the heat quantity Q, we need to use the appropriate conversion factor to convert it into the amount of standard coal.
Reply #82024-06-04
Thank you. This is just a rough estimate; it’ll have to do

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