CN110570010B - An energy management method for a distributed system with a heat storage device - Google Patents

Abstract

The invention provides an energy management method of a distributed system containing a heat storage device, which comprises the following steps: distributing electric energy, distributing heat energy, controlling a heat storage device, designing the capacity of the heat storage device and the like; the invention combines the characteristics of electric heating and cooling loads of users, realizes the optimal scheduling of system energy by controlling the storage and release of the heat energy of the distributed system through the heat storage device, improves the control method of the distributed system and the design method of the heat storage device, reduces the heat energy loss of the system, effectively improves the comprehensive energy utilization efficiency, reduces the investment cost and the recovery period of the system, has greater economic advantage and can effectively promote the application of the heat storage technology to the actual distributed system.

Description

Translated fromChinese

一种含储热装置的分布式系统的能量管理方法An energy management method for a distributed system with a heat storage device

技术领域technical field

本发明属于分布式系统领域，特别涉及一种含储热装置的分布式系统的能量管理方法。The invention belongs to the field of distributed systems, in particular to an energy management method of a distributed system including a heat storage device.

背景技术Background technique

分布式系统具有靠近用户、可再生能源接入性好、能源利用效率高等优势，是新能源和可再生能源系统发展的趋势。其中，以微小型燃气机组为供能主体的冷热电联供系统(Combined cooling,heating and power,CCHP)集供电、供热、供冷于一体，通过电冷热三联供的方式可实现能源的梯级利用，综合能源利用率可达70％以上，因此CCHP系统成为分布式系统发展的主流形式。Distributed systems have the advantages of being close to users, good access to renewable energy, and high energy utilization efficiency, which is the development trend of new energy and renewable energy systems. Among them, the combined cooling, heating and power (CCHP) system, which uses micro gas-fired units as the main energy supply, integrates power supply, heating, and cooling. The cascade utilization of integrated energy can reach more than 70%, so the CCHP system has become the mainstream form of distributed system development.

CCHP系统包含多能源输入和多负荷需求，产能和用能之间的不均衡问题较为突出，存在系统效率高但节能率低的普遍现象。储能技术是实现其能量优化调度的关键，而储热作为单位储能容量成本较低的方式，易于推广到实际系统中。但目前含储热装置的分布式系统控制方法未能有效发挥储热系统的作用，导致其利用率较低，无法有效实现系统能量的优化调度；同时储能系统的设计容量冗余较大，经济性较差。因此，设计一种适用于含储热装置的分布式系统能量管理方法及其容量优化匹配方法，对储能技术在分布式系统中的应用推广具有重要的意义。The CCHP system includes multi-energy input and multi-load demand, and the imbalance between production capacity and energy consumption is more prominent, and there is a common phenomenon that the system efficiency is high but the energy saving rate is low. Energy storage technology is the key to realize optimal energy dispatching, and heat storage, as a way of lower cost per unit energy storage capacity, is easy to be extended to practical systems. However, the current distributed system control method including heat storage devices fails to effectively play the role of the heat storage system, resulting in a low utilization rate and the inability to effectively achieve optimal scheduling of system energy; at the same time, the design capacity of the energy storage system is relatively large. The economy is poor. Therefore, it is of great significance to design an energy management method suitable for distributed systems with heat storage devices and its capacity optimization matching method for the application and promotion of energy storage technology in distributed systems.

发明内容Contents of the invention

本发明的目的在于提供一种含储热装置的分布式系统的能量管理方法，以实现分布式系统能量的优化调度，提高系统综合能源利用率和节能率，并减少系统投入成本。The purpose of the present invention is to provide an energy management method for a distributed system including a heat storage device, so as to realize optimal dispatching of distributed system energy, improve system comprehensive energy utilization rate and energy saving rate, and reduce system input cost.

为实现上述目的，本发明采取的技术方案是：For realizing above-mentioned object, the technical scheme that the present invention takes is:

该分布式系统包括：燃气机组、余热回收装置、燃气锅炉、吸收式制冷机、热交换器、储热装置。燃气机组连接燃气管网获取天然气、产生电能并由余热回收装置将余热进行回收；燃气锅炉连接燃气管网获取天然气，产生热能。燃气机组和电网提供电能满足用户电负荷需求；余热回收装置和燃气锅炉提供热能，且通过吸收式制冷机转换为合适的冷能满足用户冷负荷需求、通过热交换器转换为合适的热能满足用户热负荷需求。提供用户的热能可通过储热装置进行调节。The distributed system includes: gas-fired unit, waste heat recovery device, gas-fired boiler, absorption refrigerator, heat exchanger, and heat storage device. The gas-fired unit is connected to the gas pipeline network to obtain natural gas, generate electric energy, and the waste heat is recovered by the waste heat recovery device; the gas-fired boiler is connected to the gas pipeline network to obtain natural gas and generate heat energy. The gas-fired unit and the power grid provide electric energy to meet the user's electrical load demand; the waste heat recovery device and gas-fired boiler provide heat energy, which is converted into appropriate cold energy through the absorption chiller to meet the cooling load demand of the user, and converted into appropriate heat energy through the heat exchanger to meet the user's demand. Heat load demand. The thermal energy supplied to the user can be regulated by means of thermal storage.

其特征在于，包括以下步骤：It is characterized in that, comprising the following steps:

步骤1、电能分配的步骤；Step 1, the steps of electric energy distribution;

步骤2、热能分配的步骤；Step 2, the step of thermal energy distribution;

步骤3、控制储热装置的步骤；Step 3, the step of controlling the heat storage device;

步骤4、设计储热装置容量的步骤；Step 4, the step of designing the capacity of the heat storage device;

所述分布式系统能量管理方法根据用户日逐时电负荷、冷负荷和热负荷，依次确定电能分配和热能分配。The energy management method of the distributed system determines the distribution of electric energy and heat energy in sequence according to the user's daily and hourly electric load, cooling load and thermal load.

所述步骤1具体为：判断用户日逐时电负荷E_load(t)与燃气机组额定电功率E_max的大小，若燃气机组t时段E_load(t)大于E_max，则燃气机组以额定状态运行、输出电功率E_cchp(t)＝E_max，燃气机组t时段E_load(t)超出E_max的电能由电网提供；若燃气机组t时段输出电功率E_cchp(t)＝E_load(t)，则不需要电网提供电能。Thestep 1 is specifically: judging the size of the user's daily hourly electric load E_load (t) and the rated electric power E_max of the gas-fired unit, and if the gas-fired unit E_load (t) is greater than E_max during the period t, the gas-fired unit operates in a rated state , Output electric power E_cchp (t)=E_max , the electric energy of the gas-_fired unit_whose E_load (t) exceeds E_max during the period t is provided by the grid; Does not require electricity from the grid.

所述步骤2具体为：根据上述电能分配的步骤得到的燃气机组逐时输出电功率，计算得到逐时负载率

结合燃气机组电效率得到燃气机组的逐时一次能源消耗量G_cchp(t)，并结合燃气机组热效率和余热回收装置效率，得到逐时可回收利用的热能Q_re(t)，结合吸收式制冷机和热交换器的转换效率，将逐时冷负荷和热负荷转换为等效热能需求，并求和得到用户逐时热能需求Q_load(t)，则t时段，余热回收装置向用户提供的热能为Q_reuse(t)＝min(Q_re(t),Q_load(t))，若余热回收装置向用户提供的热能大于或小于用户逐时热能需求Q_load(t)时，则按照步骤3中的方法分配热能。Thestep 2 is specifically: according to the hourly output electric power of the gas-fired unit obtained in the above steps of electric energy distribution, the hourly load rate is calculated

Combined with the electrical efficiency of the gas-fired unit, the hourly primary energy consumption G_cchp (t) of the gas-fired unit is obtained, and combined with the thermal efficiency of the gas-fired unit and the efficiency of the waste heat recovery device, the hourly recyclable heat energy Q_re (t) is obtained, combined with absorption refrigeration Convert the hourly cooling load and heat load into the equivalent heat energy demand, and get the user’s hourly heat energy demand Q_load (t) by summing up the conversion efficiency of the machine and the heat exchanger. Then, during the t period, the waste heat recovery device provides the user with The heat energy is Q_reuse (t)=min(Q_re (t), Q_load (t)), if the heat energy provided by the waste heat recovery device to the user is greater than or less than the hourly heat energy demand of the user Q_load (t), follow the steps The method in 3 distributes thermal energy.

所述步骤3包括以下步骤：Describedstep 3 comprises the following steps:

步骤a、计算各个时段Q_re(t)与Q_load(t)的差值，得到逐时热差值Q_diff(t)；Step a, calculate the difference between Q_re (t) and Q_load (t) in each period, and obtain the hourly heat difference Q_diff (t);

步骤b、储能阶段；选取所有Q_diff(t)大于0的时段，并按时间顺序进行排序，根据以下方法确定储热装置的储能功率：取逐时热差值Q_diff(t)、储热装置最大功率Q_{max_TESS}、储热装置未存满容量C_TESS-C_TESS(t)三个参数中绝对值的最小值作为储能装置实时功率Q_TESS(t)，将回收热能的富余部分存储到储热装置中；若仍有多余，则多余部分热能释放到周围环境中，若Q_diff(t)＞0，则储能装置的实时功率为：Step b, energy storage stage; select all time periods with Q_diff (t) greater than 0, and sort them in chronological order, and determine the energy storage power of the heat storage device according to the following method: take the hourly heat difference Q_diff (t), The maximum power Q_{max_TESS} of the heat storage device, the minimum absolute value of the three parameters C_TESS -C_TESS (t) of the heat storage device's unfulfilled capacity are used as the real-time power Q_TESS (t) of the energy storage device, and the surplus part of the heat energy will be recovered Stored in the heat storage device; if there is still excess, the excess heat energy is released to the surrounding environment. If Q_diff (t)>0, the real-time power of the energy storage device is:

Q_TESS(t)＝min{|Q_diff(t)|,|Q_{max_TESS}|,|C_TESS-C_TESS(t)|}，Q_TESS (t)=min{|Q_diff (t)|,|Q_{max_TESS} |,|C_TESS -C_TESS (t)|},

其中，C_TESS为储热装置设计的容量；C_TESS(t)为储热装置当前容量，表示截止当前时刻存储的热能总和，并认为在储能阶段的初始时刻为0；同时，将所有Q_diff(t)＞0时段Q_TESS(t)相加，得到储热装置存储的热能总和为Q_sto；Among them, C_TESS is the designed capacity of the heat storage device; C_TESS (t) is the current capacity of the heat storage device, indicating the sum of the thermal energy stored up to the current moment, and it is considered to be 0 at the initial moment of the energy storage stage; at the same time, all Q_diff (t)>0 period Q_TESS (t) is added, and the sum of the thermal energy stored in the heat storage device is Q_sto ;

步骤c、释能阶段；选取所有Q_diff(t)小于0的时段、按时间顺序进行排序，并根据以下方法确定储热装置的释能功率：取逐时热差值Q_diff(t)、储热装置最大功率Q_{max_TESS}、储热装置当前容量C_TESS(t)三个参数中绝对值的最小值、取相反数作为储能装置实时功率Q_TESS(t)，将储热装置中存储的热能进行释放以提供用户热能需求，若仍然未达到用户的热能需求，则用户未被满足的热能需求由燃气锅炉提供，即：若Q_diff(t)＜0，则储能装置实时功率为：Step c, energy release stage; select all time periods with Q_diff (t) less than 0, sort them in chronological order, and determine the energy release power of the heat storage device according to the following method: take the hourly heat difference Q_diff (t), The maximum power Q_{max_TESS} of the heat storage device and the current capacity C_TESS (t) of the heat storage device are the minimum absolute values of the three parameters, and the opposite number is taken as the real-time power Q_TESS (t) of the energy storage device. The heat energy is released to provide the heat energy demand of the user. If the heat energy demand of the user is still not met, the unsatisfied heat energy demand of the user is provided by the gas boiler, that is, if Q_diff (t)<0, the real-time power of the energy storage device is:

Q_TESS(t)＝-min{|Q_diff(t)|,|Q_{max_TESS}|,|C_TESS(t)|}，Q_TESS (t)＝-min{|Q_diff (t)|,|Q_{max_TESS} |,|C_TESS (t)|},

把释能阶段的初始时刻C_TESS(t)设置为Q_sto，其当前值通过上一时刻剩余容量减去上一时刻释能能量得到，同时，将所有Q_diff(t)＜0时段Q_TESS(t)相加并取相反数，得到储热装置释放的热能总和为Q_rel；Set the initial time C_TESS (t) of the energy release phase as Q_sto , and its current value is obtained by subtracting the energy release energy at the previous time from the remaining capacity at the previous time. At the same time, set all Q_diff (t)<0 period Q_TESS (t) add up and take the opposite number to obtain the heat energy summation released by the heat storage device as Q_rel ;

步骤d、比较Q_sto与Q_rel的大小，若Q_sto＞Q_rel，选取所有Q_diff(t)大于0的时段、按时间倒序进行排序，逐时减少储热装置存储的储热功率、并同步减少存储的热能总和Q_sto，当Q_sto＝Q_rel时，执行步骤e；Step d, compare the size of Q_sto and Q_rel , if Q_sto > Q_rel , select all time periods when Q_diff (t) is greater than 0, sort them in reverse order of time, reduce the heat storage power stored in the heat storage device hour by hour, and Synchronously reduce the sum of stored thermal energy Q_sto , when Q_sto =Q_rel , execute step e;

步骤e、获得对储热装置的控制：在Q_diff(t)＜0时段，燃气锅炉的功率由Q_GB(t)＝|Q_diff(t)-Q_TESS(t)|得到；在Q_diff(t)＞0时段，释放到周围环境中的损失热能由Q_loss(t)＝Q_diff(t)-Q_TESS(t)获得。Step e. Obtain the control of the heat storage device: during the period of Q_diff (t)<0, the power of the gas boiler is obtained by Q_GB (t)=|Q_diff (t)-Q_TESS (t)|_; (t)>0 period, the loss heat energy released to the surrounding environment is obtained by Q_loss (t)=Q_diff (t)-Q_TESS (t).

在步骤3中，对储热装置的控制以日为计量周期，储热装置存储和释放的热能相等，即：

Instep 3, the control of the heat storage device takes daily as the metering cycle, and the heat energy stored and released by the heat storage device is equal, that is:

在步骤3中，储热装置容量的设计方法为：在N个典型用户设计日工况下，分别计算各个时段回收热能

与用户热需求

的差值得到逐时热差值

将全部时段的

求和得到总和

将

小于0的时段进行求和得到不足时段的总和

将

大于0的时段进行求和，得到富余时段的总和

对于工况X，若

则：

若

则

X＝1，……，N，储热装置的容量为：Instep 3, the design method for the capacity of the heat storage device is as follows: under the daily working conditions of N typical user designs, calculate the recovered heat energy for each time period

and user thermal requirements

The hourly thermal difference

all time period

sum to get sum

Will

Sum the periods less than 0 to get the sum of the insufficient periods

Will

Sum the periods greater than 0 to get the sum of the surplus periods

For working condition X, if

but:

like

but

X=1,...,N, the capacity of the heat storage device is:

相对于现有技术，本发明具有以下的优点和有益效果：Compared with the prior art, the present invention has the following advantages and beneficial effects:

本发明设计了一种含储热装置的分布式系统的能量管理方法，该能量管理方法通过储热装置能量的存储释放实现系统热能的优化调度，减少了系统耗散到周围环境中的损失热能，有效提高了系统综合能源利用效率；同时，提供了储热装置容量的优化设计方法，减少了系统的初始投资成本，并提高了储热装置的利用率、减少了投资回收期，具有较大的经济优势。The present invention designs an energy management method for a distributed system including a heat storage device. The energy management method realizes the optimal scheduling of system heat energy through the storage and release of heat storage device energy, and reduces the loss of heat energy dissipated by the system into the surrounding environment. , which effectively improves the comprehensive energy utilization efficiency of the system; at the same time, it provides an optimal design method for the capacity of the heat storage device, which reduces the initial investment cost of the system, improves the utilization rate of the heat storage device, and reduces the investment payback period. economic advantage.

附图说明Description of drawings

图1是本发明的整体拓扑结构示意图。FIG. 1 is a schematic diagram of the overall topology of the present invention.

图2是夏季工况的逐时电负荷和等效逐时热需求。Figure 2 shows the hourly electrical load and equivalent hourly heat demand in summer conditions.

图3是冬季工况的逐时电负荷和等效逐时热需求。Figure 3 shows the hourly electrical load and equivalent hourly heat demand in winter conditions.

图4为夏季工况的电能分配图。Figure 4 is the distribution diagram of electric energy in summer working conditions.

图5为冬季工况的电能分配图。Figure 5 is the distribution diagram of electric energy in winter conditions.

图6为夏季工况的储热装置控制图。Figure 6 is the control diagram of the heat storage device in summer working conditions.

图7为冬季工况的储热装置控制图。Figure 7 is a control diagram of the heat storage device in winter conditions.

具体实施方式detailed description

实施例Example

以下结合附图和具体实施方式对本发明的技术方案做进一步的说明。The technical solutions of the present invention will be further described below in conjunction with the drawings and specific embodiments.

如图1所示，一种含储热装置的分布式系统的能量管理方法，该分布式系统包括燃气机组10、余热回收装置20、燃气锅炉30、吸收式制冷机40、热交换器50、储热装置60。燃气机组10连接燃气管网100获取天然气、产生电能并由余热回收装置20将余热进行回收；燃气锅炉30连接燃气管网100获取天然气，产生热能。燃气机组10和电网200提供电能满足用户电负荷需求；余热回收装置20和燃气锅炉30提供热能，且通过吸收式制冷机40转换为合适的冷能满足用户冷负荷需求、通过热交换器50转换为合适的热能满足用户热负荷需求。通过储热装置60的存储释放实现系统热能的调节。As shown in Figure 1, an energy management method for a distributed system including a heat storage device, the distributed system includes a gas-firedunit 10, a wasteheat recovery device 20, a gas-firedboiler 30, anabsorption refrigerator 40, aheat exchanger 50,Heat storage device 60. The gas-firedunit 10 is connected to thegas pipeline network 100 to obtain natural gas and generate electric energy, and the wasteheat recovery device 20 recovers the waste heat; the gas-firedboiler 30 is connected to thegas pipeline network 100 to obtain natural gas and generate heat energy. The gas-firedunit 10 and thepower grid 200 provide electric energy to meet the user's electrical load demand; the wasteheat recovery device 20 and the gas-firedboiler 30 provide heat energy, which is converted into appropriate cold energy by theabsorption refrigerator 40 to meet the user's cooling load demand, and converted by theheat exchanger 50 To meet the user's heat load demand for appropriate heat energy. The adjustment of the thermal energy of the system is realized through the storage and release of theheat storage device 60 .

实施对象包括夏季和冬季两种典型用户工况，图2和图3分别为夏季工况和冬季工况对应的逐时电负荷和等效逐时热需求。其中，等效逐时热需求根据用户的冷负荷和热负荷，结合吸收式制冷机的COP和热交换器的效率计算得到，本实施例中，吸收式制冷机的COP为1.2，热交换器的效率为0.9。The implementation objects include two typical user working conditions in summer and winter. Figure 2 and Figure 3 show the hourly electric load and equivalent hourly heat demand corresponding to the summer working condition and winter working condition respectively. Among them, the equivalent hourly heat demand is calculated according to the cooling load and heating load of the user, combined with the COP of the absorption chiller and the efficiency of the heat exchanger. In this embodiment, the COP of the absorption chiller is 1.2, and the heat exchanger The efficiency is 0.9.

实施对象的燃气机组为内燃机，额定功率67kw，其电效率和热效率可通过美国采暖、制冷与空调工程师学会制定的自然吸气式小型机组标准特性数据拟合得到。The gas-fired unit to be implemented is an internal combustion engine with a rated power of 67kw. Its electrical efficiency and thermal efficiency can be obtained by fitting the standard characteristic data of naturally aspirated small units formulated by the American Society of Heating, Refrigerating and Air-Conditioning Engineers.

根据电能分配的步骤，将24个小时的逐时电负荷E_load(t)与燃气机组额定功率值67kw做比较，得到的燃气机组逐时输出电功率：当E_load(t)＞67，燃气机组输出电功率为67kw，不足部分由电网提供；如夏季工况t＝10的时段，E_load(10)＝82.59kW，则燃气机组以额定状态运行，输出功率为67kW，此时电网向用户提供电能为15.59kW。当E_load(t)≤67，燃气机组输出电功率为E_cchp(t)＝E_load(t)，电网不提供电能；如夏季t＝5的时段，E_load(5)＝40.42kW，则燃气机组输出功率为40.42kW，此时电网不提供电能。图4和图5分别为夏季工况和冬季工况对应的电能分配方法。According to the steps of electric energy distribution, compare the 24-hour hourly electric load E_load (t) with the rated power value of the gas unit 67kw, and obtain the hourly output electric power of the gas unit: when E_load (t) > 67, the gas unit The output electric power is 67kw, and the insufficient part is provided by the power grid; for example, during the period of summer working condition t=10, E_load (10)=82.59kW, the gas-fired unit operates at the rated state, and the output power is 67kW. At this time, the power grid provides electric energy to the user It is 15.59kW. When E_load (t)_≤67 , the output electric power of the gas-fired unit is E_cchp (t)=E_load (t), and the grid does not provide electric energy; The output power of the unit is 40.42kW, and the grid does not provide electric energy at this time. Figure 4 and Figure 5 respectively show the power distribution methods corresponding to summer and winter conditions.

根据上述得到的逐时E_cchp(t)以及

计算得到逐时负载率PLR(t)；结合燃气机组电效率、热效率和余热回收装置效率，得到逐时可回收利用的热能Q_re(t)。如夏季工况t＝10的时段，E_cchp(10)＝67kW且E_max＝67kW，则PLR(10)＝1，此时电效率为0.2641，实时热效率为0.5479，余热回收装置效率为0.8，则Q_re(10)＝111.2kW。结合用户的等效逐时热需求，可知t时段余热回收装置向用户提供的热能为Q_reuse(t)＝min(Q_re(t),Q_load(t))。如夏季工况t＝10的时段，Q_re(10)＝111.2kw且Q_load(10)＝75.74kW，可则Q_reuse(10)＝75.74kW；夏季工况t＝16的时段，Q_re(16)＝111.2kW且Q_load(16)＝130.13kW，可则Q_reuse(16)＝111.2kW。According to the hourly E_cchp (t) obtained above and

The hourly load rate PLR(t) is calculated; combined with the gas unit electrical efficiency, thermal efficiency and waste heat recovery device efficiency, the hourly recoverable heat energy Q_re (t) is obtained. For example, in the period of summer working condition t=10, E_cchp (10)=67kW and E_max =67kW, then PLR(10)=1, at this time, the electrical efficiency is 0.2641, the real-time thermal efficiency is 0.5479, and the efficiency of the waste heat recovery device is 0.8, Then Q_re (10) = 111.2kW. Combined with the user's equivalent hourly heat demand, it can be known that the heat energy provided by the waste heat recovery device to the user during the period t is Q_reuse (t) = min(Q_re (t), Q_load (t)). For example, during the period of summer working condition t=10, Q_re (10)=111.2kw and Q_load (10)=75.74kW, then Q_reuse (10)=75.74kW; during the period of summer working condition t=16, Q_re (16)=111.2kW and Q_load (16)=130.13kW, then Q_reuse (16)=111.2kW.

该分布式系统的储热装置容量设计如下：分别计算夏季工况和冬季工况24个时段回收热能Q_re(t)与用户热需求Q_load(t)的差值得到逐时热差值Q_diff(t)。分别将夏季工况和冬季工况全部时段的Q_diff(t)求和得到总和Q_Tdiff，可知夏季工况Q_Tdiff＝292.27kWh，冬季工况Q_Tdiff＝-1430.68kWh。若Q_Tdiff＞0，则将Q_diff(t)小于0的时段进行求和得到不足时段的总和Q_insuf，并取绝对值作为预选容量值；如夏季工况大于0，计算得到不足时段的总和为-266kWh，则将266kWh作为预选容量值。若Q_Tdiff＜0，则将Q_diff(t)大于0的时段进行求和得到富余时段的总和Q_surplus，并取绝对值作为预选容量值；如冬季工况小于0，计算得到富余时段的总和为57kWh，则将57kWh作为预选容量值。比较所有的预选容量值，其中的最大值即作为储热装置的设计容量，则该储热装置设计容量为266kWh。The heat storage device capacity design of the distributed system is as follows: Calculate the difference between the recovered heat energy Q_re (t) and the user’s heat demand Q_load (t) in 24 periods of summer and winter conditions respectively to obtain the hourly heat difference Q_diff (t). The total Q_Tdiff is obtained by summing the Q_diff (t) in all periods of the summer and winter conditions respectively. It can be known that the summer condition Q_Tdiff =292.27kWh, and the winter condition Q_Tdiff =-1430.68kWh. If Q_Tdiff > 0, then sum the periods of Q_diff (t) less than 0 to obtain the sum Q_insuf of the insufficient period, and take the absolute value as the pre-selected capacity value; if the summer operating condition is greater than 0, calculate the sum of the insufficient period If it is -266kWh, then 266kWh will be used as the pre-selected capacity value. If Q_Tdiff < 0, then sum the periods of Q_diff (t) greater than 0 to obtain the sum Q_surplus of the surplus period, and take the absolute value as the pre-selected capacity value; if the winter condition is less than 0, calculate the sum of the surplus period is 57kWh, then 57kWh will be used as the pre-selected capacity value. Comparing all the pre-selected capacity values, the maximum value is taken as the design capacity of the heat storage device, then the design capacity of the heat storage device is 266kWh.

根据上述余热回收装置向用户提供的热能以及储热装置设计容量，采用以下方法设计储热装置的储能和释能控制方法：According to the heat energy provided by the above-mentioned waste heat recovery device to the user and the design capacity of the heat storage device, the following methods are used to design the energy storage and energy release control method of the heat storage device:

在储能阶段，选取所有Q_diff(t)大于0的时段、按时间顺序进行排序，并根据以下原则确定储热装置的储能功率：取逐时热差值Q_diff(t)、储热装置最大功率Q_{max_TESS}、储热装置未存满容量C_TESS-C_TESS(t)三个参数中绝对值的最小值作为储能装置实时功率Q_TESS(t)，将回收热能的富余部分存储到储热装置中；若仍有多余，则多余部分热能释放到周围环境中。如夏季工况t＝0时段，Q_diff(0)＝27.5，Q_{max_TESS}＝66.5，C_TESS-C_TESS(0)＝266，则Q_TESS(0)＝27.5kW，储热装置存储的热能总和为Q_sto＝27.5kWh；又如夏季工况t＝7时段，Q_diff(7)＝45.49，Q_{max_TESS}＝66.5，C_TESS-C_TESS(7)＝5.31，则Q_TESS(7)＝5.31kW，储热装置存储的热能总和为Q_sto＝266kWh。In the energy storage stage, select all time periods with Q_diff (t) greater than 0, sort them in time order, and determine the energy storage power of the heat storage device according to the following principles: take the hourly heat difference Q_diff (t), heat storage The minimum value of the absolute value among the three parameters of the maximum power Q_{max_TESS} of the device and the unfulfilled capacity of the heat storage device C_TESS -C_TESS (t) is used as the real-time power Q_TESS (t) of the energy storage device, and the surplus part of the recovered heat energy is stored in In the heat storage device; if there is still excess, the excess heat energy is released to the surrounding environment. For example, during summer working condition t=0, Q_diff (0)=27.5, Q_{max_TESS} =66.5, C_TESS -C_TESS (0)=266, then Q_TESS (0)=27.5kW, the total heat energy stored by the heat storage device Q_sto ＝27.5kWh; for another example in summer working condition t=7 period, Q_diff (7)＝45.49, Q_{max_TESS} ＝66.5, C_TESS -C_TESS (7)＝5.31, then Q_TESS (7)＝5.31kW , the total heat energy stored by the heat storage device is Q_sto =266kWh.

在释能阶段，选取所有Q_diff(t)小于0的时段、按时间顺序进行排序，并根据以下原则确定储热装置的释能功率：取逐时热差值Q_diff(t)、储热装置最大功率Q_{max_TESS}、储热装置当前容量C_TESS(t)三个参数中绝对值的最小值、取相反数作为储能装置实时功率Q_TESS(t)，将储热装置中存储的热能进行释放以提供用户热能需求；若仍然不足，则不足部分热能由燃气锅炉提供。如夏季工况t＝12时段，Q_diff(12)＝-17.66，Q_{max_TESS}＝66.5，C_TESS(12)＝266，则Q_TESS(12)＝-27.66kW，储热装置释放的热能总和为Q_rel＝27.66kWh；又如冬季工况t＝19时段，Q_diff(19)＝-80.49，Q_{max_TESS}＝66.5，C_TESS(19)＝0，则Q_TESS(19)＝0kW，储热装置释放的热能总和为Q_rel＝57.11kWh。In the energy release stage, select all time periods with Q_diff (t) less than 0, sort them in chronological order, and determine the energy release power of the heat storage device according to the following principles: take the hourly heat difference Q_diff (t), heat storage The maximum power Q_{max_TESS} of the device, the minimum absolute value of the three parameters of the current capacity C_TESS (t) of the heat storage device, and the opposite number is taken as the real-time power Q_TESS (t) of the energy storage device, and the thermal energy stored in the heat storage device is Released to meet the user's heat energy demand; if it is still insufficient, the insufficient heat energy will be provided by the gas boiler. For example, in the summer working condition t=12 period, Q_diff (12)=-17.66, Q_{max_TESS} =66.5, C_TESS (12)=266, then Q_TESS (12)=-27.66kW, the total heat energy released by the heat storage device is Q_rel ＝27.66kWh; another example is the winter working condition t＝19 period, Q_diff (19)＝-80.49, Q_{max_TESS} ＝66.5, C_TESS (19)＝0, then Q_TESS (19)＝0kW, heat storage device The sum of released thermal energy is Q_rel =57.11 kWh.

在储热装置存储和释放平衡校验阶段，比较Q_sto与Q_rel的大小，若Q_sto＞Q_rel，选取所有Q_diff(t)大于0的时段、按时间倒序进行排序，逐时减少储热装置存储的储热功率、并同步减少存储的热能总和Q_sto，直至Q_sto＝Q_rel时跳出该步骤。In the storage and release balance verification stage of the heat storage device, compare the sizes of Q_sto and Q_rel . If Q_sto > Q_rel , select all periods when Q_diff (t) is greater than 0, sort them in reverse order of time, and reduce the storage time by time. The heat storage power stored by the thermal device, and the sum of stored heat energy Q_sto is reduced synchronously until Q_sto =Q_rel and this step is skipped.

根据上述步骤，可以得到储热装置的控制方法，图6和图7分别为夏季工况和冬季工况对应的储热装置控制方法。则在Q_diff(t)＜0时段，燃气锅炉的功率可由Q_GB(t)＝|Q_diff(t)-Q_TESS(t)|得到；如冬季工况t＝19时段，燃气锅炉提供的热能为Q_GB(19)＝80.49kW。在Q_diff(t)＞0时段，释放到周围环境中的损失热能可由Q_loss(t)＝Q_diff(t)-Q_TESS(t)得到；如夏季工况t＝7时段，系统有40.18kWh的多余热能释放到了周围环境中。According to the above steps, the control method of the heat storage device can be obtained, and Fig. 6 and Fig. 7 respectively show the control methods of the heat storage device corresponding to the summer working condition and the winter working condition. Then in the period of Q_diff (t)<0, the power of the gas boiler can be obtained by Q_GB (t)=|Q_diff (t)-Q_TESS (t)| The heat energy is Q_GB (19) = 80.49kW. In the period of Q_diff (t) > 0, the loss heat energy released to the surrounding environment can be obtained by Q_loss (t) = Q_diff (t) - Q_TESS (t); for example, in the summer operating condition t = 7 period, the system has 40.18 kWh of excess heat energy is released into the surrounding environment.

上列详细说明是针对本发明可行实施例的具体说明，该实施例并非用以限制本发明的专利范围，凡未脱离本发明所为的等效实施或变更，均应包含于本案的专利范围中。The above detailed description is a specific description of the feasible embodiment of the present invention. This embodiment is not used to limit the patent scope of the present invention. Any equivalent implementation or change that does not deviate from the present invention should be included in the patent scope of this case. middle.

Claims (3)

Translated fromChinese

1.一种含储热装置的分布式系统的能量管理方法，基于分布式系统，所述分布式系统包括燃气机组[10]、余热回收装置[20]、燃气锅炉[30]、吸收式制冷机[40]、热交换器[50]和储热装置[60]；所述燃气机组[10]连接燃气管网[100]以获取天然气和产生电能，并由余热回收装置[20]将余热进行回收；燃气锅炉[30]连接燃气管网[100]以获取天然气，产生热能；燃气机组[10]和电网[200]提供满足用户电负荷需求的电能；余热回收装置[20]和燃气锅炉[30]用于提供热能，所述吸收式制冷机[40]用于转换合适的冷能满足用户冷负荷需求，所述热交换器[50]用于转换合适的热能满足用户热负荷需求，提供给用户的热能通过储热装置[60]进行调节；其特征在于，包括以下步骤：1. An energy management method for a distributed system with a heat storage device, based on a distributed system, the distributed system includes a gas-fired unit [10], a waste heat recovery device [20], a gas-fired boiler [30], an absorption refrigeration machine [40], heat exchanger [50] and heat storage device [60]; the gas unit [10] is connected to the gas pipeline network [100] to obtain natural gas and generate electric energy, and the waste heat recovery device [20] converts the waste heat recovery; the gas boiler [30] is connected to the gas pipeline network [100] to obtain natural gas and generate heat energy; the gas unit [10] and the power grid [200] provide electric energy to meet the user's electrical load demand; the waste heat recovery device [20] and the gas boiler [30] is used to provide thermal energy, the absorption refrigerating machine [40] is used to convert appropriate cold energy to meet user cooling load requirements, and the heat exchanger [50] is used to convert appropriate thermal energy to meet user thermal load requirements, The heat energy provided to the user is regulated through the heat storage device [60]; it is characterized in that it includes the following steps:步骤1、电能分配的步骤；Step 1, the steps of electric energy distribution;步骤2、热能分配的步骤；Step 2, the step of thermal energy distribution;步骤3、控制储热装置的步骤；Step 3, the step of controlling the heat storage device;步骤4、设计储热装置容量的步骤；Step 4, the step of designing the capacity of the heat storage device;所述步骤1具体为：判断用户日逐时电负荷E_load(t)与燃气机组额定电功率E_max的大小，在t时段E_load(t)大于E_max，则燃气机组以额定状态运行，输出电功率E_cchp(t)＝E_max，且t时段E_load(t)超出E_max的电能由电网提供；The step 1 is specifically: judging the size of the user's daily hourly electric loadE_load(t) and the rated electric powerE_max of the gas-fired unit, and ifE_load(t) is greater thanE_max during the period t, the gas-fired unit operates in a rated state and outputs The electric powerE_cchp(t)=E_max , and the electric energy thatE_load(t) exceedsE_max during the period t is provided by the grid;所述步骤2具体为：根据上述电能分配的步骤得到的燃气机组逐时输出电功率，计算得到逐时负载率

，结合燃气机组电效率得到燃气机组的逐时一次能源消耗量G_cchp(t)，并结合燃气机组热效率和余热回收装置效率，得到逐时可回收利用的热能Q_re(t)，结合吸收式制冷机和热交换器的转换效率，将逐时冷负荷和热负荷转换为等效热能需求，并求和得到用户逐时热能需求Q_load(t)，则t时段，余热回收装置向用户提供的热能为Q_reuse(t)＝min(Q_re(t),Q_load(t))，若余热回收装置向用户提供的热能大于或小于用户逐时热能需求Q_load(t)时，则按照步骤3中的方法分配热能；The step 2 is specifically: according to the hourly output electric power of the gas-fired unit obtained in the above steps of electric energy distribution, the hourly load rate is calculated

, combined with the electrical efficiency of the gas-fired unit to obtain the hourly primary energy consumptionG_cchp(t) of the gas-fired unit, and combined with the thermal efficiency of the gas-fired unit and the efficiency of the waste heat recovery device, the hourly recyclable heat energyQ_re(t) is obtained, combined with the absorption The conversion efficiency of the refrigerator and the heat exchanger converts the hourly cooling load and heat load into the equivalent heat energy demand, and sums up the user’s hourly heat energy demandQ_load(t) , then the waste heat recovery device provides the user with The heat energy isQ_reuse(t)＝min(Q_re(t), Q_load(t)) , if the heat energy provided by the waste heat recovery device to the user is greater than or less than the user’s hourly heat energy demandQ_load(t) , then according to The method in step 3 distributes thermal energy;所述步骤3包括以下步骤：Described step 3 comprises the following steps:步骤a、计算各个时段Q_re(t)与Q_load(t)的差值，得到逐时热差值Q_diff(t)；Step a, calculate the difference between each periodQ_re(t) andQ_load(t) , obtain the hourly heat differenceQ_diff(t) ;步骤b、储能阶段；选取所有Q_diff(t)大于0的时段，并按时间顺序进行排序，根据以下方法确定储热装置的储能功率：取逐时热差值Q_diff(t)、储热装置最大功率Q_{max_TESS}、储热装置未存满容量C_TESS-C_TESS(t)三个参数中绝对值的最小值作为储能装置实时功率Q_TESS(t)，将回收热能的富余部分存储到储热装置中；若储热装置存满后仍有多余，则多余部分热能释放到周围环境中，若Q_diff(t)＞0，则储能装置的实时功率为：Step b, energy storage stage; select all time periods withQ_diff(t) greater than 0, sort them in chronological order, and determine the energy storage power of the heat storage device according to the following method: take the hourly heat differenceQ_diff(t) , The maximum powerQ_{max_TESS} of the heat storage device, the minimum absolute value of the three parametersC_TESS-C_TESS(t) of the unfulfilled capacity of the heat storage device are used as the real-time powerQ_TESS(t) of the energy storage device, and the surplus part of the heat energy will be recovered Stored in the heat storage device; if the heat storage device is full and there is still excess heat, then the excess heat energy is released to the surrounding environment. IfQ_diff(t) > 0, the real-time power of the energy storage device is:Q_TESS(t)＝min{|Q_diff(t)|,|Q_{max_TESS}|,|C_TESS-C_TESS(t)|}，Q_TESS(t) =min {|Q_diff(t)|,|Q_{max_TESS}|,|C_TESS-C_TESS(t)| },其中，C_TESS为储热装置设计的容量；C_TESS(t)为储热装置当前容量，表示截止当前时刻存储的热能总和，并认为在储能阶段的初始时刻为0；同时，将所有Q_diff(t)＞0时段Q_TESS(t)相加，得到储热装置存储的热能总和为Q_sto；Among them,C_TESS is the designed capacity of the heat storage device;C_TESS(t) is the current capacity of the heat storage device, indicating the sum of the thermal energy stored up to the current moment, and it is considered to be 0 at the initial moment of the energy storage stage; at the same time, allQ_diff(t) > 0 periodQ_TESS(t) is added, and the sum of the thermal energy stored in the heat storage device isQ_sto ;步骤c、释能阶段；选取所有Q_diff(t)小于0的时段、按时间顺序进行排序，并根据以下方法确定储热装置的释能功率：取逐时热差值Q_diff(t)，储热装置最大功率Q_{max_TESS}、储热装置当前容量C_TESS(t)三个参数中绝对值的最小值，取该最小值的相反数作为储能装置实时功率Q_TESS(t)，将储热装置中存储的热能进行释放以提供用户热能需求，若仍然未达到用户的热能需求，则用户未被满足的热能需求由燃气锅炉提供，即：若Q_diff(t)＜0，则储能装置实时功率为：Step c, energy release stage; select all time periods withQ_diff(t) less than 0, sort them in chronological order, and determine the energy release power of the heat storage device according to the following method: take the hourly heat differenceQ_diff(t) , The minimum value of the absolute value among the three parameters of the maximum powerQ_{max_TESS} of the heat storage device and the current capacityC_TESS(t) of the heat storage device, the inverse of the minimum value is taken as the real-time powerQ_TESS(t) of the energy storage device, and the heat storage The thermal energy stored in the device is released to provide the thermal energy demand of the user. If the thermal energy demand of the user is still not met, the unsatisfied thermal energy demand of the user is provided by the gas boiler, that is: ifQ_diff(t) <0, the energy storage device The real-time power is:Q_TESS(t)＝-min{|Q_diff(t)|,|Q_{max_TESS}|,|C_TESS(t)|}，Q_TESS(t) = -min {|Q_diff(t) |,|Q_{max_TESS} |,|C_TESS(t) |},把释能阶段的初始时刻C_TESS(t)设置为Q_sto，则Q_sto当前值通过上一时刻剩余容量减去上一时刻释能能量得到，同时，将所有Q_diff(t)＜0时段Q_TESS(t)相加并取相反数，得到储热装置释放的热能总和为Q_rel；Set the initial timeC_TESS(t) of the energy release phase asQ_sto , then the current value ofQ_sto is obtained by subtracting the energy released at the previous time from the remaining capacity at the previous time, and at the same time, set allQ_diff(t) <0 periods Add Q_TESS(t) and take the inverse number to obtain the sum of the heat energy released by the heat storage device asQ_rel ;步骤d、比较Q_sto与Q_rel的大小，若Q_sto＞Q_rel，选取所有Q_diff(t)大于0的时段，按时间倒序进行排序，逐时减少储热装置存储的储热功率，并同步减少存储的热能总和Q_sto，当Q_sto＝Q_rel时，执行步骤e；Step d, compare the size ofQ_sto andQ_rel , ifQ_sto >Q_rel , select all time periods withQ_diff(t) greater than 0, sort them in reverse chronological order, reduce the heat storage power stored in the heat storage device hour by hour, and Synchronously reduce the stored thermal energy sumQ_sto , whenQ_sto =Q_rel , execute step e;步骤e、获得对储热装置的控制：在Q_diff(t)＜0时段，燃气锅炉的功率由Q_GB(t)＝|Q_diff(t)-Q_TESS(t)|得到；在Q_diff(t)＞0时段，释放到周围环境中的损失热能由Q_loss(t)＝Q_diff(t)-Q_TESS(t)获得。Step e, obtain the control of the heat storage device: during the period ofQ_diff(t) <0, the power of the gas boiler is obtained byQ_GB(t) = |Q_diff(t) -Q_TESS(t) |; inQ_diff(t) >0 period, the loss heat energy released to the surrounding environment is obtained byQ_loss(t) =Q_diff(t) -Q_TESS(t) .2.根据权利要求1所述的能量管理方法，其特征在于，在步骤3中，对储热装置的控制以日为计量周期，储热装置存储和释放的热能相等，即：2. The energy management method according to claim 1, characterized in that, in step 3, the control of the heat storage device takes daily as the metering cycle, and the heat energy stored and released by the heat storage device is equal, that is:

=0。

=0.3.根据权利要求1所述的能量管理方法，其特征在于，在步骤3中，储热装置容量的设计方法为：在N个典型用户设计日工况下，分别计算各个时段回收热能Q^X_re(t)与用户热需求Q^X_load(t)的差值得到逐时热差值Q^X_diff(t)，将全部时段的Q^X_diff(t)求和得到总和热差值Q^X T_diff(t)，将Q^X_diff(t)小于0的时段进行求和得到热能不足的时段的总和热差值Q^X_insuf，将Q^X_diff(t)大于0的时段进行求和，得到热能富余的时段的总和热差值Q^X_surplus，对于工况X，若

＞0则

；若

＜0，则

，X=1,...,N，储热装置的容量为：3. The energy management method according to claim 1, characterized in that, in step 3, the design method for the capacity of the heat storage device is: under the daily working conditions of N typical user designs, respectively calculate the recovered heat energyQ^X in each time period The difference between_re(t) and the user's heat demandQ^X_load(t) is obtained as the hourly heat differenceQ^X_diff(t) , and the sum ofQ^X_diff(t ) in all periods is obtained to obtain the total heat differenceQ^XT_diff(t) , sum the period whenQ^X_diff(t) is less than 0 to get the total heat differenceQ^X_insuf of the period when the thermal energy is insufficient, and sum the period whenQ^X_diff(t) is greater than 0 to get the thermal energy The sum heat differenceQ^X_surplus of the surplus period, for the working condition X, if

＞0

;like

<0, then

, X=1,...,N, the capacity of the heat storage device is:

。

.

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