河南郑州空压机知名厂

I. principle and structure of air source heat pump unit 

 

空气源热泵冷暖机组系统概述空气源热泵，除具备制取出采暖用热水的功能外，空气源热泵机组还能切换到制冷工况制取冷冻水。空气源热泵的基本原理是基于压缩式制冷循环，利用冷媒做为载体，通过风机的强制换热，从大气中吸取热量或者排放热量，以达到制冷或者制热的需求。

The air source heat pump system is an overview of the air source heat pump. Besides the function of making the heating and heating water, the air source heat pump unit can also switch to the refrigeration condition to make the frozen water. Air source heat pump is based on the basic principle of compression refrigeration circulation, cold media as a carrier, through the fan of forced heat transfer, emissions from the atmosphere absorb heat or heat, to achieve cooling or heating requirements. 

 

 

 

按照逆卡诺循环原理，该系统主要空气源热泵主机和末端两大部分组成。空气源热泵机组与末端共同使用，前者提供冷水或热水，后者将冷水或热水，通过热交换，提供冷气或采暖。空气源热泵机组是采暖系统中的主机，由于采用空气源冷凝器不需要冷却塔；而蒸发器是水冷的，夏天制冷时提供冷水，冬季制热时提供热水，风机盘管是空调系统的末端装置，装在室内如同把水从低处提升到高处而采用水泵那样，采用热泵可以把热量从低温抽吸到高温。所以热泵实质上是一种热量提升装置，热泵的作用是从周围环境中吸取热量，并把它传递给被加热的对象（温度较高的物体）。

According to the inverse carnot cycle principle, the main air source heat pump host and the terminal two main components. The air source heat pump unit is used in joint use with the end. The former provides cold water or hot water, the latter provides cold water or hot water, and provides cold air or heating through heat exchange. The air source heat pump unit is the host in the heating system, because the air source condenser does not need the cooling tower; And evaporator is water-cooled, provide cold water when refrigeration in summer, winter heating with hot water, fan coil air conditioning system is at the end of the device, installed indoors as from lower to higher ground and use the water pump, the heat pump can take heat from low temperature pump to high temperatures. So the heat pump is essentially a heat lifting device that pumps heat from the surrounding environment and passes it to the heated object (the higher temperature object). 

 

 

二、产品结构：

Ii. Product structure: 

 

空气源热泵顶出风、侧出风结构

The air source heat pump has a wind and lateral wind structure 

 

一

one 

 

空调负荷计算

Air conditioning load calculation 

 

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hebnt2015 

 

1.空调负荷计算的组成(QL)

1. Composition of air-conditioning load calculation (QL) 

 

(1)由于室内外温差和太阳辐射作用，通过建筑物围护结构传入室内的热量形成的冷负荷；

(1) due to the temperature difference and solar radiation effect of the indoor and outdoor temperature, the cooling load of heat formed in the indoor heat is introduced through the building envelope. 

 

(2)人体散热、散湿形成的冷负荷；

(2) cold load of heat dissipation and moisture formation of human body; 

 

(3)灯光照明散热形成的冷负荷；

(3) the cooling load formed by the heat dissipation of the lamplight; 

 

(4)其他设备散热形成的冷负荷；

(4) cooling load of other equipment cooling; 

 

(5)渗透空气所形成的冷负荷

(5) the cooling load formed by penetrating air 

 

(6)新风量负荷

(6) new air volume load 

 

2.空调负荷计算方法简单介绍

2. Simple introduction of air conditioning load calculation method 

 

空调动态负荷的计算显得比较繁琐，即便是采用一些简化手段，计算工作量也是比较大的。估算最简便，捷径行路，人之通性，慢慢的被它取而代之了。

The calculation of dynamic load of air conditioning is cumbersome, even with some simplifying means, the calculation workload is relatively large. The easiest way to estimate, the shortcut, the human connectedness, slowly replaced by it. 

 

但是估算的根据并不坚定，偏于保守是不可避免的，总是顾虑怕估算的小了，这也是可以理解的。估算法也要注意与实际相符合，要根据实际的经验以及不同建筑的各自不同的情况。目前空调负荷的计算还是以估算为主。

But it is understandable that the estimates are based on infirmness, which is inevitable, and always scruple to be calculated. The estimation method should also be paid attention to the actual situation, according to the actual experience and the different conditions of different buildings. The current calculation of air conditioning load is based on estimation. 

 

3.民用建筑空调单位面积冷负荷（qL）

3. Cold load (qL) for residential air conditioning unit area 

 

 

4.负荷计算——单位面积冷负荷法

4. Load calculation -- unit area cooling load method 

 

QL=qL×S

QL = QL * S 

 

式中：QL——建筑物空调房间总冷负荷 (W)

In the formula: QL -- total cold load (W) of air conditioning room of buildings 

 

QL—— 冷负荷 (W/m2 )

QL - cold load (W/m2) 

 

S—— 空调房间面积 (m2)

S -- air conditioning room area (m2) 

 

二

two 

 

末端（风盘）计算选择

End (winddisk) calculation selection 

 

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hebnt2015 

 

（1）根据风量：

(1) according to the air volume: 

 

房间面积、层高（吊顶后）和房间气体循环次数三者的乘积即为房间的循环风量。其对应的风机盘管高速风量，即可确定风机盘管型号。

The product of room area, layer height (after ceiling) and room gas cycle is the circulation air volume of the room. Its corresponding fan coil high-speed air volume can determine the fan coil model. 

 

（2）根据冷负荷：

(2) according to the cold load: 

 

根据单位面积负荷和房间面积，可得到房间所需的冷负荷值。利用房间冷负荷对应风机盘管的中速风量时的制冷量即可确定风机盘管型号一般采用第二种方法——根据冷负荷选择风机盘管，在特殊场合如对噪音要求较高的场所，可用（暖通风向标 ID:hebnt2015）第一种方法进行校核。

According to the unit area load and the room area, can get the cold load value of the room. Using the corresponding fan coil cooling load of the refrigerating capacity of the medium speed airflow when determined fan coil models generally adopt the second method, according to choose fan coil cooling load, higher requirements on special occasions such as the noise of the place, available (hvac vane ID: hebnt2015) the first method for checking. 

 

确定型号以后，还需确定风机盘管的安装方式（明装或安装），送回风方式（底送底回，侧送底回等）以及水管连接位置（左或右）等条件。

Determine model, still need to determine the installation way of fan coil units (surface mounted or installation), back to the way of the wind (bottom to send back, side to send the bottom back, etc.) and the pipe connection position (left or right), and other conditions. 

 

房间面积较大时应考虑使用多个风机盘管，房间单位面积负荷较大，对噪音要求不高时可考虑使用风量和制冷量较大的风机盘管。注意：对于风管超过一定长度的风盘，应采用中、高静压的风盘，且出风管道上不宜多于两个出风口。

When the room area is larger, should consider the use of more than one fan coil, the room unit area load is bigger, can consider to use the air quantity and the fan coil that makes a large amount of refrigerant when the noise requirement is not high. Note: for wind disks that have more than a certain length of air pipe, the wind disk should be used in the middle and high static pressure, and there should be no more than two air outlets on the air duct. 

 

三

three 

 

采暖负荷计算

Calculation of heating load 

 

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hebnt2015 

 

 

1.采暖负荷计算的组成（Qn）

1. Composition of heating load calculation (Qn) 

 

冬季采暖通风系统的热负荷，应根据建筑物下列散失和获得的热量确定：

The heat load of the winter heating and ventilation system shall be determined according to the following loss and heat loss of the building: 

 

1）围护结构的耗热量，包括基本耗热量和附加耗热量，

1) the heat consumption of the containment structure, including the basic consumption of heat and added heat, 

 

2）加热由门窗缝隙渗入室内的冷空气的耗热量

2) heating of cold air from doors and Windows into the indoor air 

 

3）加热由门、孔沿及相邻房间浸入的冷空气的耗热量；

3) heat consumption of cold air which is immersed by doors, holes and adjacent rooms; 

 

4）建筑内部设备得热；

4) heating of internal equipment; 

 

5）通过其他途径散失或获得的热量。

5) loss or gain of heat through other means. 

 

对于一般民用住宅层高在3m 以下工程上可采用面积热负荷法进行概算。

The thermal load method can be applied to the general residential building in the following works. 

 

单位面积热负荷法：Qn=K×qn×S

Unit area heat load method: Qn = K * Qn * S 

 

式中：Qn—— 建筑物的采暖设计热负荷，W

In the formula: Qn -- the heating load of the building's heating design, W 

 

S —— 建筑物的建筑面积，m2；

S -- the building area of the building, m2; 

 

qn —— 建筑物的采暖单位面积热负荷，W/m2，

Qn -- heat load of heating unit area of buildings, W/m2, 

 

K —— 附加系数

K -- additional coefficients 

 

建筑各个区域的围护结构、冷空气渗透情况均有差别，如果需要计算的较为准确，应根据各个区域在建筑中的位置（如：是否靠近外墙、外墙上的门窗）和门窗（是否有冷空气渗透）进行分别计算。

In different regions in building palisade structure also have difference, cold air infiltration rate, if you need to calculate more accurate, should according to each area on top of the building (such as: whether the exterior walls, doors and Windows on the wall) and door window (if there is a cold air permeability) were calculated. 

 

2. 室内采暖单位面积热负荷计算（qn）

2. Calculation of thermal load of indoor heating unit area (qn) 

 

1)一般原则

1) general principles 

 

别墅的负荷一般要比住宅的大一些。

The load of the villa is generally bigger than the house. 

 

别墅的顶层负荷要大于中间层或底层。

The top level of the villa is greater than the middle or bottom. 

 

普通卫生间根据面积提供500～1000W的定值来计算。

The average toilet can be calculated according to the area of 500 ~ 1000W. 

 

别墅地下室一般不配。

The villa basement doesn't fit. 

 

客卧一般负荷相对较大。

The guest bedroom general load is relatively large. 

 

对于外墙较大或玻璃面积较大的，建议做负荷计算

For larger walls or larger glass area, it is recommended to do load calculation 

 

2）室内采暖单位面积热负荷估算表（qn）

2) thermal load estimation table of indoor heating unit area (qn) 

 

 

3. 附加系数

3. Additional coefficients 

 

附加系数为采暖面积与全房间面积的比值，根据下表进行选择：

The additional coefficient is the ratio between the area of the heating area and the total room area, according to the following table: 

 

 

上表的附加系数为标准推荐数值，在实际工程中应根据实际情况做出具体调整。

The additional coefficient of the above table is the standard recommended value, which should be adjusted according to the actual situation. 

 

房间进深大于6 米时，以距外墙6 米为界分区当作不同的单独房间，分别计算供暖热负荷。

When the room is more than 6 meters deep, it is used as a separate room of 6 meters from the outer wall to calculate heating load respectively. 

 

4.另一种采暖热负荷的估算办法

4. Another estimation method for heating load 

 

Qn=a×Rn×V×（tn-tw）

Qn = a * V * Rn (tn - tw) 

 

Qn —— 采暖热负荷 W

Qn -- heating load W 

 

tn —— 室内空气温度 ℃

Tn - ℃ of indoor air temperature 

 

tw —— 室外供暖计算温度 ℃

Tw - outdoor heating calculation ℃ temperature 

 

V —— 建筑的体积 m3

V -- the volume of the building m3 

 

Rn —— 体积热指标 根据建筑的保温情况宜取0.4-0.7

Rn -- the thermal index of volume is 0.4 to 0.7 according to the insulation of the building 

 

a —— 修正系数。

A -- the coefficient of correction.