Indirect Evaporating Cooling In The Passenger Car Air Conditioning System

Heat engineering calculations of a passenger car

In order to determine the main characteristics of the air conditioning system based on the water indirect evaporating, it is essential to find the total amount of heat entering the car (Emelyanov et al., 2010). For this purpose, a heat engineering calculation has been carried out. The calculation is reduced to determine the total heat infiltrations into the car interior. The calculation scheme is shown in Figure 1.

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₁ – the amount of energy entering the car through the body rails;₂ – heat infiltration occurred in the result of exposure to solar radiation;₃ – the amount of energy entering the car through leaks in the body;₄ – the amount of energy entering the car with outdoor air in the result of cargo and passengers ventilation;₅ – heat infiltration in the result of the fan motor operation;₆ – heat infiltration from the passengers

Calculations of heat infiltrations are based on the following method:

The heat infiltration that occurs under the influence of the temperature difference is determined by the formula (2015):

$Q_{\mathrm{1}}=k\cdot F_{\mathrm{1}}\cdot (t_o-t_i)+k\cdot F_{\mathrm{2}}\cdot (t_v-t_i)$,

where $t_o$ – temperature of outdoor air, in summer $t_o=30^{\circ }С$;

$t_i$ – temperature of air inside the car, in summer; $t_i=24^{\circ }С$

$t_v$ – temperature of air in a car vestibule, in summer $t_v=27^{\circ }С$;

$F_1$ – the area of the body rail which comes into contact with the outdoor air, $F_1=311\hspace{0.33em}{m}^2$;

$F_2$ – the area of the body rail which comes into contact with a car vestibule rail, $F_2=\mathrm{16,8}\hspace{0.33em}{m}^2$.

The heat infiltration from solar radiation is determined by the formula:

$Q_{\mathrm{2}}=K_P\cdot F\cdot \frac{\mathrm{А}\cdot q}{\alpha }$,

where $F-$the outer surface of the sunlit part of the car ( $F=\mathrm{204,4}\hspace{0.33em}{m}^2$);

$А$ – solar energy absorption coefficient, accepted as 0.7;

$q$ – the average daily intensity of solar radiation, accepted as $200\hspace{0.33em}\frac{\text{W}}{m^2\cdot K}$;

$\alpha$ – heat transfer coefficient of the outer surface, $30\hspace{0.33em}\frac{\text{W}}{m^2\cdot K}$.

The heat infiltration through leaks in the doors, windows, etc. is determined by the formula:

$Q_{\mathrm{3}}=\frac{V\cdot \rho }{\mathrm{3,6}}\cdot (i_{\mathrm{1}}-i_{\mathrm{2}})$,

where $V$ – the volume of air entering through the leaks, $V=15\hspace{0.33em}\frac{m^3}{\text{hour}}$;

$\rho$ – outdoor air density, in summer $\rho =\mathrm{1,17}\hspace{0.33em}\frac{\text{kg}}{m^3}$;

₁,₂ – heat content of the air outside and inside the car $i_1=68\hspace{0.33em}\frac{\text{kJ}}{\text{kg}}$, $i_2=51\hspace{0.33em}\frac{\text{kJ}}{\text{kg}}$;

The heat infiltration during the ventilation of the car is determined by the formula:

$Q_{\mathrm{4}}=\frac{n\cdot V_r}{\mathrm{3,6}}\cdot (\mathrm{1,3}\cdot (t_{\mathrm{н}}-t_{\mathrm{в}})+r\cdot ({\varphi }^{\mathrm{1}}\cdot f^{\mathrm{1}}-{\varphi }^{\mathrm{2}}\cdot f^{\mathrm{2}}\left)\right)$,

where $n$ – ventilation rate, 4 volume/hour;

$V_r$ – volume of air to be replaced, $V_R=190\hspace{0.33em}{м}^3$;

$\mathrm{1,3}$ – air heat capacity, $\text{kJ/}{\text{m}}^3$;

$\varphi \cdot f$ – absolute humidity of the air, $\text{g/kg}$;

$⥂r$ – latent heat of steam generation, $2\text{,55}\hspace{0.33em}\text{kJ/kg}$.

The heat infiltration which is equivalent to the operation of the fans is determined by the formula:

$Q_{\mathrm{5}}=\mathrm{1000}\cdot N\cdot n\cdot \eta \cdot \frac{\tau }{\mathrm{24}}$,

where $N$ – fan motor power, $N=2\hspace{0.33em}\text{kW}$;

$n$ – the number of motors, $n=2$;

$\eta$ – efficiency of motors, $\eta =0\text{,95}$;

$\tau$ – duration of electric motor operation, $\tau =7\hspace{0.33em}\text{hours}$;

Heat infiltrations from the transported cargo and containers during cooling in the car:

$Q_{\mathrm{6}}=\frac{q_{\text{bio}}\cdot n}{\mathrm{3,6}}$,

where $q_{\text{bio}}$ – the amount of energy released by one passenger, W;

$n$ – the number of passengers, $n=38$;

The calculation results are summarized in Table 1.

The total heat infiltration to the passenger car interior is determined:

$Q_0^{н}=\mathrm{1689,2}+\mathrm{929,04}+\text{1074,1}+\mathrm{4122,9}+\mathrm{1108,33}+\mathrm{3040}=\mathrm{11863,6}\hspace{0.33em}\text{W}$

To perform calculations of an air conditioning system with indirect evaporating cooling, it is essential to find out the maximum values of heat infiltration – «gross»:

$Q_{\mathrm{0}}^{\text{gr}}=\beta \cdot Q_{\mathrm{0}}^{\mathrm{н}}$,

where $\beta$ – a coefficient that takes into account inaccuracy of the calculation, 1,15.

$Q_{\mathrm{0}}^{\text{gr}}=\mathrm{1,15}\cdot \mathrm{11863,6}=\mathrm{13643,1}\hspace{0.33em}\text{W}$

After calculating the total heat infiltrations, it is possible to calculate the system of indirect evaporating cooling of the passenger car air.

Calculation of indirect evaporating cooling

Direct evaporating cooling has different design schemes:

- indirect evaporating cooling system without water circuit (Figure 2a);

- indirect evaporating cooling system with a water circuit (Figure 2b).

In an indirect evaporating cooling system, the ambient air is blown into the system by a fan and is cooled to a constant moisture content in the indirect evaporating heat exchanger. After the heat exchanger, the main air flow is divided into two parts: secondary and operating, which is directed towards absorbing part of the system (Emelyanov & Gorbatov, 2013).

The secondary flow is a cooler and a cooling flow at the same time - after the heat exchanger, it is directed back towards the main flow. In this case, the water is supplied to the secondary flow ducts. Relevance of water supply is to slow down the growth of air temperature due to its parallel humidification: the same change in thermal energy is known to be achieved by changing the temperature as well as by simultaneously changing the temperature and humidity. When the secondary flow is humidified, the same heat exchange is achieved with a smaller temperature change.

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In the case of indirect evaporating heat exchangers of a different type, the secondary flow is not directed to the heat exchanger, but to the cooling tower, where it cools the water circulating through the indirect evaporating heat exchanger: the water is heated by the main flow, flows in the flow and is cooled in the cooling tower by the secondary air flow. A circulation pump helps the water go along the circuit.

To calculate the cycle of an indirect evaporating cooling system with circulating water, the following initial data are used:

–_a – relative humidity of the ambient air, 55%;

–_a – temperature of the ambient air, 30° С;

– – temperature difference at the cold end of the heat exchanger, 3° С;

–_w – temperature difference at the warm end of the heat exchanger, 5° С;

–_ct – the difference between the temperature of the water coming out of the cooling tower and the temperature of the air supplied to it by a wet bulb thermometer, 1° С;

– – minimum temperature difference (temperature head) between the flows in the cooling tower (∆t_min <∆t_wct), 2° С;

–_р – mass air consumption required by an absorbing part of the system, 9,89 kg/s;

–_f – Fan efficiency, 0,7%;

–_f – pressure loss in the system mechanism and lines (required fan head), 680 Pa.

The calculation method is based on the following assumptions:

– The processes of heat-mass transfer are balanced,

– There is no external heat infiltration on all areas of the system,

– The air pressure in the system is equal to atmospheric (local changes in air pressure due to air being moved by a fan or due to the fact that its circulation through aerodynamic drag is small, which makes it possible to use of I-d diagram moist air to atmospheric pressure throughout the whole circulating system).

Engineering calculation

The sequence of engineering calculations of the system is as follows (Figure 3) (2012).

Additional parameters of the ambient air (point "0" in Figure 3) are determined using the I-d diagram or using the humid air calculation program: the specific enthalpy of the air i₀, J / kg and the moisture content d₀, kg/kg.

The increment of the specific enthalpy of air in the fan (J/kg) depends on the type of fan. If the fan motor is not blown (not cooled) by the main air flow, then:

$\Delta i_a=\frac{\Delta P_a}{{\rho }_{\mathrm{0}}{\eta }_m}=\mathrm{0,84}\mathrm{}\text{kJ/kg};$

where_m – motor efficiency;

 – air density before entering the fan, kg/m³

 – barometric pressure of the environment, Pa;

_a – the gas constant of the air is equal to 287 J/(kg•K).

The increment of the specific enthalpy of air in the fan

Specific enthalpy of air which comes out of the fan (point «1», Figure 3), J/kg.

i₁= i₀+∆i_a $=\mathrm{68,84}\text{kJ/kg}$

Since the process (0 – 1, Figure 3) occurs at a constant moisture content (), it is possible to determine the air temperature that comes out of the fan (point 1, Figure 3) from the known values of.

The dew point of the ambient air is determined by the known value of,.

Psychrometric temperature difference of the main flow air that comes out of the heat exchanger (point 2, Figure 3) is, °С

∆t_2–4=∆t_x+∆t_wct $=\mathrm{2,5}\mathrm{}°\mathrm{С}$

where ∆t_х is assigned based on the specific operating conditions in the range (0.5...5.0), °C. At the same time, small values of ∆t_х will entail a relatively large size of the heat exchange unit. To ensure low values of ∆t_х, it is crucial to use highly efficient heat transfer surfaces;

_ct is selected in the range (0.8 - 3.0), °C; lower values of_гр  should be taken if it is necessary to obtain the lowest possible temperature of cold water in the cooling tower.

The humidifying process of the secondary air flow in the cooling tower from state 2-4 (Figure 3), with sufficient accuracy for engineering calculations, should go along the line.

In this case, with the value ∆t_2–4, the temperatures andcould be determined, points 2 and 4, respectively, °C. To do this, a line such that the temperature difference between point 2 and point 4 corresponds to the foundshould be found. Point 2 is located at the intersection of the lines and the constant moisture content. Point 4 is located at the intersection of the line and the curve = 100 % of relative humidity (2021).

Thus, with the help of the above diagrams, it is possible to determine the remaining parameters at points 2 and 4 (Figure 3).

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It is possible to determine the temperature of the water that comes out, at the point 1w, °C. In the calculations, the heating of the water in the pump can be neglected, therefore, before entering the heat exchanger point 1w (Figure 3) the water will have the same temperature

t_1w=t₄+∆t_wct $=\mathrm{23}\mathrm{}°\mathrm{C}$

It is necessary to determine– water temperature after heat exchanger, point 2w, °С

t_2w=t₁-∆t_h= $\mathrm{26}\mathrm{}°\mathrm{C}$

The temperature of the air discharged into the water tank point 5 is determined semigraphically using the I-d diagram. The specified method is as follows (Figure 3).

The point 1w, which characterizes the state of water at the entrance to the indirect evaporating heat exchanger, with the value of the specific enthalpy of point (4) is placed on the isotherm t_1w, which is separated from the isotherm t4 at a distance of_ct.

From the point 1w along the isentalpa, the segment 1w — is set so that. According to the fact that the process of heating the air in the cooling tower occurs100 %, from the point «p» a tangent to_пр=1 is drawn so that the point of tangency is now found out.

From the point of tangency along the isentalpa (adiabat,) the segment - is drawn so that. Thus, the minimum temperature difference between the cooled water and the secondary flow air in the cooling tower is provided. This temperature difference guarantees the operation of the cooling tower in the design regime. Now it is necessary to draw a straight line from point 1w through point n to the intersection with the straight line so that the point 2w is known. From point 2w, a straight line to the intersection with_пр100% is drawn. Point 5 which characterizes the state of the air coming out of the cooling tower is known.

According to the diagram (Figure 3), the required temperature t5 and the other parameters of point 5 can be found.

The air parameters obtained in the result of calculations and drawing:

– air temperature after the fan (32 °C);

– secondary flow temperature before entering to the cooling tower (26°C);

– secondary flow temperature in the cooling tower (20.5°C);

– temperature of the air that comes out of the cooling tower (25.5°C).

Thus, it is possible to compose a system of equations to find unknown mass flow rates of air and water. Thermal load of the cooling tower on the secondary air flow, W:

Q_ct=G_a ( i₅ – i₂) $=\mathrm{3,3}\text{W}$

Where_a – secondary air mass flow rate, kg/s

Thermal load of the cooling tower by water flow:

Q_wct=G_ow C_pw ( t_2w – t_1w) = $\mathrm{0,254}\text{kW}$

Where_p specific heat of water, 4200J/(kg•K)

– material balance by air flow

Heat load of the heat exchanger in the main air flow, W:

Q_hl=G_o( i₁ – i₂) $=\mathrm{1,6}\text{kW}$

Heat load of the heat exchanger in the water flow, W:

Q_wht=G_ow C_pw ( t_2w – t_1w)

Cooling tower heat balance:

_{ct ct}

Combined heat balance of the cooling tower and the water heat exchanger:

Q_wct=Q_whl

Air mass flow rate in the main air stream, kg/s:

G_o=G_r

Water mass flow rate through the cooling tower along the main stream, kg/s:

$G_{ww}=G_a\frac{i_{\mathrm{5}}-i_{\mathrm{2}}}{C_{pw}\left(i_{\mathrm{2}w}-i_{\mathrm{1}w}\right)}=\mathrm{2,5}\text{kg/s};$

The amount of water required to recharge the cooling tower water circuit, kg/s

G_wr=(d₅-d₂)G_w $=\mathrm{1,3}\text{kg/s}$

Thus, all the required parameters for the design calculations of the elements of the indirect evaporating air cooling system are found.

The working flow of cooled air supplied to the absorbing part of the system which is point 2 (Figure 3) can be additionally cooled, for instance, by adiabatic humidification or by any other method. As an example, Figure 3 shows the point 3* corresponding to adiabatic moisture. In this case, the points 3* and 4 match (Figure 3).