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Processing & Handling :: Thermal & Energy Management :: Heat Exchangers, Condensers & Coolers

February 1, 2012

Effective Thermal Design Of Cooling Towers

A step-by-step approach to cooling-tower design, with an example calculation to make it clear

Jonny Goyal Air Liquide Engineering and Construction, Lurgi India

Various misconceptions arise when it comes to the thermal design of cooling towers. Sometimes related parameters, such as range, approach, effectiveness, liquid-to-gas ratio ( L/G), wet-bulb temperature, cooling water temperature, relative humidity, number of transfer units (NTU) and other terms create a confusion for the designer in effectively sizing, selecting and evaluating a particular cooling tower. This leads to inadequate design.

The objective of this article is to present a stepwise understanding of how to calculate the NTU for a cooling tower, and thus to understand the basis of thermal design of counter-flow cooling towers for optimizing cost and performance.
  

Definitions

First, let’s look at some of the basic terms and briefly describe their significance and role in cooling tower design and performance.

Dry-bulb temperature. Dry-bulb temperature (tdb) — usually referred to as the air temperature — is the property of air that is most commonly used. When people refer to the temperature of the air, they are normally referring to its dry-bulb temperature. The dry-bulb temperature is an indicator of heat content and is shown along the bottom axis of a psychometric chart. The vertical lines extending upward from this axis are constant-temperature lines.

Wet-bulb temperature. Wet-bulb temperature (twb) is the reading when the bulb of a thermometer is covered with a wet cloth, and the instrument is whirled around in a sling. The wet-bulb temperature is the lowest temperature that can be reached by evaporation of water only.

Relative humidity (RH). RH is the ratio of the partial pressure of water vapor in air over the saturation vapor pressure at a given temperature. When the relative humidity is 100%, the air is saturated and therefore, water will not evaporate further. Therefore, when the RH is 100% the wet-bulb temperature is the same as the dry-bulb temperature, because the water cannot evaporate any more.

Range. The range is the difference in temperature of inlet hot water (t2) and outlet cold water (t1), t2t1. A high cooling-tower range means that the cooling tower has been able to reduce the water temperature effectively.

Approach. The approach is the difference in temperature of outlet cold water and ambient wet-bulb temperature, t1tw. The lower the approach, the better the cooling tower performance. Although both range and approach should be monitored, the approach is a better indicator of cooling tower performance.

Cooling tower capability. The capability of the cooling tower is a measure of how close the tower can bring the hot water temperature to the wet-bulb temperature of the entering air. A larger cooling tower (that is, more air or more fill) will produce a closer approach (colder outlet water) for a given heat load, flowrate and entering air condition. The lower the wet-bulb temperature, which indicates either cool air, low humidity or a combination of the two, the lower the cooling tower can cool the water. Capability tests are conducted per the ATC-105 Code of the Cooling Tower Institute (CTI; Houston; www.cti.org).

The thermal performance of the cooling tower is thus affected by the entering wet-bulb temperature. The entering air dry-bulb temperature has an insignificant effect on thermal performance.

Effectiveness. A cooling towers effectiveness is quantified by the ratio of the actual range to the ideal range, that is, the difference between cooling water inlet temperature and ambient wet-bulb temperature. It is defined in terms of percentage.
 

 

(1)

Liquid-to-gas ratio (L/G). The L/G ratio of a cooling tower is the ratio of the liquid (water) mass flowrate (L) to gas (air) mass flowrate (G). Cooling towers have certain design values, but seasonal variations require adjustment and tuning of water and air flowrates to get the best cooling tower effectiveness.

Number of transfer units (NTU). Also called the tower coefficient, the NTU is a numerical value that results from theoretical calculations based on a set of performance characteristics. The value of NTU is also representative of the degree of difficulty for the cooling process. The NTU corresponding to a set of hypothetical conditions is called the required coefficient and is an evaluation of the problem. The same calculations applied to a set of test conditions is called the available coefficient of the tower involved. The available coefficient is not a constant but varies with operating conditions. The operating characteristic of a cooling tower is developed from an empirical correlation that shows how the available coefficient varies with operating conditions.

Cooling capacity. The cooling capacity of a tower is the heat rejected [kcal/h or TR (refrigeration tons; 1 TR = 12,000 Btu/h = 3,025.9 kcal/h)], and is determined by the product of mass flowrate of water, times the specific heat times the temperature difference.
 

Theory — the Merkel equation

In a cooling tower operating in counter current flow, there are two basic principles involved for removing heat by the cooling water:

1.sensible heat transfer due to a difference in temperature levels

2. latent heat equivalent of the mass transfer resulting from the evaporation of a portion of the circulating water

 Figure 1. The Merkel equation is derived by considering
a falling water droplet surrounded by saturated air
Merkel developed the basic equation based on the above principles. The Merkel model is universally accepted for designing and rating of counter-flow cooling towers. The model is based on a drop of water falling through an upstream flow of unsaturated air at a wet-bulb temperature of twb with enthalpy hA (Figure 1), in a counter-flow cooling tower. The drop of water is assumed to be surrounded by a film of saturated air at the water temperature WT with saturation enthalpy hW. As the drop travels downward, heat and mass transfer takes place from the interface air film to the upstream air, thereby cooling the water from hot temperature to a cold temperature.

The main assumptions of Merkel theory are the following:

1. The saturated air film is at the temperature of the bulk water.

2. The saturated air film offers no resistance to heat transfer.

3. The vapor content of the air is proportional to the partial pressure of the water vapor.

4. The heat transferred from the air to the film by convection is proportional to the heat transferred from the film to the ambient air by evaporation.

5. The specific heat of the air-water vapor mixture and the heat of vaporization are constant.

6. The loss of water by evaporation is neglected.

7. The force driving heat transfer is the differential enthalpy between the saturated and bulk air.

Figure 2. This plot, known as the driving force diagram, shows the enthalpy versus
temperature for water and air
Figure 3. Solving the Merkel equation (Equation 2), is usually done graphically,
where the integral is equal to the area under the curve
 
This cooling process can best be explained on a psychometric chart, which plots enthalpy versus temperature. The process is illustrated in the so-called driving-force diagram shown in Figure 2. The air film is represented by the water operating line on the saturation curve. The main air is represented by the air operating line, the slope of which is the ratio of liquid (water) to air (L/G). The cooling characteristic, a degree of difficulty to cooling is represented by the Merkel equation:

 
(2)

Where:

K = overall enthalpy transfer coefficient, lb/h-ft2

a = Surface area per unit tower volume, ft2/ft3

V = Effective tower volume, ft3

L = Water mass flowrate, lb/h

Equation 2 basically says that at any point in the tower, heat and water vapor are transferred into the air due (approximately) to the difference in the enthalpy of the air at the surface of the water and the main stream of the air. Thus, the driving force at any point is the vertical distance between the two operating lines. And therefore, the performance demanded from the cooling tower is the inverse of this difference. The solution of the Merkel equation can be represented by the performance demand diagram shown in Figure 3. The KaV/L value is equal to the area under the curve, and represents the sum of NTUs defined for a cooling tower range.

An increase in the entering twb moves the air operating line towards the right and upward to establish equilibrium. Both the cold water temperature (CWT) and hot water temperature (HWT) increases, while the approach decreases. The curvature of the saturation line is such that the approach decreases at a progressively slower rate as the twb increases. An increase in the heat load increases the cooling ranges and increases the length of the air operating line. To maintain equilibrium, the line shifts to the right increasing the HWT, CWT, and approach. The increase causes the hot water temperature to increase considerably faster than does the cold water temperature. In both these cases, the KaV/L should remain constant. However, a change in L/G will change the KaV/L value.
 

 

Cooling tower design

On the basis of the above discussion, it is clear that there are five parameters that, in combination, dictate and define the performance of a cooling tower, namely:

1. Hot water temperature, HWT

2. Cold water temperature, CWT

3. Wet bulb temperature, t wb

4. Water mass flowrate, L

5. Air mass flowrate, G

The first four parameters are determined by the user of the cooling tower. It is the fifth quantity, G, that is selected by the designer of the cooling tower. Once these five quantities are available, the tower characteristic ( KaV/L), can be calculated through the Merkel equation.

The first step in designing a cooling tower is the generation of a demand curve. In this curve, the KaV/L values are plotted against varying L/G ratios. The next step is to superimpose fill-characteristic curves and demand curves. The Cooling Technology Institute has tested a variety of fill configurations and generated fill characteristic curves for each type; CTI’s Technical Paper TP88_05 can be referred to in this regard.

Cooling tower design is basically an iterative process. The factors that effect the selection of design L/G and consequently the fill height are: cell dimensions, water loading, air velocities across various cooling tower sections and pressure drops, and fan selection.

The classical method of thermal rating of cooling towers is to estimate the ratio of liquid to gas first and then find the proper tower volume by the means of trial and error using the tower performance curve. The L/G is the most important factor in designing the cooling tower and related to the construction and operating cost of cooling tower.

Finally we can summarize the importance of the L/G ratio with the following points.

A high L/G ratio means:

• More water to less air

• Air is more saturated — driving force is reduced

• More residence time of water needed

• Less cooling in given time

• Increase in required fan power

• Decrease in height of tower

• Low evaporation loss (under same water flowrate)
 

 

An example makes it clear

As an example, let us design a cooling tower with the following data:

Capacity ( F): 3,000 m3/h

Wet bulb temperature (twb): 29°C Relative humidity () 92%

Cooling water inlet (t2): 43°C

Cooling water outlet (t1): 33°C Altitude ( Z): 10 m

Step I. This step involves heat load calculations as follows:

1. Range = (t2 – t1) = 43 – 33 = 10°C

2. Approach = (t1twb) = 33 – 29 = 4°C

3. Heat load, Q = mCp(t2t1)

= 998.13 x F x Range

= 998.13 x 3,000 x 10

= 29,943,900 kcal/h

Step II.This step involves total psychometric calculations as follows:

1. Barometric pressure (p) at the given altitude (Z) is calculated by using the following equation:

(3)

For an altitute of 10 m, this becomes

p = 101.2 kPa

2. Assume a dry bulb temperature of say, tdb = 32°C

3. Calculate water vapor saturation pressure (pws) at the assumed tdb for the temperature range of 0 to 200°C using the equation:

(4)

Where:

C 1 = –5.8002206 x 103

C 2 = 1.3914993 x 100

C 3 = – 4.8640239 x 10–2

C 4 = 4.1764768 x 10–5

C 5 =–1.4452093 x 10–8

C 6 =6.5459673 x 100

and T represents the dry bulb temperature in Kelvin. This results in the value:

pws = 4.7585 kPa

4. The partial pressure of water (pw) at given relative humidity is found using the following equation:

(5)

pw = 4.3779 kPa

5. The partial pressure (pws) is again calculated using Equation 4. This time T represents the wet bulb temperature in Kelvin, which calculates to:

pws = 4.0083 kPa

6. Using pws calculated in Step 5 we recalculate twb using the Carrier equation:

(6)
which gives the result:

t wb = 37.7°C

7. This step is an iterative process, whereby the assumed value of tdb in Step 2 is varied in such a way that the calculated twb in Step 6 equals the actual (real) twb.

8. After a number of iterations, the calculated t db value converges to 30.12°C.

Step III.This step involves the calculation of the inlet air enthalpy (h1) as follows:

1. The humidity ratio (W) for dry air is calculated using the following equation:

(7)

W = 0.02515 kg water/kg dry air

2. The specific volume (v) for dry air is calculated using the following equation:

(8)

v = 0.89511 m3/kg, dry air

3. Calculate the enthalpy of inlet air (h1) using the following equation:

(9)

h 1 = 94.750 kJ/kg

4. Calculate the humidity ratio at saturation (Ws) for wet air using same Equation 7. Here we now use pws:

Ws = 0.02743 kg water/kg moist air

5. Calculate the specific volume (v) for wet air using Equation 8 with Ws.

v = 0.89827 m3/kg moist air

Step IV.This step involves the calculation of the exit air properties, as follows:

1. Assume some value of the L/G ratio, say 1.575, and calculate h2 for exit air using the following equation:

(10)

h 2 = 160.50 kJ/kg

2. Assume that the exit air has a relative humidity of 97–99% (design RH at the outlet), and also assume some value of exit air dry-bulb temperature.

3. Use the same partial pressure and humidity equations as discussed in Step II and Step III, to calculate the enthalpy of exit air at these assumed values in Point 2 above. At an assumed RH of 98.5% and exit air t db of 37°C, we recalculate

h2 = 141.18 kJ/kg

4. This is again an iterative process. Next, assume the value of exit air tdb in Point 2 (at same relative humidity) in such a way that the calculated h2 in Point 3 equals the h2 calculated in Point 1.

5. After a number of iterations, the calculated exit air tdb value converge to 39.55°C.

6. Now that the dry-bulb temperature and RH are known values, recalculate the wet-bulb temperature using Equation 6.

twb = 39.31°C

7. Calculate the dry and wet specific volume of exit air using Equation 8. Also calculate the density of dry air and wet air (for inlet and exit).

v = 0.9540 m3/kg, dry air

v = 0.9551 m3/kg, moist air

Average density dry = 1.0827 kg/m3

Average density wet = 1.0801 kg/m3

Step V. This step is to help draw the driving force diagram as follows:

1. Take different temperature ranges (covering cooling water inlet and outlet temperature) and calculate the enthalpy of air using Equation 9 and psychometric calculations discussed above. Plot the air saturation curve (enthalpy versus temperature) as shown in Figure 2.

2. Take cooling water outlet temperature and calculate the enthalpy of fin (h’) using Equation 9. Here the partial pressure, saturation pressure and humidity ratio are calculated for the corresponding temperature taken (Table 1).

h’ at 33°C = 116.569 kJ/kg

This enthalpy specifies Point B on the graph of Figure 2, and is the starting point of the water operating line.

3. Similarly calculate enthalpy of air ha, at wet-bulb temperature (Table 1).

ha = 94.667 kJ/kg

This enthalpy specifies Point A on the graph of Figure 2 and is the starting point of the air operating line.

4. Take incremental change in temperature (say 0.5 or 1.0) up to the hot water temperature and calculate the h' and ha. The ending points are shown as C and D in Figure 2 on the water operating and air operating line respectively.

5. The difference between h’ and ha will give you the enthalpy driving force for incremental change in temperature.

6. Take the inverse of enthalpy difference in each incremental step (Table 1).

7. Calculate NTU = 4.18 x ∆t x (Average of incremental increase in inverse of enthalpy difference).

Or for 0.5°C increment in temperature, calculated NTU = 0.096

8. Similarly, calculate the NTU for each step and add to get the total NTU for the particular assumed L/G ratio (Table 1). Or, for an assumed L/G of 1.575, Total NTU = 1.7533 — this is KaV/L.

9. Now to plot the tower characteristic curve, first we vary the L/G ratio and repeat all calculations discussed above to generate the data for various NTU to plot. The curve represents “Design NTU” on the graph, shown in Figure 4.

10. Take the design L/G ratio and plot the tower characteristic curve by assuming the slope of the line (m), which usually varies between –0.5 to –0.8. One can also consult with vendors for this value as it also depends on the type of fins used.

11. Calculate the value of the constant C, related to cooling tower design using equation:

NTU = C x ( L/G) m

C = 2.522

 Figure 4. The intersection of the tower characteristic curve and the design NTU curve gives the design L/G ratio

Table 1.
Water Temperature, t Water Vapor Saturation Pressure, pws Partial Press. of H2O Vapor, pw Humidity Ratio, W Enthalpy of Film, h' Enthalpy of Air, ha Enthalpy Difference, (h'ha) 1/(h'ha) ∆t NTU ∑NTU Cumulative Cooling Range
°C K kPa kPa kg water/kg dry air kJ/kg kJ/kg kJ/kg kg/kJ        
33 306.15 5.0343 5.0343 0.0326 116.5686 94.6668 21.9018 0.0457        
33.5 306.65 5.1774 5.1774 0.0335 119.5982 97.9585 21.6397 0.0462 0.5 0.096 0.096 0.5
34 307.15 5.3239 5.3239 0.0345 122.6986 101.2503 21.4483 0.0466 0.5 0.097 0.193 1
34.5 307.65 5.4740 5.4740 0.0356 125.8718 104.5420 21.3298 0.0469 0.5 0.098 0.291 1.5
35 308.15 5.6278 5.6278 0.0366 129.1197 107.8338 21.2859 0.0470 0.5 0.098 0.389 2
35.5 308.65 5.7853 5.7853 0.0377 132.4444 111.1255 21.3189 0.0469 0.5 0.098 0.487 2.5
36 309.15 5.9466 5.9466 0.0388 135.8480 114.4173 21.4307 0.0467 0.5 0.098 0.585 3
37 310.15 6.2810 6.2810 0.0412 142.9006 121.0008 21.8999 0.0457 1.0 0.193 0.778 4
38 311.15 6.6315 6.6315 0.0436 150.2958 127.5843 22.7116 0.0440 1.0 0.187 0.965 5
39 312.15 6.9987 6.9987 0.0462 158.0530 134.1678 23.8852 0.0419 1.0 0.180 1.145 6
40 313.15 7.3835 7.3835 0.0489 166.1928 140.7513 25.4415 0.0393 1.0 0.170 1.314 7
41 314.15 7.7863 7.7863 0.0518 174.7371 147.3348 27.4023 0.0365 1.0 0.158 1.473 8
42 315.15 8.2080 8.2080 0.0549 183.7094 153.9183 29.7911 0.0336 1.0 0.146 1.619 9
43 316.15 8.6492 8.6492 0.0581 193.1348 160.5018 32.6330 0.0306 1.0 0.134 1.753 10
  ∑NTU 1.75334674  
Table 2. TYPICAL THERMAL CALCULATIONS OF COUNTER-FLOW COOLING TOWER
Flowrate 3,000 m3/h   Hot water temperature 43 °C
Wet-bulb temperature 29 °C Cold water temperature 33 °C
Approach 4 °C Range 10 °C
Assumed dry-bulb temperature 30.12 °C Heat load 29943888.32 kcal/h
Assumed L/G 1.575   Barometric pressure (p) 101.2 kPa
No. of cells 3        
Cell length 14 m Water vapor saturation pressure (pws), at tdb 4.2754 kPa
Cell width 14 m Partial pressure of water vapor (pw) 3.9333 kPa
Air inlet height 5.5 m Water vapor saturation pressure (pws), at twb 4.0083 kPa
Design RH 92%   Recalculating twb 29.0 °C
Density of water 1,000 kg/m3 Difference 0.0 °C
Altitude 10 m Dew point temperature (td) of moist air (for temperature between 0 and 93°C) 28.66 °C
   
Inlet Air Properties Exit Air Properties
Inlet twb 29 °C Enthalpy 160.50 kJ/kg
RH 92%   Exit air temperature (tdb) 39.55 °C
Inlet tdb at above RH 30.12 °C Water vapor saturation pressure (pws), at tdb 7.2097 kPa
Humidity ratio (W) 0.0252 kg water/kg dry air Put some value of exit RH (Assuming it is in between 97 and 99%) 98.5 %
Specific volume (v) 0.8951 m3/kg, dry air Partial pressure of water vapor (pw) 7.1016 kPa
Density 1.1172 kg/m3, dry air Humidity ratio (W) at above tdb 0.04692 kg water/kg dry air
Humidity ratio at saturation (Ws) 0.0274 kg water/kg moist air Enthalpy of exit air 160.50 kJ/kg
Specific volume, (v) at saturation 0.8983 m3/kg, moist air Difference 0.00  
Density 1.1133 kg/m3, moist air Exit twb 39.31 °C
Enthalpy of moist air 94.5702 kJ/kg   Water vapor saturation pressure (pws) at exit twb 7.1170 kPa
Humidity Ratio at twb 0.0256 kg water/kg moist air   Recalculating twb 39.31 °C
Enthalpy at twb 94.6668 kJ/kg   Difference 0.00  
    Dew point temperature (td) of moist air (for temperature between 0 and 93°C) 39.24 °C
Air Flow   Specific volume (v) 0.9540 m3/kg, dry air
Average density dry 1.0827 kg/m3   Density 1.0482 kg/m3, dry air
Average density wet 1.0801 kg/m3   Humidity ratio (Ws) at saturation 0.0477 kg water/kg moist air
Air flowrate at fin 162.90 m3/s, per cell   Specific volume (v) at saturation 0.9551 m3/kg, moist air
Air flow at inlet, at rain zone 157.87 m3/s, per cell   Density 1.0470 kg/m3, moist air
Air flowrate at fan 168.26 m3/s, per cell   KaV/L 1.7533  
    Constant related to tower design (C) 2.522  

The significance of these calculations is that now we can directly calculate the cooling tower characteristic by using our equations and can compare with the vendor data if the provided height of the cell is adequate to meet the calculated NTU. Obviously the height also depends on the type of packing, but along with vendor input we can create a complete economical design of our cooling tower. Further, we can develop a calculation sheet in Microsoft Excel, which gives all the results of the psychometric chart as well as the cooling tower design. Typical thermal calculations for a counter-flow cooling tower can be seen in Table 2.

Edited by Gerald Ondrey

 

Reference:

1) Cooling Tower Thermal Design Manual,” Daeil Aqua Co., Ltd., available as download at: che.sharif.ir/~heatlab/Lab/Benefit%20Book%20&%20Journal/Benefit%20book/Cooling%20Tower%20Thermal%20Design%20Manual.pdf

2) Rosaler, Robert C., The Standard Handbook of Plant Engineering, 2nd Edition, McGraw-Hill, New York, 1995.

3) Green, Don W., others, Perrys Chemical Engineers Handbook, 6th Edition, McGraw-Hill, New York, 1984.

4) Psychometric Data from ASAE D271.2 December 1994.

5) ASHRAE Handbook — Fundamentals, Chapter 1, American Soc. of Heating, Refrigerating and Air-Conditioning Engineers, Atlanta, Ga.

Author

Jonny Goyal is an assistant lead engineer at Air Liquide Engineering and Construction, c/o Lurgi India Co. Pvt. Ltd. (A-22/10, Mohan Cooperative Industrial Estate, Mathura Road, New Delhi 110044 India; Phone: +91-4259-5400; Fax: +91-4259-5051: Email: jonny.goyal@airliquide.com), where he is involved in process designing activities related to basic and detailed engineering projects. On the technology and business development front, he is consistently involved in working with and developing Lurgi Fixed-Bed, Dry-Bottom gasification technology and downstream acid-gas removal. Prior to joing Lurgi in 2007, Goyal had two years work experience in Mojj Engineering Pune in the design and commissioning of bioethanol plants from molasses or grains as feedstock, and also has experience in plant operation of distillation columns in the solvent recovery plant of Nectar Lifesciences Ltd. Goyal has a B.Tech.Ch.E. from SBSCET, Ferozepur, which is affilated with PTU Jalandhar, Punjab.


In the original, print article (February 2012, pp. 24–29), there were mistakes in Equations (4) and (6). The correct equations are as follows:

 

     (4)

   

     (6)

 
 

These typos do not affect any of the numbers in the Tables or the example problem because the correct equations (not those printed) were used for the calculations. Theonline version of the article has been co


 

Comments (3) for Effective Thermal Design Of Cooling Towers
1.
Many thanks for publishing this article. Please note my new email ID <jonny.goyal@lanzatech.com> for further email communication.

Regards,
Jonny
Posted by Jonny Goyal on Monday, December 3, 2012 @ 12:02 AM
2.
Dear sir,
thanks for your nice detail but while design any cooling tower the Air flow rate is differ by
1.volume of the tank
2.Inlet temp of tower
3.atmospheric pressure
4.Inlet water flow rate

so how we consider the air flow rate required for cooling tower.is there is any thumb rule for L/G ratio or air flow requirement.
please let me know..

thank in advance..
Posted by vishwa on Saturday, May 25, 2013 @ 07:20 AM
3.
many many thanks for giving such a clear details.
Posted by Vikas kumar on Thursday, July 3, 2014 @ 09:22 PM

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