Capillary Tube Length and Heat Transfer Dynamics in Air Conditioners: A Comparative Analysis of R-12 and Its Alternatives

Air conditioning and refrigeration systems are integral components in a diverse range of sectors, including industries such as food, textiles, chemicals, printing, and transportation, as well as in various infrastructural settings like banks, restaurants, schools, hotels, and recreational facilities. Consequently, the effective installation, repair, and maintenance of these systems are crucial for the seamless operation of the associated activities. Within these systems, the capillary tube, typically fabricated from copper with diameters ranging from 0.5 mm to 5 mm and lengths between 0.5 m to 5 m, plays a vital role. The selection of the capillary tube's size is contingent upon the power and load capacity of the system, and it is particularly suited for systems with small or fluctuating cooling loads, such as refrigerators, water coolers, and small air conditioners. This study focuses primarily on straight capillary tubes, often necessitated by their extensive length requirements.

The capillary tube, serving as a throttling device in refrigeration and air conditioning systems, offers several advantages: Firstly, its simplicity allows for easy and cost-effective manufacturing. Secondly, the capillary tube limits the maximum refrigerant charge in the system, obviating the need for a receiver. Thirdly, upon cessation of the refrigeration plant, the pressure across the capillary tube equalizes, ensuring constant pressure throughout the refrigeration cycle. This feature significantly reduces the load on the compressor upon restart, as it does not need to counteract high pressures.

Distinguished from thermostatic expansion valves, capillary tubes lack moving components, contributing to their widespread application as expansion devices in air-conditioning and refrigeration machines. Their appeal has grown in tandem with the development of more adaptable compressors, advanced refrigerants, and near leak-proof systems [1]. Most small refrigeration and air conditioning systems employ capillary tubes as expansion devices due to their simplicity and low cost [2].

3.1 Replacements for R-12

As mentioned earlier due to environmental concerns for R12, it has unusually high potential to cause the depletion of the ozone layer in the upper layers of the atmosphere that causes the greenhouse effect. R-134a is considered the finest R12 substitute. R134a's chemical name is tetrafluoromethane, and its chemical formula is (CF3CH2F). It is a hydrofluorocarbon (HFC) with little greenhouse effect and does not contribute to ozone depletion. R-134a is nonflammable and nonexplosive, with high chemical stability despite a little affinity for moisture. R-401a and R-401b refrigerants are considered good replacements that may be used instead of R12. R-401A is a 34% R-124, 13% R-152a, and 53% R-22 combination by weight. It is a suitable substitute for R12 when the temperature of evaporator is -23℃  or above. It may be utilized in applications such as home refrigerators, dairy display cases, food, walk-in coolers, vending machines and beverage dispensers. R-401b is a 61% R22, 28% R-124 and 11% R-152a combination by weight. This refrigerant is appropriate for home and commercial freezers and transportation refrigeration [11].

3.2 Assumptions used to analyze the flow through the tube

1- The capillary tube inner diameter and surface roughness are constant.

2- The capillary tube is completely insulated.

3- Pure refrigerant (oil-free) is fill the capillary tube.

4- Steady state, one-dimensional flow.

5- The flow is homogenous in the two-phase.

6- The entrance point lies on saturated liquid or sub-cooled liquid.

7- The capillary tube is horizontal.

8- The capillary tube is adiabatic.

3.3 Govering equations

The governing equations used to describe the physical flow behaviour through all capillary tube forms depend principally on the basic equations of the conservation of energy, momentum and mass and the basic fundamentals of fluid flow through the tubes, which are explained by considering an infinitesimal control volume as shown in Figure 1 and applying the conservation equations as follows.

- Mass conservation:

$\begin{gathered}\frac{d(\rho \mathrm{u})}{d x}=0 \\ \frac{\dot{m}}{A}=\frac{u_1}{v_1}=\frac{u_2}{v_2} \quad \mathrm{~A}_1=\mathrm{A}_2\end{gathered}$      (1)

- Momentum conservation in x-direction:

$\frac{d\left(\rho \mathrm{u}^2\right)}{d x}=-\frac{\mathrm{dp}}{\mathrm{dx}}-\frac{f}{d} \frac{1}{2} \rho \mathrm{u}^2$         (2)

$\begin{gathered}\rho u A d u=p \cdot A-(p+d p) A-\tau_w d A \\ \dot{m} d u=-d p \cdot A-\tau_w \cdot \pi \cdot d \cdot \Delta \mathrm{L} \\ \tau_w=\frac{f \rho u^2}{8}{ Darcy \, equation }\end{gathered}$       (3)

Substitute shear stress and divide on A in the momentum equation:

$\begin{gathered}\frac{\dot{m}}{A}\left(u_2-u_1\right)=\left(p_1-p_2\right) \frac{A}{A}-\frac{f \cdot \rho \cdot u^2 \cdot \pi \cdot d \cdot \Delta \mathrm{L}}{8 \cdot A} \\ \frac{\dot{m}}{A}\left(u_2-u_1\right)=\left(p_1-p_2\right)-\frac{4 \cdot f \cdot \rho \cdot u^2 \cdot \pi \cdot d \cdot \Delta \mathrm{L}}{8 \cdot \pi \cdot d^2} \\ \frac{\dot{m}}{A}\left(u_2-u_1\right)=\left(p_1-p_2\right)-\frac{f \cdot \rho \cdot u^2 \cdot \Delta \mathrm{L}}{2 \cdot d}\end{gathered}$         (4)

Simplify the last part of the equation:

$f \frac{\rho \cdot u^2 \cdot \Delta \mathrm{L}}{d .2}=f \frac{\Delta \mathrm{L} \cdot u}{d \cdot 2} \frac{u}{v}=f \frac{\Delta \mathrm{L} \cdot u}{d .2} \frac{\dot{m}}{A}$

For Reynolds's number of turbulent regions, an applicable equation for the friction factor f  is:

$\begin{aligned} & f=\frac{0.33}{R e^{0.25}} \\ & \operatorname{Re}=\frac{\rho d u}{\mu}\end{aligned}$      (5)

The viscosity of the two-phase refrigerant at a given position in the tube is a function of the vapour fraction x:

$\begin{aligned} & \mu=\mu_f(1-x)+\mu_g x \\ & \mu=\mu_f+x\left(\mu_g-\mu_f\right)\end{aligned}$     (6)

For the increase of length $1-2$, we can apply mean friction $f_m$ and mean velocity $u_m$.

$u_m=\left[\frac{u_1+u_2}{2}\right]$       (7)

$f_m=\left[\frac{f_1+f_2}{2}\right]$    (8)

Substitute in momentum Eq. (4):

$\begin{aligned} & \frac{\dot{m}}{A}\left(u_2-u_1\right)=\left(p_1-p_2\right)-f_m \frac{\Delta \mathrm{L} \cdot u_m}{d \cdot 2} \frac{\dot{m}}{A} \\ & \Delta \mathrm{L}=\left[\left(p_1-p_2\right)-\frac{\dot{m}}{A}\left(u_2-u_1\right)\right] \frac{2 \cdot d \cdot A}{f_m \cdot u_m \cdot \dot{m}}\end{aligned}$       (9)

- Energy conservation:

$\rho \mathrm{u} \frac{d h}{d x}+\frac{\rho \mathrm{ud}\left(\frac{1}{2 \mathrm{u}^2}\right)}{d x}=0$    (10)

By applying the first law of thermodynamics to the control volume:

$Q-W=g \cdot d z+u d u+d h$

From assumptions:

$\begin{gathered}u d u+d h=0 \\ h_2+\frac{u_2^2}{2}=h_1+\frac{u_1^2}{2}\end{gathered}$      (11)

$\begin{aligned} & h=h_f(1-x)+h_g x \\ & h=h_f+x\left(h_g-h_f\right)\end{aligned}$ (12)

$\begin{aligned} & v=v_f(1-x)+v_g x \\ & v=v_f+x\left(v_g-v_f\right)\end{aligned}$   (13)

Combine the continuity Eq. (1) and the energy Eq. (11):

$h_2+\frac{v_2^2}{2}\left(\frac{\dot{m}}{A}\right)^2=h_1+\frac{u_1^2}{2}$      (14)

Substitute Eqs. (12) and (13) into Eq. (14):

$\mathrm{h}_{\mathrm{f} 2}+\left(\mathrm{h}_{\mathrm{g} 2}-\mathrm{h}_{\mathrm{f} 2}\right) \mathrm{x}+\frac{\left[\left(\mathrm{v}_{\mathrm{f} 2}+\left(\mathrm{v}_{\mathrm{g} 2}-\mathrm{v}_{\mathrm{f} 2}\right) \mathrm{x}\right)\right]^2}{2}\left(\frac{\dot{\mathrm{m}}}{\mathrm{A}}\right)^2=\mathrm{h}_1+\frac{\mathrm{u}_1^2}{2}$  (15)

Everything in Eq. (15) is known except x, which can be solved by the quadratic equation:

$\begin{gathered}x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\ a=\frac{1}{2}\left(v_{g 2}-v_{f 2}\right)^2\left(\frac{\dot{m}}{A}\right)^2 \\ b=\left(h_{g 2}-h_{f 2}\right)+v_{f 2}\left(v_{g 2}-v_{f 2}\right)\left(\frac{\dot{m}}{A}\right)^2 \\ c=\left(h_{f 2}-h_1\right)+\frac{1}{2}\left(\frac{\dot{m}}{A}\right)^2 v_{f 2}^2-\frac{u_1^2}{2}\end{gathered}$

1.png

Figure 1. The control volume (C.V) for refrigerant flow in capillary tube

3.4 Calculating the length of an increment

1-Select t_2.

2-Compute $h_{g 2}, h_{f 2}, v_{f 2}, v_{g 2}, p_2, \mu_{g 2}$ and $\mu_{f 2}$, all of which are functions of $t \, h_g, h_f, v_f, v_g, p, \mu_g$, $\mu_f$=f(t)$, \phi=a_0+a_1 t+a_2 t^2+a_3 t^3+a_4 t^4+a_5 t^5$.

3-Compute vapour fraction (x) from continuity Eq. (1) and energy Eq. (15).

4-With the value of $x$ known $h_2, v_2$ and $u_2$ can be computed.

5-Compute the Re at point 2 using the viscosity from Eq. (6), the friction factor at point 2 from Eq. (5), and the mean friction factor for the increase from Eq. (8).

6-Finally, substitute Eq. (7) into Eq. (9) to calculate $\Delta \mathrm{L}$ and Solving by EES Program.

1-The operating conditions at the capillary tube's inlet and the refrigerant's thermophysical characteristics, particularly the viscosity, affect the capillary tube's length.

2-The capillary tube's diameter and length are directly proportional. The length is shorter if the diameter is smaller and the length is longer if the diameter is greater.

3- Both refrigerants' performances show the same degree of variation when the conditions remain is the same.

4-In the two-phase region, it is quite different in quality because the viscosity of R-12 is less than R-134a, and length of capillary tube of R-12 is longer than that of R-134a under the same conditions.