This question pertains to the change of state of liquid using a Metering Device as the refrigerant enters an Evaporator Coil of an HVAC system:

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The Metering Device takes liquid and then reduces the pressure and then spits out 75 percent liquid and 25 percent vapor. I don't understand why decreasing pressure, thus decreasing temperature will create a saturation point where liquid is both water and vapor. Shouldn't it be the other way around? Decreasing pressure should change saturation point so the liquid is both water and ice.

This is my rationale behind what I say:

In terms of changing of state of a liquid, as you add more heat content (molecules in motion or BTU), such as 340 BTU all the way up to 1310 BTU, all the liquid is converting into gas. This is latent heat. So adding heat changes liquid to gas. Conversely, when we remove heat it changes state of liquid to ice. At 0 degrees F and 0 BTU, you have ice.

In terms of gas laws, if you compress a gas to reduce its volume, both the temperature and pressure will increase. If you expand a gas to increase its volume, both the temperature and pressure will decrease.

We can use PSI or PSIA (absolute) to measure pressure per square inch of area. If you look at a Pressure Temperature Chart for Water, 212 Degrees F and 14.7 PSIA is at saturation where there is both liquid and vapor. If we add any heat, it will change state from liquid to vapor.

If we reduce the atmospheric pressure (e.g. from 14.7 PSIA to 0.33 PSIA) and if we put the water into a vacuum (e.g. from 29.6 inHg to 0.68 inHg), then we can change the state of the water at 70 degrees F. Why? Because decrease in pressure means decrease in temperature.

Consequently, how is reducing pressure creating vapor in the case of the Metering Device?


I'm a bit confused by your question -- your diagram shows the path of refrigerant through the cooling system, but you seem to be discussing water's state changes.

Liquid refrigerant under pressure enters the metering device. As it leaves the device at a lower pressure, it boils, producing some vapor, but not evaporating completely. The rest of the liquid evaporates as it passes through the evaporator coil, absorbing heat as it does so.


The refrigerant in the HVAC system is not water. In this case, apparently, it is R22 (chlorodifluoromethane).

(Note that use of the units °F, psi, inHg, BTU is deprecated. You might want to consider using proper SI units. Furthermore, you should not use gauge pressures when making thermodynamic calculations since such values depend on the ambient pressure; you should use absolute pressures instead.)

The R22 enters the metering device at a temperature of $T_0=105\ \mathrm{^\circ F}=40.6\ \mathrm{^\circ C}$ and a pressure of $p_0=278\ \mathrm{psi(g)}=20.2\ \mathrm{bar}$, which corresponds to a specific enthalpy of $h_0=250.26\ \mathrm{kJ/kg}$ (IIR reference state). Since the equilibrium pressure of R22 at this temperature is $p=211\ \mathrm{psi(g)}=15.5\ \mathrm{bar}$, the refrigerant is liquid at this point.

When passing through the so-called metering device (i.e. through an orifice), the pressure of the refrigerant is relaxed to $p_1=69\ \mathrm{psi(g)}=5.8\ \mathrm{bar}$. The equilibrium temperature at this pressure is $T_1=40.3\ \mathrm{^\circ F}=4.6\ \mathrm{^\circ C}$. The corresponding specific enthalpy is $h_{1,\mathrm l}=205.44\ \mathrm{kJ/kg}$ for liquid R22 and $h_{1,\mathrm g}=406.71\ \mathrm{kJ/kg}$ for gaseous R22.

Assuming a fast and adiabatic process, the evaporated fraction $x$ of R22 can be estimated using a simple enthalpy balance as follows:

$$\begin{align} h_0&=x\cdot h_{1,\mathrm g}+(1-x) \cdot h_{1,\mathrm l}\\[6pt] x&= \frac{h_0-h_{1,\mathrm l}}{h_{1,\mathrm g}-h_{1,\mathrm l}}\\[6 pt] &= \frac{250.26\ \mathrm{kJ/kg}-205.44\ \mathrm{kJ/kg}}{406.71\ \mathrm{kJ/kg}-205.44\ \mathrm{kJ/kg}}\\[6pt] &=0.223\\[6pt] &=22.3\ \% \end{align}$$ which is quite close to the given $x=25\ \%$.


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