# Thermo: Principles behind temperature drop across a valve

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1. Jul 10, 2016

### Aledrus

Hi everyone

Would like some help here.

If I draw a control volume across a valve that I can control the flow area to, and I know the temperature and pressure upstream of the valve, I will know its enthalpy:in.

High temperature liquid going through this valve supposedly changes to steam due to a drop of pressure because of the restriction in the valve. Ok, but why does the temperature drop too?

On the internet, websites say that high temperature fluid just drops in temperature across a valve because the pressure drops. But why does it?

If I were to attempt to explain it, I would say:
1. Pressure drops because of the restriction in flow but is being pumped out the other end or going into larger volume
2. Because pressure is lower at this point, more of the liquid turns into steam via flashing
3. Temperature drops as pressure drops according to the vapour-liquid line on the steam tables

But if temperature and pressure at outlet of the valve is lower than temperature and pressure at inlet of the valve, then is energy not conserved? It seems that heat is dumped somewhere out so that it becomes cold, but where goes the energy:in? Where goes the kJ/kg that was sent to the valve?

For example of this : an aerosol can - it gets cold although it starts at the same temperature. The only device in there is the restriction/valve, there are no condensers or heat sinks or anything to take the heat away.. but it gets cold.

2. Jul 10, 2016

### Nidum

Conversion of water into steam requires energy . Think about where that energy comes from in your problem .

3. Jul 10, 2016

### Staff: Mentor

If the valve is insulated, what does the open system version of the first law tell you about the change in enthalpy per unit mass of the water/steam passing through the valve?

4. Jul 10, 2016

### Aledrus

I actually don't know. I don't have a background in this field.

"Conversion of water into steam requires energy"

I thought that's what temperature is.. a joule is the energy required to 1g of water by 1 Kelvin.. so temperature is a tank of joules I suppose. If temperature drops, where does the joule go? How do I account for the missing joules?

"what does the open system version of the first law tell you about the change in enthalpy per unit mass of the water/steam passing through the valve?"

I'll give it a try.. 0 = Q - W + w1(u1 + V1/2 + gz1) - w2(u2 + V2/2 + gz2)... assuming z1=z2 and no heat input then:

W = 0.5 [ w1 (2 x u1 + V1^2 ) - w2 (2 x u2 + V2^2) ]

H = u + pv..

W = external work + flow work = W (ext) + Fv = W (ext) + (P/A)v = W (ext) + Pv/A = W (ext) + (H - u)/A

W (ext) = 0

H-u = 0.5 A [ w1 (2 x u1 + V1^2) - w2 (2 x u2 + V2^2) ]

Not sure if this equation is correct, but from what I understand is if flow area is small then final velocity should be higher than initial velocity. So if V2 > V1 then H-u should increase by about V2^2 - V1^2? I think I've messed up somewhere.

5. Jul 10, 2016

### Staff: Mentor

If you neglect the change in kinetic energy, the flow version of the first law tells you the enthalpy of the fluid does not change in passing through the throttle. So, since the fluid pressure drops in the throttle, $\Delta (Pv)$ is negative, the internal energy of the liquid must increase (a little). This means that the liquid comes out of the throttle valve a little hotter than when it went in. Now it is released from the throttle into a region where the pressure is much lower. In fact, the pressure is lower than the equilibrium vapor pressure at the temperature of the fluid coming out of the valve. This means that there is a driving force for evaporation to occur. The heat of vaporization is supplied by the liquid (and by the subsequent vapor) itself. This operation is called a Flash vaporization. Some of the liquid evaporates to produce some vapor, and the liquid and vapor leave the flash unit with a temperature lower than the liquid entering the throttle. But the combined enthalpy of the liquid and vapor, even at the lower temperature, is the same as that of the liquid entering the throttle. How can this be? It can be because the enthalpy of the vapor at a given temperature is much higher than the enthalpy of the liquid. So no energy is gained and none is lost (assuming adiabatic operation).

6. Jul 12, 2016

### Aledrus

Thank you for taking the time to write that, I truly appreciate this.

I have some very naive / amateur questions based on your reply. I hope you don't mind.

"since the fluid pressure drops in the throttle, the internal energy of the liquid must increase (a little)."

How can this be? What provided energy to the liquid?

"The heat of vaporization is supplied by the liquid (and by the subsequent vapor) itself."

This is the H(fg) right? Initially there was a big H(fg) but because the pressure has dropped, the H(fg) has become lower? But what did you mean that the liquid is supplying the heat of vaporization? Does this mean a similar thing that the heat of vaporization at this lower pressure has been supplied by the liquid that preexisted at the prior higher pressure?

" and the liquid and vapor leave the flash unit with a temperature lower than the liquid entering the throttle"

This part I don't quite understand.. there isn't a delta-T term in the equations that I can manipulate?

I understand the last line though, thank you.
If I were to write that in my own words, would you accept this: the enthalpy of the liquid at temperature T2 is equal to the enthalpy of liquid at temperature T1 (where T1<T2) plus enthalpy of vapour at temperature T1 according to mass quality?

7. Jul 12, 2016

### Staff: Mentor

Viscous dissipation of mechanical energy within the valve. Since h is constant and $v\Delta P$ is negative, $\Delta U$ is positive. This is typically a very small effect, and can usually be neglected.
The heat of vaporization is a very weak function of the saturation temperature and pressure. So the heat of vaporization is about constant in this operation. But the liquid comes in with a high temperature, and, any remaining liquid and vapor leave the unit at a lower temperature. This means that the sensible heat of the liquid and vapor have dropped. This decrease in sensible heat provides the energy necessary to vaporize whatever amount of liquid is evaporated.
Are you saying that you would like to actually see what the heat balance equation for the evaporator looks like?
This sounds something like the heat balance equation, but your terminology is a little confusing to me.

8. Jul 12, 2016

### Aledrus

Thank you. Sorry about the terminology, I am trying to understand it and it could be because English isn't my primary language.

I actually want to understand this part:

"and, any remaining liquid and vapor leave the unit at a lower temperature."

This is what I don't understand. Why does it leave at a lower temperature?

I am not sure what a heat balance equation is, but I think it might help me understand better.

9. Jul 12, 2016

### Staff: Mentor

Let the subscript "in" refer to the conditions of the inlet stream and "out" represent the conditions of the outlet streams. Let the subscript L refer to the liquid and the subscript V refer to the vapor:

The mass balance gives: $$m_{L,in}=m_{L,out}+m_{V,out}$$
The enthalpy balance gives:
$$H_{in}=H_{out}$$where $H_{in}=H_{L,in}$ and $H_{out}=H_{L,out}+H_{V,out}$, with
$$H_{L,in}=m_{L,in}h_{L,in}$$
$$H_{L,out}=m_{L,out}h_{L,out}$$
and $$H_{V,out}=m_{V,out}h_{V,out}$$
where $h_{L,in}$, $h_{L,out}$, and $h_{V,out}$ are the enthalpies per unit mass of the streams.

If we takes as the datum of enthalpy per unit mass the liquid at the inlet temperature $T_{in}$ and $P_{in}$, we have:
$$h_{L,in}=0$$
$$h_{L,out}=C(T_{out}-T_{in})+v_L(P_{out}-P_{in})\tag{1}$$and
$$h_{V,out}=C(T_{out}-T_{in})+v_L(P_{out}-P_{in})+\lambda_{out}\tag{2}$$
where C is the heat capacity of the liquid (assumed constant), $v_L$ is the specific volume of the liquid, and $\lambda_{out}$ is the heat of vaporization at the outlet temperature and pressure. Note that Eqns. 1 and 2 neglect the thermal expansion of the liquid.
If we combine the previous equations, we obtain:
$$m_{L,in}\left[C(T_{out}-T_{in})+v_L(P_{out}-P_{in})\right]=-m_{V,out}\lambda_{out}\tag{3}$$
The pressure term in Eqn. 3 is typically much smaller than the temperature term. If we neglect the pressure term, we obtain:
$$(T_{out}-T_{in})=-\left(\frac{m_{V,out}}{m_{L,in}}\right)\left(\frac{\lambda_{out}}{C}\right)$$

10. Jul 12, 2016

### Aledrus

Thank you so much for preparing that.

I can follow it better, and I have been able to reproduce equation (3).

Question: Where does equation (1) come from?

Is that from h = u + Pv? therefore Δh = Δu + ΔPv? Where Δu = CpΔT?

11. Jul 12, 2016

### Staff: Mentor

It comes from the equation:$$dh=C_pdT+v(1-\alpha T)dP\tag{1}$$where $\alpha$ is the coefficient of volumetric thermal expansion. This is a general equation for a single phase substance, derived from the starting equation dh=Tds+vdP. The derivation is in every Thermo book. If the 2nd term in parenthesis is neglected (since the coefficient of volumetric thermal expansion of the liquid is small), we obtain dh = C_p dT + vdP.