I Maximizing Flow: Understanding the Impact of Fittings on Water Velocity

[physea]

Hello!
When water flows through fittings like U bends, elbows, etc, is kinetic energy lost and velocity reduced?​

 

[BvU]

No

 

[physea]

No

How no?
Aren't there collisions in the bend that decrease the velocity of molecules?

 

[BvU]

Yes there are. However:
Water is practically incompressible, so it can't accumulate and speed has to be maintained. Something else has to give: the driving force.

 

[physea]

Yes there are. However:
Water is practically incompressible, so it can't accumulate and speed has to be maintained. Something else has to give: the driving force.

Why speed has to be maintained?
I am referring to a driving force that sends water at the beginning of a pipe and the end of the pipe is free. Will the liquid exit the pipe at the same velocity? Given diameter is constant, but there are fittings like bends and friction.

 

[russ_watters]

Why speed has to be maintained?
I am referring to a driving force that sends water at the beginning of a pipe and the end of the pipe is free. Will the liquid exit the pipe at the same velocity? Given diameter is constant, but there are fittings like bends and friction.

If the velocity dropped, volumetric flow rate would drop and you would have more water going into a pipe than coming out the other end. What would happen to this missing water?

 

[Baluncore]

I am referring to a driving force that sends water at the beginning of a pipe and the end of the pipe is free.

Potential energy is lost as the pressure reduces towards the open end. Conservation of diameter requires velocity be fixed and so KE is fixed.

 

[physea]

Lets have a tube connected to a tap and that tube has bends etc and ends freely so that the water drops.

Just after the tap, we have a flow meter that gives a specific velocity of the water.

Is that velocity the same across the whole tube up to the end including the bends?

 

[BvU]

Yes

 

[physea]

Yes

And when the tap generates a flow of 100kg/s, which corresponds to 1m/s (let's say) for 1m pipe diameter, will that velocity of the fluid be the average velocity across the whole pipe, if there are bends etc along the pipe?

 

[Chestermiller]

And when the tap generates a flow of 100kg/s, which corresponds to 1m/s (let's say) for 1m pipe diameter, will that velocity of the fluid be the average velocity across the whole pipe, if there are bends etc along the pipe?

Do you believe in conservation of mass? Do you believe that liquid water is very nearly incompressible?

 

[BvU]

And when the tap generates a flow of 100kg/s, which corresponds to 1m/s (let's say) for 1m pipe diameter, will that velocity of the fluid be the average velocity across the whole pipe, if there are bends etc along the pipe?

100 kg/s is 0.1 m³/s. A pipe diameter of 1 m is $\pi/4$ m². I get [edit] $0.4 / \pi$ m/s (ahem...after thinking ).
And: yes, that velocity will be the average velocity all over the pipe (provided the diameter is constant).

Can you imagine a very (very!) long pipe, so long that the velocity drops to zero before the end of the pipe is reached ?

 

[physea]

So turbulence does not affect mean velocity?

 

[sophiecentaur]

The important thing to remember about water flow in situations as described is that things take time to establish themselves and for the system to reach a steady state after the tap is opened. The pressure drops across the various fittings and pipe runs will always add up to the pump pressure (same as Kirchof’s second law for electrical circuits). Mass flow rate has to be the same all the way round if there is no reservoir / accumulator in circuit.

 

[Chestermiller]

So turbulence does not affect mean velocity?

Of course not.

 

[Andy Resnick]

Hello!
When water flows through fittings like U bends, elbows, etc, is kinetic energy lost and velocity reduced?​

I think this discussion went off in the wrong direction. It is true that fittings and bends can create a pressure drop that will reduce the volume flux- an extreme example is what happens when you kink a hose. I only have some third-hand ancient engineering books that cover the topic, but there are tables that incorporate pressure drops due to, for example, sudden changes in pipe diameter. The reason, IIRC, is viscous dissipation.

 

[russ_watters]

I think this discussion went off in the wrong direction. It is true that fittings and bends can create a pressure drop that will reduce the volume flux- an extreme example is what happens when you kink a hose. I only have some third-hand ancient engineering books that cover the topic, but there are tables that incorporate pressure drops due to, for example, sudden changes in pipe diameter. The reason, IIRC, is viscous dissipation.

I'm not following: reduce the volume flux vs what?

Versus before you put the kink in the hose? Sure: it's trivially true that different systems may have different flow rates.

Upstream vs downstream of the kink? No.

The difference between these two scenarios is often misapplied, but I don't think we're talking about two different scenarios...though frankly it sounds like you switched back and forth between them in your answer.

 

[gmax137]

I think this discussion went off in the wrong direction. It is true that fittings and bends can create a pressure drop that will reduce the volume flux- an extreme example is what happens when you kink a hose. I only have some third-hand ancient engineering books that cover the topic, but there are tables that incorporate pressure drops due to, for example, sudden changes in pipe diameter. The reason, IIRC, is viscous dissipation.

Yes, the fittings and bends cause pressure drop, and the overall pressure drop affects the flow rate, but it affects the flow rate all along the hose. The velocity downstream of the kink is the same as the velocity upstream, assuming the hose diameters are equal up- and down-stream.

 

[physea]

Yes, the fittings and bends cause pressure drop, and the overall pressure drop affects the flow rate, but it affects the flow rate all along the hose. The velocity downstream of the kink is the same as the velocity upstream, assuming the hose diameters are equal up- and down-stream.

The velocity downstream and upstream a bend may be equal, but IN the bend? Won't it get affected?
And the turbulence created after the bend, won't affect the mean velocity nearly after the bend?

So, I understand that any losses inside a pipe will be exhibited as pressure drops and not velocity drops? (again I mean average velocity).

 

[Chestermiller]

The velocity downstream and upstream a bend may be equal, but IN the bend? Won't it get affected?
And the turbulence created after the bend, won't affect the mean velocity nearly after the bend?

So, I understand that any losses inside a pipe will be exhibited as pressure drops and not velocity drops? (again I mean average velocity).

See my post #11.

 

[boneh3ad]

The velocity downstream and upstream a bend may be equal, but IN the bend? Won't it get affected?
And the turbulence created after the bend, won't affect the mean velocity nearly after the bend?

So, I understand that any losses inside a pipe will be exhibited as pressure drops and not velocity drops? (again I mean average velocity).

Think of it this way. Let's say you have a pipe network with no branches (one inlet, one outlet) and you are pumping in water at a steady volumetric flow rate of $Q$. Along the entire length of that pipe network, regardless of the fittings, the volumetric flow rate must still be $Q$, otherwise you would have mass missing (total mass flow out has to equal total mass flow in).

Now, volumetric flow rate is related to cross-sectional area, $A$, and the average velocity, $V_{avg}$, over that area by $Q = V_{avg}A$. So, regardless of fittings or turbulence or anything else, in order to make sure mass is conserved, the average velocity is always $V_{avg} = Q/A$. If you pick two positions with the same cross-section, then the average velocity will be the same, even if there is a fitting between them, since $Q$ is constant.

Now, across any cross-section of the pipe, the velocity will not be constant. it will vary from zero at the wall to some maximum at the centerline (or at least near the centerline), and this fact is not reflected in the use of $V_{avg}$. Perhaps this is where you are getting confused. The relationship between the velocity at a specific point at that average velocity is
V_{avg} = \dfrac{\oint_A V(r,\theta)\;dA}{\oint_A dA} = \dfrac{1}{A}\oint_A V(r,\theta)\;dA.
So, in short, fittings and turbulence can affect the distribution of $V(r,\theta)$, but they won't change the value of $V_{avg}$ because that would break the law of conservation of mass.

For example, a laminar flow in a straight pipe will have a parabolic $V(r,\theta)$ profile centered at the centerline with no $|theta$ dependence. A turbulent flow would have a smaller maximum $V(r,\theta)$ but the profile would be "fuller", i.e. it would have a sharper gradient near the wall and a flatter profile near the centerline. A bend would tend to move the maximum point off of the centerline. Fittings could do any number of things depending on the type. The one universal law here is that $Q$ is constant but $V(r,\theta)$ is not.

 

[sophiecentaur]

The velocity downstream and upstream a bend may be equal, but IN the bend? Won't it get affected?

It is true to say that, if the volume flow rate is the same and the cross sectional area reduces then the velocity of the water must be higher (Area times speed = flow rate).
But the flow rate has to be the same all the way round - where else could water go to or come from?

 

[russ_watters]

The velocity downstream and upstream a bend may be equal, but IN the bend? Won't it get affected?

If there is a diameter change, yes. This the premise of the Venturi Effect.

And the turbulence created after the bend, won't affect the mean velocity nearly after the bend?

Correct. The mean volumetric flow rate has to be the same everywhere, everywhere and everywhere along the pipe.

So, I understand that any losses inside a pipe will be exhibited as pressure drops and not velocity drops? (again I mean average velocity).

Correct.

 

[physea]

But the cross sectional area shouldn't be always perpendicular to the flow? In a bend, it's not, so maybe there is always a change in the cross sectional area in bends, even though the bore of the pipe is the same?

 

[BvU]

So don't bother looking there. Before and after the bend, if the pipe diameter is the same then the v are also the same.

 

[physea]

What happens with rotameters in turbulent flow? Do they over-estimate velocity? Is there any data on this?

 

[BvU]

Brooks has plenty white papers I expect.

 

[russ_watters]

But the cross sectional area shouldn't be always perpendicular to the flow? In a bend, it's not, so maybe there is always a change in the cross sectional area in bends, even though the bore of the pipe is the same?

No. "Cross" is short for "across". As in "perpendicular". Now you're trying to break a definition in order to support a nonsensical and pointless idea you don't want to let go of.

 

[russ_watters]

What happens with rotameters in turbulent flow? Do they over-estimate velocity? Is there any data on this?

Flow in pipes is turbulent. Flow meters work. Yes, there's data on it (any flow meter spec sheet provides its accuracy). Please stop trying to argue your way out of a reality you don't like and just accept it.

 

[Chestermiller]

@physea You still haven't answered this: If you have water flowing continuously into your piping system at a constant flow rate and water coming out of your piping system continuously at a lower constant flow rate, what happens to all that extra water that is continuously building up within your piping system? For example, if you have water going in at 30 gallons per minute (gpm) and water coming out at 20 gpm, then the rate of buildup is 10 gpm. So after 1 minute, you would have an extra 10 gallons in your system. No big problem yet (maybe). But, after a day, you would have an extra 14400 gallons in your piping system. This would be enough to fill my entire living room. Where did it all go?

 

[Andy Resnick]

The difference between these two scenarios is often misapplied, but I don't think we're talking about two different scenarios...though frankly it sounds like you switched back and forth between them in your answer.

That could be. My understanding is that when unsteady flow is created (for example, just after a discontinuous expansion or flow through an orifice in a pipe, either one resulting in a vena contracta), the flow field that exists prior to resumption of steady flow is associated with viscous energy losses. My (ancient) book spends a chapter or so on hydraulic grade lines and energy grade lines. It describes "minor losses" and 'head losses', but isn't very specific on *what* is being lost.

 

[russ_watters]

That could be. My understanding is that when unsteady flow is created (for example, just after a discontinuous expansion or flow through an orifice in a pipe, either one resulting in a vena contracta), the flow field that exists prior to resumption of steady flow is associated with viscous energy losses. My (ancient) book spends a chapter or so on hydraulic grade lines and energy grade lines. It describes "minor losses" and 'head losses', but isn't very specific on *what* is being lost.

Unsteady flow is what you have when the flow field is in the process of changing, caused by the system setup changing. E.G., for the second it takes to bend a hose into a kink or few seconds it takes to actuate a valve or variable orifice. Durign this time, the volumetric flow rate is changing and is indeed also not constant along the pipe.

The OP asks about flow through fittings, which are welded/soldered in place before the system is even turned on. So I don't see a way to apply an unsteady flow to it. And I don't think it is really what you mean:

It sounds like you are describing undeveloped or developing flow. This is where the velocity profile across the pipe is changing along the pipe due to a discontinuity (in the image in the wiki, an entrance). After the discontinuity, where the pipe becomes uniform again, the velocity profile is still changing, but the average velocity is not.
https://en.wikipedia.org/wiki/Entrance_length#Hydrodynamic_entrance_length

 

[Asymptotic]

What happens with rotameters in turbulent flow? Do they over-estimate velocity? Is there any data on this?

Do a search on "flow meter installation guidelines".

How much turbulence affects flow meters depends on how they operate, and ranges from 'practically no effect' to 'significant'. Turbine (paddlewheel) sensors are ubiquitous, but sensitive to turbulent flow, and require (sometimes significant; up to 50, although usually no less than 10 upstream/5 downstream pipe diameters) of straight run to operate well.

Read sensor manufacturer literature (and post #21 and #32) for clues to the 'whys', but changes in average flow rate isn't one of them.

 

[boneh3ad]

That could be. My understanding is that when unsteady flow is created (for example, just after a discontinuous expansion or flow through an orifice in a pipe, either one resulting in a vena contracta), the flow field that exists prior to resumption of steady flow is associated with viscous energy losses. My (ancient) book spends a chapter or so on hydraulic grade lines and energy grade lines. It describes "minor losses" and 'head losses', but isn't very specific on *what* is being lost.

What is being lost is "total head", which is a code word for total pressure. If you think about it in terms of the modified Bernoulli equation, the total head would be
h_{total}=\dfrac{p}{\rho g} + \dfrac{V^2}{2g} + z,
where $z$ is a height. The relationship between two points in the flow is then
\dfrac{p_1}{\rho g} + \dfrac{V_1^2}{2g} + z_1 = \dfrac{p_2}{\rho g} + \dfrac{V_2^2}{2g} + z_2 + h_{loss}.
Generally, $V$ is set by conservation of mass through the system, so the losses are manifested as a loss in either pressure or elevation.

 

[Andy Resnick]

What is being lost is "total head", which is a code word for total pressure.

It sounds like you are describing undeveloped or developing flow. This is where the velocity profile across the pipe is changing along the pipe due to a discontinuity (in the image in the wiki, an entrance). After the discontinuity, where the pipe becomes uniform again, the velocity profile is still changing, but the average velocity is not.

I guess this is the source of my confusion: if there is a pressure loss associated with, for example, a sudden constriction, how is this manifested in the flow?

http://images.slideplayer.com/24/7380603/slides/slide_94.jpg

The figure above mentions that velocity increases but the kinetic energy decreases...??

 

[boneh3ad]

I guess this is the source of my confusion: if there is a pressure loss associated with, for example, a sudden constriction, how is this manifested in the flow?

http://images.slideplayer.com/24/7380603/slides/slide_94.jpg

The figure above mentions that velocity increases but the kinetic energy decreases...??

A sudden constriction generally causes separation bubbles to form in the corners, and these recirculating regions lead to very high viscous dissipation. Physically, that is what is causing the extra head loss in a sudden contraction (or expansion, for that matter). The slide here is fairly vague in what they are describing as energy. They don't mean kinetic energy, as that is clearly increasing with the velocity increase. They are referencing the total available pool of energy available to the flow, e.g. the sum of kinetic energy, gravitational potential energy, and "pressure energy".

In the picture they drew, the height isn't changing so we can ignore gravity. If you ignore losses, the velocity increase predicted by conservation of mass would result in a pressure drop predicted by the Bernoulli equation. If you then include losses due to, for example, the contraction, the velocity change would be the same (conservation of mass still applies), but the pressure after the contraction would be lower than that predicted by Bernoulli alone. That is the nature of the equation I provided earlier. It basically says the sum of all of the forms of energy at point 1 equals the sum of all of those terms plus terms representing losses at point 2.

 

[russ_watters]

I guess this is the source of my confusion: if there is a pressure loss associated with, for example, a sudden constriction, how is this manifested in the flow?

http://images.slideplayer.com/24/7380603/slides/slide_94.jpg

The figure above mentions that velocity increases but the kinetic energy decreases...??

That slide is very poorly worded: there are two sources of pressure drop, but only one source of pressure loss.

One drop is due to conservation of energy, e.g., the Venturi Effect. It is recoverable if the pipe expands later.

The other drop is due to drag, which is complicated for an orifice as described above. This is a true loss.

But you asked "how is this manifested", and I'm not sure what you mean. How is the energy loss when you push a box across the floor "manifested"?

 

[boneh3ad]

But you asked "how is this manifested", and I'm not sure what you mean. How is the energy loss when you push a box across the floor "manifested"?

When you push a box along the floor, the friction dissipates some of that energy as heat. Loosely speaking, the same general idea applies in fluids, and any process that results in higher viscous losses, such as separation, is going to result in greater energy dissipation.

 

[Andy Resnick]

The other drop is due to drag, which is complicated for an orifice as described above. This is a true loss.

But you asked "how is this manifested", and I'm not sure what you mean. How is the energy loss when you push a box across the floor "manifested"?

A sudden constriction generally causes separation bubbles to form in the corners, and these recirculating regions lead to very high viscous dissipation. Physically, that is what is causing the extra head loss in a sudden contraction (or expansion, for that matter). The slide here is fairly vague in what they are describing as energy. They don't mean kinetic energy, as that is clearly increasing with the velocity increase. They are referencing the total available pool of energy available to the flow, e.g. the sum of kinetic energy, gravitational potential energy, and "pressure energy".

Hopefully you have anticipated my next question- let me thread a horizontal pipe through a black box, and set the inlet pressure to be constant. At the outlet where the pipe opens to the outside air, fluid comes out at a constant rate. We may be tempted to measure that flow rate and compare it with, for example, a prediction assuming Poiseuille flow.

However, unbeknownst to you, inside the black box I welded two fittings; one is a constrictor that decreases the diameter and then a second fitting that increases the diameter back to the original diameter. I guess I could be even more simple and insert an orifice rather than weld two fittings. Maybe I left the pipe diameter constant but roughened the interior pipe walls- you have no idea.

But regardless of what I did, the inlet pressure is unchanged and the outlet pressure is unchanged. I suppose the volume flow (say, m^3/s) changes, so our predicted flow rate won't match. We would ascribe that to some sort of 'pressure loss' ('minor loss') within the hidden section of pipe, right?

The OP asked if there is a loss of kinetic energy or velocity through a fitting. Here, volume flow is reduced in comparison to Poiseuille flow due to a pressure loss. It would seem that would relate to loss of kinetic energy and/or velocity, right?

 

[physea]

So how can I explain that the difference between Darcy losses and experimentally calculated, is consistently increasing in a exponential-like way?
The set up is just a pump that pumps water inside a pipe with bends and the end of the pipe let's the water flow out freely.

The only factor I found that causes such systematic error is velocity!

 

[boneh3ad]

In the black box example, you are correct. The losses inside would mean that, for a constant inlet and outlet pressure, the flow rate would drop relative to the case of Poiseuille flow. That's not how I interpreted the OP's question, though, nor did anyone else.

If that's what he meant then it illustrates the importance of being as exact as possible when describing you problem. We all believed he was asking whether a fitting or other loss will slow down the downstream flow relative to the upstream flow without area change

 

[russ_watters]

Hopefully you have anticipated my next question- let me thread a horizontal pipe through a black box, and set the inlet pressure to be constant. At the outlet where the pipe opens to the outside air, fluid comes out at a constant rate. We may be tempted to measure that flow rate and compare it with, for example, a prediction assuming Poiseuille flow.

However, unbeknownst to you, inside the black box I welded two fittings; one is a constrictor that decreases the diameter and then a second fitting that increases the diameter back to the original diameter. I guess I could be even more simple and insert an orifice rather than weld two fittings. Maybe I left the pipe diameter constant but roughened the interior pipe walls- you have no idea.

So it is a box with a balancing valve in it, of unknown setting. Fair enough.

But regardless of what I did, the inlet pressure is unchanged and the outlet pressure is unchanged. I suppose the volume flow (say, m^3/s) changes, so our predicted flow rate won't match. We would ascribe that to some sort of 'pressure loss' ('minor loss') within the hidden section of pipe, right?

The OP asked if there is a loss of kinetic energy or velocity through a fitting. Here, volume flow is reduced in comparison to Poiseuille flow due to a pressure loss. It would seem that would relate to loss of kinetic energy and/or velocity, right?

You're making the same mistake the OP did, and maybe it means that your previous question was asked backwards. But I'll come back to that.

First off, if the flow rate into and out of the box are the same and the pressure on the inlet and outlet of the box are the same, then you haven't lost anything! To "lose" means you don't have something anymore. To "lose" it in the box means it's still in the box! So what did we lose in the box? We lost *pressure*. The *pressure* at the inlet of the box is higher than the pressure at the outlet of the box.

Now, depending on what is causing the flow to happen in the first place, odds are good you also will have a lower flow with the second scenario than the first, but that isn't necessarily true. We don't know enough about the scenarios (namely, what is causing the flow to happen -- a pump? under control? What kind) to say whether the flow rates will be different.

So you are indeed mixing and matching what happens in one scenario vs what happens between two different scenarios. "Loss" is not referring to how one scenario differs from another, it is about what happens in a particular scenario.

And back to "manifest". You asked how pressure drop manifests, but I think that's the answer to the question you really needed to ask: how does energy loss manifest? Via permanent pressure drop (loss) across a fitting.

 

[Chestermiller]

So how can I explain that the difference between Darcy losses and experimentally calculated, is consistently increasing in a exponential-like way?
The set up is just a pump that pumps water inside a pipe with bends and the end of the pipe let's the water flow out freely.

The only factor I found that causes such systematic error is velocity!

In my judgment, based on your prior responses, the reason the calculation does not match the observations adequately is that you did the calculation incorrectly. Since you have not seen fit to share the details of the system design and operating parameters with us, we can't help you find your error. It seems to me you are just continuing to waste everyone's time.

 

[russ_watters]

Or, like I said in your other thread, maybe it's a feature, not a bug; two similar quadratic functions with different constants will diverge.

But as said, unless we see the data and calculations, we're only guessing at what the problem might be.

 

[Andy Resnick]

So it is a box with a balancing valve in it, of unknown setting. Fair enough.

You're making the same mistake the OP did, and maybe it means that your previous question was asked backwards. But I'll come back to that.

First off, if the flow rate into and out of the box are the same and the pressure on the inlet and outlet of the box are the same, then you haven't lost anything! To "lose" means you don't have something anymore. To "lose" it in the box means it's still in the box! So what did we lose in the box? We lost *pressure*. The *pressure* at the inlet of the box is higher than the pressure at the outlet of the box.

Now, depending on what is causing the flow to happen in the first place, odds are good you also will have a lower flow with the second scenario than the first, but that isn't necessarily true. We don't know enough about the scenarios (namely, what is causing the flow to happen -- a pump? under control? What kind) to say whether the flow rates will be different.

So you are indeed mixing and matching what happens in one scenario vs what happens between two different scenarios. "Loss" is not referring to how one scenario differs from another, it is about what happens in a particular scenario.

And back to "manifest". You asked how pressure drop manifests, but I think that's the answer to the question you really needed to ask: how does energy loss manifest? Via permanent pressure drop (loss) across a fitting.

I wonder if we are speaking past each other.

As I set up the problem, it should have been clear that I am considering pressure-driven flow: Poiseuille flow applies for axisymmetric laminar viscous flow through a straight circular impermable tube of radius R, resulting in a flow rate Q= π/8μ dP/dz R⁴. I can hold the inlet and outlet pressures constant.

As for: "So you are indeed mixing and matching what happens in one scenario vs what happens between two different scenarios. "Loss" is not referring to how one scenario differs from another, it is about what happens in a particular scenario.", I guess that makes sense. For example, in Poiseuille flow there is a pressure loss between inlet and outlet. But isn't the point of "head loss coefficients" to compare to some ideal configuration?

In the end, we seem to be agreeing with each other: fittings can introduce a permanent pressure drop. There should also be a loss of momentum in addition to energy, because there is an interaction between the fluid and fitting, resulting in a transfer of energy and momentum from the fluid to the fitting.

 

[Andy Resnick]

If that's what he meant then it illustrates the importance of being as exact as possible when describing you problem. We all believed he was asking whether a fitting or other loss will slow down the downstream flow relative to the upstream flow without area change

I don't know about 'we ALL believed', but I agree the OP was ambiguous.

 

[Chestermiller]

The OP has last been on site at 7am today, and still has not provided us either a diagram of the system or his calculation. How do responders feel about closing this thread?

 

[physea]

The OP has last been on site at 7am today, and still has not provided us either a diagram of the system or his calculation. How do responders feel about closing this thread?

wow

 

[russ_watters]

I wonder if we are speaking past each other.

I don't think so. I'll try to focus:

In the end, we seem to be agreeing with each other: fittings can introduce a permanent pressure drop. There should also be a loss of momentum in addition to energy, because there is an interaction between the fluid and fitting, resulting in a transfer of energy and momentum from the fluid to the fitting.

This is wrong: there is no [permanent] loss of momentum, only a loss of energy across the fitting.

 

[russ_watters]

The OP has last been on site at 7am today, and still has not provided us either a diagram of the system or his calculation. How do responders feel about closing this thread?

I'd agree. And he can pm you (Or me) the requested info to reopen the thread with.