How to Average Heat Transfer Coefficients

<1)...calculate...exit gas velocity and density.>

Apply the known inlet and outlet conditions to the gases to obtain their densities at those conditions. Apply the principle of conservation of mass, and carry out a mass balance on the system:

Σ[Input flows] = Σ[output flows] + Σ[accumulation rates].

<2)...meaningful value...U...>

What would the meaning be, i.e., what message would it convey? What needs to be done with this information?

Thanks for the reply though conservation of mass is just one requirement. By only applying this parameter I could have any number of pressures / density / velocity combinations. The question is which one is the right one? Appologies if I was unclear about the heat transfer coefficient. In my application the heat transfer coefficient varies from inlet to outlet by a factor of approx 3. Since a simple heat transfer calculation only uses a single heat tranfer coefficient value then what value do I use? The arithmetic average between inlet and outlet perhaps or is there a recognised way of dealing with this situation. Thanks Mark-G

<2)...meaningful value...U...>

What would the meaning be, i.e., what message would it convey? What needs to be done with this information?

There can be alot going on in a heat exchanger. Everything changes when you are changes flow, temperatures and such.

If this is for a customer, you can ask him what are the conditions he uses, other wise, you can use MS-Excel and make a chart with the results for some of the possible conditions. And use that as a somewhat baseline

Results can be from empirical info or pulled from a program used to size it.

p911

I smell something wrong in "the gases cool massively, the density fall massively and the velocity actually decelerates". Is not a matter of thermodynamics: a mass balance shows that the product among density, velocity and cross section (i.e., the mass flow rate) has to be constant along the HE.

Anyway, I guess that the simplest approach may be to consider a small number of sections (say, 3 or 4) for which the variation of U can be thought as small enough to allow using a plain average (between Uin and Uout) without introducing an important error. Then you can solve the problem as if it were a series of 3 or 4 heat exchangers.

Is this a homework problem?

Hi Mark-G,

Your spec doesn't add up in the real world! You talk of 0.5 bar abs;surely you mean Bar G! (unless your engine is exhausting into a condenser!!)

How do you manage to get 900C out of the engine? 550/600 yes;higher and the engine is not working in the real world, unless you are refiring into the flue gases.

You talk of exit temps of 100C this will put you below the acid dewpoint of the gases and cause ,at the very least,condensation, deposition build-up and eventually blocking of the tubes. You are using extremely small bore tubes,c10mm bore, for such an application,they will become fouled; in fact it would appear that the equipment was designed for an entirely different purpose.

Hopefully this is just a 'back of a fag packet' brain stormer and not a serious attempt at an engineering solution;or perhaps it is just homework after all!

If you want a quick answer to the velocities etc just apply the Universal Gas Laws equation and save a lot of debate.

Good luck,

Massey.

Since the gas is not condensing, the overall coefficient will be controlled by the gas side coefficient, which is quite small. While one can have hours of fun calculating parameters at each point along the tube length, for all practical purposes averaging between the inlet and outlet should be adequate.

The problem is more complex since the convection coefficient is function of Reynolds via the Nusselt criterion and the velocity is function of the temperature which depends on the quantity of energy transferred between entry and considered point along the tube thus depending on the convection coefficient. It is the snake which bites its tail! The best way is to write the differential equation and solve it making before e few simplifications as considering for the gas physical properties averaged values between the temperatures at inlet and outlet and as constant over the whole length. Before we go into details some remarks made above are pertinent especially with respect to lowest temperature. It depends on the used materials, the lower it is the higher the risk to get a condense rich is SO3H2 or even SO4H2. Under industrial conditions for long range usage the outlet is maintained around 150°C in order to avoid it. If the walls are of stainless steels and use short in time a lower temperature can be afforded. The thermal resistance of the convective layer to water and the conductive tube are very low in comparison to the one presented by the convection (forced) between gas and wall. This means that it is possible to assume the wall temperature as almost constant (it is not really true but the variations are a lot less than the gas temperature) further to above assumption for gas properties as almost constant. For forced convection the Nusselt criterion is Nu=0.023*Pr^(1/3)*Re^(0.8); Nu= h*d/kf Where: Pr= Prandl criterion = 0.7 for diatomic gases [dimensionless] Re= w*d/ν = averaged velocity of fluid * nominal length/ cinematic viscosity [dimensionless] h = convection coefficient [W/m^2°K] kf= conductivity of gas [W/m°K] In the tube the only constant is the mass flow. Gas velocity will be at a given point w(x)=M'*v(x) Where M' = dM /dt=mass flow in [kg/s] and v(x)= specific volume of gas under conditions at "x" [m^3/kg]. Since the pressure drop is low it could be considered that v(x)=Cv*T(x). Here may I make same remark as a comment above pressure cannot be 0.5 bar(abs) since atmospheric pressure is 1 bar(abs) and for a gas flow the pressure at inlet MUST be higher than at outlet! We obtain thus: h(x)*d/kf=0.023*Pr^(1/3)*(d/ν)^0.8*M'*Cv*T(x) This is based on the assumption that as well kf as ν are independent of the coordinate "x". If we group all constant values together in a global constant we obtain that h(x)=Ch*T(x)^0.8. Now let us write the power balance for a "dx" element at distance "x" from inlet: Power content P(x)= Cp*M'*T(x) and the derivative is →d(P)/dx= Cp*M'*d(T(x))/dx Power loss due to convection d(P)=- h(x)*π*d*(T(x)-Tw)*dx → dP/dx=- Ch*T(x)^0.8*π*d*(T(x)-Tw)→ dP/dx= - Ch*T(x)^0.8*π*d*(T(x)-Tw) = Cp*M'*d(T(x))/dx If we consider Tw= constant we can introduce a new variable θ(x)= T(x)-Tw → dθ/dx=dTx/dx - Ch*(θ+ Tw)^0.8*π*d*θ = Cp*M'*dθ/dx This equation, integrated gives the values for θ and thus for T(x) as function of all involved parameters: -Ch*(θ+ Tw)^0.8*π*d*θ = θ' → -(Ch*π*d /(Cp*M'))*θ *(θ+ Tw)^0.8= θ' → θ' = - Co*θ *(θ +Tw)^0.8 In a 1st approximation which is not far from exact result it can be assumed that the exponent is 1 instead of 0.8 the equation can be thus written as : → dθ / (θ^2+θ*Tw)= - Co*dx And can be then directly integrated since the variables are separated. The integral left is of the type du/(a*u^2+bu+c) with a=1, b=Tw and c=0 thus Δ= 4*a*c-b^2=-Tw^2 For Δ<0 the result is : I= 1/b*ln ((2*a*u/(2*a*u+2*b)) +Ci= 1/Tw*ln(θ/(θ+ Tw)) +Ci= - Co*x → θ/(θ+ Tw ) = exp [- (Co*Tw*x +Ci)] With boundary condition x=0 →θ(0)=(Tin –Tw) → (Tin-Tw)/(Tin-Tw+ Tw)= exp(-Ci) → Ci=-ln(1-Tw/Tin) If for x=L the computed Tout is different from the expected (wished) value the some parameters have to be adapted as tube length, diameter or number. The basic equation with exponent 0.8 can be integrated numerically with a Runge-Kutta algorithm even on an XLS sheet! This approach is not the most precise but it is VERY fast and allows a good estimation. The other approach is to integrate the equation with all variables but it is cumbersome if one does not have the right software and requires A LOT of work. Hope it will help Nick

Thanks Nick Name. You've spent a lot of offort writing this reply and It's going to take a while to digest it all. Just picking up on a couple of your point's you'll see I've mentioned these in my previous reply. The low temperature output option is just one of many operating points. The exchanger works in conjunction with a bypass pipe with the toal flow being split between the exchanger and the bypass. We need to be able to regulate the final combined exit temperature to any point between 900 and 100 deg C. This can be achieved by controling the percentage of gases going into the exchanger and the bypass and suitably controlling the exit temp from the exchanger. The strategy is to control the final temp by maintaining the exchanger exit temp at 180 deg C and varying mass flow split until we reach a point where the final temperature must fall below 180 deg C and then the exchanger too must fall below this limit. It's use is for research and an inevitable limited life will be acceptable. Thanks, Mark-G

Thank you for all the replies. Sorry for the delay in responding but I've been away a couple of says. 1) This is not a homework question. It's a project to make a heat exchanger that will be used in a test cell for engine development. I did make a mistake with my original description, I stated that as the gases cool from 900 deg C down to 100 deg C the density also falls. it does not of course, my mistake, it rises and that's why the velocity falls. 2) The target to make the exit gas temp just 100 deg C is a requirement to meet one of the engine operating points. We recognise that this will cause acidic condensation and lead to increased fouling rates but the time spent at this condition will be limited and the exchanger will be positioned to allow easy rodding out of the gas tubes via access covers. The limited running time and the small size of the exchanger is such that it is probably more reasonable to swap the exchanger for a new one than make it out of more exotic materials than stainless steel. 3) The gases are at subatmospheric pressure because the exit from the exchanger is fed back into the engine inlet manifold. Now it's time for me to study the replies in full. Many thanks once again.