Example 15.4-2: Parallel or Counter-Flow Heat Exchangers

Heather Dore and Chris Powers

CENG 402 Project

Begin with a Macroscopic Energy Balance for the whole stream.

We define Th1 and Th2 as the hot stream inlet and outlet temperatures, respectively. Similarly, Tc1 and Tc2 represent the inlet and outlet temps of the cold stream. These are defined as show in Fig 15.4-2 on page 466 of BS&L.

> restart;

We begin by computing a steady-state energy balance according to equation 15.1-4

> eq1:=Delta*(Uhat+p*Vhat+KEhat+Phihat)=Qhat-What;

We assume negligible changes in kinetic and potential energy. There is no outside work on the system, either. Also, we should note for later use that Qc=-Qh.

> KEhat:=0: Phihat:=0: What:=0:

Also, U+pV is the enthalpy per unit mass.

> Uhat:=Hhat-p*Vhat;

> eq1;

Since the enthalpy is per unit mass, we need to multiply by the mass flow rate w to get the total heat, Q.

> eq2:=w*DeltaHhat=Q;

We have this equation for both hot and cold streams, so

> eq3:=wh*DeltaHhath=Qh; eq4:=wc*DeltaHhatc=Qc;

For this situation, we can assume an incompressible liquid flowing under reasonable pressure drop or an ideal gases. With this assumption for constant Cp, DeltaHhat=Cphat*DeltaT.

> DeltaHhath:=Cphath*(Th2-Th1); DeltaHhatc:=Cphatc*(Tc2-Tc1); eq3; eq4;

We will use this result later in the analysis. At this point, we move on to consider a Macroscopic Balance applied in differential form.

A similar derivation of Eq. 15.1-4 to a segment dz yields

> eq5:=wh*Cphath*dTh=dQh;

We can now replace dQh by an equation with the overall heat transfer coefficient in it; specifically, either equation 13.1-8 or 9.6-3.

> dQh:=U0*(2*Pi*r0*dz)*(Tc-Th);

Here, r0 is the inside pipe radius, and U0 is overall heat transfer coefficient).

> eq5;

This shall be arranged for easier integration.

> eq5:=eq5/(Tc-Th)/(wh*Cphath);

We now show the corresponding equation for the cold stream.

> eq6:=-dTc/(Tc-Th) = 2*U0*Pi*r0*dz/(wc*Cphatc);

By adding these two equations, we obtain the following result, which gives the temperature difference between the two fluids at any cross-section in the exchanger.

> eq7:=eq5+eq6;

The differential Th and Tc can be combined in one differential, and some other rearrangement yields eq 15.4-13

> eq8:=-d(Th-Tc)/(Th-Tc)=U0*((wh*Cphath)^(-1)+(wc*Cphatc)^(-1))*(2*Pi*r0*dz);

We now assume that U0 is independent of z, and we integrate between the inlet and the outlet. First, the Left Hand Side of the Equation.

> LHS:=int((-1/(deltaT)),deltaT=deltaT1...deltaT2);

In order to simplify the integration for MAPLE, we set Th-Tc=deltaT as shown above.

> deltaT2:=(Th2-Tc2); deltaT1:=(Th1-Tc1); LHS;

Now, we integrate the Right Hand Side of the Equation, from z=0 to z=L.

> RHS:=int((2*U0*(1/(wh*Cphath)+1/(wc*Cphatc))*Pi*r0),z=0...L);

> eq9:=LHS=RHS;

This gives us Bird's result from Equation 15.4-14

Now we can plug in our earlier result for the steady-state whole-stream energy balances.

> wh:=solve(eq3,wh);

> wc:=solve(eq4,wc);

> eq9;

But Qh=-Qc, so

> Qh:=-Qc;

> Qc:=solve(eq9, Qc);

Realizing that 2*Pi*r0*L is the total inside surface area, we can simplify.

> L:=A0/(Pi*r0*2);

> Qc;

Note that all of the above temperature operations are commonly grouped together and called the log mean temperature difference, notated (Th-Tc)ln.