Heather Dore and Chris Powers

CENG 402 Project

Example Problem: Double-Pipe Exchanger Heat Transfer Rates

Now we solve a sample problem: The hot liquid is an oil and the cold liquid is cooling water.

Part 1: Here we will determine what the required heat exchanger area is for the following conditions in counter-current flow.

> restart;

> U0:=180*Btu/(hr*ft^2*R):

> wh:=-5000*lbm/hr: wc:=5000*lbm/hr: Cphath:=0.5*Btu/(lbm*R): Cphatc:=1.0*Btu/(lbm*R): Th2:=(190+459.67)*R: Th1:=(110+459.67)*R: Tc1:=(80+459.67)*R:

Use eq 15.4-8 and 15.4-9, and the relation that Qh=-Qc, to solve for Tc2.

> Qh:=wh*Cphath*(Th2-Th1); Qc:=wc*Cphatc*(Tc2-Tc1);

> eq1:=Qc=-Qh;

> Tc2:=solve(eq1,Tc2);

Thus, the cooling water return (water outlet) Temp should is 120 F. Now we can use eq 15.4-15 or -16 to solve for the required area.

> eq2:=Qc=U0*A0*((Th2-Th1-Tc2+Tc1)/(ln((Th2-Tc2)/(Th1-Tc1))));

> A0:=solve(eq2,A0);

Part 2: What is the minimum amount of water that can be used to achieve the desired oil temperature change?

This means Th2 and Th1 should be the same as before. Also, Tc1 must be the same as well. Thus, we should increase the area and decrease the anount of water flow to achieve this result.

This is for Part 1, counter-current flow.

> Tc2:='Tc2': wc:='wc':

Get required Qh and equation for Qc..

> Qh:=wh*Cphath*(Th2-Th1); Qc:=wc*Cphatc*(Tc2-Tc1);

Now the maximum temperature that the cooling water can be heated up to would be 190 F, which gives a 0 degree tempertaure approach (unrealistic).

> Tc2:=(190+459.67)*R:

Now we solve for the amount of water required at these conditions.

> eq:=Qc=-Qh;

> wc:=solve(eq,wc);

This is roughly 40 % of the cooling water used in the previous case.

This is for Part 1, parallel-flow.

> Th2:=(110+459.67)*R: Th1:=(190+459.67)*R: Tc2:='Tc2': wc:='wc': wh:=5000*lbm/hr:

Now follow similar computation method as above

> Qh:=wh*Cphath*(Th2-Th1); Qc:=wc*Cphatc*(Tc2-Tc1);

The outlet water stream temperature should not be any higher than the outlet oil stream; otherwise, we would be heating the oil back up with the water. Thus, the water outlet, Tc2=110 F.

> Tc2:=(110+459.67)*R:

> eq:=Qc=-Qh;

> wc:=solve(eq,wc);

To obtain the same temperature decrease in the oil, it should be noted that parallel flow requires a much larger flow rate of cooling water than does counter-current flow. This is because counter-current flow is more effecient.

Part 3: Here we will determine the required heat exchanger area as in Part 1, but with the heat-transfer coeffcient varying linearly with water temperature..

This requires the use of equation 15.E-1

> U1:=25*Btu/(hr*ft^2*R): U2:=175*Btu/(hr*ft^2*R): Tc2:='Tc2': A0:='A0':

> wh:=-5000*lbm/hr: wc:=5000*lbm/hr: Th2:=(190+459.67)*R: Th1:=(110+459.67)*R: Tc1:=(80+459.67)*R:

> Qh:=wh*Cphath*(Th2-Th1); Qc:=wc*Cphatc*(Tc2-Tc1);

> eq1:=Qc=-Qh;Tc2:=solve(eq1,Tc2);

This gives the same temperature as in Part 1.

Now we apply equation 15.E-1 in order to see the effect of linearly varying U values.

> eq2:=Qc=A0*((U1*(Th2-Tc2)-U2*(Th1-Tc1))/(ln((U1*(Th2-Tc2))/(U2*(Th1-Tc1)))));

> A0:=solve(eq2,A0);

Thus, with the heat transfer coeffecient varying linearly, the required heat exchanger area is greater.

Part 4:

What is the required area if we have parallel flow operation and the mass flow rate of water is 24500 lbm/hr, with a U0 constant at 180*Btu/(hr*ft^2*R)? Since it is parallel flow, both flow rates have the same sign.

> wh:=5000*lbm/hr: wc:=24500*lbm/hr: U0:=180*Btu/(hr*ft^2*R): Th2:=(110+459.67)*R: Th1:=(190+459.67)*R: Tc1:=(80+459.67)*R:

> Tc2:='Tc2': A0:='A0':

> Qh:=wh*Cphath*(Th2-Th1); Qc:=wc*Cphatc*(Tc2-Tc1);

> eq3:=Qc=-Qh;

> Tc2:=solve(eq3,Tc2);

> eq4:=Qc=U0*A0*((Th2-Th1-Tc2+Tc1)/(ln((Th2-Tc2)/(Th1-Tc1))));

> A0:=solve(eq4,A0);

This is similar to Part 1, but with a much higher water flow rate and the same oil flow rate. We obtain the same oil temperature difference with less required heat exchanger area when the water flow rate is increased.