Log mean temperature difference, ft applied to heat exchangers. problem 4.9-2, p351. oil flowing at the rate of 5.04 kg/s (cpm = 2.09 kj/kg. k) is cooled in a 1-2 heat exchanger from 366.5 k to 344.3 k by 2.02 kg/s of water entering at 283.2 k. the overall heat transfer coefficient u0, is 340 w/m^2 .k. calculate the area required. (hint: a heat balance must first be made to determine the outlet water temperature.)

the Area required is 12m2

Explanation:

A heat exchanger is a mechanical device used to transfer heat from one fluid to another by heating or cooling one of the fluids with the help of the other. They are of different types which include;

Shell and Tube heat exchangers, that categorically involves large number of small tubes embedded within a cylindrical shell. It may be

1 shell 1 tube pass (1-1 exchanger) which means the heat enters through one end and travels on a straight line to exit the other end.

1 shell multi-tube pass (1-2,1-4,1-6,1-8 etc exchangers) meaning the heat enters through one end, and travels in a directed path to exit a different end.

To calculate the quantity of heat exchanged for both cases, we use the same formula

Q=UA∆Tlm

Where Q= Rate of Quantity of heat exchanged

U= Heat transfer coefficient

A=Area of tube

Tlm= Logarithm mean temperature difference which signifies the average logarithm of the difference in temperature between the hot region and the cold region of both ends of the pipe exchanger.

The major difference is that,

For 1-1 exchanger, Tlm=Tm= [(Thi-Tco)-(Tho-Tci)]/ln[(Thi-Tco)/(Tho-Tci)]

But

For 1-2, 1-4, 1-6 etc exchangers, Tlm=Tm×F

Where F is called the correction factor, and it is the point on the graph (of correction factor (F) plotted against Log mean temperature difference (Tlm) for cross-flow exchangers) where the Y value and Z value intercepts.

The Y and Z values can be calculated using the Formulas below

Y=[(Tco-Tci)/(Thi-Tci)] and Z=[(Thi-Tho)/(Tco-Tci)]

For the question above, to find the Area of tube, we need to calculate the rate of quantity of heat exchanged (Q), then use the value to find the outlet temperature of cold fluid (Tco), then find the Log mean temperature difference which will require F,Y,Z and Tm, then use the results gotten to compute the Area.

Given;

Mass flow rate of hot fluid (oil), Mh=5.04kg/s

Specific heat of hot fluid, Cph=2.09kj/kg.k

Inlet temperature of hot fluid, Thi=366.5K

Mass flow rate of cold fluid (water), Mc=2.02kg/s

Specific heat of cold fluid, Cpc=4.186kj/kg.k

Inlet temperature of cold fluid, Tci=283.2K

Outlet temperature of hot fluid, Tho=344.3k

Outlet temperature of cold fluid = Tco

Overall heat transfer coefficient, U= 340W/m2.k

Area of tube =A

First we have to find the quantity of heat exchanged (Q). We use,

Qh=∆Hh=MhCph(Tho-Thi)

=5.04×2.09 × (344.3-366.5)

=-233.85kj/s

The quantity of heat lost by the hot fluid Qh is equal to the quantity of heat gained by the cold Fluid Qc.

Therefore;

Qh=Qc

Qc=233.85kj/s

To calculate Tco, we use

Qc=∆Hc=McCpc(Tco-Tci)

233.85=2.02×4.186×(Tco-283.2)

233.85=8.45572Tco-2394.66

Tco=(233.85+2394.66)/8.45572=310.86k

Next, we use the formula Q=UA∆Tlm to get our A, but for 1-2 exchanger,

Tlm=Tm×F

So we find F using Y and Z values

Y=(Tco-Tci)/(Thi-Tci)=(310.86-283.2)/(366.5-283.2)=0.3321

Z=(Thi-Tho)/(Tco-Tci)=(366.5-344.3)/(310.86-283.2)=0.8026

At point Y=0.3321 and point Z=0.8026, we have an intercept of F=0.975

Also,

Tm=[(Thi-Tco)-(Tho-Tci)]/Ln[(Thi-Tco)/(Tho-Tci)]

=[(366.5-310.86)-(344.3-283.2)]/ln[(366.5-310.86)/(344.3-283.2)]

=(-5.46)/(-0.0936)=58.33k

∆Tlm=Tm×F

=58.33×0.975

=56.872k

Now to get A, we input all calculated values into our general formula;

Q=233.85Kj= (233.85×1000)j

Q=UA∆Tlm

233.85×1000=340×A×56.872

A=(233.85×1000)/(340×56.872)=12m2

Therefore, the Area required is 12m2.

x+2 because it just doubles hon so the pattern can actually tell you the expression.

so for example,

if x=2 then 2+2 is 4

if x=4 then 4+4 is 8

and so on and so forth.

i really hope that this makes sense and you out in a way hon: )

diy

explanation: