A counter flow gas turbine cooler is to be designed to recover heat from exhaust

Question

A counter flow gas turbine cooler is to be designed to recover heat from exhaust gases. The mass flow rate for each gas stream is 2.5 kg/s, i.e. nearly balanced counter flow. The incoming air of the cold stream is 20 C and the incoming air of the high temperature gas stream is 500 C. Due to space limitations, the heat exchanger is composed of 25 channels for each gas stream and utilizes the internal fin structure with performance characteristics. The dimensions of each channel are L = 30 cm, w =30 cm, and H = 6.35 nun. Determine the heat transfer rate Q and the outlet temperatures of each gas stream. You may neglect the wall conduction resistance. What pressure drop results for each gas stream? Are they the same, and if not explain.

Explanation / Answer

solution:

1)here as only mass flow rate and inlet temperature are given hencce we have to solve by NTU method of heat exchanger as follows

2)here heat capacity ratio is given as follows

C=Cmin/Cmax=m*Cpc/m*Cph=1.006/1.093=.9204

3)here each flow we have to calculate heat transfer coefficient and hence for hot gases

mass flow rate=m/25=2.5/25=.1 kg/s per channel

m=density*A*V

A=W*H=300*6.35*10^-6 m2

so we get velocity as

v=115.11 m/s

here reynold number is

Re1=VDh/nu

Dh=2(H*W)/(H+W)=12.43 mm

Re=18228.04

Nu for cooling of gases

Nu=.023*RE^.8*Pr^.3

Nu=52.94

Nu=h1*Dh/K

h1=238.54 w/m2k

3)in same way for cold air we have

mass flow rate=m/25=2.5/25=.1 kg/s per channel

m=density*A*V

A=W*H=300*6.35*10^-6 m2

so we get velocity as

v=43.562 m/s

here reynold number is

Re1=VDh/nu

Dh=2(H*W)/(H+W)=12.43 mm

Re=36099.04

Nu for heating of gases

Nu=.023*RE^.8*Pr^.4

Nu=89.26

Nu=h2*Dh/K

h2=179.52 w/m2k

3)where overall heat transfer coefficient

U=[1/(1/h1+1/h2)]=102.43 w/m2k

total area=25*2(W*L+H*L)=4.5952 m2

4)here NTU are

NTU=UA/m*Cpc=102.43*4.5952/.1*1.006=4678.82

5)effectiveness is raio of actual heat transfer to maximum possible heat transfer and given by

ef=1-e^-[(1-C)NTU]/1-C*e^-[(1-C)NTU]=1

ef=Tc2-Tc1/Th1-Tc1

Tc2=Th1=773 k

hence from energy balance

1.093(Th1-Th2)=1.006(Tc2-Tc1)

Th2=331.2 k

6)here overall heat transfer is

Q=25*m*Cph(Th1-Th2)=1207.21 kw

7)here pressure drop is given by

p=f*(L/Dh)*density*V^2/2

Tmh=Th1+Th2/2=552.1 k

Tmc=Tc1+Tc2/2=533 k

as density is inversely proportional to temperature hence

500/552.1=density1/.0456

densityh=.04129g/m3

for cold air

533/20=1.205/densityc

densityc=.04521

friction factor=16/Re

fh=16/RE1=8.777*10^-4

fc=4.432*10^-4

hence pressure for gases

pg=5.8048 m

for air

pc=.4588 m

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