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A shell-and-tube heat exchanger is to used to heat water (in the tube side) from 30...

A shell-and-tube heat exchanger is to used to heat water (in the tube side) from 30 deg C to 40 deg C at a mass flow rate of 4 kg/s. The fluid used for heating (shell side) is water entering at 90 deg C with a mass flow rate of 2 kg/s. A 1-2 STHE is used and the overall heat transfer coefficient based on the inside area is 1390 W/m2-K. The tubes are 1.875 in diameter (inside) and require an average velocity of 0.375 m/s.

What is the effective (corrected) temperature driving force in deg C?

a. 37
b. 35
c. 42
d. 30

How many tubes are available per pass?

a. 31
b. 20
c. 39
d. 43

What is the required heat transfer area in m2?

a. 4.2
b. 5.7
c. 4.8
d. 5.2

What length of pipe will be required to accomplish the desired heat transfer?

a. 1.12 m
b. 2.12 m
c. 1.79 m
d. 2.5 m


Solutions

Expert Solution

solution :

Tube side fluid = water

Inlet temp. T1 = 30 c

Outlet temp. T2 = 40 c

Mass flow rate Mwt = 4 kg/s

Cp of water = 4.2 KJ/kg*c

Heat required to increased the temp. of tube side fluid = Mwt * Cp* (T2-T1)

                           

Q= 4*4.2*(40-30)

Q = 168 KJ/s

Shell side fluid = water

Inlet temp. (t1) = 90 c

Outlet temp. (t2) =??

Mass flowrate Mws = 2 KG/s

Heat supplied by shell side fluid = 168 kJ/s

Q= Mws * Cp * (t1-t2)

168 = 2 * 4.2 * (90- t2)

So     t2 = 70 C

for a 1-2 STHE ;

effective (corrected) temperature driving force =

effective (corrected) temperature driving force =

FOR LMTD CORRECTION FACTOR :

USING THE VALUE OF R AND S;

WE FIND LMTD CORRECTION FACTOR = 0.98

ACTUAL LMTD = FT * TEMP. DRIVING FORCE = 0.98 * 43.28 = 42.55 C

the effective (corrected) temperature driving force in deg C = 42 C

2)  

U =  1390 W/m2-K

Q = U*A * the effective (corrected) temperature driving force in deg C

168 KJ /S = (1.390 KJ/S*M^2*K) * A * 42

A = 2.88 M^2

The tubes are 1.875 in = 0.047625 M diameter (inside) and require an average velocity of 0.375 m/s.

DENSITY OF WATER 1000 KG/M^3

VISCOSITY OF WATER = 8.90* 10 ^4 KG/M*S

REYNOLD NO. =

REYNOLD NO   = 20066

SO

SO AS PER STANDARD TABLE

FOR 1 SHELL 2 TUBE PASSES ;

NO. OF TUBES =20

3)

SO ACTUAL AREA REQUIRED = 20 * AREA OF 1 TUBE

= 20* ( 2.88 /12 )

= 4.8 M^2

4) LENGTH OF TUBE :

AREA = nt * 3.14 * D * L

4.8 = 20 * 3.14 * 0.047625 * L

L = 1.60 M


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