A shell and tube heat exchanger as a kettle boiler to make s

A shell and tube heat exchanger as a kettle boiler to make steam from a stagnant pool of water on the outside of the tube bank using high-temperature heating oil flowing through the tubes. The exchanger is designed to have one shell and two tube passes. Each tube pass is 65.9 m long. The water is pressurized to 1.5233 bars and enters as saturated liquid, and exits the heat exchanger as superheated steam at 395 K. The heating oil enters at 525 K and exits at 442 K, with a flow rate of 79.5 kg/s. The exchanger is made of Nt = total number of AISI 1010 carbon steel tubes per pass, with each tube having an outside diameter of 2.76 cm and inside diameter of 2.43 cm. The heating oil has a convective heat transfer coefficient of 77 W/m2-K, and the steam has a convective heat transfer coefficient of 698 W/m2-K. The heat capacity of the heating oil at the mean temperature is 2794 J/kg-K. Assume no fouling. You may use 400 K to look up the thermal conductivity of the tubes. What temperature driving force should be used to calculate the heat transfer rate between the water and heating oil? Given the internal energy lost by the heating oil, what is the rate of steam production in the boiler? Use the LMTD method to determine the number of tubes per pass needed to achieve the rate of heat transfer required.

Solution

solution: here we have shell and tube type heat exchanger in which two tubes are place in shell

1) from given data in problem we can calculate total energy lost by hot oil flowing through tube as

Qoil=-Mh*ch*(Thi-Tho)=-79.5*2.794*(525-442)=-18436.209 kj/s

2) where wate enter at 1.5233 bar pressure at which saturation temperature as inlet temp

Tci=109.29 c or 382.29 k

Tco=395 K

where cps=1.9363 kj/kg k from steam table

so steam generation rate as

Qoil=Qwater

18436.209=Ms\'*1.9363*(395-382.29)

ms\'=749.11 kg/s

4) where by LMTD method we have driving force as

dTm=dT1-dT2/ln(dT1/dT2)

wher parralel He

dT1=Thi-Tci=525-382.29=142.70

dT2=Tho-Tco=442-395=47

hence dTm=(142.7-47)/i=ln(142.7/47)=86.1698 K

driving force is dTm=86.1698 K

5) overall heat transfer coefficeint U as

Q=UAdTm

18436.209=UA(86.1698)

UA=213.95

wher Ai=2*pi*L*Ri=5.03085

Ao=2*pi*L*Ro=5.7140

6) due to resistance achieved Uo as follows

Uo=1/[(Ao/Ai*hi)+(Ao/2*pi*l*K)*(ln(Ro/Ri))+(1/ho)]

wher for steel 1010

a=-2.0622*10^-4 per K

T=400 k

K=35.15 w/mk

hence Uo=61.6021 W/m2k

where area for outer tube

UA=213.95

UoAo\'=213.95

Ao\'=213.95/61.6021=3.4730 m2

wher our tube area outer is

Ao=5.7140 m2

hence no of tube required is

Ao\'=n*Ao

3.4730=n*5.7140

n=.6070 approx=1

n=1 is enough

hence single tube per pass is enough

7) here we are using 2 tube hence available

2*Uo\'Ao=213.95

Uo\'=18.7215 w/m2k

Uo> Uo\'

hence exchanger with two tube is more reliable