Question

A copper pipe of diameter 3.5 cm and wall thickness 4 mm is used in an antibiotic factory to convey hot water a distance of 60 m at a flow rate of 20 L/s. The inlet temperature of the water is 90°C. Due to heat losses to the atmosphere, the outlet water temperature is 84°C. The ambient temperature is 25°C. The inside of the pipe is covered with a layer of fouling with a fouling factor of 7500 W/m²°C.

1. What is the rate of heat loss from the water in kW?
2. What proportion in percentage (%) of the total resistance to heat transfer is provided by the fouling layer?
3. If the copper pipe were replaced by a clean stainless-steel pipe of the same thickness, what proportion in percentage (%) of the total resistance to heat transfer would be provided by the pipe wall? To answer this question you need to estimate this ratio and multiply by 100%: B/k1/Ux 100%   where B is the pipe-wall thickness, k is the thermal conductivity of stainless-steel

Answer 1

Diameter of copper pipe = 3.5 cm

Flow rate of water =Q= 20L/s = 0.02 m3/s

Length of pipe = L= 60 m

Crossectional area of pipe = 0.0352(3.142) /4 = 0.0009621 m2

Velocity of water = Q/A= 0.02/0.0009621 = 20.787 m/s

From Perry handbook the properties of water are computed at average temperature of (84+90)/2= 87°C

Thermal conductivity (k) = 0.671 W/mK

Density of water () = 967.29 Kg/m3

Viscosity of water () = 0.3261 mPa =

3.261 ×10-4

Cp = 3.8361 KJ/kg K

Thermal conductivity of copper = 401 W/mK

Nre = 0.035(20.787) (967.29) /(0.0003261)

= 2158071.153

The flow is turbulent , hence dittus bioler equation can be used

Npr = 1.864

Dittus bioler equation is

Substituting all values we get

h= 63224.014 W/m2°C

Fouling water = 7500 W/m2°C

Overall heat transfer coefficient(U)

1/U = 1/(h) +(1/f)

f = fouling factor

1/U = 1/(7500) +1/(63224.014)

U = 6704.654 W/m2K

Total resistance = R = resistance by convection + resistance by pipe wall = (1/U + ∆x/k )

R = (1/6704.654) +(0.0004/401) = 1.5904×10-4

Ambient temperature = 25°C

Inlet temperature difference = ∆Ti = 90-25 = 65°C

Outlet temperature difference = ∆T2 = 84-25 = 59°C

∆Tlmtd = (∆Ti-∆To) /ln(∆Ti/∆To)

∆Tlmtd =( 65-59) /ln(65/59) = 61.951°C

A) Q /A = ∆Tlmtd/R

A = 3.142(d) (L) = 3.142(0.035) (60) = 6.597 m2

Subsituting all values we get

Q = 2569.735KW

Heat lost by water = 2569.735 KW

Resistance by conduction = ∆x/K = 0.004/401 = 9.975×10-6

Resistance by fouling = 1/7500 = 1.333×10-4

Resistance by convection = 1/63224.014 = 1.5816×10-5

Total resistance R = (9.975×10-6) +(1.5816×10-5) +(1.333×10-4) = 1.5909×10-4

Percentage of resistance offered by fouling layer = (1.333×10-4) /(1.5909×10-4) = 0.8378

= 83.788%

So 83.788% of resistance is provided by fouling factor

C)

If copper pipe is replaced by clean steel pipe

Then there is no fouling factor

From perry handbook

K of stainless steel = 14.4 W/mK

Resistance from conduction or pipe wall = ∆x/k = 0.004/14.4 = 2.777×10-4

Resistance from convection = 1/63224.014 = 1.5816×10-5

Total resistance = 1.5816×10-5+(2.777×10-4) = 2.935 ×10-4

Percentage of resistance offered by pipe wall to the total resistance = (2.777×10-4) /(2.935×10-4) = 0.94611

= 94.611%

So to the total resistance, 94.611% of resistance is offered by pipe wall or conduction

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