3. Consider a 10-m-long smooth rectangular tube, with a = 50 mm and b=25 mm, that is maintained at a constant surface temperature. Liquid water enters the tube at 20C with a mass flow rate of 0.01 kg/s. Determine the tube surface temperature necessary to heat the water to the desired outlet temperature of 80C.

surface temperature is 86.31c

Explanation:

given that

l = 10m

width a = 0.05m

height b = 0.025

inlet flow temperature  T1 = 20c

mass flow rate , m = 0.01kg/s

outlet flow temperature, T2 = 80c

Average temperature

Tave = T1 + T2/2

= 20 + 80/2

= 50C

Hydraulic diameter of a rectanguar tube

Dh = 4Ac/p

= 4(a x b)/2(a +b)

4 (0.05 x 0.025)/2(0.05 + 0.025)

= 0.0333m

Average velocity

Vavg = m/ pAc

Vavg = 0.01/988.1 x ( 0.05 x 0.025)

=8.096 x 10∧-3m/s

properties of water  at 1 atm and average temperature

Density , p = 988.1kg/m³

specific heat  , cp = 4181 j/kg-k

kinematic viscosity, ∪ = 0. 547 x 10 ∧-3 kg/m-s

pranditl number , pr = 3.55

Reynolds number

Re = pVavgDh/ ∪

= 988.1 x 8.096 x 10∧-3 x0.0333/0.54 x 10 ∧-3

= 487.02.

As the obtained reynolds number is less than   the 2300 , the flow is laminar.

Hydraulic entry lenght

lh = 0.05 x Re x Dh

Lh = 0.05 x 487.02 x 0.0333

= 0.811m

therefore,

thermal  entry length =  pr x lh

= 3.55 x 0.811

= 2.879m

laminar flow now, a/b value is

a/b = 0.05/0.0025

a/b = 2

note, the nusselt number for a/b = 2 is

Nu = 3.39

Heat transfer coefficient  

h = k x Nu/Dh

h = 0.644 x3.39/0.333

= 65.56 w/m²k

The surface area =A

As = 2(a + b)L

A = 2 ( 0.05 + 0.025) 10

A = 1.5m²

the tube temperature

Te = Ts - (Ts-T1) esp [hAs/mcp]

80 = Ts- (Ts -20)esp[ 65.56 x 15/0.01 x 4181]

80 = Ts - (Ts  - 20)exp[-2.352]

by solving this we get T = 86.31c

hence the surface temperature is 86.31c

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