Answer:
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
