A 10 m long and 10 mm inner-diameter pipe made of commercial steel is used to heat a liquid in an industrial process. The liquid enters the pipe with T₁ = 25°C, V = 0.8 m/s. A uniform heat flux is maintained by an electric resistance heater wrapped around the outer surface of the pipe, so that the fluid exits at 75°C. Assuming fully developed flow and taking the average fluid properties to be
p = 1000 kg / (m ^ 3)
Cp = 4000 J/kgK,
viscosity=2*10^ -3 kg/ms ,
k=0.48 W/mK, and Pr = 10
determine:
(a) The required surface heat flux q'' produced by the heater
(b) The surface temperature at the exit Ts
Answers
Answer:
180 degrees
Explanation:
If my calculation is correct,
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Heat flux q'' produced by the heater is 40.1kW/m² and The surface temperature at the exit Ts is 53°c
Given:
length of the tube L= 10m
Inner diameter of tube D = 10mm
Initial temperature of water Ti = 20°c
Final temperature of water To = 75°c
velocity of flow, V=0.8m/s
properties of water:
p= 1000kg/m³
Cp = 4000J/KgK
μ = 2 x 10^-3kg/m.s
k = 0.8W/mK
Pr = 10
(a)
m= pAV
= p x (π/4)D² x V
= 1000 x π/4 x (0.01)² x 0.8 = 0.063kg/s
and
q x pi x DL = mCp(To-Ti)
==> q= mCp(To-Ti)/(π x DL)
= 0.063 x 4000 x (75-25)/ (π x 0.01 x 10)
q = 40.1kW/m²
(b)
let surface temperature = Ts
Reynold's number,
ReD = PVD/μ = 1000 x 0.8 x 0.01 / 2 x 10³ = 4 x 10³
L/D = 10/0.01 = 1000
SO, μD = 0.023ReD^4/5 Pr
= 0.023 x (4 x 10³)^4/5 x 10 = 175.14
hD/k = 175.14
h = 175.14 x 0./0.01 = 14011.2 W/m²K
heat flux = q = h(Ts- (To +Ti)/2)
= 40.1 x 10³ = 14011.2 x (Ts - 100/2)
= Ts-50 = 2.87
= Ts = 53°c
Hence, Heat flux q'' produced by the heater is 40.1kW/m² and The surface temperature at the exit Ts is 53°c