Question 12.2: Air at 330 K, flowing at 10 m/s, enters a pipe of inner diam......
Air at 330 K, flowing at 10 m/s, enters a pipe of inner diameter 25 mm, maintained at 415 K. The drop of static pressure along the pipe is 80 N/m² per metre length. Using the Reynolds analogy between heat transfer and fluid friction, estimate the air temperature 0.6 m along the pipe.
From equations 12.98 and 3.18:
\frac{1}{{A}t}\frac{{\bf M}}{\rho}(C_{A s}-C_{A w})=(-{N}_{A})_{y=0} (12.98)
h_{f}=\frac{-\Delta P_{f}}{\rho g}=4\frac{R}{\rho u^{2}}\frac{l}{d}\frac{u^{2}}{g} (3.18)
– ΔP = 4 (R/ρu²)(l/d)ρu² = 4 (h/C_{p}ρu)(l/d)ρu²
(using the mean pipeline velocity u in the Reynolds analogy)
Then, in SI units.
-ΔP/l = 80 = 4[h/(C_{p} x 10)] (1/0.025) x 10²
and: h = 0.05C_{p} W/m² K
In passing through a length dL of pipe, the air temperature rises from T to T + dT.
The heat taken up per unit time by the air,
\mathrm{d}Q=(\rho u C_{p})(\pi\,d^{2}/4)\,\mathrm{d}TThe density of air at 1 bar and 330 K = (M P)/(RT) = {\frac{29\times10^{5}}{8314\times330}}=1.057\,\,{\mathrm{kg/m^{3}}}
Thus; dQ = (1.057 x 10 x C_{p})[π(0.025)²/4]dT
= 0.0052C_{p} dT W (i)
The heat transferred through the pipe wall is also given by:
dQ = h(πd dL)(415 – T)
= (0.05C_{p})(π x 0.025 dL)(415 – T) (ii)
= 0.039C_{p}(415 – T)dL W
Equating (i) and (ii):
giving ln[85/(415 – T_{\widehat{O}})] = 0.45
and: 85/(415 – T_{\widehat{O}}) = e^{0.45} = 1.57
and: \underline{\underline{T_{\widehat{O}}=360\ K}}