Question #261305

Air at a pressure of 101.3 k Pa and 288.8 K enters inside a tube having an inside diameter of 12.7 mm and a length of 1.52 m with a velocity of 24.4 m/s. Condensing steam on the outside of the tube maintains the inside wall temperature at 372.1 K . Calculate the convection coefficient of the air. ( Note: This solution is trial and error. First, assume an outlet temperature of the air. )


1
Expert's answer
2021-11-05T09:36:40-0400

Solution;

Given;

Air pressure;Pa=101.3kPaP_a=101.3kPa

Air temperature,Ta=288.8KT_a=288.8K

Inside diameter,Di=12.7mmD_i=12.7mm

T ube length,L=1.52mL=1.52m

Air velocity,v=24.4m/sv=24.4m/s

Surface temperature,Ts=372.1KT_s=372.1K

Now;

As=πDL=π×12.7×103×1.52=60.65×103m2A_s=πDL=π×12.7×10^{-3}×1.52=60.65×10^{-3}m^2

Ac=π4d2=π4×(12.7×103)2=126.68×106m2A_c=\frac{π}{4}d^2=\frac{π}{4}×(12.7×10^{-3})^2=126.68×10^{-6}m^2

Assume that temperature at outlet in great than at inlet.

Take T0=300K>TaT_0=300K >T_a

According to temperature distribution;

TsToTsTa=ehAsδAcvcp\frac{T_s-T_o}{T_s-T_a}=e^{\frac{-hA_s}{\delta A_c vc_p}}

In which;

δair=1.4kg/m3\delta_{air}=1.4kg/m^3

Cp=1006J/kgKC_p=1006J/kgK

By direct substitution;

372.1300372.1288.8=eh×60.60×1031.14×24.4×126.68×106×1006\frac{372.1-300}{372.1-288.8}=e^{\frac{-h×60.60×10^{-3}}{1.14×24.4×126.68×10^{-6}×1006}}


Hence;

h=8.88W/m2Kh=8.88W/m^2K



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