Question

8.12 Dry, compressed air at Tm.i75°C, p-10 atm, with a mass flow rate of 0.001 kg/s, enters a 30-mm-diameter, 5-m-long tube whose surface is at Ts - 25°C. (a) Determine the thermal entry length, the mean temperature of the air at the tube outlet, the rate of heat transfer from the air to the tube wall, and the power required to flow the air through the tube. For these conditions the fully developed heat transfer coefficient is h- 3.58 W/m2 K. (b) In an effort to reduce the capital cost of the installation it is proposed to use a smaller, 28-mm-diameter tube. Determine the thermal entry length, the mean temperature of the air at the tube outlet, the heat transfer rate, and the required power for the smaller tube For laminar flow conditions it is known that the value of the fully developed heat transfer coefficient is inversely proportional to the tube diameter

Answer 1

Mean temperature: Ti +T75 +25 - 50 oC mem Thermal properties of air at this mean temperature: -1.94x10 N.s/m2 P 10am 10x 10.95 kg/m3 c 1008 J/kgK Pr 0.704 Determine the Reynolds number: 4x0.001 π×0.03 × 1.94 × 10-5 πdμ - 2187.7 <2300 Hence the flow is laminar Calculate the thermal entry length: em0.05dRe. Pr 0.05×0.03×21 87.7×0.704 2.31 m

Outlet temperature: T -T T-T exp|- ×n mc m,o -ex mc × 0.03 × 5 0.001 × 1 008 25-T 25-75 Im 34.38 oC Heat transfer rate expl x3.58 surfaceim AT-AT In(AT,/AT) (34.38-25)-(75-25) In (34.38-25)/(75-25) 3.58π × 0.03 × 5 = 40.95 W Power required:

Pressure drop lu 2d Friction factor for laminar flow: 64 64 21877:0.02925 e Flow velocity: 4×0.001 0.1292 m/s DA ρπd2 10.95×π×0.032 Pressure drop: AnP10.02925 10.95x(0.1292) 2x 0.03 2d 0.445 Pa Power required regd

1m Ap 0.445x 4.064×10-5 w 0.001 10.95 Given 3.58 28 mm h, 30 mm =3.835 Wirk Determine the Reynolds number: 4mm 4 x 0.001 2343.96 >2300 Hence the flow is turbulent. Calculate the thermal entry length: X. therm rurb -10x 0.028

= 0.28 m Outlet temperature: T.-T T -Tm T-T T -T 25-T 25-75 'm = 34.381°C exp mc mc π×0.028×5 × 3.835 = exp|- 0.001x 1008 Heat transfer rate: surface Im = 3.835π x 0.028x51-(34.38-25)-(75-25) In (34.38-25)/(75-25) 40.942 W

Power required: regd Pressure drop 2d Friction factor for turbulent flow: f0.316Re0.316x 2343.96-140.0454 Flow velocity: 4 x0.001 pA ρπdz 10.95 XTx0.028 0.1483 m/s Pressure drop: pu 1 0.95x(0.1483) 2x0.028 0.976 Pa Power required regd

-0.976x 0,001 10.95 - 8.91x10 W