Question

Prob#1 mainstream A 0.5-m-square duct contains air flowing from Um = 1.5 m/s entering the duct at 1 atm and 25 C, as shown in Figure below. Applying the appropriate boundary layer concept, calculate the (a) The average velocity at 2 m in the duct. (b) Pressure drop along the duct, in Pa/m unit. 0.5-m-square duct Boundary layers U mainstream U -2 m

Answer 1

Obtain the following properties of the air at 1 atm and 25°C from the air properties table. Density of the air, p=1.184 kg/m2 Viscosity of the air, u=1.849x10-5 kg, m.s Calculate the Reynolds number as follows: pVx Re u 1.184x1.5x2 Re 1.849 x 10- Re = 1.92x10 Since, the Reynolds number (1.92x10') is less than the critical Reynolds number (5x10). So, the flow is laminar flow. Calculate the displacement thickness as follows: 1.72 1x 1/2 * 1.721x2 1/2 = (1.92x105) 5* = 0.00785 m Calculate the cross-sectional area of flow at inlet as follows: 4 = w, xw 4 = 0.5 x 0.5 4 = 0.25 m

The cross-sectional area of flow is decreasing along the duct due to formation of the boundary layer. Calculate the cross-sectional area of flow at exit (x = 2 m) as follows: 4. =(w; – 28*)x(w; – 28*) Since, the area is reducing from both sides and this reduction will be equal to the two times of the displacement thickness. 4. =(0.5 - 2x0.00785)(0.5- 2x0.00785) 4. = 0.2345 m2 (a): Calculate the mean velocity in the core using continuity equation as follows: V.A = AV. V. = V mean? A 1.5x 0.25 V man 0.2345 V = 1.60 m/s 1ean

(b): Calculate the mean pressure by applying Bernoulli's equation between inlet and exit as follows: PV? PV? P. + P. 2 2 PD2 PV? CD2 i + = P. -P 2 2 1.184x1.602 1.184x1.52 P. + =101325+ 2 2 P. +1.51 = 101325 +1.33 P. = 101324.82 N/m Calculate the pressure drop as follows: dp_101324.84 -101325 2-0 dp -0.08 Pa/m dx dx