6. Flow of fluids and Bernoulli’s equation. Q , volume flow rate – volume of fluid flowing per unit time. W, weight flow rate - weight of fluid flowing M , mass flow rate – mass of fluid flowing

Volume flow rate , which calculate from Q = Av Weight flow rate , which calculate from W = Q Mass flow rate calculate from, M = pQ

Velocity of flow of fluid in a closed system depends on the principle of continuity.

then the mass flow rate can be expressed as , M1=M2 fluid flow from section 1 to section 2 at a constant rate. That is , the quantity of fluid flowing past any section in a given amount of time is constant. This is referred to as stead flow. if there is no any fluid added , removed , or stored between section 1 and section 2, then the mass flow rate can be expressed as , M1=M2

or because M = pAv, p1A1v1 = p2A2v2 so because of p1 and p2 are equal, A1v1 = A2v2

Exp :6.4 ,the inside diameters of the pipe at section 1 and 2 are 50mm and 100mm , respectively. Water at 70°C is flowing with an average velocity of 8 m/ s at section 1 . Calculate the following: Velocity at section 2 Volume flow rate Weight flow rate Mass flow rate ,

Velocity at section 2, A1v1 =A2v2 v2= v1 ( A1 / A2 ) A1 =( ¶ D2/ 4) = (¶ (50mm) 2)/ 4 = 1963 mm2 A2 =( ¶ D2/ 4) = (¶ (100mm) 2)/ 4 = 7854 mm2 v2= v1 ( A1 / A2 ) = 8.0m/s x 1963 mm 2 / 7854 mm 2 = 2.0 m/ s

b) Volume flow rate , Q=A1v1 = 1963 mm2 x 8 m/ s x 1m2/ (103mm)2 = 0.0157 m3 / s c) Weight flow rate , W = Q = 9.59kN/m3 X 0.0157 m3/s = 0.151kN/s

d) Mass flow rate ,M, M = pQ = 978 kg / m3 x 0.0157 m3/ s = 15.36 kg / s

Exp : 6. 5. At one section in an air distribution. system, air at 101 Exp : 6.5 At one section in an air distribution system, air at 101.35kPa and 40°C has an average velocity of 6.1 m/ s and the duct is 30.5 cm square . At another section ,the duct is round with a diameter of 457 mm , and the velocity is measured to be 4.57 m/ s . Calculate (a) the density of the air in the round section and (b) the weight flow rate of air in N/s . At 101.35 k Pa and 40°C , the density of air is 1.134 kg/ m3 and the specific weight is 11.14 N/m3 .

Then , we can calculate the area of the two section and solve for p2 : Solution : According to the continuity equation for gases , Eq (6-4) , we have p1A1v1 = p2A2v2 Then , we can calculate the area of the two section and solve for p2 : p2 = p1 (A1 / A2)(v1 / v2) A1 = ( 30.5 cm )( 30.5 cm ) = 930.25 A2= ¶D22 /4 = ¶(45.7 cm) 2 = 1640 cm 2

(a) Then , the density of the air in the round section is p2 = (1.134 kg/m3 )( 930.25 cm 2/1620 cm2 )(6.1 m/s /4.57 m/s) p2 = 0.859 kg/m3 (b) The weight flow rate can be found at section 1 from W= A1v1 . Then , the weight flow rate is , W= 1A1v1 y W= (11.14 N/m3 ) (930.25 cm2 )(1 m2 / 10000 cm2 )(6.1 m / s) = 6.32 N/s