1. 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

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

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

4. 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

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

6. 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 ,

7. 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

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

9. d) Mass flow rate ,M,
M = pQ = 978 kg / m3 x m3/ s
= kg / s

10. Exp : 6. 5. At one section in an air distribution. system, air at 101
Exp : At one section in an air distribution system, air at kPa 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 k Pa and 40°C , the density of air is kg/ m3 and the specific weight is N/m3 .

11. 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 ) =
A2= ¶D22 /4 = ¶(45.7 cm) 2
= 1640 cm 2

12. (a) Then , the density of the air in the round section is
p2 = (1.134 kg/m3 )( cm 2/1620 cm2 )(6.1 m/s /4.57 m/s)
p2 = kg/m (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 ) ( cm2 )(1 m2 / cm2 )(6.1 m / s)
= 6.32 N/s