Exercise 2

KJ2113 Basic Fluid Mechanics Exercise 2

1.

An open cylindrical tank as shown in Fig. E2.1 contains water at 20°C. (a) (b)

Derive the expression which represents the rate of change in water level dh/dt in terms of flowrates Q1, Q2 and Q3 as well as the tank diameter d, If V1 = 3 m/s, Q3 = 0.01 m3/s, and the water level h is constant, calculate the outlet velocity V2.

3

Q3 = 0.01 m3/s

1

D1 = 5 cm h d

2

h

D2 = 7 cm

Fig. E2.1 2.

Water flow in a pipe and then exit through a bended nozzle as shown in Fig. E2.2. The nozzle is connected to the main pipe using a flanged joint at (1). The diameter of the pipe is D1 = 10 cm and is constant, whilst the diameter at the outlet section of the nozzle (2) is D2 = 3 cm. The flowrate of the water is Q = 15 liter/and the water pressure at the flange is p1 = 230 kPa. By neglecting the weight of water and the nozzle, (a) (b)

Determine the force that the bolts have to withstand at (1), Calculate the torque exerted to the nozzle at (1).

1

Q

Flange

Water

30°

1.2 m

2 Nozzle

40°

Fig. E2.2 3.

A pump-turbine system as shown in Fig. E2.3 draws water from the upper lake during the day to supply electricity to nearby township. During the night, it pumps water from the lower lake to the upper lake so that the original water level can be recovered before the day time operation takes place. The design specification of the system requires that the volumetric flowrate through the system is 950 liter/s for both day and night operations and the total head loss for each operation is 5.2 m. Assuming that the flow is steady, inviscid and incompressible, calculate: (a) (b) (c)

Loss in power for both operations, Power generated by the turbine during the day, Power required to operate the pump during the night. Altitude 45 m Upper lake

Altitude 7.5 m Pumpturbine

Fig. E2.3

Lower lake

4.

A water jet exits through a nozzle to the atmosphere at sea level and is directed to the stationary Pitot tube as shown in Fig. E2.4. If the pressure in section (1) is 110 kPa, by neglection all kinds of losses, determine: (a) (b)

mass flow rate of the flow in kg/s, total head H in the Pitot tube.

Water 12 cm

4 cm

H

Water jet

(1)

Fig. E2.4

Answers to Exercise 2

1.

(a)

Select the water volume in the tank to be the control volume: d ρ dV + ρ (Q2 − Q1 − Q3 ) = 0 dt ∫CV πd 2 dh + ρ (Q2 − Q1 − Q3 ) = 0 ρ 4 dt dh Q1 − Q2 + Q3 Q1 − Q2 + Q3 = = dt A πd 2 4

(

(b)

)

If h is constant:

Q1 − Q2 + Q3 = 0 ⇒ Q2 = Q1 + Q3 A 1 1 ( )2 ( )2 (3.0) 4 π 0.07 V2 = 0.01 + 4 π 0.05 V2 = 4.13 m/s 2.

(a)

For the given volumetric flowrate Q:

[

]

−3 m& = ρQ = 998 15(10) = 14.97 kg/s

Q 15(10) = = 1.91 m/s A1 14 π(0.1)2 −3

V1 =

Q 15(10) V2 = = = 21.2 m/s A2 14 π(0.03)2 −3

Using the x-momentum equation:

∑ Fx = − Fbolt + p1 A1 = m& u 2 − m& u1 Fbolt = p1 A1 − m& (− V2 cos 40° − V1 ) = 230(10)

[ π(0.1) ]− 14.97(− 21.2 cos 40° − 1.91)

3 1 4

2

= 2078 N = 2.08 kN (b)

From the diagram, rO2 = −1.2j dan rO1 = 0. Moment equilibrium at O, ∑ M A = 0 : ∑ M O = TO = m& (rO 2 × V2 ) − m& (rO1 × V1 )

TO = 14.97 [− 1.2 j × 21.2(− cos 40°i − sin 40° j)] − 14.97 [0 × 1.91i ] = − 292 N ⋅ m (clockwise)

3.

(a)

Power loss:

W& f = ρgQh f = 998(9.81)(0.950)(5.2) = 48.4 kW (b)

Turbine power: p1 V12 p 2 V22 + + z1 = + + z 2 + h f + ht ρg 2 g ρg 2 g 0 + 0 + 45 = 0 + 0 + 7.5 + 5.2 + ht ⇒ ht = 32.3 m W& t = ρgQht = 998(9.81)(0.950 )(32.3) = 300 kW

(c)

Pump power:

p1 V12 p 2 V22 + + z1 = + + z2 + h f − hp ρg 2 g ρg 2 g 0 + 0 + 7.5 = 0 + 0 + 45 + 5.2 − h p ⇒ h p = 42.7 m

W& p = ρgQh p = 998(9.81)(0.950)(42.7 ) = 397 kW 4.

(a)

Using the Bernoulli and continuity equations: p1 V12 p V2 + + z1 = 2 + 2 + z 2 = H ρg 2 g ρg 2 g 3 V 2 (0.04 0.12) V22 110(10) 0 + 2 +0= + +0 998(9.81) 2(9.81) 998(9.81) 2(9.81) 4

V22 = 223.2 ⇒ V2 = 14.9 m/s

[

]

2 m& = ρA2V2 = 998 14 π(0.04 ) (14.9) = 18.7 kg/s

(b)

From (a): H=

V22 (14.9)2 = = 11.4 m 2 g 2(9.81)