• Author / Uploaded
• Vinay Gupta

• Categories
• Presión
• Ingeniería Química
• Dinámica (Mecánica)
• Mecanica clasica
• Física Aplicada e Interdisciplinaria

Citation preview

COURSEWORK: MOMENTUM PRINCIPLE

SPRING 2005

Issue date: Tuesday 15 March 2005 Hand-in date: Monday 11 April 2005 (by 4 pm)

Question 1 (11 marks) 60 mm

A contracting pipe bend turns water through 135° (in a horizontal plane) before discharging it to the atmosphere. The inlet diameter is 80 mm and the outlet diameter is 60 mm. The volumetric flow rate is 20 L s–1. (a) Neglecting energy losses, calculate the upstream pressure and the magnitude and direction of the force on the bend. (b) If the upstream pressure is actually 24 kPa, calculate the loss of head in the pipe bend.

80 mm

Q = 20 L/s

Question 2 (11 marks) A junction splits a horizontal pipe of diameter 80 mm into two horizontal pipes of diameter 60 mm and 50 mm at 30° to the original flow direction as shown. The junction is arranged so as to split the flow in the ratio 2:1 between 80 mm these two pipes. For water flow, if the upstream pressure and velocity are 10 kPa and 5 m s–1 respectively, and the outlet pipes discharge to atmosphere, calculate the force on the junction in the x-y coordinate system shown.

60 mm 2Q/3

Q y

A large square plate of mass 50 kg is pivoted from its upper edge. A high-velocity jet of diameter 60 mm carrying 30 L s–1 of water hits it dead centre. Estimate the 60 mm deflection from the vertical, .

30

o

Q/3 50 mm

θ

Q = 30 L/s

Coursework: Momentum Principle

o

x

Question 3 (8 marks)

Hydraulics 1

30

m = 50 kg

David Apsley

Answer 1 Data: D1 = 0.08 m, D2 = 0.06 m, Q = 0.02 m3 s–1. 4Q 4 × 0.02 = = 3.979 m s −1 2 D1 × 0.08 2 4Q 4 × 0.02 U2 = = = 7.074 m s −1 2 2 D2 × 0.06 By Bernoulli, with p2 = 0 (discharge to atmosphere), p1 + 12 U 12 = 0 + 12 U 22

(a)

U1 =




p1 =

1 2




(U 22 − U 12 ) = 12 × 1000 × (7.074 2 − 3.979 2 ) = 17100 Pa


Let the force on the pipe bend be (Fx, Fy). Then the force on the fluid is (– Fx, – Fy). Apply the momentum principle in the x-direction: D2 p1 1 − Fx = Q(−U 2 cos 45° − U 1 ) 4 D12 U2 7.074 × 0.08 2 Fx = p1 + Q( + U 1 ) = 17100 × + 1000 × 0.02 × ( + 3.979) 4 4 2 2 = 85.95 + 179.62 = 265.6 N 









Apply the momentum principle in the y-direction: − Fy = Q(U 2 cos 45° − 0)


Fy = − Q


U2 2

= −1000 × 0.02 ×

7.074 2

= −100.0 N

The magnitude of the force is F = Fx2 + Fy2

= 265.6 2 + 100.0 2

= 283.8 N

The direction of the force is Fy − 100.0 = tan −1 = tan −1 ( ) = −20.6° Fx 265.6

(b) The loss of head is p1actual U2 p U2 p actual − p1ideal +z+ 1 − 2 +z+ 2 = 1 g 2g g 2g g 



=



24000 − 17100 1000 × 9.81

= 0.7034 m

Answer: (a) 17.1 kPa and 284 N at 20.6° (below) inflow direction; (b) 0.70 m

Hydraulics 1

Coursework: Momentum Principle

David Apsley

Answer 2

Data: D1 = 0.08 m, D2 = 0.06 m, D3 = 0.05 m, U1 = 5 m s–1. The cross-sectional areas of the pipes are D12 0.08 2 A1 = = × = 0.005027 m 2 4 4 D22 0.06 2 A2 = = × = 0.002827 m 2 4 4 2 D3 0.05 2 A3 = = × = 0.001963 m 2 4 4 The total quantity of flow is that in pipe 1, i.e. Q1 = U 1 A1 = 5 × 0.005027 = 0.025135 m 3 s −1 This is split in the ratio 2:1, so 2 2 Q2 = Q1 = × 0.025135 = 0.01676 m 3 s −1 3 3 1 1 Q3 = Q1 = × 0.025135 = 0.008378 m 3 s −1 3 3 The velocities in these pipes are then Q 0.01676 = 5.929 m s −1 U2 = 2 = A2 0.002827 Q 0.008378 U3 = 3 = = 4.268 m s −1 A3 0.001963

Let the force on the junction be (Fx, Fy). Then the force on the fluid is (– Fx, – Fy). Apply the momentum principle in the x-direction: p1 A1 − Fx = Q2U 2 cos 30° + Q3U 3 cos 30° − Q1U 1 Hence Fx = p1 A1 − [(Q2U 2 + Q3U 3 ) cos 30° − Q1U 1 ] 








= 10000 × 0.005027 − 1000 × [(0.01676 × 5.929 + 0.008378 × 4.268) × cos 30° − 0.025135 × 5) = 50.27 + 8.65 = 58.92 N Apply the momentum principle in the y-direction: − Fy = Q2U 2 sin 30° − Q3U 3 sin 30° − 0 Hence Fy = (Q3U 3 − Q2U 2 ) sin 30°







= 1000 × (0.008378 × 4.268 − 0.01676 × 5.929) ×

1 2

= −31.8 N

Answer: (59, – 32) N

Hydraulics 1

Coursework: Momentum Principle

David Apsley

Answer 3

At equilibrium the weight of the plate and the fluid force on the plate must have equal moments about the pivot. Let the component of fluid force normal to the plate be Fn and the length of a side be L. Then, taking moments about the pivot, Fn ( L / 2) = mg ( L / 2) sin Hence F (*) sin = n mg


The force on the plate is equal and opposite to the force of the plate on the jet. For the jet we have, upstream, D = 0.06 m, Q = 0.03 m3 s–1. Hence, 4Q 4 × 0.03 U= = 10.61 m s −1 = 2 2 D × 0.06 



Applying the momentum principle normal to the plate, − Fn = Q(0 − U cos ) Hence, Fn = QU cos





Substituting this in (*) gives QU cos sin = mg Hence, QU 1000 × 0.03 × 10.61 tan = = = 0.6489 mg 50 × 9.81 whence = 32.98°







Answer: 33°

Hydraulics 1

Coursework: Momentum Principle

David Apsley