Aula Prática 5. Fluxes (Problem 1.07) Consider the flow in a rectangular duct, formed by two paralell plates (width b=1m and height 2h= 30cm) where air.

Apresentação em tema: "Aula Prática 5. Fluxes (Problem 1.07) Consider the flow in a rectangular duct, formed by two paralell plates (width b=1m and height 2h= 30cm) where air."— Transcrição da apresentação:

1 Aula Prática 5

2 Fluxes (Problem 1.07) Consider the flow in a rectangular duct, formed by two paralell plates (width b=1m and height 2h= 30cm) where air (=1.2 kg/m3) is flowing. Assuming stationary flow and a situation 1 with uniform velocity profile and another with a parabolic profile (represented on the figure), 1 –In case of uniform profile with 5 m/s velocity, compute : a)The volumetric discharge, the mass flow rate and the average velocity in the cross section. b) The momentum and kinetic energy fluxes.

3 2 –Assuming the parabolic profile given by the equation: Compute the maximum velocity to get the same discharge and the momentum and kinetic energy fluxes. Show that they are higher than in case of uniform velocity

4 2 – Repita o cálculo admitindo um perfil parabólico dado pela equação: Calcule o valor da velocidade máxima para que o caudal fosse o mesmo da alínea anterior. Calcule os fluxos de quantidade de movimento e de energia cinética nessas condições.

5 Fluxos Considere o escoamento numa conduta de secção rectangular, formada por duas placas planas paralelas, de largura b=1m afastadas de uma distância 2h= 30cm, na qual se escoa ar (=1.2 kg/m3). Vamos considerar uma situação com perfil de velocidade do tipo tampão e outra com perfil parabólico (representado na figura). 1 – Calcule, se o escoamento fosse do tipo tampão com velocidade de 5 m/s: a)O caudal volúmico e o caudal mássico e a velocidade média na secção de passagem. b) Os fluxos de Quantidade de movimento e de energia cinética.

6 Resolution Q=uA =5ms -1 *2*0.3*5m 2 =3m 3 s -1 m=1.2kg*m -3 *3m 3 s -1 =3.6 kgs -1

7 Momentum an Kinetic Energy

8 Parabolic Profile To get Q =3m 3 s -1 the maximal velocity must be 7.5 ms -1. The average velocity is:

9 Momentum flux is higher

10 Kinetic energy flux is higher as well

11 Problem 1.08 The shear stress on the wall of a pipe can be computed knowing the average velocity (U) and the friction coefficient (f). For a water flow in a pipe 20 m long with a diameter D=2.5cm, if the average velocity was U=1.5 m/s and f=0.002 compute: The water discharge, The force necessary to keep the pipe in place (equal to the drag force on the pipe) The pressure drop assuming that the velocity profile at the inlet and outlet are identical. The energy consumption.

12 Resolution A) The average velocity is the discharge per unity of area: B) The flow exerts friction force on the pipe (tangential to the pipe) and pressure (normal to the pipe). The pressure force tends to inflate the pipe, but not to move it. The only force that could move the pipe is the friction force.

13 c) If the velocity profiles are identical at the entrance and exit, there is no acceleration and consequently the pressure drop has to balance the friction force: The pressure drop in a pipe is proportional to the square of the discharge in to the inverse of the fifth power of D. In ordinary domestic installations the pressure drop is the supply pressure (provided by the municipality about 5 Pa) and thus the discharge is very much dependent on the pipe diameter. Small diameters allow (very) small discharges.

14 d) The energy consumption would be the work of the pressure force. The friction force is applied on the wall where velocity is null and thus doesn’t produce any work. The work of the pressure force to transport the fluid along the pipe is the force times the displacement: The power is the work per unit of time the work per unit of time: Where T is the time to cross the whole pipe. The velocity is the length of the pipe divided by that time. The power is proportional exchanged with a fluid in a flow is the pressure drop/increase in the flow time the discharge.

15 Estimate the power generated in a 70 m high dam when the flow discharged is 5 m3/s. The pressure at the entrance of the turbine is the pressure at the bottom of the dam and the pressure at the exit is the atmospheric pressure. Thus the power is: Looking carefully to the equation one can see that the energy provided is the potential energy of the top water per unit of volume: times the discharge that is the volume per unit of time.

16 Why are dams as high as possible? Because like that the potential energy per unit of volume is higher and also because being high they can store more water. Anyhow, whatever is the volume stored, only creating potential energy it is possible to extract energy.