Question

For a particular casting setup, the top of the sprue has a diameter of 0.030 m, and its length is 0.200 m. The volume flow rate of metal into the mold is 0.03 m3 /min. What is the resultant velocity at the bottom of the sprue? Also, what diameter should be specified at the bottom of the sprue to prevent air aspiration? What is the Reynolds number at the bottom of the sprue. Is the flow laminar or turbulent? Assume the pressure at the top and bottom of the sprue is atmospheric pressure. Viscosity of the metal used is 0.004 N.s / m2. Density of the metal is 2700 kg/ m3 .

Answer 1

Answer with Explanation:

We are given that

Diameter=0.030 m

Length of sprue=
`h_1`=0.200 m

Metal volume flow rate,Q=0.03
`m^3/min`

Q=
`(0.03)/(60)=5* 10^(-4)m^3/s` because 1 minute=60 seconds

Let 1 for the top and 2 for the bottom

`d_=0.030 m`

`h_2=0`

`A_1=(\pi d^2)/(4)=(3.14* (0.030)^2)/(4)`

`A_1=7.065* 10^(-4) m^2`

`v_1=(Q)/(A_1)=(5* 10^(-4))/(7.065* 10^(-4))`

`v_1=0.708 m/s`

Pressure at the top and bottom of the sprue is atmospheric

`h_1+(v^2_1)/(2g)=h_2+(v^2_2)/(2g)`

Substitute the values

`0.2+((0.708)^2)/(2\cdot 9.8)=0+\frac[v^2_2}{2\cdot 9.8}`

`v^2_2=2\cdot 9.8\cdot (0.2\cdot 9.8\cdot 2+0.501264)/(2\cdot 9.8)=4.421264`

`v_2=√(4.421264)=2.1 m/s`

`Q=A_2v_2`

`5* 10^(-4)=A_2* 2.1`

`A_2=(5* 10^(-4))/(2.1)=2.381* 10^(-4) m^2`

Reynolds number=
`(v_2D\rho)/(\eta)`

`\eta=0.004 N.s/m^2`

`\rho=2700 kg/m^3`

Substitute the values then we get

Reynolds number=
`(2.1* 0.03* 2700)/(0.004)`

Reynolds number=42525

The Reynolds number is greater than 4000 .Therefore, the flow is turbulent.