Chapter 5, Problem 5.71P

To determine

The profile of sprue diameter as function of its height.

Answer to Problem 5.71P

The profile of sprue diameter as function of its height is shown in Figure-A.

Explanation of Solution

Given data:

The diameter at top of the sprue is $d_t=100\text{mm}$ .

The height of the sprue is $h=300\text{mm}$ .

The diameter at bottom of the sprue is $d_b=300\text{mm}$ .

Formula used:

The expression for the area is given as,

  $A=\frac{\pi }{4}{d}^2$

Here, $d$ is the diameter.

Calculation:

The relation between area and height can be given as,

  $\frac{A_1}{A_2}=\sqrt{\frac{h}{h_1}}$

Here, $A_1$ is the area of top of the sprue at the height $h_1$ , and $A$ is the area of the top surface at height $h$ .

Evaluate the above equation further,

  $\begin{array}{c}\frac{A_1}{A_2}=\sqrt{\frac{h}{h_1}}\\ \frac{\left(\frac{\pi }{4}{d}_1^2\right)}{\left(\frac{\pi }{4}{d}^2\right)}=\sqrt{\frac{h}{h_1}}\\ d^2=d_1^2\sqrt{\frac{h}{h_1}}\\ d=d_1{h}_1^{0.25}{h}^{0.25}\end{array}$ ...... (1)

In equation 1, take $d_1{h}_1^{0.25}$ as constant and replace with $k$ .

Here, equation 1 becomes,

  $\begin{array}{l}d=d_1{h}_1^{0.25}{h}^{0.25}\\ d=k{h}^{0.25}\end{array}$ ...... (2)

As the reference location for the height is not known, take $h_0$ as the location of reference.

So, the location of surface at height from top surface is $h_0+h$ .

Substitute the value in equation 2,

  $d=k{\left(h_0+h\right)}^{0.25}$ ...... (3)

Further applying the boundary conditions,

At $h=0$ and $d=0.1\text{m}$ .

Substitute the values in equation 3,

  $\begin{array}{c}d=k{\left(h_0+h\right)}^{0.25}\\ 0.1\text{m}=k{\left(h_0\right)}^{0.25}\\ k=0.1{\left(h_0\right)}^{-0.25}\end{array}$ ...... (4)

At $h=0.3\text{m}$ and $d=0.\text{025}\text{m}$ ,

Substitute the values in equation 3,

  $\begin{array}{c}d=k{\left(h_0+h\right)}^{0.25}\\ \mathrm{0.0.25}\text{m}=k{\left(0.3+h_0\right)}^{0.25}\\ k=0.1{\left(0.3+h_0\right)}^{-0.25}\end{array}$ ...... (5)

Solve equation 4 and 5,

  $\begin{array}{c}0.1{\left(h_0\right)}^{0.25}=0.025{\left(0.3+h_0\right)}^{0.25}\\ \frac{{\left(0.3+h_0\right)}^{0.25}}{{\left(h_0\right)}^{0.25}}=4\\ 256h_0=h_0+0.3\\ h_0=1.176\times {10}^{-3}\text{m}\end{array}$

Substitute the value in equation 4,

  $\begin{array}{l}k=0.1{\left(1.176\times {10}^{-3}\right)}^{0.25}\\ k=0.0185\end{array}$

Substitute the values in equation 3,

  $\begin{array}{l}d=k{\left(h_0+h\right)}^{0.25}\\ d=\left(0.0185\right){\left[\left(1.176\times {10}^{-3}\right)+h\right]}^{-0.25}\end{array}$

The graph further can be plotted between the diameter and sprue height using the above equation.

  

Figure- A Graph between diameter and height of the sprue

Conclusion:

Therefore, the profile of sprue diameter as function of its height is shown in Figure-A.