Question 8 of 8: Hydrostatic Casting Forces on a Spherical Die-Cast Mold with a Fill Riser

Question 8: Hydrostatic Casting Forces on a Spherical Die-Cast Mold with a Fill Riser (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Hollow spherical mold cavity, radius $R$, diameter $D=2R$, filled with molten zinc alloy of density $\rho$, through a small cylindrical fill hole/riser of length $L$ set at the top of the mold and standing full of metal to height $L$ above the mold's own top pole; $z=0$ at the mold's top pole (per the hint), $z=D$ at the bottom; gauge pressure referenced to zero at the free surface at the top of the riser (height $L$ above $z=0$).

Find. $F_{top}(L)$ and $F_{bottom}(L)$ (vertical forces on the top and bottom hemispherical wall halves) as functions of the riser length $L$, each expressed as a multiple of the sphere's own metal weight $W=\rho g\left(\tfrac43\pi R^3\right)$; and whether increasing $L$ without bound is an appropriate way to improve mold–metal contact.

Approach. Because the riser is filled and connected to the cavity, Pascal's principle applies: the gauge pressure at any point in the mold equals $\rho g$ times its full vertical distance below the riser's free surface, i.e. $\rho g(L+z)$ — a uniform head $\rho gL$ superimposed on the same $z$-dependent hydrostatic profile as a hole-free mold. Parametrize the sphere by polar angle $\theta$ from the top ($z(\theta)=R(1-\cos\theta)$), project the local pressure onto the vertical direction, and integrate over each hemisphere.

1. Geometry and pressure with the riser head included. At polar angle $\theta$ (0 at top, $\pi$ at bottom): $z(\theta)=R(1-\cos\theta)$, surface element $dA=2\pi R^2\sin\theta\,d\theta$, outward-normal vertical component $n_z=-\cos\theta$ (positive $z$ downward), and gauge pressure $$P(\theta)=\rho g\big[L+R(1-\cos\theta)\big]$$
2. Vertical force element. $$dF_z=P(\theta)(-\cos\theta)(2\pi R^2\sin\theta)\,d\theta=-2\pi\rho gR^2\big[L+R(1-\cos\theta)\big]\cos\theta\sin\theta\,d\theta$$
3. Top half, $\theta:0\to\pi/2$. Split the integral into the riser-head term and the no-riser term (the latter already evaluated to $\tfrac16$ for a hole-free mold): with $u=\cos\theta$, $$\int_0^{\pi/2}L\cos\theta\sin\theta\,d\theta=\dfrac{L}{2},\qquad \int_0^{\pi/2}R(1-\cos\theta)\cos\theta\sin\theta\,d\theta=\dfrac{R}{6}$$ $$F_{top}(L)=-2\pi\rho gR^2\left(\dfrac{L}{2}+\dfrac{R}{6}\right)=\boxed{-\pi\rho gR^2\left(L+\dfrac{R}{3}\right)}$$ (negative $\Rightarrow$ net force is upward, growing in magnitude the taller the riser stands.)
4. Bottom half, $\theta:\pi/2\to\pi$. Similarly, $$\int_{\pi/2}^{\pi}L\cos\theta\sin\theta\,d\theta=-\dfrac{L}{2},\qquad \int_{\pi/2}^{\pi}R(1-\cos\theta)\cos\theta\sin\theta\,d\theta=-\dfrac{5R}{6}$$ $$F_{bottom}(L)=-2\pi\rho gR^2\left(-\dfrac{L}{2}-\dfrac{5R}{6}\right)=\boxed{\pi\rho gR^2\left(L+\dfrac{5R}{3}\right)}$$ (positive $\Rightarrow$ net force is downward, also growing with $L$.)
5. Baseline check ($L=0$) and expression relative to the sphere's own weight. Setting $L=0$ recovers the hole-free result exactly: $F_{top}(0)=-\pi\rho gR^3/3=-W/4$, $F_{bottom}(0)=5\pi\rho gR^3/3=5W/4$ of this exam (which posed the same geometry with no riser). In general, $$\dfrac{F_{top}(L)}{W}=-\dfrac{3L}{4R}-\dfrac14,\qquad\dfrac{F_{bottom}(L)}{W}=\dfrac{3L}{4R}+\dfrac54$$
6. The riser's leverage — a second hydrostatic paradox. $F_{bottom}(L)-F_{top}(L)=2\pi\rho gR^2(L+R)$, which exceeds the sphere's own metal weight $W=\tfrac43\pi\rho gR^3$ by $2\pi\rho gR^2L+\tfrac23\pi\rho gR^3$ — a term that grows with the mold's cross-sectional area $R^2$, not with the (small) volume of metal actually sitting in the thin riser. A riser of modest bore therefore adds force to the WHOLE mold surface out of all proportion to its own weight, exactly the mechanism (Pascal's principle acting on a wide vessel through a narrow, elevated connection) that makes the natural hydrostatic pressure profile tend to lift the upper mold segment in the first place, and why that tendency grows, not shrinks, as $L$ increases.
7. Is "increase $L$ as far as possible" appropriate? A modest riser IS beneficial: it raises the gauge pressure at every point in the cavity by $\rho gL$, exactly the same passive mechanism the external-pressure die-casting systems used elsewhere in this exam series achieve actively — better mold–metal conformity and reduced porosity risk, especially near the top of the casting where the hole-free pressure was lowest. But growing $L$ without bound is not appropriate: (i) both $F_{top}(L)$ and $F_{bottom}(L)$ — and hence the required mold-clamping force — increase linearly and without limit in $L$, while any real clamp has a finite rated force; (ii) the metal standing in the riser above the casting is trim that must be sheared off and re-melted, a yield loss that grows linearly with $L$; (iii) a taller riser carries more thermal mass and stays liquid longer, which usefully continues to feed the casting's own solidification shrinkage for a while — the genuine reason to keep a non-trivial $L$ — but once the riser comfortably outlives the casting's own freezing time, additional height adds cycle time and cost with no further feeding benefit. $$\boxed{\text{A modest, non-zero }L\text{ improves contact pressure passively; "as far as possible" is not appropriate once clamping-force, yield-loss and cycle-time costs are weighed against the saturating feeding benefit}}$$