Question

1. measure the correct amount of part b in a separate cup. if the cup has not been used before, add 2.5 grams (when pouring part b into a, not all material will transfer).
2. quickly pour part b into the cup with part a. keep pouring until the liquid starts dripping.
3. stir for 25 - 30 seconds. do not stir longer or foaming will be underway prior to sealing.
4. immediately pour the mixture into the cavity of the mold half with pins pointing up. quickly put the top half on the mold and clamp the mold halves together (the notch on the side of each mold half should be lined up). when clamping, be sure the vent is near the top and open to allow reaction gases to escape.
5. de - mold part after 30 minutes or more, or after the residue in the cup is no longer tacky.
6. check the ball temperature with the infrared thermometer. clean up mold and work area.
7. show the ball to the instructor and be \checked off\ for completion.
8. make sure your actual process parameters are recorded (temperatures, times, pressures, scrap mass).

reference video: https://youtu.be/y7vuc67o0pk
discussion questions:

1. explain the difference between open cell and closed cell foams. open cells are interconnected whilst closed cells are completely sealed off from one another
2. determine the masses of part a and part b for a total mass of 60 grams, assuming a 0.575:1 mass ratio. (m_a + m_b=60) (m_b=\frac{60}{1 + 0.575}=38.095g) (60 - 38.095=) (m_a = 21.905)
3. develop the equation for calculating the density (si) of a foam ball in terms of its diameter. diameter should be the only variable in your equations. show all intermediate steps. include all conversion factors. note: your final equation should be fully reduced, similar to density = mass/(5.17d³). no fractions, no pi, just list a decimal number for the coefficient with your final units. (

ho=\frac{m}{v}) (6/pi = 1.9099) density=(\frac{1.91cdot m}{d³})

1. develop the equation for calculating the specific weight (in us customary units) of a foam ball in terms of its diameter. diameter should be the only variable in your equations. show all intermediate steps. include all conversion factors. note: your final equation should be fully reduced, similar to specific weight=(2.18*weight)/(d³). no fractions, no pi, just list a decimal number for the coefficient with your final units.

Explanation:

Step 1: Recall density formula

The density formula is $
ho=\frac{m}{V}$. For a sphere (foam - ball), the volume formula is $V = \frac{4}{3}\pi r^{3}$, and since $r=\frac{D}{2}$, then $V=\frac{4}{3}\pi(\frac{D}{2})^{3}=\frac{\pi D^{3}}{6}$.

Step 2: Substitute volume into density formula

Substitute $V=\frac{\pi D^{3}}{6}$ into $
ho=\frac{m}{V}$, we get $
ho=\frac{m}{\frac{\pi D^{3}}{6}}=\frac{6m}{\pi D^{3}}$. Given $\frac{6}{\pi}\approx1.91$, so $
ho=\frac{1.91m}{D^{3}}$ (in SI units, where mass $m$ is in kg and diameter $D$ is in m, and density $
ho$ is in $kg/m^{3}$).

Step 3: Recall specific - weight formula

The specific weight $\gamma$ is related to density by $\gamma=
ho g$, where $g = 9.81m/s^{2}$ in SI units. First, convert SI units to US customary units. $1kg = 2.205lb$ and $1m=3.281ft$.

The density in SI is $
ho=\frac{1.91m}{D^{3}}$ ($m$ in kg, $D$ in m). Convert mass $m$ from kg to lb and diameter $D$ from m to ft.

$m(kg)\to m(lb)=2.205m(kg)$ and $D(m)\to D(ft)=3.281D(m)$

$
ho=\frac{1.91\times2.205m_{lb}}{(3.281D_{ft})^{3}}=\frac{4.20155m_{lb}}{35.3147D_{ft}^{3}}$.

Since $\gamma=
ho g$, and $g = 32.2ft/s^{2}$ in US customary units.

$\gamma=\frac{4.20155m_{lb}}{35.3147D_{ft}^{3}}\times32.2ft/s^{2}=\frac{135.29091m_{lb}}{35.3147D_{ft}^{3}}\approx\frac{3.83m_{lb}}{D_{ft}^{3}}$

Answer:

1. Density formula: $

ho=\frac{1.91m}{D^{3}}$ (where $
ho$ is in $kg/m^{3}$, $m$ is in kg, and $D$ is in m)

1. Specific - weight formula: $\gamma=\frac{3.83m}{D^{3}}$ (where $\gamma$ is in $lb/ft^{3}$, $m$ is in lb, and $D$ is in ft)