Temat: Metalurgia proszków
Przed spiekaniem | Po spiekaniu |
---|---|
Ciśnienie (MPa) | wym. wysokość h (mm) |
100 | 6,672 |
100 | 6,68 |
200 | 5,775 |
200 | 5,745 |
350 | 4,99 |
350 | 5 |
Wzór na objętość walca Πr2h
V= Πr2h = Π ∙ (7,484)2 · 6,572 = 1156,41 mm3
V= Πr2h = Π ∙ (7,191)2 · 6,072 = 986,4 mm3
V= Πr2h = Π ∙ (7,462)2 · 5,685 = 994,33 mm3
V= Πr2h = Π ∙ (7,167)2 · 5,338 = 861,52 mm3
V= Πr2h = Π ∙ (7,464)2 · 4,899 = 857,43 mm3
V= Πr2h = Π ∙ (7,238)2 · 4,713 = 775,68 mm3
Gęstość właściwa $\ \rho = \frac{m}{V}$ [$\frac{g}{\text{mm}3}\rbrack$
ρ 1 = $\frac{5,85}{1156,41\ }\ $= 5,05·10-3 [$\frac{g}{\text{mm}3}\rbrack$
ρ 2 = $\ \frac{5,81}{986,4\ }\ $= 5,89·10-3 [$\frac{g}{\text{mm}3}\rbrack$
ρ 3 = $\frac{5,82}{994,33\ }\ $= 5,88·10-3 [$\frac{g}{\text{mm}3}\rbrack$
ρ 4 = $\frac{5,86}{861,52}\ $= 6,8·10-3 [$\frac{g}{\text{mm}3}\rbrack$
ρ 5 = $\frac{5,86}{857,43\ }\ $= 6,83·10-3 [$\frac{g}{\text{mm}3}\rbrack$
ρ 6 = $\frac{5,85}{775,68\ \ }$ = 7,54·10-3 [$\frac{g}{\text{mm}3}\rbrack$
Gęstość względna ρw $= \frac{\rho}{\rho t}\ $·100 % ρt = 0,00897 [$\frac{g}{mm3}\rbrack$
ρw1 = $\frac{5,05 10 - 3}{8,97 10 - 3}\ $·100 % = 56,29 %
ρw2 = $\frac{5,89 10 - 3}{8,97 10 - 3}\ $·100 % = 65,66 %
ρw3 = $\frac{5,88 10 - 3}{8,97 10 - 3}\ $·100 % = 65,55 %
ρw4 = $\frac{6,8 10 - 3}{8,97 10 - 3}\ $·100 % = 75,8 %
ρw5 = $\frac{6,83 10 - 3}{8,97 10 - 3}\ $·100 % = 76,14 %
ρw6 = $\frac{7,54 10 - 3}{8,97 10 - 3}\ $·100 % = 84,05 %