To cite: The reason for greater drying shrinkage in slip or hydro plastic casted products that contain smaller size clay particles.

Question

Want to see the full answer?

Answer 1

Question

Chapter 13, Problem 23QAP

To determine

To cite:

The reason for greater drying shrinkage in slip or hydro plastic casted products that contain smaller size clay particles.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

Qu 4 Silver has FCC crystal structure at room temperature, and a lattice constant, a, of 0.407 nm. Draw a reduced sphere silver unit cell in the grids provided below, clearly label the lattice dimensions. Within the unit cell you drew, shade the (1 0 0) plane. How many atoms are contained within the (1 0 0) plane? Calculate the area of (1 0 0) plane in [nm?]. Express your answer in [nm?] to three significant figures. Calculate the planar density of the (1 0 0) plane in [atoms/nm?]. Express the answer in atoms/nm to three significant figures. show all work step by step

Crystallographic planes Crystallographic planes are denoted by Miller indices. 5b Algorithm for Miller indices 1. Read off intercepts of plane with axes in terms of a, b, c 2. Take reciprocals of intercepts 3. Reduce to smallest integer values 4. Enclose in parentheses, no commas. 353 1/3 1/5 1/3 535 (535) In the cubic system, a plane and a direction with the same indices are orthogonal. E.g. [100] direction is perpendicular to (100) plane. Correspondingly, [123] direction is perpendicular to (123) plane. [2,3,3] Plane intercepts axes at 3a, 2b, 2c 2 11 1 Reciprocal numbers are: 3'2'2 b. Indices of the plane (Miller): (2,3,3) 2 a Indices of the direction: [2,3,3] X (200) (100) (110) (111) (100) Indices of crystallographic plane can be found from cross product of indices of any two non-parallel directions in this plane.

Crystallographic positions Crystallographic position is denoted by three numbers, which are coefficients of the position vector, e.g. ½½½ for the red atom. Here the 'new' atom is at a/2 + b/2 + c/2 Silicon crystal has so-called "diamond type lattice". Each Si atom has 4 nearest neighbors. Diamond lattice starts with a FCC lattice and then adds four additional INTERNAL atoms at locations r = a/4+b/4+c/4 away from each of the atoms. In other words, diamond lattice is formed by two FCC lattices sifted by the vector r.