On Thu, Jul 14, 2011 at 5:52 PM, David Jonsson <[email protected] > wrote:
> Hi > > Can someone help me to derive the spring constant between water molecules > based on the bulk modulus of water? It seems simple but i just > can't figure it out. > > How does spring constant between water molecules in > F = - k x > relate to the bulk modulus > K = - V dp / dV > > k = spring constant > F = force between molecules > x = elongation of spring or displacement of molecules relative each other > K = bulk modulus of water = 2.2 MPa. It describes the pressure needed to > make a relative volume change. 1 % compression requires 22 kPa, 10 % > compression requires 220 kPa, etc. > V = Volume of the fluid > dV = volume change due to pressure > dp = pressure change causing volume change > > This is a general question for all fluids and not only water. It has vortex > and rotation applications. I will show you later. > > David > > David Jonsson, Sweden, phone callto:+46703000370 > > I don't know that a simple model like springs connecting the molecules works for a liquid, but here's a way to connect the concepts: If you think of piston in a rigid cylinder with area A and length x, then the volume is V = Ax. The pressure is P = F/A. So, dp/dV = (dF/dx) / A^2. But dF/dx is k, so K = (V/A^2) * k or k = K (A/x) Here, x is not the displacement, but the length of the cylinder, and k represents the spring constant of the entire cylinder. This is how Young's modulus is connected to the spring constant for a solid. But Young's modulus is not normally assigned to a liquid. To get a ball park estimate of the spring constant between two molecules, you could try to estimate the cross-section and length of cylinder that contains 2 molecules. But this doesn't take account of couplings to other molecules, so I don't know how valid it is. But again, it should give a ball park figure.
