Pressure in seepage

[fonseh]

Homework Statement

Can anyone explain why the pore pressure at C is given by( H1 + z + (h/ H2)(z) ) (y_w) ?

Homework Equations

The Attempt at a Solution

Shouldnt it be ( H1 + z + ) (y_w) only ?
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[Chestermiller]

The quantity $(H+z)\gamma_w$ is the hydrostatic contribution to the pore pressure at C. The additional contribution of seepage flow to the pore pressure at C is $\frac{k}{\mu}vz$, where z is the pore flow distance between C and A, k is the permeability, $\mu$ is the water viscosity, and v is the seepage velocity. For point B, the contribution of seepage flow to the pore pressure at B is $\frac{k}{\mu}vH_2=h\gamma_w$, where $H_2$ is the pore flow distance between B and A and h is the additional head above the water table as a result of seepage flow. So, from the relationship at B, we have:
$$\frac{k}{\mu}v=\frac{h}{H_2}\gamma_w$$Therefore, substituting this into the additional contribution of seepage flow to the pore pressure at C, we obtain $\frac{h}{H_2}z\gamma_w$. Therefore, the total pore pressure at C is $$(H+z)\gamma_w+\frac{h}{H_2}z\gamma_w=\left(H+z+\frac{h}{H_2}z\right)\gamma_w$$

 

[fonseh]

The quantity $(H+z)\gamma_w$ is the hydrostatic contribution to the pore pressure at C. The additional contribution of seepage flow to the pore pressure at C is $\frac{k}{\mu}vz$, where z is the pore flow distance between C and A, k is the permeability, $\mu$ is the water viscosity, and v is the seepage velocity. For point B, the contribution of seepage flow to the pore pressure at B is $\frac{k}{\mu}vH_2=h\gamma_w$, where $H_2$ is the pore flow distance between B and A and h is the additional head above the water table as a result of seepage flow. So, from the relationship at B, we have:
$$\frac{k}{\mu}v=\frac{h}{H_2}\gamma_w$$Therefore, substituting this into the additional contribution of seepage flow to the pore pressure at C, we obtain $\frac{h}{H_2}z\gamma_w$. Therefore, the total pore pressure at C is $$(H+z)\gamma_w+\frac{h}{H_2}z\gamma_w=\left(H+z+\frac{h}{H_2}z\right)\gamma_w$$

I have another example here . In this case , it's downwards seepage ... Why for this case , the Pressure at B is (H1 + z -iz )yw ? Shouldn't the pressure increases with the depth ?

 

[Chestermiller]

The hydrostatic portion of the pressure variation does increase with depth. But, if the viscous seepage flow is downward, its contribution to the pressure variation must involve a pressure gradient component that can drive the fluid downward.

 

[fonseh]

The hydrostatic portion of the pressure variation does increase with depth. But, if the viscous seepage flow is downward, its contribution to the pressure variation must involve a pressure gradient component that can drive the fluid downward.

so , do you mean as the water flow from top to the bottom , so the water is saying to be flow from higher pressure to low pressure ? So , in the case of downwards seepage , the pressure at A > C >B ?

 

[Chestermiller]

so , do you mean as the water flow from top to the bottom , so the water is saying to be flow from higher pressure to low pressure ? So , in the case of downwards seepage , the pressure at A > C >B ?

Only the viscous seepage portion of the pressure variation, which superimposes linearly upon the hydrostatic portion of the pressure variation, to give the overall total pressure variation.

 

[fonseh]

Only the viscous seepage portion of the pressure variation, which superimposes linearly upon the hydrostatic portion of the pressure variation, to give the overall total pressure variation.

So , the pressure due to seepage variation is A > C >B ??

 

[Chestermiller]

So , the pressure due to seepage variation is A > C >B ??

Yes, if the flow is downward.

 

[fonseh]

The quantity $(H+z)\gamma_w$ is the hydrostatic contribution to the pore pressure at C. The additional contribution of seepage flow to the pore pressure at C is $\frac{k}{\mu}vz$, where z is the pore flow distance between C and A, k is the permeability, $\mu$ is the water viscosity, and v is the seepage velocity. For point B, the contribution of seepage flow to the pore pressure at B is $\frac{k}{\mu}vH_2=h\gamma_w$, where $H_2$ is the pore flow distance between B and A and h is the additional head above the water table as a result of seepage flow. So, from the relationship at B, we have:
$$\frac{k}{\mu}v=\frac{h}{H_2}\gamma_w$$Therefore, substituting this into the additional contribution of seepage flow to the pore pressure at C, we obtain $\frac{h}{H_2}z\gamma_w$. Therefore, the total pore pressure at C is $$(H+z)\gamma_w+\frac{h}{H_2}z\gamma_w=\left(H+z+\frac{h}{H_2}z\right)\gamma_w$$

Can you explain what causes The additional contribution of seepage flow to the pore pressure at C is $\frac{k}{\mu}vz$ ?? Is there any name for the term ?

 

[Chestermiller]

Can you explain what causes The additional contribution of seepage flow to the pore pressure at C is $\frac{k}{\mu}vz$ ?? Is there any name for the term ?

The differential equation for the variation of pressure in a porous medium (in the vertical direction) is $$\frac{dp}{dz}+\gamma=-\frac{k}{\mu}v$$ where, in this equation, z is the elevation and v is the superficial upward seepage velocity. This is Darcy's Law.