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) ?

The Attempt at a Solution

Shouldnt it be ( H1 + z + ) (y_w) only ?
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Mentor
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
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 ?

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Mentor
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
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 ?

Mentor
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
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 ??

Mentor
So , the pressure due to seepage variation is A > C >B ??
Yes, if the flow is downward.

fonseh
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 ?

Mentor
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.

fonseh