Question

4.11

4.11 An embankment content constructed at a 10% moisture content. of the be the water and to be unit is WolLlll. M ate be placed at the rate of 250 ccy/hr. The specified dry weight compacted fill is 2,850 lb/cy. How many gallons of water must be from supplied each hour to increase the moisture content of the material 5 to 10% by weight? (4,277 gal/hr) aamnacted measure The

Answer 1

ANSWER

From theory we know that water content will be:

From theory we know that water content will be: w = W_w/W_s x100 So, W_w is the weight of water and W_s is the weight of solids. W_w = W_s(w/100) W_w = gamma_dV_s(w_2 - w_1/100) Rate of compacted soil per hour is V_s = 250 ccy W_w = gamma_d V_s(w_2- w_1/100) W_w = (2850)(250)(10-5/100) W_w = 35626 lb/hr W_w = 35625/8.33 gallons/hr W_w = 4276.71 gallons/hr W_w = 4277 gallons/hr Therefore, the supplied water per hour to increase the water content is 4277 gallons/hr.

So, $W_{w}$ is the weight of water and $W_{s}$ is the weight of solids.

$W_{w}=W_{s}\left ( \frac{w}{100} \right )$

$W_{w}=\gamma _{d}V_{s}\left ( \frac{w_{2}-w_{1}}{100} \right )$

Rate of compacted soil per hour is $V_{s}=250ccy$

$W_{w}=\gamma _{d}V_{s}\left ( \frac{w_{2}-w_{1}}{100} \right )$

$W_{w}=(2850)(250)\left ( \frac{10-5}{100} \right )$

$W_{w}=35625lb/hr$

$W_{w}=\frac{35625}{8.33}gallons/hr$

$W_{w}=4276.71gallons/hr$

$\mathbf{W_{w}=4277gallons/hr}$    ANS.

Therefore, the supplied water per hour to increase the water content is $\mathbf{4277gallons/hr}$ .

Regards!!!