Answer:
a) benzene = 910 days
b) toluene = 1612.67 days
Explanation:
Given:
Kd = 1.8 L/kg (benzene)
Kd = 3.3 L/kg (toluene)
psolid = solids density = 2.6 kg/L
K = 2.9x10⁻⁵m/s
pores = n = 0.37
water table = 0.4 m
ground water = 15 m
u = K/n = (2.9x10⁻⁵ * (0.4/15)) / 0.37 = 2.09x10⁻⁶m/s
a) For benzene:

The time will take will be:

b) For toluene:


Answer:
The Euler buckling load of a 160-cm-long column will be 1.33 times the Euler buckling load of an equivalent 120-cm-long column.
Explanation:
160 - 120 = 40
120 = 100
40 = X
40 x 100 / 120 = X
4000 / 120 = X
33.333 = X
120 = 100
160 = X
160 x 100 /120 = X
16000 / 120 = X
133.333 = X
Answer:
1) 
2) 
Explanation:
For isothermal process n =1

![V_o = \frac{5}{[\frac{72}{80}]^{1/1} -[\frac{72}{180}]^{1/1}}](https://tex.z-dn.net/?f=V_o%20%20%3D%20%5Cfrac%7B5%7D%7B%5B%5Cfrac%7B72%7D%7B80%7D%5D%5E%7B1%2F1%7D%20-%5B%5Cfrac%7B72%7D%7B180%7D%5D%5E%7B1%2F1%7D%7D)

calculate pressure ratio to determine correction factor

correction factor for calculate dpressure ration for isothermal process is
c1 = 1.03

b) for adiabatic process
n =1.4
volume of hydraulic accumulator is given as
![V_o =\frac{\Delta V}{[\frac{p_o}{p_1}]^{1/n} -[\frac{p_o}{p_2}]^{1/n}}](https://tex.z-dn.net/?f=V_o%20%3D%5Cfrac%7B%5CDelta%20V%7D%7B%5B%5Cfrac%7Bp_o%7D%7Bp_1%7D%5D%5E%7B1%2Fn%7D%20-%5B%5Cfrac%7Bp_o%7D%7Bp_2%7D%5D%5E%7B1%2Fn%7D%7D)
![V_o = \frac{5}{[\frac{72}{80}]^{1/1.4} -[\frac{72}{180}]^{1/1.4}}](https://tex.z-dn.net/?f=V_o%20%20%3D%20%5Cfrac%7B5%7D%7B%5B%5Cfrac%7B72%7D%7B80%7D%5D%5E%7B1%2F1.4%7D%20-%5B%5Cfrac%7B72%7D%7B180%7D%5D%5E%7B1%2F1.4%7D%7D)

calculate pressure ratio to determine correction factor

correction factor for calculate dpressure ration for isothermal process is
c1 = 1.15

Answer:
S = 5.7209 M
Explanation:
Given data:
B = 20.1 m
conductivity ( K ) = 14.9 m/day
Storativity ( s ) = 0.0051
1 gpm = 5.451 m^3/day
calculate the Transmissibility ( T ) = K * B
= 14.9 * 20.1 = 299.5 m^2/day
Note :
t = 1
U = ( r^2* S ) / (4*T*<em> t </em>)
= ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4
Applying the thesis method
W(u) = -0.5772 - In(U)
= 7.9
next we calculate the pumping rate from well ( Q ) in m^3/day
= 500 * 5.451 m^3 /day
= 2725.5 m^3 /day
Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping
S = 
where : Q = 2725.5
T = 299.5
W(u) = 7.9
substitute the given values into equation above
S = 5.7209 M