Answer:
A) approximate alkalinity = 123.361 mg/l
B) exact alkalinity = 124.708 mg/l
Explanation:
Given data :
A) determine approximate alkalinity first
Bicarbonate ion = 120 mg/l
carbonate ion = 15 mg/l
Approximate alkalinity = ( carbonate ion ) * 50/30 + ( bicarbonate ion ) * 50/61
= 15 * (50/30) + 120*( 50/61 ) = 123.361 mg/l as CaCO3
B) calculate the exact alkalinity of the water if the pH = 9.43
pH + pOH = 14
9.43 + pOH = 14. therefore pOH = 14 - 9.43 = 4.57
[OH^- ] = 10^-4.57 = 2.692*10^-5 moles/l
[ OH^- ] = 2.692*10^-5 * 179/mole * 10^3 mg/g = 0.458 mg/l
[ H^+ ] = 10^-9.43 * 1 * 10^3 = 3.7154 * 10^-7 mg/l
therefore the exact alkalinity can be calculated as
= ( approximate alkalinity ) + ( [ OH^- ] * 50/17 ) - ( [ H^+ ] * 50/1 )
= 123.361 + ( 0.458 * 50/17 ) - ( 3.7154 * 10^-7 * 50/1 )
= 124.708 mg/l
The rainfall run off model HEC-HMS is combined with river routing model. They are used for simulating the rainfall process.
Explanation:
The HEC - HMS rainfall model is used for simulating the rainfall runoff process. In this study the soil conservation service and curve number method is used to calculate the sub basin loss in basin module.
It provides various options for providing the rainfall distributions in the basin. It has the control specification module used to control the time interval for the simulations.
The one dimensional continuity equation is
бA / бT + бQ / бx= 0
Who doesn’t play war zone
Answer:
a) 152000 slugs
b) 2220000 kg or 2220 metric tons
Explanation:
A body with a weight of 4.9*10^6 lbf has a mass of
4.9*10^6 lbm * 1 lbf/lbm = 4.9*10^6 lbm
This mass value can then be converted to other mass values.
1 slug is 32.17 lbm
Therefore:
4.9*10^6 lbm * 1 slug / (32.17 lbm) = 152000 slugs
1 lb is 0.453 kg
Therefore:
4.9*10^6 lbm / (1/0.453) * kg/lbm = 2220000 kg
