Question

4.4 A water storage tank initially contains 400 m of water. The average daily usage is 40 m. If water is added to the tank at an average rate of 20lexp(-1/20)] m per day, where t is time in days, for how many days will the tank contain water?

Answer 1

dv/dt = Quantity added - Quantity used because dv/dt = 20 exp (-t/20) m^3/day - 40 m^3/day integral^0_v_i dv = 20 integral^t_0 e^-t/20 dt = 40 integral^t_0 dt -V_i = -20 times 20 e^-t/20^t_0 - 40 t \r\n V_i = 400 (e^-t/20 - d) + 40 t Gives, Initial volume V_i = 400 m^3 therefore 400 = 400 (e^-t/20) - 400 + 40 t 800 = 400 e^-t/20 + 40 t 2 = e^-t/20 + t/10 (2 - t/10) = e^-t/20 ln (2 - t/10) = -t/20 t = -20 ln (2 - t/10) This is an transcendental equation and can only be solved using some plotting software. This will given the time �t� when all the water is exhausted in the tank