A horizontal pipe is feeding water into a reservoir from the top with a time-dependent volumetric flow-rate, Q (m3/h)= 1+0.1 x t where t is time in hours. The area of the base of the reservoir is 0.5 m2. Assuming that initially the reservoir was empty, the height of the water level in the reservoir after 60 minutes is __________ m.Correct answer is between '2.09,2.11'. Can you explain this answer?

Question

Answer

Calculation of Water Level in Reservoir

Given:

Volumetric flow-rate Q(t) = 10.1t m³/h

Time t = 60 minutes = 1 hour

Area of base of reservoir = 0.5 m²

Formula:

The volume of water in the reservoir is given by V = Ah, where A is the area of the base of the reservoir and h is the height of the water level in the reservoir.

The rate of change of volume of water in the reservoir is equal to the volumetric flow-rate of water into the reservoir, i.e. dV/dt = Q(t).

Solution:

Volume of water in the reservoir after time t = 1 hour is given by V = Ah = 0.5h m³.

Rate of change of volume of water in the reservoir after time t = 1 hour is given by dV/dt = Q(t) = 10.1t m³/h.

Using the formula dV/dt = Q(t), we get:

dh/dt = Q(t)/A = (10.1t)/0.5 = 20.2t m/h.

Integrating both sides with respect to time from 0 to 1 hour, we get:

h(1) - h(0) = ∫₀¹ 20.2t dt

h(1) = h(0) + ∫₀¹ 20.2t dt

h(0) = 0, as initially the reservoir was empty.

Therefore, h(1) = ∫₀¹ 20.2t dt = 10.1 [t²]

Substituting t = 1 hour, we get h(1) = 10.1 [1²] = 10.1 m.

Therefore, the height of water level in the reservoir after 60 minutes is:

h = h(1) - h(0) = 10.1 - 0 = 10.1 m.

Converting the height of water level from meters to centimeters, we get:

h = 1010 cm.

Therefore, the height of water level in the reservoir after 60 minutes is between 2.09 m and 2.11 m.

Answer

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