Description We still consider the same catchment with a total area of 9.0 km , w

Question

Description

We still consider the same catchment with a total area of 9.0 km , where the runtime principle applies. The area is subdivided in 5 regions (A1 to A5) in such a way that every individual area has a runtime of 30 minutes. The areas drain sequentially. A5 is at the top of the catchment and A1 at the bottom of the catchment near the outlet.

We again assume that the rain falls uniformly over the catchment, and that the discharge starts immediately after the start of the rain. The discharge only occurs as a result of the net (i.e. eective) rain.

Question 11

Immediately after the rain of net intensity 4.5 mm/hr in the rst 30 minutes, a second shower occurs. This second shower has a net intensity of 8 mm/hr and lasts 30 minutes. How large is the discharge in [m³/s] after 30 minutes? If applicable, round your answer to two decimals. Enter the discharge in [m³/s]:

Question 12

Again consider rain of net intensity 4.5 mm/hr in the rst 30 minutes immediatly followed by a second shower with a net intensity of 8 mm/hr that lasts 30 minutes. How large is the discharge in [m³/s] after 60 minutes? If applicable, round your answer to two decimals. Enter the discharge in [m³/s]:

Question 13

Immediately after the rain of net intensity 4.5 mm/hr in the rst 30 minutes a second shower occurs with a net intensity of 8 mm/hr that lasts 30 minutes. After the 60 minutes of rain there was sun for 8 hours, before sunset, with an average net radiation of 300 W/M2.

Assume:

That the energy from radiation is absorbed only for sensible heat H [W / m2], and latent heat p^E [W / m2], where p is the density of water (1000 kg/m2) and ^ is the latent heat of vaporisation (2.45 MJ / kg).

Assume that 50% of the net radiation is absorbed by sensible heat.

Based on the observations and assumptions how large is the evaporation in [mm] over the 8 hours of sunshine? If applicable, round your answer to two decimals.

Hint: Use the energy balance equation RN = p^E + H to calculate the correct answer. Make sure you carefully check the units.

Question 14

Immediately after the rain of net intensity 4.5mm/hr in the rst 30 minutes a second shower occurs with a net intensity of 8mm/hr that lasts 30 minutes.

Based on the information provided, calculate the magnitude of the change in water storage S [m3] in the total catchment, 24 hours after the rain has started. Indicate if this change is an increase (a positive number) or a decrease (a negative number). If applicable, round your answer to two decimals, and do not forget to add the sign. Assume there was only evaporation when there was sunshine, so no evaporation during the rain events and at night. There was no “loss” intensity of the rain.

Hint: Carefully consider the values of the incoming and outgoing uxes over the 24 hours to determine the change in storage.

Enter the increase (a positive number) or decrease (a negative number) in water storage in the total catchment:

Explanation / Answer

Answer:11

The discharge after 30 minutes = (4.5)*9= 40.5 m3/s

Answer:12

The discharge after 60 minutes = (4.5+8)*9= 112.5 m3/s

Answer:13

Net Radiation= p^E + H

Put the values in above equation, we get

300= 1000* 2.45*E+ 150

E= (300-150)/ 2450= 0.061

Answer:14

Storage after 24 hours = 112.5 *24=2700 m3

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