Skip to main content

Interception and Interception losses | Numerical on capacity determination of impounded reservoir

 INTERCEPTION 

When rainfall occurs over catchment areas some of the amount of water might be lost because of vegetation cover and then evaporation occurs. The amount of water which is evaporation due to vegetation cover is called interception. About 10 to 20% of the total rainfall are lost due to interception. When the water retained by vegetation cover evaporated and reaches to atmosphere, then the process is called interception losses. It’s very difficult to measure the amount of water intercepted during rainfall and storms. It mostly depends upon the extent of vegetation cover and also on the velocity of wind during the rainfall (Storms). If the area is experiencing a large number of storms, then interception losses is quite large due to the forest losses and accounts for about 20% of the total annual precipitation. Interception losses are observed to be larger for the small storms as shown in fig below and smaller as the rainfall precipitation increases. 




Ad:



https://happyshirtsnp.com/



amount of interception losses is estimated by

Ii = Si +Ki Et

Here, Ii = interception losses in mm, Si=interception storage whose values varies from 0.25 to 1.25 whose values depends upon the extent of vegetation cover, Ki= ratio of vegetation cover area to its projected area, E= evaporation rate in mm/hr during the process of precipitation and t= time of rainfall in hours


Coniferous trees, dense grasses have more interception losses than that of deciduous and thin grasses and accounts for nearly 20% of the total precipitation losses. Interception process has greater roles in balancing the water in the region and in in-situ water harvesting


Numerical on capacity determination of impounded reservoir




The capacity of the impounded reservoir is calculated using an analytical method or mass curve method

Analytical method

IR capacity= Maximum cumulative surplus + maximum cumulative deficit-total inflow+ total demand.

Procedure
  • Calculate cumulative hourly demand (outflow) and cumulative hourly supply (inflow) for 24hrs.
  • Find hourly excess of demand or outflow (i.e. deficit), excess of supply or inflow (i.e. surplus), total outflow (TO), and total inflow(TI).
  • Obtain the maximum cumulative surplus (MCS) and maximum cumulative deficit (MCD).
          Then the capacity of the impounded reservoir (CR) be
           CR=MCS+MCD-TI+TO, if TI>TO
           CR=MCS+MCD, if TI< or = TO

Procedure
  1. Denote monthly runoff as Q, and monthly draft or demand as D.
  2. Calculate deficiency for each value i.e. (D-Q), (-ve values is surplus and +ve values is deficit)
  3. Calculate the cumulative deficiency SUM(D-Q), by starting from the beginning of the maximum dry period.
  4. The required maximum storage is the highest numerical values in sum(D-Q) values or the maximum cumulative deficiency at the end of the dry period.

Mass curve method/Graphical method

Mass curve method is a plot between the cumulative inflow in the reservoir with time
Capacity depends on the rates of inflow, losses (evaporation, absorption, seepage etc) and demand or outflow.
The inflow and outflow values are to be determined for various month of years.

Procedure

  1. Find cumulative run-off adding run-off values for the period considered.
  2. Find the cumulative demand or the draft values for the period considered (adding of demands or draft values) 
  3. Plot the 'time periods' versus 'cumulative values'.
  4. Draw parallel lines to the cumulative demand lines, tangent to the mass curve at the starting and the ending of the dry periods.
  5. Required storage capacity of the reservoir is the maximum of the vertices intercepts ordinates between the two tangents drawn to a peak and trough consecutively.

Interception losses

• When it rains over a catchment, not all the precipitation falls directly onto the ground.

• Before it reaches the ground, a part of it may be caught by the vegetation and subsequently

evaporated.

• The volume of water so caught is called interception.

• The intercepted precipitation may follow one of the three possible routes:

(a) It may be retained by the vegetation as surface storage and returned to the atmosphere by

evaporation; a process termed interception loss:

(b) It can drip off the plant leaves to join the ground surface or the surface flow; this is known as

through fall; and

(c) The rainwater may run along the leaves and branches and down the stem to reach the ground

surface. This part is called stemflow.

Interception loss is solely due to evaporation and does not include transpiration, through fall or streamflow.

Fig: Interception losses vs rainfall 





• The amount of water intercepted in a given area is extremely difficult to measure.

• It is estimated that of the total rainfall in an area during a plant-growing season the interception

loss is about 10 to 20%.

• Interception is satisfied during the first part of a storm and if an area experiences a large number of

small storms, the annual interception loss due to forests in such cases will be high, amounting to

greater than 25% of the annual precipitation.


• It is seen that the interception loss is large for a small rainfall and levels off to a constant value for

larger storms.

• For a given storm, the interception loss is estimated as: Ii= Si+ Ki E t

• where Ii =interception loss in mm, Si= interception storage whose value varies from 0.25 to 1.25 mm

depending on the nature of vegetation, Ki= ratio of vegetal surface area to its projected area ,

E=evaporation rate in mm/h during the precipitation & t= duration of rainfall in hours.


Ad:





                                                              Read more:
Location: Kathmandu 44600, Nepal

Comments

Post a Comment

Popular posts from this blog

Hydrodynamically smooth and rough boundaries | Velocity distribution for turbulent flow

 Hydrodynamically smooth and rough boundaries Fig: Definition of smooth and rough boundaries In general, a boundary with irregularities of large average height 'K' on its surface is considered to  be a rough boundary and one with smaller 'K' value is considered as a smooth boundary. ✓ However, for a proper classification of smooth and rough boundaries, the flow and fluid characteristics are required to be considered in addition to the boundary characteristics. ✓ As the flow outside the laminar sub-layer is turbulent, eddies of various sizes are present which try  to penetrate through the laminar sublayer. But, due to greater thickness of laminar sub-layer ,   eddies can't reach the surface irregularities and thus the boundary acts as a smooth boundary. Such  a boundary is termed as " Hydro-dynamically Smooth Boundary" ✓ With the increase in Reynold's no (Re), the thickness of the laminar sub-layer  decreases and it's  can even become much smaller
Post a Comment
Read more

Water hammer and unsteady flows in pipes

             Ad: https://happyshirtsnp.com/                                                                             UNSTEADY FLOW IN PIPES When water is flowing through a pipe is suddenly stopped by the use of valve then we can see sudden rise in pressure inside the pipe and this pressure waves transmitted along the length of the pipe and it creates knocking like sound effects knows as 'water hammer'. In doing so, there are rapid pressure oscillation in the pipe, often accompanied by a hammering like sound, this phenomenon is called as 'Water hammer effect'. Sometimes the pressure may rise to the greater extent and may causes serious damage to the pipe. Therefore the 'water hammer' effects should be properly studied in the pipes. The magnitude of pressure rise depends on:   i)    Speed at which the valve is closed   (ii)     The velocity of flow (V) (iii)     The length of the pipe (L) (iv)      Elastic properties of pipe material (E) and flowing fluid (K).
Post a Comment
Read more

THREE RESERVOIRS AND PIPE NETWORK PROBLEMS

 Ad: https://happyshirtsnp.com/         THREE RESERVOIRS PROBLEM AND PIPE NETWORK   In practice, several pipes are interconnected forming various loops (or circuit) in municipal water distribution systems. There are a number of pipes connected either in series or in parallel or a combination of both. A group of interconnected pipes forming several loops (or circuit) in complex manner is called pipe networks, The complexity in a pipe networking is to determine the solution Of pipe problem (flow distribution) in such pipe network. Branched pipes: Three Reservoirs Problem Fig: Branched pipes: three inter connected Reservoir     In water supply system, Often a number of reservoirs are required to be interconnected by means Of a pipe system consisting of a number of pipes namely main and branches which meet at a junction. Figure shows three reservoirs 'B' and 'C' are interconnected by pipes 'a', 'b' and 'c' which meet at a junction 'D'. In the
Post a Comment
Read more