GATE-2018 (24) :Wyllie – Time average equation

GATE 2017

82. Consider a vertical seismic profile (VSP) data acquisition experiment as shown in the figure below. The subsurface consists of a horizontal layer of 2 km thickness underlain by a semi-infinite half- space. The P-wave velocities (Vp) in the first layer and the half – space are 2.0 km/sec and 2.5 km/s, respectively. The vertical well has a string of receivers (denoted by inverted triangles) spaced 10 m a part, with the shallowest receiver at a depth of 0.5 km and the deepest receiver at a depth of 1.5km. The source (denoted by star) is placed 0.5 km from the well head. The travel time of the primary reflection event at the deepest receiver is…………….. s (Thanks to Alisha form ANU)

 

 Solution:

TRAVEL TIME OF THE SOURCE TO SHALLOWEST RECEIVER:

First i want to calculate the travel time from the Source to first shallowest receiver  (0.5km ).

For that we need travel time of the ABG or GBA.

The triangle ABC and BFG ,we get the distances AB and BG

 Triangle ABC:

$AB^{2} = BC^{2} + CA^{2}$

$AB^{2} = 1.5^{2} + 0.25^{2}$

$AB =\sqrt{ 1.5^{2} + 0.25^{2}}$

$AB = 1.5206km$

Triangle BFG:

$BG^{2} = BF^{2} + FG^{2}$

$BG^{2} = 2^{2} + 0.25^{2}$

$BG =\sqrt{ 2^{2} + 0.25^{2}}$

$BG = 2.1055km$

Therefore total distance ABG = AB+ BG

                                      ABG = 1.5206 + 2.0155

                                      ABG = 3.5361 km 

Formula for travel time t= Distance / velocity

Velocity of the first layer is given by Vp = 2km/sec

t = 3.5361 / 2

$t = 1.768 seconds$

Therefore time taken by P-wave to travel the distance from the source to shallow  receiver  is 1.768 seconds.

TRAVEL TIME OF THE SOURCE TO DEEPEST RECEIVER:

First we need to calculate the travel time between the Source to deepest receiver KBG

For that we need distance of KB and BG.

From the triangles KBG and BFG we get the distances of KB and BG.

Triangle KBD:

$KB^{2} = KD^{2} + BD^{2}$

$KB^{2} = 0.25^{2} + 0.5^{2}$

$KB =\sqrt{ 0.25^{2} + 0.5^{2}}$

$KB= 0.559km$ 

Triangle BFG:

$BG^{2} = BF^{2} + FG^{2}$

$BG^{2} = 2^{2} + 0.25^{2}$

$BG =\sqrt{ 2^{2} + 0.25^{2}}$

$BG= 2.0155 km$ 

Therefore the total distance is KBG = KB+ BG

                                              KBG = 0.559 +2.0155

                                              KBG = 2.5745 km

Formula for travel time t is 
$t=\frac{Distance}{Velocity}$

Velocity of the first layer is given by Vp = 2km/sec

$t=\frac{2.5745}{2}$

$t = 1.2877 seconds$

Therefore time taken by P-wave to travel the distance from the source to deepest receiver  is 1.2877 seconds.