Thrust Block Calculations:
3
y
x
4
135
2
In order to determine whether or not a thrust block is needed at the y-connection, we used
the momentum equation for fluid flow to calculate a reactionary force. We assumed the
weight of water to be negligible compared to the forces generated by the pressure and
momentum. To find the reactionary force we first calculated the pressures just before and
after the junction using the energy equation. The equations and values for these pressures
are shown below:
p2  (z1  z2  hl )  435,000Pa
v 22
v 32
p3  (  
  )  436,000Pa

2g
2g
P2
v 22
v 42
p4  (  
  )  434,000Pa

2g
2g
P2

Next we used the component form of the momentum equation to calculate the magnitude
and direction of the reactionary force needed to hold the piping in place. The force
summation equations are given below.
F
x
 Rx  p4 A4  p3 A3 cos 45  p2 A2 cos 45  mÝ3v 3  mÝ4 v 4  mÝ2v 2  0
Q22
Q32
Q42

p
A

p
A
cos
45

p
A
cos
45




4 4
3 3
2 2
A22
A32
A42
 Rx  4627N
 Rx  
F
y
 Ry  p3 A3 sin 45  p2 A2 sin 45  mÝ3v 3  mÝ2v 2
 Ry   p3 A3 sin 45  p2 A2 sin 45  
 Ry  2655N

Q32
Q2
 2
A3
A2
Finally, we combined these two forces to create a resultant reactionary force and find an
angle at which this force is applied. The angle theta is
Rtotal  Rx2  Ry2  5335N
  tan1 (
Ry
)  28.9
Rx

Rx
Ry


Rtotal

