$d_{2}=0.3m, d_{1}=0.6m, p_{1}=171.675 kN/m^{2}$

Magnitude and direction of force extend on bend

i) Case 1) When no flow i.e Q=o

Consider section 1-1 at inclet and section 2-2 at exit

$V_{1}=V_{2}$=0

$A_{1}=\frac{\pi}{4}\times d_{1}^{2}=\frac{\pi}{4}\times (0.6)^{2}=0.2827m^{2}$

$A_{2}=\frac{\pi}{4}\times d_{2}^{2}=\frac{\pi}{4}\times (0.3)^{2}=0.0707 m^{2}$

By Bernoullis equation

$\frac{P_{1}}{y}+\frac{V_{1}^{2}}{2g}=\frac{p_{2}}{y}+\frac{V_{2}^{2}}{2g}$

$\frac{171.675}{981}=\frac{P_{2}}{9.81}=P_{2}=171.675 kN/M^{2}$

Force exerted by bend x-direction

$P_{1}A_{1}-P_{2}A_{2}cose\theta+Fx=PQ(V_{2}cos\theta -V_{1})$ (Q=0)

$171.675\times 0.2827-171.675\times 0.0707cos 60+fs=0$

x=42.464 kN

Force exterted by bend in y-direction

$F_{y}-P_{2}A_{2}sin\theta=171.675\times$ 0.07607 sin60

=10.511 kN

F$\sqrt{F_{x}^{2}+f^{2}_{y}}=\sqrt{(42.464)^{2}+(10.511)^{2}}$

=43.746 kN

$\Phi=\tan^{-1}(\frac{Fy}{Fx})=\tan^{-1}(\frac{10.511}{43.746})$=13.g

ii) Case ii-When flow is 876 l/s i.e Q=0.879 $m^{3}/s$

V_{1}=$\frac{Q}{A_{1}}=\frac{0.876}{0.2807}$=3.12m/s

V_{2}=$\frac{Q}{A_{2}}=\frac{0.876}{0.0707}$=12.48m/s

By Bernoulis equation

$\frac{P_{1}}{y}+\frac{V_{1}^{2}}{2g}=\frac{P_{2}}{y}+\frac{V_{2}^{2}}{2g}$

$\frac{171.678}{9.81}+\frac{3.12^{2}}{2\times 9.81}+\frac{(12.48)^{2}}{2\times 9.81}$

$P_{2}=127.87 kN/m^{2}$

Force Exerted in x-direction

$P_{1}A_{1}=P_{2}A_{2}cos\theta+Fx=PQ(V_{2}sin\theta)\times 10^{ -3}$

=$171.675\times 0.2828-127.87\times 0.0707cos 60+fx==1000\times 0.876(12.48 cos 60-3.12)$

Fx=41.343 kN

Force exerted in y-direction

$P_{1}A_{1}-P_{2}A_{2}sin\theta=P.Q(V_{2}sin\theta )\times 10^{-3}$

$F_{y}-127.87\times 0.0707 sin 60=1000\times 0.876\times12.48 sin 60\times 10^{-3}$

17.301kN

Resultant Force F=$\sqrt{f_{x}^{2}+f_{y}^{2}}$

=$\sqrt{(41.343)^{2}+(17.301)^{2}}$

=44.817 kN

Direction $\Phi=tan^{-1}(\frac{fy}{fx})$

=$tan^{-1}(\frac{17.301}{44.817})$

=21.1$^{\circ}$