Question 5.15: A fireman must reach a window 40 m above the ground (Fig. 5....

Let u_{1} be the velocity of discharge from the nozzle,

then, u_{1}=\frac{\dot{m}}{\rho A}=\frac{30}{1000 \times(\pi / 4) \times(0.03)^{2}}=42.44 m/s

(Note that u_{1}^{2} / 2 g should be more than the height of the window for the jet to reach at all. In this case u_{1}^{2} / 2 g = 91.80 m which is greater than 40 m).

Let α be the angle of the nozzle with horizontal. Considering the time taken by a fluid particle in the jet to reach the window at point 2 from the discharge point 1 as t, we can write

x=42.44 \cos \alpha t (5.110a)

and 38=42.44 \sin \alpha t-1 / 2 g t^{2} (5.110b)

where, x is the horizontal distance between the nozzle and window. Eliminating t from Eqs (5.110a)and (5.110b), we have

38=x \tan \alpha-\frac{9.81}{2 \times(42.44)^{2}} \frac{x^{2}}{\cos ^{2} \alpha} (5.111)

The value of x depends upon the value of α. For maximum value of x we require an optimization with α.

Differentiating each term of Eq. (5.111) with respect to α, we get

For maximum x, \frac{ d x}{ d \alpha}=0

Hence, x \sec ^{2} \alpha-2 \times 0.0027 x^{2} \sec ^{2} \alpha \tan \alpha=0

x \tan \alpha=\frac{1}{2 \times 0.0027}=185.2 m (5.112)

Solving for x and α from Eq. (5.111) and (5.112), we get,

which gives x/cos α = 233.5 m

Again, x tan α = 185.2 m

or α = 52.5º

and x = 142 m