Question 3.16: A 10-cm fire hose with a 3-cm nozzle discharges 1.5 m^3/min ...

We use Bernoulli’s equation and continuity to find the pressure p_1 upstream of the nozzle, and then we use a control volume momentum analysis to compute the bolt force, as in Fig. E3.16.

The flow from 1 to 2 is a constriction exactly similar in effect to the venturi in Example 3.15, for which Eq. (1) gave

p_1 = p_2 + \frac{1}{2} \rho (V_2^2 – V^2_1)                      (1)

The velocities are found from the known flow rate Q = 1.5 m^3/min or 0.025 m^3/s:

V_2 = \frac{Q}{A_2} = \frac{0.025  m^3/s}{(\pi/4)(0.03  m)^2} = 35.4 m/s

V_1 = \frac{Q}{A_1} = \frac{0.025  m^3/s}{(\pi/4)(0.1  m)^2} = 3.2 m/s

We are given p_2 = p_a = 0 gage pressure. Then Eq. (1) becomes

p_1 = \frac{1}{2}(1000  kg/m^3) [(35.4^2 – 3.2^2)m^2/s^2] = 620,000  kg/(m \cdot s^2) = 620,000 Pa gage

The control volume force balance is shown in Fig. E3.16b:

and the zero gage pressure on all other surfaces contributes no force. The x-momentum flux is +\dot{m}V_2 at the outlet and -\dot{m}V_1 at the inlet. The steady flow momentum relation (3.40) thus gives

\sum{F} = \frac{d}{dt} \left(\int_{CV}{V \rho  d^\circ \mathcal{V}}\right) + \sum{(\dot{m}_iV_i)_{out}} – \sum{(\dot{m}_iV_i)_{in}}             (3.40)

or                    F_B = p_1A_1 – \dot{m}(V_2 – V_1)               (2)

Substituting the given numerical values, we find

\dot{m} = \rho Q = (1000  kg/m^3)(0.025  m^3/s) = 25 kg/s

F_B = (620,000  N/m^2)(0.00785  m^2) – (25  kg/s)[(35.4 – 3.2)m/s] = 4872  N – 805 (kg\cdot m)/s^2 = 4067  N (915  lbf)                 Ans.