Forces on a Fire Hose Nozzle

Contents

1 Introduction

Recently, I tutored a first-year engineering student from a London university in fluid mechanics. His lecturer had derived the reaction force on a fire hose nozzle using the steady flow momentum equation, continuity, and Bernoulli’s principle. The lecturer concluded that the firefighters are applying a backwards force on the nozzle to “stop it getting pulled towards the fire”. Now, one does not need to be a trained firefighter to realise that this is not right. Anybody who has ever operated a garden hose or pressure washer knows that in fact the opposite is true, namely that there is a “recoil” in the opposite direction of the flow when the nozzle is opened, which needs to be resisted by the operator.

In Sec. 2 we will derive the equilibriating force on a nozzle, and in Sec. 3 we will determine the force that firefighters need to exert on the nozzle to keep it stationary.

2 Derivation of force on nozzle

Let’s start with the Steady Flow Momentum Equation: $$F+\underset{\text{CV}}{\iint }\rho \mathbf{VV}\bullet \mathbf{dA}+\underset{\text{CV}}{\iint }p\mathbf{dA}=0,$$ (1)where $F$ is the sum of all external forces on the control volume and $\mathbf{dA}$ is positive into the control volume⁠[1][1]: Commonly, $\mathbf{dA}$ is taken positive out of the control volume, but I prefer to define it the other way around. That way all forces (momentum, pressure, and external) are taken to act on the CV and less care needs to be taken getting signs right when deriving the equation.. In our specific situation, we have $$-F=\rho A_1{V}_1^2-\rho A_2{V}_2^2+p_1{A}_1$$ (2) when using gauge pressures ($p_2$ is atmospheric pressure). By continuity, we have $$V_1{A}_1=V_2{A}_2.$$ (3)Neglecting hydrostatic terms, Bernoulli gives $$p_1=\frac{1}{2}\rho (V_2^2-V_1^2).$$ (4)Substituting Eqns. 3 and 4 into Eqn. 2, we get $$\begin{array}{rl}-F& =\rho A_1\left[V_1^2-V_1{V}_2+\frac{1}{2}(V_2^2-V_1^2)\right]\\ & =\frac{1}{2}\rho A_1(V_1^2-2V_1{V}_2+V_2^2)\\ & =\frac{1}{2}\rho A_1(V_1-V_2{)}^2\\ & =\frac{1}{2}\rho A_1{V}_1^2{\left(1-\frac{V_2}{V_1}\right)}^2.\end{array}$$ (5)But from Eqn. 3, we have $V_2/V_1=A_1/A_2$. Making the term in parentheses positive, we obtain the classic result $$-F=\frac{1}{2}\rho A_1{V}_1^2{\left(\frac{A_1}{A_2}-1\right)}^2.$$ (6)Equivalently, if $V_2$ is known instead of $V_1$: $$-F=\frac{1}{2}\rho A_1{V}_2^2{\left(1-\frac{A_2}{A_1}\right)}^2.$$ (7)

Since the sum of external forces on the control volume is in the negative x-direction (i.e. to the left), the lecturer concluded that the firefighters are pulling back on the nozzle (to “stop it getting pulled torwards the fire”). In the following section, we shall see why this is incorrect.

3 Force exerted by the firefighters

By choosing a control volume whose boundaries go across the walls of the nozzle (or the hose), we are effectively slicing through the nozzle (or hose) at the location of the boundary. In doing so, we need to take account of the longitudinal stress ${\sigma }_l$ in the wall of the nozzle (or hose). Designating the force that the firefighters are exerting on the nozzle as $F_f$, we then have $$F=F_f-F_w,$$ (8)where $F_w$ is the force caused by ${\sigma }_l$.

To find out the sign of $F_f$ (which will tell us if the firefighters are pushing or pulling on the nozzle), we need to determine the magnitude of $F_w$. This is straightforward if we assume that the nozzle remains stationary after first being opened (and therefore wall stresses are constant, since strains have not changed), and that the firefighters are not exerting any force on the nozzle before it is opened. Before the nozzle is opened, the sum of pressure forces must therefore be balanced by the longitudinal wall stresses: $$F_w=p_0{A}_1,$$ (9)where $p_0$ is the pressure in the hose before the nozzle is opened. If we assume that the total pressure remains unchanged after the nozzle is opened⁠[2][2]: When drawing water from a fire truck, one might expect a drop in total pressure once the water starts flowing, as the pump now needs to work harder. This should not be an issue when a fire hydrant is supplying the water, however frictional losses inside the hose should lead to some loss of total pressure in both cases., we have $$p_0=p_1+\frac{1}{2}\rho V_1^2$$ (10) and thus, using Eqn. 4, $$F_w=\frac{1}{2}\rho A_1{V}_2^2.$$ (11)This may now be substituted into Eqn. 8, along with Eqn. 7, to find $F_f$: $$\begin{array}{rl}{F}_f& =F+F_w\\ & =-\frac{1}{2}\rho A_1{V}_2^2{\left(1-\frac{A_2}{A_1}\right)}^2+\frac{1}{2}\rho A_1{V}_2^2\\ & =\frac{1}{2}\rho A_1{V}_2^2{\left(\frac{A_2}{A_1}\right)}^2.\end{array}$$ (12)A rather neat result! $F_f$ is a function of the hose diameter (via $A_1$), the exit velocity, and the nozzle contraction ratio only. Furthermore, $F_w$ is positive, which confirms our intuition that the firefighters are pushing on the nozzle in the flow direction, rather than pulling on it.