Answer:
-0.25 KN
Since, the force transmitted by the coupling between the nozzle and hose is on the left due to the negative sign.
Then the coupling must be in TENSION
Explanation:
Applying continuity and x component of momentum to inertia and using gage pressure to cancel [tex]P_{atm}[/tex]
Δ = [tex]\frac{\delta}{\delta t} \int\limits_{dv} \, pdv + \int\limits_{cs} upV.dA[/tex]
[tex]F_{sx} + F_{cx} =\frac{\delta}{\delta t} \int\limits_{cv} \, pdv + \int\limits_{cs} upV.dA[/tex]
Then;
[tex]0 = {-|pV_1A_1}+{|pV_2A_2|}\\=-pV_1A_1+pV_2A_2[/tex]
[tex]V_1=V_2\frac{A_2}{A_1}[/tex]
[tex]=V_2(\frac{D_2}{D_1} )^2[/tex]
= [tex]32 \frac{m}{s}*(\frac{25mm}{35mm})^2[/tex]
= 32 m/s × 0.5102040816
= 16.33 m/s
and
[tex]R_x+p_1gA_1=u_1{-|fV_1A_1}+u_2{|pV_2A_2|}[/tex]
where; [tex]u_1=v_1[/tex] and [tex]u_2=v_2[/tex]
[tex]R_x=-p_1gA_1-V_1fV_1A_1+V_2fV_2A_2[/tex]
[tex]= -p_1gA_1 + fV_2A_2(V_2-V_1)[/tex]
[tex]= - 510*10^3\frac{N}{m^2} * \frac{\pi}{4}(0.035)^2 m^2 + 999 \frac{kg}{m^3}*32 \frac{m}{s} *\frac{\pi}{4}(0.025)^2m^2(32.0-16.33)\frac{m}{s} *\frac{Ns^2}{kg.m}[/tex]
[tex]R_x = -244.78 N\\R_x= -0.25 KN[/tex]
Since, the force transmitted by the coupling between the nozzle and hose is on the left due to the negative sign.
Then the coupling must be in TENSION
