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1962348829's Question
Math
Posted 7 months ago
calculate wrong

Firefighters are holding a nozzle at the end of a hose while trying to extinguish a fire. If the hose inner diameter is 10 cm , the nozzle exit diameter is 5 cm and the water flow rate is 12 m3/min12 \mathrm{~m}^{3} / \mathrm{min}, determine the horizontal resistance force required of the firefighters to hold the nozzle. Express your answer in kN .
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Answer from Sia
Posted 7 months ago
Solution by Steps
step 1
Calculate the cross-sectional area of the hose at the nozzle exit: Anozzle=π(dnozzle2)2=π(5 cm2)2=π(2.5 cm)2=6.25π cm2A_{\text{nozzle}} = \pi \left(\frac{d_{\text{nozzle}}}{2}\right)^2 = \pi \left(\frac{5 \text{ cm}}{2}\right)^2 = \pi \left(2.5 \text{ cm}\right)^2 = 6.25\pi \text{ cm}^2
step 2
Calculate the cross-sectional area of the hose at the inner diameter: Ahose=π(dhose2)2=π(10 cm2)2=π(5 cm)2=25π cm2A_{\text{hose}} = \pi \left(\frac{d_{\text{hose}}}{2}\right)^2 = \pi \left(\frac{10 \text{ cm}}{2}\right)^2 = \pi \left(5 \text{ cm}\right)^2 = 25\pi \text{ cm}^2
step 3
Calculate the velocity of water at the nozzle exit using the flow rate: Q=Ahose⋅vhose=Anozzle⋅vnozzleQ = A_{\text{hose}} \cdot v_{\text{hose}} = A_{\text{nozzle}} \cdot v_{\text{nozzle}}
step 4
Given the flow rate Q=12 m3/min=12×106 cm3/minQ = 12 \text{ m}^3/\text{min} = 12 \times 10^6 \text{ cm}^3/\text{min}, convert to seconds: Q=12×106 cm360 s=2×105 cm3/sQ = \frac{12 \times 10^6 \text{ cm}^3}{60 \text{ s}} = 2 \times 10^5 \text{ cm}^3/\text{s}
step 5
Calculate the velocity of water at the nozzle exit: vnozzle=QAnozzle=2×105 cm3/s6.25π cm2=2×1056.25π≈1018.59 cm/s=10.1859 m/sv_{\text{nozzle}} = \frac{Q}{A_{\text{nozzle}}} = \frac{2 \times 10^5 \text{ cm}^3/\text{s}}{6.25\pi \text{ cm}^2} = \frac{2 \times 10^5}{6.25\pi} \approx 1018.59 \text{ cm/s} = 10.1859 \text{ m/s}
step 6
Calculate the velocity of water at the hose inner diameter: vhose=QAhose=2×105 cm3/s25π cm2=2×10525π≈254.65 cm/s=2.5465 m/sv_{\text{hose}} = \frac{Q}{A_{\text{hose}}} = \frac{2 \times 10^5 \text{ cm}^3/\text{s}}{25\pi \text{ cm}^2} = \frac{2 \times 10^5}{25\pi} \approx 254.65 \text{ cm/s} = 2.5465 \text{ m/s}
step 7
Apply the Bernoulli equation to find the horizontal resistance force: ΔP=12ρ(vnozzle2−vhose2)\Delta P = \frac{1}{2} \rho (v_{\text{nozzle}}^2 - v_{\text{hose}}^2)
step 8
Given ρ=1000 kg/m3\rho = 1000 \text{ kg/m}^3, calculate ΔP\Delta P: ΔP=12×1000 kg/m3×(10.18592−2.54652) m2/s2=500×(103.75−6.48)≈500×97.27=48635 Pa\Delta P = \frac{1}{2} \times 1000 \text{ kg/m}^3 \times (10.1859^2 - 2.5465^2) \text{ m}^2/\text{s}^2 = 500 \times (103.75 - 6.48) \approx 500 \times 97.27 = 48635 \text{ Pa}
step 9
Calculate the force: F=ΔP×Anozzle=48635 Pa×6.25π×10−4 m2≈95.6 NF = \Delta P \times A_{\text{nozzle}} = 48635 \text{ Pa} \times 6.25\pi \times 10^{-4} \text{ m}^2 \approx 95.6 \text{ N}
step 10
Convert the force to kN: F=95.6 N×1 kN1000 N=0.0956 kNF = 95.6 \text{ N} \times \frac{1 \text{ kN}}{1000 \text{ N}} = 0.0956 \text{ kN}
Answer
0.0956 kN
Key Concept
Bernoulli's Equation
Explanation
Bernoulli's equation relates the pressure, velocity, and height at two points in a fluid flow. In this problem, it helps determine the pressure difference and thus the force required to hold the nozzle.

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