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Citrus2011 [14]
3 years ago
6

Water flows through a nozzle at the end of a fire hose. If the nozzle exit velocity must be 20 m/s and the exit diameter is 40 m

m, determine the minimum flow rate that the pump must provide.
Engineering
1 answer:
lara31 [8.8K]3 years ago
5 0

Answer:

minimum flow rate provided by pump is 0.02513 m^3/s

Explanation:

Given data:

Exit velocity of nozzle = 20m/s

Exit diameter = 40 mm

We know that flow rate Q is given as

Q = A \times V

where A is Area

A =\frac{\pi}{4} \times (40\times 10^{-3})^2 = 1.256\times 10^{-3} m^2

Q = 1.256\times 10^{-3} \times 20 = 0.02513 m^3/s

minimum flow rate provided by pump is 0.02513 m^3/s

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3 years ago
The solid cylinders AB and BC are bonded together at B and are attached to fixed supports at A and C. The modulus of rigidity is
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Answer:

a) 0.697*10³ lb.in

b) 6.352 ksi

Explanation:

a)

For cylinder AB:

Let Length of AB = 12 in

c=\frac{1}{2}d=\frac{1}{2} *1.1=0.55in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}0.55^4=0.1437\ in^4\\

\phi_B=\frac{T_{AB}L}{GJ}=\frac{T_{AB}*12}{3.3*10^6*0.1437}  =2.53*10^{-5}T_{AB}

For cylinder BC:

Let Length of BC = 18 in

c=\frac{1}{2}d=\frac{1}{2} *2.2=1.1in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}1.1^4=2.2998\ in^4\\

\phi_B=\frac{T_{BC}L}{GJ}=\frac{T_{BC}*18}{5.9*10^6*2.2998}  =1.3266*10^{-6}T_{BC}

2.53*10^{-5}T_{AB}=1.3266*10^{-6}T_{BC}\\T_{BC}=19.0717T_{AB}

T_{AB}+T_{BC}-T=0\\T_{AB}+T_{BC}=T\\T_{AB}+T_{BC}=14*10^3\ lb.in\\but\ T_{BC}=19.0717T_{AB}\\T_{AB}+19.0717T_{AB}=14*10^3\\20.0717T_{AB}=14*10^3\\T_{AB}=0.697*10^3\ lb.in\\T_{BC}=13.302*10^3\ lb.in

b) Maximum shear stress in BC

\tau_{BC}=\frac{T_{BC}}{J}c=13.302*10^3*1.1/2.2998=6.352\ ksi

Maximum shear stress in AB

\tau_{AB}=\frac{T_{AB}}{J}c=0.697*10^3*0.55/0.1437=2.667\ ksi

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