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Zolol [24]
2 years ago
12

Explain why failure of this garden hose occurred near its end and why the tear occurred along its length. Use numerical values t

o explain your result. Assume the water pressure is 30 psistr

Engineering
1 answer:
alukav5142 [94]2 years ago
8 0

Answer:

  • hoop stress
  • longitudinal stress
  • material used

all this could led to the failure of the garden hose and the tear along the length

Explanation:

For the flow of water to occur in any equipment, water has to flow from a high pressure to a low pressure. considering the pipe, water is flowing at a constant pressure of 30 psi inside the pipe which is assumed to be higher than the allowable operating pressure of the pipe. but the greatest change in pressure will occur at the end of the hose because at that point the water is trying to leave the hose into the atmosphere, therefore the great change in pressure along the length of the hose closest to the end of the hose will cause a tear there. also the other factors that might lead to the failure of the garden hose includes :

hoop stress ( which acts along the circumference of the pipe):

αh = \frac{PD}{2T}     EQUATION 1

and Longitudinal stress ( acting along the length of the pipe )

αl = \frac{PD}{4T}       EQUATION 2

where p = water pressure inside the hose

          d = diameter of hose, T = thickness of hose

we can as well attribute the failure of the hose to the material used in making the hose .

assume for a thin cylindrical pipe material used to be

\frac{D}{T} ≥  20

insert this value into equation 1

αh = \frac{20 *30}{2}  = 60/2 = 30 psi

the allowable hoop stress was developed by the material which could have also led to the failure of the garden hose

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Answer:

P = 4.745 kips

Explanation:

Given

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Then we use the formula

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For AB:

ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)

⇒  ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB

For AC:

ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)

⇒  ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC

Now, we use the condition

ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in

⇒  ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in

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(2.107*10⁻⁶ in/lbf)*P = 0.01 in

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The pic shown can help to understand the question.

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Answer:

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Explanation:

given data:

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WE KNOW THAT mass flow rate is given as

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for pressure 1 MPa, the density of steam is = 4.068 kg/m3

therefore we have

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Answer:

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Solution:

As per the question:

Temperature at inlet, T_{i} = 20^{\circ}C = 273 + 20 = 293 K

Temperature at the outlet, T_{o} = 200{\circ}C = 273 + 200 = 473 K

Pressure at inlet, P_{i} = 80 kPa = 80\times 10^{3} Pa

Pressure at outlet, P_{o} = 800 kPa = 800\times 10^{3} Pa

Speed at the outlet, v_{o} = 20 m/s

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Input power, P_{i} = 400 kW = 400\times 10^{3} W

Now,

To calculate the heat transfer, Q, we make use of the steady flow eqn:

h_{i} + \frac{v_{i}^{2}}{2} + gH  + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH' + p_{s}

where

h_{i} = specific enthalpy at inlet

h_{o} = specific enthalpy at outlet

v_{i} = air speed at inlet

p_{s} = specific power input

H and H' = Elevation of inlet and outlet

Now, if

v_{i} = 0 and H = H'

Then the above eqn reduces to:

h_{i} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH + p_{s}

Q = h_{o} - h_{i} + \frac{v_{o}^{2}}{2} + p_{s}                (1)

Also,

p_{s} = \frac{P_{i}}{ mass, m}

Area of cross-section, A = \frac{\pi D^{2}}{4} =\frac{\pi 0.1^{2}}{4} = 7.85\times 10^{- 3} m^{2}

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p_{s} = \frac{400\times 10^{3}}{0.925} = 432.432 kJ/kg

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\Delta h = h_{o} - h_{i} = c_{p}\Delta T =c_{p}(T_{o} - T_{i}) = 1.005(200 - 20) = 180.9 kJ/kg

Now, using these values in eqn (1):

Q = 180.9 + \frac{20^{2}}{2} + 432.432 = 813.33 kW

Now, rate of heat transfer, q:

q = mQ = 0.925\times 813.33 = 752.33 kW

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