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nasty-shy [4]
4 years ago
8

Water flows at a rate of 10 gallons per minute in a new horizontal 0.75?in. diameter galvanized iron pipe. Determine the pressur

e gradient, ?P/L, along the pipe.
Engineering
1 answer:
ruslelena [56]4 years ago
3 0

Answer:

\frac{\delta p }{l} = 30.4 lb/ft^3

Explanation:

Given data:

flow rate = 10 gallon per  minute = 0.0223 ft^3/sec

diameter = 0.75 inch

we know discharge is given as

Q =  VA

solve for velocity V = \frac{Q}{A}[/tex]

V = \frac{0.223}{\frac{\pi}{4} \frac{0.75}{12}}

V = 7.27 ft/sec

we know that Reynold number

Re = \frac{VD}{\nu}

Re = \frac{7.27 \times \frac{0.75}{12}}{1.21\times 10^{-5}}

Re = 3.76 \times 10^4

calculate the \frac{\epsilon }{D}ratio to determine the fanning friction f

\frac{\epsilon }{D} = \frac{0.0005}{\frac{0.75}{12}} = 0.008

from moody diagram f value corresonding to Re and \frac{\epsilon }{D}is 0.037

for horizontal pipe

\delta p = \frac{f l \rho v^2}{2D}

\frac{\delta p }{l} = \frac{1 \times 0.037 \times 1.94 \times 7.27}{\frac{0.75}{12}}

where 1.94 slug/ft^3is density of  water

\frac{\delta p }{l} = 30.4 lb/ft^3

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astra-53 [7]

Answer:

0.024 m = 24.07 mm

Explanation:

1) Notation

\sigma_c = tensile stress = 200 Mpa

K = plane strain fracture toughness= 55 Mpa\sqrt{m}

\lambda= length of a surface crack (Variable of interest)

2) Definition and Formulas

The Tensile strength is the ability of a material to withstand a pulling force. It is customarily measured in units (F/A), like the pressure. Is an important concept in engineering, especially in the fields of materials and structural engineering.

By definition we have the following formula for the tensile stress:

\sigma_c=\frac{K}{Y\sqrt{\pi\lambda}}   (1)

We are interested on the minimum length of a surface that will lead to a fracture, so we need to solve for \lambda

Multiplying both sides of equation (1) by Y\sqrt{\pi\lambda}

\sigma_c Y\sqrt{\pi\lambda}=K   (2)

Sequaring both sides of equation (2):

(\sigma_c Y\sqrt{\pi\lambda})^2=(K)^2  

\sigma^2_c Y^2 \pi\lambda=K^2   (3)

Dividing both sides by \sigma^2_c Y^2 \pi we got:

\lambda=\frac{1}{\pi}[\frac{K}{Y\sigma_c}]^2   (4)

Replacing the values into equation (4) we got:

\lambda=\frac{1}{\pi}[\frac{55 Mpa\sqrt{m}}{1.0(200Mpa)}]^2 =0.02407m

3) Final solution

So the minimum length of a surface crack that will lead to fracture, would be 24.07 mm or more.

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