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lukranit [14]
3 years ago
10

Using the tables for water, determine the specified property data at the indicated states. In each case, locate the state on ske

tches of the p–υ and T–υ diagrams.a. At p = 3 bar, υ = 0.5 m3 /kg, find T in °C and u in kJ/kg.b. At T = 320°C, υ = 0.03 m3 /kg, find p in MPa and u in kJ/kg.c. At p = 28 MPa, T = 520°C, find υ in m3 /kg and h in kJ/kg.d. At T = 10°C, υ = 100 m3 /kg, find p in kPa and h in kJ/kg.e. At p = 4 MPa, T = 160°C, find υ in m3 /kg and u in kJ/kg.
Engineering
1 answer:
Anarel [89]3 years ago
5 0

heres a link! ;)

http://www.econsulat.ro/

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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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3 years ago
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