I will put other link in comments

A company specification calls for a steel component to have a minimum tensile strength of 1240 MPa. Tension tests are conducted

Pavlova-9 [17]

Answer:

a) The minimum acceptable value is 387.5 HV using Vickers hardness test.

b) The minimum acceptable value is 39.4 HRC using Rockwell C hardness test.

Explanation:

To get the tensile strength of a material from its hardness, we multiply it by an empirical constant that depends on things like yield strength, work-hardening, Poisson's ratio and geometrical factors. The incidence of cold-work varies this relationship.

According to DIN 50150 (a conversion table for hardness), the constant for Vickers hardness is ≈ 3.2 (an empirical approximate):

$\mbox{Tensile strength}=HV*3.2\\\\HV = \frac{\mbox{Tensile strength}}{3.2} =\frac{1240}{3.2}=387.5$

According to DIN 50150, the constant for Rockwell C hardness test is ≈31.5 around this values of tensile strength:

$\mbox{Tensile strength}=HRC*31.5\\\\HRC = \frac{\mbox{Tensile strength}}{31.5} =\frac{1240}{31.5}=39.4$

8 0

3 years ago

1. A cylindrical casting is 0.3 m in diameter and 0.5 m in length. Another casting has the same metal is rectangular in cross-se

Lorico [155]

Based on the Chvorinov's rule, the diference in the solidification times of the two castings is 14.092 times the solidification time of the prism casting.

<h3>How to apply the Chvorinov's rule for casting processes</h3>

The Chvorinov's rule is an empirical method to estimate the cooling time of a casting in terms of a reference time. This rule states that cooling time (t) is directly proportional to the square of the volume (V), in cubic meters, divided to the surface area (A), in square meters. Now we proceed to model each casting:

<h3>Cylindrical casting</h3>

t = C · [0.25π · D² · L/(0.5π · D² + π · D · L)]²

t = C · [0.25 · D · L/(0.5 · D + L)]² (1)

<h3>Prism casting</h3>

t' = C · [3 · T² · L/(6 · T · L + 2 · T · L + 6 · T²)]²

t' = C · [3 · T · L/(8 · L + 6 · T)]² (2)

<h3>Relationship between the cross sections of both castings</h3>

3 · T² = 0.25π · D² (3)

Where:

t - Cooling time of the cylindrical casting, in time unit.
t' - Cooling time of the prism casting, in time unit.
C - Cooling factor, in time unit per square meter.
D - Diameter of the cylinder, in meters.
L - Length of the casting, in meters.
T - Width of the cross section of the prism casting, in meters.

If we know that D = 0.3 m, then the thickness of the prism casting is:

$T = \sqrt{\frac{\pi}{12} }\cdot D$

T ≈ 0.153 m

And (1) and (2) simplified into these forms:

<h3>Cylindrical casting</h3>

t = C · {0.25π · (0.3 m) · (0.5 m)/[0.5 · (0.3 m) + 0.5 m]}²

t = 0.0329 · C (1b)

<h3>Prism casting</h3>

t' = C · {3 · (0.153 m) · (0.5 m)/[8 · (0.5 m) + 6 · (0.153 m)]}²

t' = 0.00218 · C (2b)

Lastly we find the percentual difference in the solidification times of the two castings by using the following expression:

r = (1 - t'/t) × 100 %

r = (1 - 0.00218/0.0329) × 100 %

r = 93.374 %

The cooling time of the prism casting is 6.626 % of the solidification time of the cylindrical casting. The diference in the solidification times of the two castings is 14.092 times the solidification time of the prism casting. $\blacksquare$

To learn more on solidification times, we kindly invite to check this verified question: brainly.com/question/13536247

3 0

2 years ago

Water vapor at 6 MPa, 600 degrees C enters a turbine operating at steady state and expands to 10kPa. The mass flow rate is 2 kg/

kirill115 [55]

Answer:

Explanation:

Obtain the following properties at 6MPa and 600°C from the table "Superheated water".

$h_1=3658.8KL/Kg\\s_1=7.1693kJ/kg.k$

Obtain the following properties at 10kPa from the table "saturated water"

$h_{f2}=191.81KJ/Kg.K\\h_{fg2}=2392.1KJ/Kg\\s_{f2}=0.6492KJ/Kg.K\\s_{fg2}=7.4996KJ/Kg.K$

Calculate the enthalpy at exit of the turbine using the energy balance equation.

$\frac{dE}{dt}=Q-W+m(h_1-h_2)$

Since, the process is isentropic process $Q=0$

$0=0-W+m(h_1-h_2)\\h_2=h_1-\frac{W}{m}\\\\h_2=3658.8-\frac{2626}{2}\\\\=2345.8kJ/kg$

Use the isentropic relations:

$s_1=s_{2s}\\s_1=s_{f2}+x_{2s}s_{fg2}\\7.1693=6492+x_{2s}(7.4996)\\x_{2s}=87$

Calculate the enthalpy at isentropic state 2s.

$h_{2s}=h_{f2}+x_{2s}.h_{fg2}\\=191.81+0.87(2392.1)\\=2272.937kJ/kg$

a.)

Calculate the isentropic turbine efficiency.

$\eta_{turbine}=\frac{h_1-h_2}{h_1-h_{2s}}\\\\=\frac{3658.8-2345.8}{3658.8-2272.937}=0.947=94.7%$

b.)

Find the quality of the water at state 2

since $h_f$ at 10KPa < $h_2$ < $h_g$ at 10KPa

Therefore, state 2 is in two-phase region.

$h_2=h_{f2}+x_2(h_{fg2})\\2345.8=191.81+x_2(2392.1)\\x_2=0.9$

Calculate the entropy at state 2.

$s_2=s_{f2}+x_2.s_{fg2}\\=0.6492+0.9(7.4996)\\=7.398kJ/Kg.K$

Calculate the rate of entropy production.

$S=\frac{Q}{T}+m(s_2-s_1)$

since, Q = 0

$S=m(s_2-s_1)\\=2\frac{kg}{s}(7.398-7.1693)kJ/kg\\=0.4574kW/k$

6 0

3 years ago

2. There are three drawings that architects and designers use to indicate spaces. What are these drawing?

Zarrin [17]

Answer:

Architectural plans.

Explanation:

An architectural plan is called the drawings made by architects, civil engineers or designers of spaces or interiors, in which these professionals capture their building projects, organizing the distribution of the spaces to be used, the elements to be located in them and, fundamentally, to give construction planning a projection into reality. Thus, the plans help professionals to have a better understanding of the expected end result of the projects they are carrying out.

3 0

3 years ago

Infinitivo de vivia kkk xd

blagie [28]

Answer:

pls put a question not random letters

Explanation:

8 0

3 years ago