All categories

2 years ago

The flow curve for a certain metal has parameters: strain-hardening

Engineering

1 answer:

ZanzabumX [31]2 years ago

3 0

Answer:(a) $45,300.24 lb/in/^2$

(b)0.255

Explanation:

Given

Strain hardening exponent(n)=0.22

Strength coefficient(k) $=54000 lb/in/^2$

and we know

$\sigma =k\left ( \epsilon \right )^n$

where

$sigma =true\ stress$

$\epsilon =true\ strain$

(a)True strain=0.45

$\sigma =54000\times 0.45^{0.22}$

$\sigma =45,300.24 lb/in^2$

(b)true stress $=40,000 lb/in^2$

$40000=54000\times \epsilon ^{0.22}$

$\epsilon ^{0.22}=0.7407$

$\epsilon =0.7407^{4.5454}=0.255$

You might be interested in

CHALLENGE ACTIVITY 1.4.1: Basic syntax errors. Type the statements. Then, correct the one syntax error in each statement. Hints:

Lunna [17]

Answer:

Lets check each statement for the errors.

Explanation:

System.out.println("Predictions are hard.");

This statement has no syntax errors. When this statement is executed the following line will be displayed:

Predictions are hard.

System.out.print("Especially ');

This statement is missing the closing quotation mark. A single quotation mark is placed instead of double quotation mark in the statement.

The following error message will be displayed when this program statement will be compiled:

Main.java:15: error: unclosed string literal

String literals use double quotes. So to correct this syntax error, the statement should be changed as follows:

System.out.print("Especially");

The output of this corrected line is as following:

Especially

System.out.println("about the future.").

In this line a period . is placed at the end of the statement instead of a semicolon ; but according to the syntax rules statements should end in semicolons.

The error message displayed when this line is compiled is as following:

Main.java:15: error: ; expected

Main.java:15: error: not a statement

So in order to correct this syntax error the statement should be changed as following:

System.out.println("about the future.");

The output of this corrected line is as following:

about the future

System.out.println("Num is: " - userName);

There is a syntax error in this statement because of - symbol used instead of +

+ symbol is used to join together a variable and a value a variable and another variable in a single print statement.

The error message displayed when this line is compiled is as following:

Main.java:13: error: bad operand types for binary operator '-'

So in order to correct this syntax error the statement should be changed as following:

System.out.println("Num is: " + userName);

This line will print two things one is the string Num is and the other is the value stored in userName variable.

So let userName= 5 then the output of this corrected line is as following:

Num is: 5

8 0

2 years ago

Fly thermostat, an automatic temperature controller for homes, learns the patterns for raising and lowering the temperature in a

Lubov Fominskaja [6]

Answer:

The Internet of things

Explanation:

When devices embedded with a network connectivity, software or sensors, are able to exchange data with the manufacturer through the network of physical things, then we say the Internet of things (IoT) has been defined.

Internet of things is when devices like smartphones, sensors and other smart devices are connected together which allows them to exchange data through the network of physical things. By exchanging data, these devices learn new patterns of adaptation.

In this case, the fly thermostat, learns the patterns for raising and lowering the temperature in a house through the Internet of things beacause the sensor has made it possible by automatically observing the temperature pattern of the house.

4 0

2 years ago

Read 2 more answers

A thermal energy storage unit consists of a large rectangular channel, which is well insulated on its outer surface and encloses

yaroslaw [1]

Answer:

the temperature of the aluminum at this time is 456.25° C

Explanation:

Given that:

width w of the aluminium slab = 0.05 m

the initial temperature $T_1$ = 25° C

$T{\infty} =600^0C$

h = 100 W/m²

The properties of Aluminium at temperature of 600° C by considering the conditions for which the storage unit is charged; we have ;

density ρ = 2702 kg/m³

thermal conductivity k = 231 W/m.K

Specific heat c = 1033 J/Kg.K

Let's first find the Biot Number Bi which can be expressed by the equation:

$Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{h \dfrac{w}{2}}{k}$

$Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{100 \times \dfrac{0.05}{2}}{231}$

$Bi = \dfrac{2.5}{231}$

Bi = 0.0108

The time constant value $\tau_t$ is :

$\tau_t = \dfrac{pL_cc}{h} \\ \\ \tau_t = \dfrac{p \dfrac{w}{2}c}{h}$

$\tau_t = \dfrac{2702* \dfrac{0.05}{2}*1033}{100}$

$\tau_t = \dfrac{2702* 0.025*1033}{100}$

$\tau_t = 697.79$

Considering Lumped capacitance analysis since value for Bi is less than 1

Then;

$Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]$

where;

$Q = -\Delta E _{st}$ which correlates with the change in the internal energy of the solid.

So;

$Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]= -\Delta E _{st}$

The maximum value for the change in the internal energy of the solid is :

$(pVc)\theta_1 = -\Delta E _{st}max$

By equating the two previous equation together ; we have:

$\dfrac{-\Delta E _{st}}{\Delta E _{st}{max}}= \dfrac{ (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]} { (pVc)\theta_1}$

Similarly; we need to understand that the ratio of the energy storage to the maximum possible energy storage = 0.75

Thus;

$0.75= [1-e^{\dfrac {-t}{ \tau_1}}]}$

So;

$0.75= [1-e^{\dfrac {-t}{ 697.79}}]}$

$1-0.75= [e^{\dfrac {-t}{ 697.79}}]}$

$0.25 = e^{\dfrac {-t}{ 697.79}}$

$In(0.25) = {\dfrac {-t}{ 697.79}}$

$-1.386294361= \dfrac{-t}{697.79}$

t = 1.386294361 × 697.79

t = 967.34 s

Finally; the temperature of Aluminium is determined as follows;

$\dfrac{T - T _{\infty}}{T_1-T_{\infty}}= e ^ {\dfrac{-t}{\tau_t}}$

$\dfrac{T - 600}{25-600}= e ^ {\dfrac{-967.34}{697.79}$

$\dfrac{T - 600}{25-600}= 0.25$

$\dfrac{T - 600}{-575}= 0.25$

T - 600 = -575 × 0.25

T - 600 = -143.75

T = -143.75 + 600

T = 456.25° C

Hence; the temperature of the aluminum at this time is 456.25° C

3 0

2 years ago

Derive the probability that a receptor is occupied by a ligand using a model that treats the L ligands in solution as distinguis

love history [14]

that is the same thing as you are not going through this week or something

7 0

2 years ago

The outer surface of an engine is situated in a place where oil leakage can occur. When leaked oil comes in contact with ahot su

VladimirAG [237]

Answer:

TBC thickness of 4 mm is insufficient to prevent fire hazard

Explanation:

Given:

- Temperature of hot-fluid inner surface T_i = 333°C

- The convection coefficient hot-fluid h_i = 7 W/m^2K

- The thermal conductivity of engine cover k_1 = 14 W/mK

- The thickness of engine cover L_1 = 0.01 m

- The thermal conductivity of TBC layer k_2 = 1.1 W/mK ... (Typing error)

- The thickness of TBC layer L_2 = 0.004 m

- Temperature of ambient air outer surface T_o = 69°C

- The convection coefficient ambient air h_o = 7 W/m^2K

Find:

Would a TBC layer of 4 mm thickness be sufficient to keep the engine cover surface below autoignition temperature of 200°C to prevent fire hazard?

Solution:

- We will use thermal circuit analogy for the 1-D problem and steady state conduction with no heat generation in the cover or TBC layer.

The temperature at each medium interface and the Thermal resistance for each medium is given in the attachment schematic and circuit analogy.

- We will calculate the total heat flux for the entire system q:

q = ( T_i - T_o ) / R_total

- R_total is the equivalent thermal resistance of the entire circuit. Since all resistances are in series we have:

R_total = 1 / h_i + L_1 / k_1 + L_2 / k_2 + 1 / h_o

- Plug in the values and compute:

R_total = 1 / 7 + 0.01 / 14 + 0.004 / 1.1 + 1 / 7

R_total = 0.2900649351 T-m^2 / W

- Calculate the Total heat flux q:

q = ( 333 - 69 ) / 0.2900649351

q = 910.141 W / m^2

- Just like the total current in a circuit remains same, the total heat flux remains same. We will use the total heat flux q to calculate the temperature of outer engine surface T_2 as follows:

q = ( T_i - T_2 ) / R_i2

Where,

R_i2 = 1 / h_i + L_1 / k_1

R_i2 = 1 / 7 + 0.01 / 14 = 0.14357 T-m^2 / W

Hence,

( T_i - T_2 ) = q*R_i2

T_2 = T_i - q*R_i2

Plug the values in:

T_2 = 333 - 910.141*0.14357

T_2 = 202.33°C

- The outer surface of the engine cover has a temperature above T_ignition = 200°C. Hence, the TBC thickness of 4 mm is insufficient to prevent fire hazard

3 0

2 years ago