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3 years ago

A method that uses low temperature heat-treating that imparts toughness without reduction in hardness is called:_______

Engineering

1 answer:

lbvjy [14]3 years ago

5 0

Answer:

"Tempering Process" seems to be the appropriate choice.

Explanation:

Tempering seems to be a method of heat preparation which is mostly used in completely hard materials to increase consistency, strength, durability, and also some decreasing brittleness.
The tempering method is used to examine good functionality as well as flexural by reducing stiffness again after the substance has indeed been quenched towards its toughest state.

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Air flows steadily and isentropically from standard atmospheric conditions to a receiver pipe through a converging duct. The cro

Liula [17]

Answer:

The answer is "0.0728"

Explanation:

Given value:

$P_0= 14.696\ ps\\\\\ p _{0}= 0.00238 \frac{slue}{ft^{3}}\\\\\ A= 0.05 ft^2\\\\\ T_0= 59^{\circ}f = 518.67R\\\\\ air \ k= 1\\\\ \ cirtical \ pressure ( P^*)=P_0\times \frac{2}{k+1}^{\frac{k}{k-1}}\\$

$= 14.696\times (\frac{2}{1.4+1})^{\frac{1.4}{1.4-1}}\\\\=7.763 Psia\\\\$

if $P$ flow is chocked

if $P>P^{*} \to$ flow is not chocked

When P= 10 psia < $P^{*}$ $\to$ not chocked

match number:

$\ for \ P= \ 10\ G= \sqrt{\frac{2}{k-1}[(\frac{\ p_{0}}{p})^{\frac{k-1}{k}}-1]}$

$= \sqrt{\frac{2}{1.4-1}[(\frac{14.696}{10})^{\frac{1.4-1}{1.4}}-1]}$

$M_0=7.625$

$p=p_0(1+\frac{k-1}{2} M_0 r)^\frac{1}{1-k}$

$=0.00238(1+\frac{1.4-1}{2}0.7625`)^{\frac{1}{1-1.4}}\\\\\ p=0.001808 \frac{slug}{ft^3}$

$\ T= T_0(1+\frac{k-1}{2} Ma^r)^{-1}\\\\\ T=518.67(1+\frac{1.4-1}{2} 0.7625^2)^{-1}\\\\\ T=464.6R\\\\$

$\ velocity \ of \ sound \ (C)=\sqrt{KRT}\\\\$

$=\sqrt{1.4\times1716\times464.6}\\\\=1057 ft^3\\\\$

R= gas constant=1716

$m=PAV\\\\$

$=0.001808\times0.05\times(Ma.C)\\\\=0.001808\times0.05\times0.7625\times 1057\\\\=0.0728\frac{slug}{s}$

5 0

3 years ago

Problem 4 You are designing a circuit to drive LED1, using the following circuit. The datasheet for the LED specifies that VF =

HACTEHA [7]

Answer:

a) I_LED= 1/6 A b) Vf= 2.5V

Explanation:

Consider circuit in the attachment.

a) We will simplify current source in paraller with resistor to a voltage source in series with a resistor(see attachment 2)

Solving the circuit in attachment 2 using mesh analysis

-9+2I1+4(I1-I2)-4+2I1=0

8I1 - 4I2= 13 ............... eq 1

4+4(I2-I1)+ I2 + 2=0

4I1- 5I2 = 6 ............ eq 2

I1= 41/24 ; I2 = 1/6; I2= I_LED

b) Solving the circuit in attachment 2 again, this time I2=0

8I1 - 4I2= 13

8I1- 4(0)=13

I1= 13/8

Vf= 4(I1- I2) -4

I2=I_LED=0

Vf= 2.5 V

4 0

4 years ago

Air at 1 atm enters a thin-walled ( 5-mm diameter) long tube ( 2 m) at an inlet temperature of 100°C. A constant heat flux is ap

alexdok [17]

Answer:

heat rate = 7.38 W

Explanation:

Given Data:

Pressure = 1atm

diameter (D) = 5mm = 0.005m

length = 2

mass flow rate (m) = 140*10^-6 kg/s

Exit temperature = 160°C,

At 400K,

Dynamic viscosity (μ) = 22.87 *10^-6

Prandtl number (pr) = 0.688

Thermal conductivity (k) = 33.65 *10^-3 W/m-k

Specific heat (Cp) = 1.013kj/kg.K

Step 1: Calculating Reynolds number using the formula;

Re = 4m/πDμ

= (4*140*10^-6)/(π* 0.005*22.87 *10^-6)

= 5.6*10^-4/3.59*10^-7

= 1559.

Step 2: Calculating the thermal entry length using the formula

Le = 0.05*Re*Pr*D

Substituting, we have

Le = 0.05 * 1559 * 0.688 *0.005

Le = 0.268

Step 3: Calculate the heat transfer coefficient using the formula;

Nu = hD/k

h = Nu*k/D

Since Le is less than given length, Nusselt number (Nu) for fully developed flow and uniform surface heat flux is 4.36.

h = 4.36 * 33.65 *10^-3/0.005

h = 0.1467/0.005

h = 29.34 W/m²-k

Step 4: Calculating the surface area using the formula;

A = πDl

=π * 0.005 * 2

=0.0314 m²

Step 5: Calculating the temperature Tm

For energy balance,

Qc = Qh

Therefore,

H*A(Te-Tm) = MCp(Tm - Ti)

29.34* 0.0314(160-Tm) = 140 × 10-6* 1.013*10^3 (Tm-100)

0.921(160-Tm) = 0.14182(Tm-100)

147.36 -0.921Tm = 0.14182Tm - 14.182

1.06282Tm = 161.542

Tm = 161.542/1.06282

Tm = 151.99 K

Step 6: Calculate the rate of heat transferred using the formula

Q = H*A(Te-Tm)

= 29.34* 0.0314(160-151.99)

= 7.38 W

the Prandtl number using the formula

5 0

3 years ago

Two kilograms of air within a piston-cylinder assembly execute a Carnot power cycle with maximun1 and minimum temperatures of 75

ExtremeBDS [4]

Answer:

The Answer and Explanation is in the folloing attachment

Explanation:

8 0

4 years ago

When you see a street with white markings only, what kind of street is it?

Georgia [21]

Answer:

it's a one way street

3 0

4 years ago