Since no one is perfect is that a sentence fragment

A 5-mm-thick stainless steel strip (k = 21 W/m•K, rho = 8000 kg/m3, and cp = 570 J/kg•K) is being heat treated as it moves throu

Drupady [299]

Answer:

The temperature of the strip as it exits the furnace is 819.15 °C

Explanation:

The characteristic length of the strip is given by;

$L_c = \frac{V}{A} = \frac{LA}{2A} = \frac{5*10^{-3}}{2} = 0.0025 \ m$

The Biot number is given as;

$B_i = \frac{h L_c}{k}\\\\B_i = \frac{80*0.0025}{21} \\\\B_i = 0.00952$

$B_i$ < 0.1, thus apply lumped system approximation to determine the constant time for the process;

$\tau = \frac{\rho C_p V}{hA_s} = \frac{\rho C_p L_c}{h}\\\\\tau = \frac{8000* 570* 0.0025}{80}\\\\\tau = 142.5 s$

The time for the heating process is given as;

$t = \frac{d}{V} \\\\t = \frac{3 \ m}{0.01 \ m/s} = 300 s$

Apply the lumped system approximation relation to determine the temperature of the strip as it exits the furnace;

$T(t) = T_{ \infty} + (T_i -T_{\infty})e^{-t/ \tau}\\\\T(t) = 930 + (20 -930)e^{-300/ 142.5}\\\\T(t) = 930 + (-110.85)\\\\T_{(t)} = 819.15 \ ^0 C$

Therefore, the temperature of the strip as it exits the furnace is 819.15 °C

5 0

2 years ago

A piston-cylinder assembly has initially a volume of 0.3 m3 of air at 25 °C. Mass of the air is 1 kg. Weights are put on the pis

kogti [31]

Answer:

n=2.32

w= -213.9 KW

Explanation:

$V_1=0.3m^3,T_1=298 K$

$V_2=0.1m^3,T_1=1273 K$

Mass of air=1 kg

For polytropic process $pv^n=C$ ,n is the polytropic constant.

$Tv^{n-1}=C$

$T_1v^{n-1}_1=T_2v^{n-1}_2$

$298\times .3^{n-1}_1=1273\times .1^{n-1}_2$

n=2.32

Work in polytropic process given as

w= $\dfrac{P_1V_1-P_2V_2}{n-1}$

w= $mR\dfrac{T_1-T_2}{n-1}$

Now by putting the values

w= $1\times 0.287\dfrac{289-1273}{2.32-1}$

w= -213.9 KW

Negative sign indicates that work is given to the system or work is done on the system.

For T_V diagram

We can easily observe that when piston cylinder reach on new position then volume reduces and temperature increases,so we can say that this is compression process.

5 0

2 years ago

Give two causes that can result in surface cracking on extruded products.

Andreas93 [3]

Answer:

1. High friction

2. High extrusion temperature

Explanation:

Surface cracking on extruded products are defects or breakage on the surface of the extruded parts. Such cracks are inter granular.

Surface cracking defects arises from very high work piece temperature that develops cracks on the surface of the work piece. Surface cracking appears when the extrusion speed is very high, that results in high strain rates and generates heat.

Other factors include very high friction that contributes to surface cracking an d chilling of the surface of high temperature billets.

6 0

2 years ago

A 11.5 nC charge is at x = 0cm and a -1.2 nC charge is at x = 3 cm ..At what position or positions on the x-axis is the electric

diamong [38]

Answer:

Explanation:

Given

$q_1=11.5\ nC$ charge is placed at $x=0\ cm$