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3 years ago

Select the examples of Energy tasks. Check all that apply.

Engineering

2 answers:

Arada [10]3 years ago

5 0

Answer:

B. maintaining equipment

C. performing inspections

E. analyzing information

G. designing systems

Explanation:

or 2,3,5,7

Edge 2021!

Natali5045456 [20]3 years ago

4 0

Answer:

b c e g

Explanation:

right on edge 2020

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A direct contact heat exchanger (where the fluid mixes completely) has three inlets and one outlet. The mass flow rates of the i

lara31 [8.8K]

Answer:

Enthalpy at outlet=284.44 KJ

Explanation:

$m_1=1 Kg/s,m_2=1.5 Kg/s,m_3=22 Kg/s$

$h_1=100 KJ/Kg,h_2=120 KJ/Kg,h_3=500 KJ/Kg$

We need to Find enthalpy of outlet.

Lets take the outlet mass m and outlet enthalpy h.

So from mass conservation

$m_1+m_2+m_3=m$

m=1+1.5+2 Kg/s

m=4.5 Kg/s

Now from energy conservation

$m_1h_1+m_2h_2+m_3h_3=mh$

By putting the values

$1\times 100+1.5\times 120+2\times 500=4.5\times h$

So h=284.44 KJ

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3 years ago

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Read 2 more answers

Find values of the intrinsic carrier concentration n for silicon at –70° 0° 20° C, 100° C, and C. At 125° each temperature, what

Dominik [7]

Answer:

Part (i) at –70° C, intrinsic carrier concentration of silicon is 2.865 x 10⁵ carriers/cm³ and fraction of the atoms ionized is 5.37 x 10⁻¹⁸

Part (ii) at 0° C, intrinsic carrier concentration of silicon is 1.533 x 10⁹ carriers/cm³ and fraction of the atoms ionized is 3.067 x 10⁻¹⁴

Part (iii) at 20° C, intrinsic carrier concentration of silicon is 8.652 x 10⁹ carriers/cm³ and fraction of the atoms ionized is 1.731 x 10⁻¹³

Part (iv) at 100° C, intrinsic carrier concentration of silicon is 1.444 x 10¹² carriers/cm³ and fraction of the atoms ionized is 2.889 x 10⁻¹¹

Part (iv) at 125° C, intrinsic carrier concentration of silicon is 4.754 x 10¹² carriers/cm³ and fraction of the atoms ionized is 9.508 x 10⁻¹¹

Explanation:

$ni^2 = BT^3(e^{\frac{-E_g}{KT}})\\\\ni = \sqrt{ BT^3(e^{\frac{-E_g}{KT}})}$

where;

B = 5.4 x 10⁻³¹

Eg = 1.12 ev

K = 8.62 x 10⁻⁵ eV/K

T = (273 + ⁰C) K

Number of atoms in silicon crystal = 5 x 10²² atoms/cm³

Part (i) For –70° C, T = (273 -70 ⁰C)K = 203 K

$ni = \sqrt{ 5.4*10^{31}*203^3(e^ \ {\frac{-1.12}{8.62*10^{-5}*203}})}} \ =2.685*10^5 \ carriers/cm^3$

$Fraction \ of \ atoms \ ionized = \frac{2.685*10^5}{5 *10^{22}} = 5.370 *10^{-18}$

Part (ii) For 0° C, T = (273 +0 ⁰C)K = 273 K

$ni = \sqrt{ 5.4*10^{31}*273^3(e^ \ {\frac{-1.12}{8.62*10^{-5}*273}})}} \ =1.533*10^9 \ carriers/cm^3$

$Fraction \ of \ atoms \ ionized = \frac{1.533*10^9}{5 *10^{22}} = 3.067 *10^{-14}$

Part (iii) For 20° C, T = (273 + 20 ⁰C)K = 293 K

$ni = \sqrt{ 5.4*10^{31}*293^3(e^ \ {\frac{-1.12}{8.62*10^{-5}*293}})}} \ =8.652*10^9 \ carriers/cm^3$

$Fraction \ of \ atoms \ ionized = \frac{8.652*10^9}{5 *10^{22}} = 1.731 *10^{-13}$

Part (iv) For 100° C, T = (273 + 100 ⁰C)K = 373 K

$ni = \sqrt{ 5.4*10^{31}*373^3(e^ \ {\frac{-1.12}{8.62*10^{-5}*373}})}} \ =1.444*10^{12} \ carriers/cm^3$

$Fraction \ of \ atoms \ ionized = \frac{1.444*10^{12}}{5 *10^{22}} = 2.889 *10^{-11}$

Part (v) For 125° C, T = (273 + 125 ⁰C)K = 398 K

$ni = \sqrt{ 5.4*10^{31}*398^3(e^ \ {\frac{-1.12}{8.62*10^{-5}*398}})}} \ =4.754*10^{12} \ carriers/cm^3$

$Fraction \ of \ atoms \ ionized = \frac{4.754*10^{12}}{5 *10^{22}} = 9.508 *10^{-11}$

7 0

3 years ago