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4 years ago

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A 5-cm-diameter shaft rotates at 4500 rpm in a 15-cmlong, 8-cm-outer-diameter cast iron bearing (k = 70 W/m·K) with a uniform cl

earance of 0.6 mm filled with lubricating oil (μ = 0.03 N·s/m2 and k = 0.14 W/m·K). The bearing is cooled externally by a liquid, and its outer surface is maintained at 40°C. Disregarding heat conduction through the shaft and assuming one-dimensional heat transfer, determine (a) the rate of heat transfer to the coolant, (b) the surface temperature of the shaft, and (c) the mechanical power wasted by the viscous dissipation in oil.

1 answer:

-BARSIC- [3]4 years ago

5 0

Answer:

(a) the rate of heat transfer to the coolant is Q = 139.71W

(b) the surface temperature of the shaft T = 40.97°C

(c) the mechanical power wasted by the viscous dissipation in oil 22.2kW

Explanation:

See explanation in the attached files

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Consider a cubic crystal with the lattice constant a. Complete the parts (a)-(c) below. (a) Sketch the crystallographic planes w

Anna [14]

Answer:

(a) See attachment

(b) The two planes are parallel because the intercepts for plane [220] are X = 0,5 and Y = 0,5 and for plane [110] are X = 1 and Y = 1. When the planes are drawn, they keep the same slope in a 2D plane.

(c) $d = \frac{a}{\sqrt{h^{2} + k^{2} + l^{2}}} = \frac{1}{\sqrt{2}} = 0,707$

Explanation:

(a) To determine the intercepts for an specific set of Miller indices, the reciprocal intercepts are taken as follows:

For [110]

$X = \frac{1}{1} = 1; Y = \frac{1}{1} = 1; Z = \frac{1}{0} = \inf.$

For [220]

$X = \frac{1}{2} = 0,5;Y = \frac{1}{2} = 0,5;Z = \frac{1}{0} = \inf.$

The drawn of the planes is shown in the attachments.

(b) Considering the planes as two sets of 2D straight lines with no intersection to Z axis, then the slope for these two sets are:

For (1,1):

$K_1 = \frac{1}{1} = 1$

For (0.5, 0.5):

$K_2 = \frac{0.5}{0.5} = 1$

As shown above, the slopes are exactly equal, then, the two straight lines are considered parallel and for instance, the two planes are parallel also.

(c) To calculate the d-spacing between these two planes, the distance is calculated as follows:

The Miller indices are already given in the statement. Then, the distance is:

$\frac{1}{d^{2}} = \frac{h^{2} + k^{2} + l^{2}}{a^{2}}$

$d = \frac{a}{\sqrt{h^{2} + k^{2} + l^{2}}} = \frac{1}{\sqrt{2}} = 0,707$

7 0

3 years ago

A masonry facade consisting of 3,800 square feet is to be constructed for a building. The total cost per worker hour is estimate

lana66690 [7]

Answer:

Days: 6.9444 days

Production rate: 547.2035 ft²/s

Explanation:

the solution is attached in the Word file

6 0

3 years ago

KVL holds for the supermesh, so we can write a KVL equation to generate the second equation we need to solve for the two unknown

kaheart [24]

Answer:

The values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA while that of V_Δ is -25 V

Explanation:

As the complete question is not given the complete question is found online and is attached herewith.

By applying KCL at node 1

$i_x+50mA=i_y\\i_x-i_y=0.05A$

Also

$V_{\Delta}=1K*i_y$

Now applying KVL on loop 1 as indicated in the attached figure

$1K*i_y+5K(i_y-i_z)+3K*i_x=0\\3i_x+6 i_y-5i_z=0$

Similarly for loop 2

$2V_{\Delta}+5K(i_z-i_y)=0\\2*1K*i_y+5K(i_z-i_y)=0\\2K*i_y+5K(i_z-i_y)=0\\3i_y-5i_z=0$

So the system of equations become

$i_x-i_y+0i_z=0.05\\3i_x+6i_y-5i_z=0\\0i_x-3i_y+5i_z=0$

Solving these give the values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA. Also the value of voltage is given as

$V_{\Delta}=1K*i_y\\V_{\Delta}=1K*-25 mA\\V_{\Delta}=-25 V$

The values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA while that of V_Δ is -25 V

8 0

4 years ago

lorasvet [3.4K]

4-ways tell me if I’m wrong

4 0

3 years ago

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