Which of the following is not an electronic device ?

You are an engineer at company XYZ, and you are dealing with the need to determine the maximum load you can apply to a set of bo

qwelly [4]

Answer:

a)

$F_i = 28.8 kN$

b)

Alternating and mean forces on the bolt are 0.85 kN and 29.65 kN respectively.

Mean and alternating stresses in the bulk of the bolt are 378 MPa and 10 MPa.

c)

Safety factor = 5.5

Explanation:

Basic Dimension of Isometric Screw thread" is considered for analysis.

At Nominal diameter, d = 12 mm , $A_t = 84.3mm^{2}$ , $F_i = 0.9A_tS_p$ , $S_p = 380 MPa$

Rolled threads; $K_f = 2.2$

8 0

2 years ago

Air enters the compressor of a cold air-standard Brayton cycle with regeneration and reheat at 100 kPa, 300 K, with a mass flow

yanalaym [24]

Answer:

a. 47.48%

b. 35.58%

c. 2957.715 KW

Explanation:

$T_2 =T_1 + \dfrac{T_{2s} - T_1}{\eta _c}$

T₁ = 300 K

$\dfrac{T_{2s}}{T_1} = \left( \dfrac{P_{2}}{P_1} \right)^{\dfrac{k-1}{k} }$

$T_{2s} = 300 \times (10) ^{\dfrac{0.4}{1.4} }$

$T_{2s}$ = 579.21 K

T₂ = 300+ (579.21 - 300)/0.8 = 649.01 K

T₃ = T₂ + $\epsilon _{regen}$ (T₅ - T₂)

T₄ = 1400 K

Given that the pressure ratios across each turbine stage are equal, we have;

$\dfrac{T_{5s}}{T_4} = \left( \dfrac{P_{5}}{P_4} \right)^{\dfrac{k-1}{k} }$

$T_{5s}$ = 1400× $\left( 1/\sqrt{10} \right)^{\dfrac{0.4}{1.4} }$ = 1007.6 K

T₅ = T₄ + ( $T_{5s}$ - T₄)/ $\eta _t$ = 1400 + (1007.6- 1400)/0.8 = 909.5 K

T₃ = T₂ + $\epsilon _{regen}$ (T₅ - T₂)

T₃ = 649.01 + 0.8*(909.5 - 649.01 ) = 857.402 K

T₆ = 1400 K

$\dfrac{T_{7s}}{T_6} = \left( \dfrac{P_{7}}{P_6} \right)^{\dfrac{k-1}{k} }$

$T_{7s}$ = 1400× $\left( 1/\sqrt{10} \right)^{\dfrac{0.4}{1.4} }$ = 1007.6 K

T₇ = T₆ + ( $T_{7s}$ - T₆)/ $\eta _t$ = 1400 + (1007.6 - 1400)/0.8 = 909.5 K

a. $W_{net \ out}$ = cp(T₆ -T₇) = 1.005 * (1400 - 909.5) = 492.9525 KJ/kg

Heat supplied is given by the relation

cp(T₄ - T₃) + cp(T₆ - T₅) = 1.005*((1400 - 857.402) + (1400 - 909.5)) = 1038.26349 kJ/kg

Thermal efficiency of the cycle = (Net work output)/(Heat supplied)

Thermal efficiency of the cycle = (492.9525 )/(1038.26349 ) =0.4748 = 47.48%

b. $bwr = \dfrac{W_{c,in}}{W_{t,out}}$

bwr = (T₂ -T₁)/[(T₄ - T₅) +(T₆ -T₇)] = (649.01 - 300)/((1400 - 909.5) + (1400 - 909.5)) = 35.58%

c. Power = 6 kg *492.9525 KJ/kg = 2957.715 KW

3 0

3 years ago

25 In differential aeration corrosion, rich oxygenated parts are

sleet_krkn [62]

Answer:what are you trying to say

Explanation:

6 0

2 years ago

For each function , sketch the Bode asymptotic magnitude and asymptotic phase plots.

horrorfan [7]

Answer:

attached below

Explanation:

a) G(s) = 1 / s( s+2)(s + 4 )

Bode asymptotic magnitude and asymptotic phase plots

attached below

b) G(s) = (s+5)/(s+2)(s+4)

phase angles = tan^-1 w/s , -tan^-1 w/s , tan^-1 w/4

attached below

c) G(s)= (s+3)(s+5)/s(s+2)(s+4)

solution attached below

5 0

2 years ago

What is clearance? What is backlash? What is interference? Explain briefly.

Anton [14]

Explanation:

Clearance:

For easy matching and dis matching of hole and shaft we use size of hole little bit more than the size of shaft and this difference in size is called clearance.

Backlash:

It is the clearance between the two mating gear to avoids failure of gears.Actually when temperature of gears increases then at the same time the size of gear also increases ,due this there is a possibility foe jamming of gears so to avoids this backlash is provides.

Interference:

When two gears are matting then addendum of one gear inters into the deddendum of another gear and due to this gears get jam .This phenomenon is called interference.

5 0

2 years ago