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3 years ago

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Compute the electrical resistivity of a cylindrical silicon specimen 7.0 mm (0.28 in.) diameter and 57 mm (2.25 in.) in length i

n which a current of 0.25 A passes in an axial direction. A voltage of 24 V is measured across two probes that are separated by 45 mm (1.75 in.).

1 answer:

snow_tiger [21]3 years ago

8 0

No no no no no no no no no no

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Air enters the compressor of an air-standard Brayton cycle with a volumetric flow rate of 60 m3/s at 0.8 bar, 280 K. The compres

natima [27]

Answer:

a) The Net power developed in this air-standard Brayton cycle is 43.8MW

b) The rate of heat addition in the combustor is 84.2MW

c) The thermal efficiency of the cycle is 52%

Explanation:

To solve this cycle we need to determinate the enthalpy of each work point of it. If we consider the cycle starts in 1, the air is compressed until 2, is heated until 3 and go throw the turbine until 4.

Considering this:

$h_{i} =T_{i}C_{pair}=T_{i}1.005\frac{KJ}{Kg K}$

$\mu_{comp}=\frac{h_{2S}-h_{1}}{h_{2}-h_{1}}$

$\mu_{comp}=\frac{h_{3}-h_{4}}{h_{3}-h_{4S}}$

$G_{m} =\frac{PMG_{v}}{TR} =59.73\frac{Kg}{s}$

Now we can calculate the enthalpy of each work point:

h₁=281.4KJ/Kg

h₂=695.41KJ/Kg

h₃=2105KJ/Kg

h₄=957.14KJ/Kg

The net power developed:

$P_{net}=P_{Tur}-P_{Comp}=G_{m}((h_{3}-h_{4})-(h_{2}-h_{1}))$

The rate of heat:

$Q=G_{m}(h_{3}-h_{2})$

The thermal efficiency:

$\mu_{ther}=\frac{P_{net}}{Q}$

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2 years ago

Question 7.1: Two possible overhead valve combustion chambers are being considered – the first has two valves; the second has fo

AleksandrR [38]

Answer:

1) The adoption of the second design we can see that the total valve perimeter is increased by 60.8%

2) Increase in flow are : 29%

3) Additional benefits in using 4 valves per cylinder:

a)For the purpose of controlling the combustion process, the inlet valves will give more flexibility

b) There is a larger valve throat areas for the flow of gas

Explanation:

1) Perimeter of the first possible overhead valve combustion chamber with two valves:

P₂ = πd = π × 23 = 72.26mm

Perimeter of the second possible overhead valve combustion chamber with four valves:

P₄ = π2d = π × 18.5 × 2 = 116.24 mm

If second design is adopted, percentage increase = ((P₄ - P₂)/P₂)×100

= ((116.24 - 72.26)/72.26)×100 = 0.6086 ×100 = 60.86%

Therefore, the total valve perimeter is shown to have increased by 60.8%

2) Formula for flow Area (A) = P × L = πkd²

Area of the first possible overhead valve combustion chamber with two valves: A₂ = πkd² = πk(23)² = 1662k mm²

Area of the first possible overhead valve combustion chamber with four valves: A₄ = πkd² = 2πk(18.5)² = 2150k mm²

The percentage increase in flow area: ((A₄ - A₂)/A₄)×100 = ((2150 - 1662)/2150)×100 = 29%

3) The additional benefits of using are:

a) For the purpose of controlling the combustion process, the inlet valves will give more flexibility

b) There is a larger valve throat areas for the flow of gas

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2 years ago

The acceleration of a particle is given by a = 2t − 10, where a is in meters per second squared and t is in seconds. Determine t

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Answer

given,

a = 2 t - 10

velocity function

we know,

$\dfrac{dv}{dt}=a$

$\dfrac{dv}{dt}=(2t-10)$

integrating both side

$\int dv =\int (2t -10) dt$

v = t² - 10 t + C

at t = 0 v = 3

so, 3 = 0 - 0 + C

C = 3

Velocity function is equal to v = t² - 10 t + 3

Again we know,

$\dfrac{dx}{dt}=v$

$\dfrac{dx}{dt}=(t^2-10t + 3)$

integrating both side

$\int dx =\int (t^2-10t + 3)dt$

$x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t + C$

now, at t= 0 s = -4

$-4 = \dfrac{0^3}{3}- 10\dfrac{0^2}{2} + 0 + C$

C = -4

So,

$x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t-4$

Position function is equal to $x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t-4$

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Answer:

risk = probability x loss

Explanation:

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