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3 years ago

STEM! Edge 2021!

Engineering

1 answer:

jeyben [28]3 years ago

3 0

Answer:

Explanation:

requires U.S. government programs to use SI units - Treaty of the Meter

advisory panel to consider the feasibility of adopting the metric system - Metric Conversion Act

established a uniform and standardized metric system - U.S Metric Study

I might be wrong though

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A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16

abruzzese [7]

Answer:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The net work per cycle is 845.88 kJ/kg

The power developed in horsepower ≈ 45374 hP

Explanation:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The dimension of the cylinder bore diameter = 3.7 in. = 0.09398 m

Stroke length = 3.4 in. = 0.08636 m.

The volume of the cylinder v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³

The clearance volume = 16% of cylinder volume = 0.16×5.99×10⁻⁴ m³

The clearance volume, v₂ = 9.59 × 10⁻⁵ m³

p₁ = 14.5 lbf/in.² = 99973.981 Pa

T₁ = 60 F = 288.706 K

$\dfrac{T_{2}}{T_{1}} = \left (\dfrac{v_{1}}{v_{2}} \right )^{K-1}$

Otto cycle T-S diagram

T₂ = 288.706* $6.25^{0.393}$ = 592.984 K

The maximum temperature = T₃ = 5200 R = 2888.89 K

$\dfrac{T_{3}}{T_{4}} = \left (\dfrac{v_{4}}{v_{3}} \right )^{K-1}$

T₄ = 2888.89 / $6.25^{0.393}$ = 1406.5 K

Work done, W = $c_v$ ×(T₃ - T₂) - $c_v$ ×(T₄ - T₁)

0.718×(2888.89 - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg

The power developed in an Otto cycle = W×Cycle per second

= 845.88 × 2400 / 60 = 33,835.377 kW = 45373.99 ≈ 45374 hP.

8 0

4 years ago

Air at 400kPa, 970 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occu

Sonja [21]

Answer:

The rate of work developed is $\frac{\r W}{\r m}= 300kJ/kg$

The rate of entropy produced within the turbine is $\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K$

Explanation:

From the question we are told

The rate at which heat is transferred is $\frac{\r Q}{\r m } = - 30KJ/kg$

the negative sign because the heat is transferred from the turbine

The specific heat capacity of air is $c_p = 1.1KJ/kg \cdot K$

The inlet temperature is $T_1 = 970K$

The outlet temperature is $T_2 = 670K$

The pressure at the inlet of the turbine is $p_1 = 400 kPa$

The pressure at the exist of the turbine is $p_2 = 100kPa$

The temperature at outer surface is $T_s = 315K$

The individual gas constant of air R with a constant value $R = 0.287kJ/kg \cdot K$

The general equation for the turbine operating at steady state is \

$\r Q - \r W + \r m (h_1 - h_2) = 0$

h is the enthalpy of the turbine and it is mathematically represented as

$h = c_p T$

The above equation becomes

$\r Q - \r W + \r m c_p(T_1 - T_2) = 0$

$\frac{\r W}{\r m} = \frac{\r Q}{\r m} + c_p (T_1 -T_2)$

Where $\r Q$ is the heat transfer from the turbine

$\r W$ is the work output from the turbine

$\r m$ is the mass flow rate of air

$\frac{\r W}{\r m}$ is the rate of work developed

Substituting values

$\frac{\r W}{\r m} = (-30)+1.1(970-670)$

$\frac{\r W}{\r m}= 300kJ/kg$

The general balance equation for an entropy rate is represented mathematically as

$\frac{\r Q}{T_s} + \r m (s_1 -s_2) + \sigma = 0$

=> $\frac{\sigma}{\r m} = - \frac{\r Q}{\r m T_s} + (s_1 -s_2)$

generally $(s_1 -s_2) = \Delta s = c_p\ ln[\frac{T_2}{T_1} ] + R \ ln[\frac{v_2}{v_1} ]$

substituting for $(s_1 -s_2)$

$\frac{\sigma}{\r m} = \frac{-\r Q}{\r m} * \frac{1}{T_s} + c_p\ ln[\frac{T_2}{T_1} ] - R \ ln[\frac{p_2}{p_1} ]$

Where $\frac{\sigma}{\r m}$ is the rate of entropy produced within the turbine

substituting values

$\frac{\sigma}{\r m} = - (-30) * \frac{1}{315} + 1.1 * ln\frac{670}{970} - 0.287 * ln [\frac{100kPa}{400kPa} ]$

$\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K$

5 0

4 years ago

Is a water proof material used around tubs and

Leviafan [203]

I think it’s hard board

4 0

3 years ago

Read 2 more answers

A flow field is characterized by the stream function ψ= 3x2y−y3. Demonstrate thatthe flow field represents a two-dimensional inc

boyakko [2]

Answer:

δu/δx+δu/δy = 6x-6x =0

9r^2

Explanation:

The flow is obviously two-dimensional, since the stream function depends only on the x and y coordinate. We can find the x and y velocity components by using the following relations:

u =δψ/δy = 3x^2-3y^2

v =-δψ/δx = -6xy

Now, since:

δu/δx+δu/δy = 6x-6x =0

we conclude that this flow satisfies the continuity equation for a 2D incompressible flow. Therefore, the flow is indeed a two-dimensional incompressible one.

The magnitude of velocity is given by:

|V| = u^2+v^2

=(3x^2-3y^2)^2+(-6xy)^2

=9x^4+18x^2y^2+9y^2

=(3x^2+3y^2)^2

=9r^2

where r is the distance from the origin of the coordinates, and we have used that r^2 = x^2 + y^2.

The streamline ψ = 2 is given by the following equation:

3x^2y — y^3 = 2,

which is most easily plotted by solving it for x:

x =±√2-y^3/y

Plot of the streamline is given in the graph below.

Explanation for the plot: the two x(y) functions (with minus and plus signs) given in the equation above were plotted as functions of y, after which the graph was rotated to obtain a standard coordinate diagram. The "+" and "-" parts are given in different colors, but keep in mind that these are actually "parts" of the same streamline.

8 0

4 years ago

Determine the resistance values for a voltage divider that must meet the following specifications:_______.

Nikolay [14]

Answer:

i) when circuit is unloaded : R1 + R2 = 2kΩ.

ii) when 5V output voltage is applied : R1 = 1 kΩ , R2 = 1 kΩ

iii) when 2.5 v output voltage is applied : R1 = 1500 Ω, R2 = 500 Ω

iv) when: R1 = 1 kΩ , R2 = 1 kΩ is connected in parallel output voltage < 5 V

When : R1 = 1500 Ω, R2 = 500 Ω is connected in parallel output voltage > 2.5V

Explanation:

Current drawn from source under loaded condition ≤ 5 mA

source voltage = 10 v , required output = 5 v , 2.5 v

attached below is the sketch of the circuit

Resistance values

i) when the circuit is unloaded

Req = R1 + R2 = 2 kΩ ( Req = Vs / I = 10 / 5*10^-3 = 2 kΩ )

ii) when output voltage = 5 v

we will apply voltage divider rule

R1 = 1 kΩ ,

R2 = 1 kΩ

iii) When the output voltage = 2.5 v

applying voltage divider rule

R1 = 1500 Ω

R2 = 500 Ω

iv) when the load is connected to each tap one at a time

i.e. when the resistance are in parallel

when: R1 = 1 kΩ , R2 = 1 kΩ is connected in parallel output voltage < 5 V

When : R1 = 1500 Ω, R2 = 500 Ω is connected in parallel output voltage > 2.5V

attached below is the detailed solution to the given problem

8 0

3 years ago