An ideal Diesel cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the maximum air temperature and the rate of heat addition to this cycle when it produces 150 hp of power, the cycle is repeated 1200 times per minute, and the state of the air at the beginning of the compression is 95 kPa and 17°C. Use constant specific heats at room temperature. The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4.

Answer:

  radiative heat loss substantially increases as the wall temperature declines

Explanation:

The body's heat loss due to convection is ...

  (2 W/m^2·K)((32 -20)K) = 24 W/m^2

__

The body's heat loss due to radiation in the summer is ...

  [tex]\epsilon\sigma(T_b^4-T_w^4)\quad\text{where $T_b$ and $T_w$ are body and wall temperatures ($^\circ$K)}\\\\0.90\cdot 5.6703\cdot 10^{-8}(305.15^4-300.15^4)\,\text{W/m$^2$}\\\\\approx 28.3\,\text{W/m$^2$}[/tex]

The corresponding heat loss in the winter is ...

  [tex]0.90\cdot 5.6703\cdot 10^{-8}(305.15^4-287.15^4)\,\text{W/m$^2$}\\\\\approx 95.5\,\text{W/m$^2$}[/tex]

Then the total of body heat losses to surroundings from convection and radiation are ...

  summer: 24 +28.3 = 52.3 . . . W/m^2

  winter: 24 +95.5 = 119.5 . . . W/m^2

__

It is reasonable that a person would feel chilled in the winter due to the additional radiative loss to the walls in the winter time. Total heat loss is more than doubled as the wall temperature declines.

Answer:

B- Fluorinated gas

Explanation:

Answer:

B.) fluorinated gas

Explanation:

Answer:

The answer is 920 kJ

Explanation:

Solution

Given that:

Mass = 5kg

Pressure = 600 kPa

Temperature = 80° C

Liquid vapor mixture state (quality) = 0.3

Now we find out the amount of heat extracted in the process

Thus

Properties of  RI34a at:

P₁ = 600 kPa

T₁ = 80° C

h₁ = 320 kJ/kg

So,

P₁ = P₂ = 600 kPa

X₂ =0.3

h₂ = 136 kJ/kg

Now

The heat removed Q = m(h₁ -h₂)

Q = 5 (320 - 136)

Q= 5 (184)

Q = 920 kJ

Therefore the amount of heat extracted in the process is 920 kJ

Answer:

Yes, it is true

Explanation:

The computation of the time required to cool the water is shown below:

GIven that

Average cop = 2.8

Therefore the cooling effect is

= 2.8 × 400 W

= 1,120 W

ANd, the specific heat of water is 4.2 kj/kg°C i.e 4,200 j/kg°C

Now we assume the time is t

As we know that

[tex]1,120 \times t = 10\times 4,200 \times (40 - 10)\\\\ therefore\\\\ t = \frac{10\times4,200\times30}{1,120}[/tex]

t = 1,125 seconds

So, it would be 19 minutes

Therefore it is true

Answer:

zsxdcffffusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernamev

Explanation:

The criteria for a guard having to be used on a machine is;

As a safety measure If the operation exposes you to an injury.

When operating a machine, there are possibilities that the operator could be injured or exposed to injury.

Due to the possible safety issues when operating a machine, the Occupational Safety and Health Administration (OSHA) in their 29 code mandated that a safeguard must be put at each machine to ensure that there is adequate safety that prevents or minimizes the risk of getting injured.

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Answer:

gh fjh,vx j ahj ds djv dk

Explanation:

doing for points

The Cost of 1 kilowatt hour of energy at the rate of $0.75 per 1.5 watt hour is $500 which is 5000 times greater than the cost of energy at $0.10 per kilowatt hour.

Battery power = Current × Voltage

Cost of 1.5 Watt-hour = $0.75

Converting Energy to Watt - hour :

1 kilowatt = 1000 watt

1 kilowatt hour = 1000 watt - hour

Hence,

Cost of 1 kilowatt - hour = 1000 watt - hour can be calculated thus :

1.5 Watt-hour = $0.75

1000 Watt-hour = c

Cross multiply :

1.5c = $0.75 × 1000

1.5c = 750

c = 750 / 1.5

c = 500

Therefore, cost of 1 kilowatt - hour of energy will be $500

Comparing the cost of Energy at $500 per Kilo-Watt hour to Cost at $0.10:

$500 / $0.10 = 5000

Therefore, the cost of energy at $500 per kilowatt hour is 5000 times greater than cost at $0.10 per kilowatt hour.

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Answer:

Technician B is correct

Explanation:

To resize a connecting rod’s big end, the bolts are removed and the cap and rod mating surfaces will be ground to an approximate measurement. Thereafter, the rod cap will be bolted back onto the rod while the bore is machined back to it's proper size. However, the challenge with powdered metal rods is that due to the fact that the fractured surface is critical to their alignment, they cannot tolerate having those surfaces ground smooth. Hence, they cannot be resized, except bearing inserts of larger diameters are available for that particular engine.

Thus, technician B is correct.

Answer:

Tech A is correct.

Explanation:

Gears are toothed wheels that can be used to transmit power. When two or more gears are in tandem, a gear train is formed.

Gear ratio = [tex]\frac{number of teeth of the driven gear}{Number of teeth of the driving gear}[/tex]

                = [tex]\frac{27}{9}[/tex]

                = [tex]\frac{3}{1}[/tex]

Gear ratio = 3:1

The driver gear is called the input gear since it transfers its power to the driven gear. While the driven gear is called the output gear because it produces an effect due to both gears.

Tech A is correct.

Answer:

The critical buckling load is [tex]\mathbf{P_o = 141.61 \ kN}[/tex]

Explanation:

Given that:

the width of the rectangular steel = 37.5 mm = 0.0375 m

the thickness = 50 mm  = 0.05 m

the length = 1.75 m

modulus of elasticity = 200 Gpa = 200 10⁹ × Mpa

We are to calculate the critical buckling load  [tex]P_o[/tex]

Using the formula:

[tex]P_o = \dfrac{\pi ^2 E I}{L^2}[/tex]

where;

[tex]I = \dfrac{0.0375^3*0.05}{12}[/tex]

[tex]I = 2.197 * 10^{-7}[/tex]

[tex]P_o = \dfrac{\pi ^2 *200*10^9 * 2.197*10^{-7}}{1.75^2}[/tex]

[tex]P_o = 141606.66 \ N[/tex]

[tex]\mathbf{P_o = 141.61 \ kN}[/tex]

The critical buckling load is [tex]\mathbf{P_o = 141.61 \ kN}[/tex]

The negative mark is balanced by a positive mark on the set key scale while the jaws are closed.

It is common practice to shut the jaws or faces of the system before taking some reading to guarantee a zero reading. If not, please take care of the read. This read is referred to as "zero defect."

There are two forms of zero error:

zero-mistake positive; and

Non-null mistake.

----------------------------

Hope this helps!

Brainliest would be great!

----------------------------

With all care,

07x12!

Answer:

1) twelve

Explanation:

The dual voltage motors are used in day to day operations. The wye is connected with 9 lead motors. Maximum resistance can be obtained if the resistance are connected in series. To check resistance of dual voltage wye motor there must be twelve resistance readings of 1 ohm each.

Answer:

Explanation:

neither of the technicians is correct

Answer:

Explanation:

i) In the attached truth table, TRUE means the respective key owner is present and their keys are inserted at the same time. The "Opens" column is TRUE when two owners are present, not including the case where the only two owners present are B and C.

__

ii) The second attachment is a Karnaugh map of the truth table. The circled terms are ...

  Opens = AB +AC

Answer:

A TVS screen works when the pixels are switched on electronically using liquid crystals to rotate polarized light.

Explanation:

Answer:

Attached is the complete question but the weight of the mailbox and cross bar differs from the given values which are : weight of mail box = 3.2 Ib, weight of the uniform cross member = 10.3 Ib

Answer : moment of inertia = 186.7 Ib - in

Explanation:

Given data

weight of the mailbox = 3.2 Ib

weight of the uniform cross member = 10.3 Ib

The origin is of mailbox and cross member is 0

The perpendicular distance from Y axis of centroid of the mailbox

= 4 + (25/2) = 16.5"

The centroid of the bar =( ( 1 + 25 + 4 + 4 ) / 2 ) - 4  = 13"

therefore The moment of Inertia( Mo) = (3.2 * 16.5) + ( 10.3 * 13)

                                     = 52.8 + 133.9 = 186.7 Ib-in

The combined moment about O due to the weight of the mailbox and the cross member AB is; M_o = 122.4 lb.in (ccw)

We are given;

Weight of mailbox; W_m = 3.2 lb

Weight of uniform cross member; W_c = 10.3 lb

Now, from the attached diagram, let us calculate the geometric location of the mailbox and uniform cross section from point O.

Geometric location of mailbox from point O; g_m = 3 + (19/2) = 12.5 in

Geometric location of cross member from point O;

g_c = (¹/₂(1 + 19 + 3 + 7)) - 7

g_c = 8 in

Thus. combined moment about point O is;

M_o = (W_m × g_m) + (W_c × g_c)

M_o = (3.2 × 12.5) + (10.3 × 8)

M_o = 122.4 lb.in

Since positive then it is counterclockwise. Thus;

M_o = 122.4 lb.in (ccw)

The image of this question is missing and so i have attached it.

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[tex]

<html>

<head>

<meta name="viewport" content="width=device-width, initial-scale=1">

<title>Time Picker</title>

</head>

<body>

<!--24 Hours format-->

<input type="time" placeholder="Enter Time" />

<input type="date">

</body>

</html>

[/tex]

Answer:

a) Δd(change in wood diameter) = 5%

b) The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter

C) new diameter (D2) = 10.5 in

Explanation:

Wood pole diameter = 10 inches

moisture content = 5%

FSP = 30%

A) The percentage change in the wood's diameter

note : moisture fluctuations from 5% to 30% causes dimensional changes in the wood but above 30% up to 55% causes no change. hence this formula can be used to calculate percentage change in the wood's diameter

Δd/d = 1/5(30 - 5)

Δd/d = 5%  

Δd = 5%

B) would the wood swell or shrink

The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter

C) The new diameter of the wood

D2 = D + D( [tex]\frac{M1}{100}[/tex] )

D = initial diameter= 10 in , M1 = initial moisture content = 5%

therefore D2 = 10 + 10( 5/100 )

new diameter (D2) = 10.5 in

The change in the diameter of the wood would be 5%

the new diameter would be 10.5 inches

Wood pole diameter = 10 inches

Moisture content = 5%

Fiber saturation point = 30 %

[tex]\frac{1}{5} (30-5)[/tex]

= 25/5

= 5%

The percentage change in the diameter of the wood would be 5%

b. This wood is going to rise up instead of shrinking. This is due to the fact that the moisture content that it has has gone up by 55%

c. The new diameter that this wood would have

diameter = 10

moisture = 5%

D = D+D(m)

= 10 + 10(5%)

= 10.5 inches

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Answer:

The speed will be "3.58 ft/s". The further explanation is given below.

Explanation:

Number of knots

= 15

For the similarity of Froude number:

⇒  [tex]\frac{V_{m}}{\sqrt{g_{m}l_{m}} }=\frac{V}{\sqrt{gl} }[/tex]

Here,

[tex]l = length[/tex]

[tex]g_{m}=g[/tex]

⇒  [tex]\frac{V_{m}}{V}=\sqrt{\frac{l_{m}}{l} }[/tex]

    [tex]V_{m}=\sqrt{\frac{1}{50} }\times number \ of \ knots[/tex]

         [tex]=\sqrt{\frac{1}{50}}\times 15[/tex]

         [tex]=2.12 \ knots[/tex]

Now,

⇒  [tex]1 \ knots=0.514\times 3.281[/tex]

                 [tex]=1.69 \ ft/s[/tex]

So that,

⇒  [tex]V_{m}=2.12\times 1.69[/tex]

          [tex]=3.58 \ ft/s[/tex]

Answer:

Space velocity = 30 hr⁻¹

Explanation:

Space velocity for reactors express how much reactor volume of feed or reactants can be treated per unit time. For example, a space velocity of 3 hr⁻¹ means the reactor can process 3 times its volume per hour.

It is given mathematically as

Space velocity = (volumetric flow rate of the reactants)/(the reactor volume)

Volumetric flowrate of the reeactants

= (molar flow rate)/(concentration)

Molar flowrate of the reactants = 300 millimol/hr

Concentration of the reactants = 100 millimol/liter

Volumetric flowrate of the reactants = (300/100) = 3 liters/hr

Reactor volume = 0.1 liter

Space velocity = (3/0.1) = 30 /hr = 30 hr⁻¹

Hope this Helps!!!

Answer:

Reynolds number = 654350.92

Explanation:

Given data:

Cross section of  rectangular cross section = 1.8ft * 8 in  ( 8 in = 2/3 ft )

Flow rate of air = 5400 cfm = 90 ft^3 / sec

v ( kinematic viscosity of air ) = 1.8*10^-4 ft^2/s

Reynolds number

Re = VDn / v

Dn ( hydraulic diameter ) = 4A / P

where A = area, P = perimeter

a = 1.8 ft  ( length )

b = 2/3 ft ( width )

hence Dn = [tex]\frac{4(ab)}{2(a+b)}[/tex]  = [tex]\frac{4(1.8*0.6667}{2(1.8+0.6667)}[/tex]  =   0.9729 ft

V ( velocity of air flow ) = [tex]\frac{Q}{\pi /4 * Dn^2 }[/tex]  = [tex]\frac{90}{\pi /4 * 0.9729^2 }[/tex] = 121.064 ft/sec

back to Reynolds equation

Re = VDn / v  -------------- equation 1

V = 121.064 ft/sec

Dn = 0.9729 ft

v = 1.8*10^-4 ft^2/s

insert the given values into equation 1

Re = (121.064 * 0.9729 ) / 1.8*10^-4

     = 654350.92

Answer: (0,0)+ (1,0)= 1 lines upwards( suggesting that this is a line graph not saying it is but as an example) an (1,1) and (0,-1) all make a small square ( as this is a 2 dimensional graph that it has a negative side too,(below the positive side)) i hope this helps and is what you are looking for

Explanation:

Answer:

a. The magnitude of the electric field is [tex]E = -\dfrac{K}{e \cdot S}[/tex]

b. 1. In the direction of the electron's motion

c. The ratio of the magnitude of the electric field the ions travel through to the magnitude of electric field for the electrons is M:m or 34634.6:1

Explanation:

The change in kinetic energy of electron = [tex]\dfrac{1}{2} \cdot m \cdot v^2_f - \dfrac{1}{2} \cdot m \cdot v^2_i[/tex]

Given that the work done, W = the change in kinetic energy of electron, we have;

[tex]W = \dfrac{1}{2} \cdot m \cdot v^2_f - \dfrac{1}{2} \cdot m \cdot v^2_i[/tex]

Given that the final velocity is 0, we have;

[tex]W =- \dfrac{1}{2} \cdot m \cdot v^2_i = -K[/tex]

However, we have;

W = E·q·S

Where:

q = Particle charge = Electron charge = e

S = Particle displacement

E = Electric field strength

Therefore;

W = -K = E·q·S

Which gives;

[tex]E = -\dfrac{K}{e \cdot S}[/tex]

b. The direction of the electric field is the direction of the force exerted by the electric field on a positive charge

Given that the particle is an electron, its direction is opposite to that of a positive charge, however whereby the electron is brought to rest the direction of the field is in the same direction of as that of the electron's motion

The correct option is therefore, 1. In the direction of the electron's motion

c. Given that the mass of fluoride ion, Fe⁻, M = 3.155×10⁻²⁶ kg

The mass of an electron, m = 9.1094×10⁻³¹ kg

We have;

x × 0.5 × m × v² = 0.5×M×v²

∴ x = M/m

Therefore. the kinetic energy of the fluoride ion is M/m multiplied by the kinetic energy of the electron

x × 0.5 × 9.1094×10⁻³¹ × v² = 0.5×3.155×10⁻²⁶×v²

Therefore, x = (3.155×10⁻²⁶)÷(9.1094×10⁻³¹) = 34634.6

Therefore, the kinetic energy of the Fluoride ion is 34634.6 times that of the kinetic energy of the electron

Whereby the electric field strength is directly proportional to the kinetic energy, we have;

[tex]E_{(electron)} = -\dfrac{K_{(electron)}}{e \cdot S}[/tex]

[tex]E_{(fluoride \ ion)} = -\dfrac{K_{(fluoride \ ion)}}{e \cdot S}[/tex]

Which gives;

[tex]\dfrac{E_{(fluoride \ ion)} }{E_{(electron)}} = \dfrac{-\dfrac{K_{(fluoride \ ion)}}{e \cdot S}}{ -\dfrac{K_{(electron)}}{e \cdot S}} = \dfrac{{K_{(fluoride \ ion)}}}{ {K_{(electron)}}}[/tex]

[tex]\dfrac{E_{(fluoride \ ion)} }{E_{(electron)}} = \dfrac{\dfrac{M}{m} \times {K_{(electron)}}}{ {K_{(electron)}}} = \dfrac{M}{m}[/tex]

[tex]\dfrac{E_{(fluoride \ ion)} }{E_{(electron)}} = \dfrac{34634.6 \times {K_{(electron)}}}{ {K_{(electron)}}} = 34634.6[/tex]

The magnitude of the uniform electric field in the fluoride ion path is 34634.6 times that of the magnitude of the electric field in the electron path.

The ratio of the magnitude of the electric field the ions travel through to the magnitude of electric field for the electrons is M:m or 34634.6:1

Answer:

See explanation

Explanation:

Solution:-

- The shell and tube heat exchanger are designated by the order of tube and shell passes.

- A single tube pass: The fluid enters from inlet, exchange of heat, the fluid exits.

- A multiple tube pass: The fluid enters from inlet, exchange of heat, U bend of the fluid, exchange of heat, .... ( nth order of pass ), and then exits.

- By increasing the number of passes we have increased the "retention time" of a specific volume of tube fluid; hence, providing sufficient time for the fluid to exchange heat with the shell fluid.

- By making more U-turns we are allowing greater length for the fluid flow to develop with " constriction and turns " into turbulence. This turbulence usually at the final passes allows mixing of fluid and increases the heat transfer coefficient by:

                                U ∝ v^( 0.8 )    .... ( turbulence )

- The higher the velocity of the fluids the greater the heat transfer coefficient. The increase in the heat transfer coefficient will allow less heat energy carried by either of the fluids to be wasted ; hence, reduced losses.

Thereby, increases the thermal efficiency of the heat exchanger ( higher NTU units ).

Answer:

863 K

Explanation:

See the attachment

Answer:

a) The amount of heat transfer in the regenerator, q = 114.12 kJ/kg

b) Thermal efficiency = 35.9%

Explanation:

The calculations are neatly handwritten and attached as files to this solution for easiness of expression and clarity. The cycle is also drawn on the T - S diagram and included in the attached files. Check the files below for the complete calculation.

I hope this helps!

Answer:

a) T₃ = 1818.8 K

b) η = 0.614 = 61.4%

c) MEP = 660.4 kPa

Explanation:

a) According to Table A-2 of The ideal gas specific heat of gases, the properties of air are as following:

At 300K

The specific heat capacity at constant pressure = [tex]c_{p}[/tex] = 1.005 kJ/kg.K,

The specific heat capacity at constant volume = [tex]c_{v}[/tex] = 0.718 kJ/kg.K

Gas constant R for air = 0.2870 kJ/kg·K

Ratio of specific heat  k = 1.4

Isentropic Compression :

[tex]T_{2}[/tex] =  [tex]T_{1}[/tex]  [tex](v1/v2)^{k-1}[/tex]

   = 300K ([tex]16^{0.4}[/tex])

[tex]T_{2}[/tex]    = 909.4K

P = Constant heat Addition:

[tex]P_{3}v_{3} / T_{3} = P_{2} v_{2} /T_{2}[/tex]

[tex]T_{3}=v_{3}/v_{2}T_{2}[/tex]

2[tex]T_{2}[/tex] = 2(909.4K)

      = 1818.8 K

b) [tex]q_{in}[/tex] = [tex]h_{3}-h_{2}[/tex]

         =  [tex]c_{p}[/tex] ([tex]T_{3}[/tex] - [tex]T_{2}[/tex])

         = (1.005 kJ/kg.K)(1818.8 - 909.4)K

         = 913.9 kJ/kg

Isentropic Expansion:

[tex]T_{4}[/tex] =  [tex]T_{3}[/tex]  [tex](v3/v4)^{k-1}[/tex]

    =  [tex]T_{3}[/tex] [tex](2v_{2} /v_{4} )^{k-1}[/tex]

    = 1818.8 K (2 / 16[tex])^{0.4}[/tex]

    = 791.7K

v = Constant heat rejection

[tex]q_{out}[/tex] = μ₄ - μ₁

      = [tex]c_{v} ( T_{4} - T_{1} )[/tex]

      = 0.718 kJ/kg.K (791.7 - 300)K

      = 353 kJ/kg

 η[tex]_{th}[/tex] = 1 - [tex]q_{out}[/tex] / [tex]q_{in}[/tex]

       = 1 - 353 kJ/kg / 913.9 kJ/kg

       = 1 - 0.38625670

       = 0.6137

       = 0.614

      = 61.4%

c) [tex]w_{net}._{out}[/tex] = [tex]q_{in}[/tex] - [tex]q_{out}[/tex]

                = 913.9 kJ/kg - 353 kJ/kg

                = 560.9 kJ/kg

[tex]v_{1} = RT_{1} /P_{1}[/tex]

   = (0.287 kPa.m³/kg/K)*(300 K) / 95 kPa

   =  86.1 / 95

   = 0.9063 m³/kg = v[tex]_{max}[/tex]

[tex]v_{min} =v_{2} = v_{max} /r[/tex]

Mean Effective Pressure = MEP =   [tex]w_{net,out}/v_{1} -v_{2}[/tex]

                                                    = [tex]w_{net,out}/v_{1}(1-1)/r[/tex]

                                                    = 560.9 kJ/kg / (0.9063 m³/kg)*(1-1)/16

                                                    = (560.9 kJ / 0.8493m³) (kPa.m³/kJ)

                                                    = 660.426 kPa

Mean Effective Pressure = MEP = 660.4 kPa

The temperature after the addition process is 1724.8k, the thermal efficiency of the engine is 56.3% and the mean effective pressure is 65.87kPa

Assumptions made:

a) The temperature after the addition process:

Considering the process 1-2, Isentropic expansion

at

[tex]T_1=300k\\u_1=214.07kJ/kg\\v_o_1=621.3\\v_o_2=\frac{v_2}{v_1} *v_o_1[C.R=16]=v_2/v_1\\v_o_2=(v_2/v_1)v_o_1=1/16*621.2=38.825[/tex]

From using this value, v[tex]_o_2[/tex]=38.825, solve for state point 2;

[tex]T_2=862.4k\\h_2=890.9kJ/kg[/tex]

Considering the process 2-3 (state of constant heat addition)

[tex]\frac{p_3v_3}{t_3}=\frac{p_2v_2}{t_2} \\\\T_3=\frac{P_3V_3T_2}{V_2} \\T_3=(\frac{V_3}{V_2}) T_2\\\frac{v_3}{v_2}=2\\T_3=2(862.4)=1724.8k\\[/tex]

NB: p[tex]_3[/tex]≈p[tex]_2[/tex]

b) The thermal efficiency of the engine is

Q[tex]_i_n[/tex]=h[tex]_3-h_2[/tex] = 1910.6-890.9=1019.7kJ/kg

Considering process 3-4,

[tex]v_o_4=\frac{v_A}{v_2}\\ v_o_3 =\frac{V_a}{V_2}*\frac{v_2}{v_3}\\v_o_3=\frac{16}{2}*4.546\\v_o_3=36.37;v_4=659.7kJ/kg[/tex]

Q[tex]_o_u_t=v_4-u_1=659.7-214.07=445.3kJ/kg[/tex]

nth = [tex]1-\frac{Q_o_u_t}{Q_i_n}=1-\frac{445.63}{1019.7}=0.5629*100=56.3%[/tex]%

The thermal efficiency is 56.3%

W[tex]_n_e_t[/tex]=[tex]Q_i_n-Q_o_u_t=574.07kJ/kg[/tex]

[tex]v_1=\frac{RT_1}{p_1}=\frac{0.287*300}{95}=0.906m^3/kg\\v_2=v_1/16=0.05662m^3/kg\\[/tex]

Therefore, the mean effective pressure of the system engine is

[tex]\frac{W_n_e_t}{v_1-v_2}=675.87kPa[/tex]

The mean effective pressure is 65.87kPa as calculated above

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Answer:

a) 397.89 ohm

b) attached below

c) 0.707  as a ratio

Gain in dB = 20 log 0.707 = -3 dB

d)  0.8087

Gain in dB = 20 log [tex]|\frac{Vout}{Vin}|[/tex] = -1.844 dB

Explanation:

A) Find the appropriate resistor

c = 0.01 uf

fc = 40 kHz

cut-off frequency ; fo = [tex]\frac{1}{2\pi RC }[/tex]

from the above equation  R = [tex]\frac{1}{2\pi foC}[/tex]  = 397.89 ohm

B) sketch of the circuit  is attached

C) The gain of the filter at the cutoff frequency

fc = 40 kHz,  

C = 0.01 uF ⇒ [tex]\frac{-j}{2\pi foC }[/tex] =  -j 397.89

Vout = Vin * ( R / R- C )

Vout = Vin * ( 397.89 / (397.89 - j 397.89))

Vout = [tex]\frac{1}{\sqrt{2} }[/tex]  Vin ∠45⁰

therefore gain = |[tex]\frac{Vout}{Vin }[/tex]| = [tex]\frac{1}{\sqrt{2} }[/tex] = 0.707  as a ratio

Gain in dB = 20 log 0.707 = -3 dB

D) Gain of filter at 55 kHz

c = 0.01 uF =  [tex]\frac{-J}{2\pi foC }[/tex] =  -j 289.373 ohms

Vout = Vin * [tex]\frac{R}{R-C}[/tex]  

        = Vin * ( 397.89 / ( 397.89 - j 289.373))

Gain in ratio [tex]|\frac{Vout}{Vin}|[/tex] = 0.8087 ∠ 36.03⁰

therefore gain in ratio = 0.8087

Gain in dB = 20 log [tex]|\frac{Vout}{Vin}|[/tex] = -1.844 dB

Answer:

L= 312.75 mm

Explanation:

given data

elastic modulus E = 106 GPa

cross-sectional diameter d = 3.9 mm

tensile load F = 1660 N

maximum allowable elongation ΔL = 0.41 mm

to find out

maximum length of the specimen before deformation

solution

we will apply here allowable elongation equation that is express as

ΔL =     [tex]\dfrac{FL}{AE}[/tex]     ....................1

put here value and we get L

L   =    [tex]\dfrac{0.41\times 10^{-3}\times \dfrac{\pi}{4}\times (3.9\times 10^{-3})^2\times 106\times 10^9}{1660}[/tex]

solve it we get

L = 0.312752 m

L= 312.75 mm