A single-phase transformer has 480 turns on the primary and 90 turns on the secondary. The mean length of the flux path in the core is 1.8 m and the joints are equivalent to an airgap of 0.1 mm. The value of the magnetic field strength for 1.1 T in the core is 400 A/m, the corresponding core loss is 1.7 W/kg at 50 Hz and the density of the core is 7800 kg/m3. If the maximum value of the flux density is to be 1.1 T when a p.D. Of 2200 V at 50 Hz is applied to the primary, calculate: a. The cross-sectional area of the core; b. The secondary voltage on no load; c. The primary current and power factor on no load.

Answer:

a) This cycle observes the First and Second Laws of Thermodynamics, b) This cycle observes the First and Second Laws of Thermodynamics, c) This cycle does not observe the First Law of Thermodynamics, d) This cycle does not observe the Second Law of Thermodynamics.

Explanation:

The Carnot cycle offers a reliable criterion to determine the maximum theoretical efficiency ([tex]\eta_{th,max}[/tex]) for a power cycle in term of the temperatures of hot and cold reservoirs and expressed in percentage:

[tex]\eta_{th, max} = \left(1-\frac{T_{L}}{T_{H}} \right)\times 100\,\%[/tex]

Where:

[tex]T_{L}[/tex], [tex]T_{H}[/tex] - Temperatures of cold and hot reservoirs, measured in kelvins.

If [tex]T_{L} = 400\,K[/tex] and [tex]T_{H} = 1000\,K[/tex], the maximum theoretical thermal efficiency is:

[tex]\eta_{th, max} = \left(1-\frac{400\,K}{1000\,K} \right)\times 100\,\%[/tex]

[tex]\eta_{th,max} = 60\,\%[/tex]

In addition, the real efficiency of the heat engine is described by the following formula:

[tex]\eta_{th} = \left(1-\frac{Q_{L}}{Q_{H}} \right)\times 100\,\%[/tex]

Where:

[tex]Q_{H}[/tex] - Heat absorbed by the heat engine from hot reservoir, measured in kilojoules.

[tex]Q_{L}[/tex] - Heat released by the heat engine to the cold reservoir, measured in kilojoules.

The following conditions must be observed by all heat engines:

First Law of Thermodynamics

[tex]W = Q_{H}-Q_{L}[/tex]

Second Law of Thermodynamics

[tex]\eta_{th} \leq \eta_{th,max}[/tex]

Now, each cycle is checked:

a) [tex]Q_{H} = 300\,kJ[/tex], [tex]Q_{L} = 140\,kJ[/tex] and [tex]W = 160\,kJ[/tex]

First Law of Thermodynamics

[tex]W = 300\,kJ-140\,kJ[/tex]

[tex]W = 160\,kJ[/tex]

This cycle observes the First Law of Thermodynamics.

Second Law of Thermodynamics

[tex]\eta_{th} = \left(1-\frac{140\,kJ}{300\,kJ} \right)\times 100\,\%[/tex]

[tex]\eta_{th} = 53.333\,\%[/tex]

This cycle observes the Second Law of Thermodynamics.

b) [tex]Q_{H} = 300\,kJ[/tex], [tex]Q_{L} = 120\,kJ[/tex] and [tex]W = 180\,kJ[/tex]

First Law of Thermodynamics

[tex]W = 300\,kJ-120\,kJ[/tex]

[tex]W = 180\,kJ[/tex]

This cycle observes the First Law of Thermodynamics.

Second Law of Thermodynamics

[tex]\eta_{th} = \left(1-\frac{120\,kJ}{300\,kJ} \right)\times 100\,\%[/tex]

[tex]\eta_{th} = 60\,\%[/tex]

This cycle observes the Second Law of Thermodynamics.

c) [tex]Q_{H} = 300\,kJ[/tex], [tex]Q_{L} = 140\,kJ[/tex] and [tex]W = 170\,kJ[/tex]

First Law of Thermodynamics

[tex]W = 300\,kJ-140\,kJ[/tex]

[tex]W = 160\,kJ[/tex]

This cycle does not observe the First Law of Thermodynamics.

Second Law of Thermodynamics

[tex]\eta_{th} = \left(1-\frac{140\,kJ}{300\,kJ} \right)\times 100\,\%[/tex]

[tex]\eta_{th} = 53.333\,\%[/tex]

This cycle observes the Second Law of Thermodynamics.

d) [tex]Q_{H} = 300\,kJ[/tex], [tex]Q_{L} = 200\,kJ[/tex] and [tex]W = 100\,kJ[/tex]

First Law of Thermodynamics

[tex]W = 300\,kJ-100\,kJ[/tex]

[tex]W = 200\,kJ[/tex]

This cycle observes the First Law of Thermodynamics.

Second Law of Thermodynamics

[tex]\eta_{th} = \left(1-\frac{100\,kJ}{300\,kJ} \right)\times 100\,\%[/tex]

[tex]\eta_{th} = 66.667\,\%[/tex]

This cycle does not observe the Second Law of Thermodynamics.

Based on the first parameters, the power cycle obeys both the First and Second Law of Thermodynamics.

Given the following data:

In order to verify a power cycle obeys the first and second laws of thermodynamics, we would use the Carnot cycle.

Mathematically, the maximum theoretical efficiency for a power cycle in terms of the temperature is given by this formula:

[tex]\eta_{th, max}=(1-\frac{T_c}{T_h} ) \times 100[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]\eta_{th, max}=(1-\frac{400}{1000} ) \times 100\\\\\eta_{th, max}=60\%[/tex]

Similarly, the real efficiency of a power cycle, we have:

[tex]\eta_{th}=(1-\frac{Q_c}{Q_h} ) \times 100[/tex]

According to the First Law of Thermodynamics, the following condition must be met:

[tex]W_{cycle}=Q_h-Q_c[/tex]

According to the Second Law of Thermodynamics, the following condition must be met:

[tex]\eta_{th }\leq \eta_{th, max }[/tex]

Next, we would determine whether or not each obeys the first and second laws of thermodynamics:

When [tex]Q_h=300\; kJ, \;W_{(cycle)}=160kJ, \;and\;Q_c=140\; kJ[/tex]

For the First Law:

[tex]W_{cycle}=Q_h-Q_c\\\\160=300-140\\\\160=160[/tex]

Therefore, the power cycle obeys the First Law of Thermodynamics.

For the Second Law:

[tex]\eta_{th}=(1-\frac{Q_c}{Q_h} ) \times 100\\\\\eta_{th}=(1-\frac{140}{300} ) \times 100\\\\\eta_{th}=53.33\%\\\\\eta_{th }\leq \eta_{th, max } =53.33\%\leq 60\%[/tex]

Therefore, the power cycle obeys the Second Law of Thermodynamics.

When [tex]Q_h=300\; kJ, \;W_{(cycle)}=180kJ, \;and\;Q_c=120\; kJ[/tex]

For the First Law:

[tex]W_{cycle}=Q_h-Q_c\\\\180=300-120\\\\180=180[/tex]

Therefore, the power cycle obeys the First Law of Thermodynamics.

For the Second Law:

[tex]\eta_{th}=(1-\frac{Q_c}{Q_h} ) \times 100\\\\\eta_{th}=(1-\frac{120}{300} ) \times 100\\\\\eta_{th}=60\%\\\\\eta_{th }\leq \eta_{th, max } =60\%\leq 60\%[/tex]

Therefore, the power cycle obeys the Second Law of Thermodynamics.

When [tex]Q_h=300\; kJ, \;W_{(cycle)}=170kJ, \;and\;Q_c=140\; kJ[/tex]

For the First Law:

[tex]W_{cycle}=Q_h-Q_c\\\\170=300-140\\\\170\neq 160[/tex]

Therefore, the power cycle doesn't obey the First Law of Thermodynamics.

For the Second Law:

[tex]\eta_{th}=(1-\frac{Q_c}{Q_h} ) \times 100\\\\\eta_{th}=(1-\frac{140}{300} ) \times 100\\\\\eta_{th}=53.33\%\\\\\eta_{th }\leq \eta_{th, max } =53.33\%\leq 60\%[/tex]

Therefore, the power cycle obeys the Second Law of Thermodynamics.

When [tex]Q_h=300\; kJ, \;W_{(cycle)}=200kJ, \;and\;Q_c=100\; kJ[/tex]

For the First Law:

[tex]W_{cycle}=Q_h-Q_c\\\\200=300-100\\\\200=200[/tex]

Therefore, the power cycle obeys the First Law of Thermodynamics.

For the Second Law:

[tex]\eta_{th}=(1-\frac{Q_c}{Q_h} ) \times 100\\\\\eta_{th}=(1-\frac{100}{300} ) \times 100\\\\\eta_{th}=66.67\%\\\\\eta_{th }\leq \eta_{th, max } \neq 63.33\%\geq 60\%[/tex]

Therefore, the power cycle doesn't obey the Second Law of Thermodynamics.

Read more on thermodynamics here: brainly.com/question/11628413

Answer:

Both technician A and technician B are correct.

Explanation: Vehicle manufacturers always specify the size of the tires required for a given vehiclefor optimal efficiency,this will ensure that the speedometer is accurate and the level of the vehicle is good enough to ensure the vehicle works efficiently.

It is also a known fact that an increase in a vehicle's rpm(revolution per minute) will eventually lead to increased fuel consumption which means the fuel economy of the vehicle will be reduced making the vehicle less efficient in its fuel consumption.

Answer:

Enthalpy is a function of pressure hence normalized enthalpy departure values will approach zero with reduced pressure approaching zero

Explanation:

On the generalized enthalpy departure chart, the normalized enthalpy departure values seem to approach zero as the reduced pressure PR approaches zero. this is because enthalpy is a function of pressure therefore as the Pressure is reducing towards the zero value, the gas associated with the pressure tends to behave more like an Ideal gas.

For an Ideal gas the Normalized enthalpy departure value will be approaching the zero value.

Answer:

A) slip(s1) = (1800 - 1750) / 1800 = 0.0277

B) 1800 rpm, 1350 rpm,  900 rpm, 450 rpm

Explanation:

Given data

frequency = 60 Hz,  

Line - line rms = 440 - V

3 phase induction-motor drive

number of poles = 4

Full-load rated speed = 1750

rated torque = 40 Nm

A) The plot of torque-speed characteristics for the following values of the frequency f : 60 Hz, 45 Hz, 30 Hz, and 15 Hz is attached below

first we calculate the rated speed:

Ns = [tex]\frac{120f}{p}[/tex]

f = 60 Hz .  p (number of poles) = 4

Ns[tex]_{1}[/tex] = [tex]\frac{120(60)}{4}[/tex] = 1800

full loaded rated value = 1750.

slip(s1) = (1800 - 1750) / 1800 = 0.0277

considering a linearly condition the slip is low

[tex]\frac{T1}{T2} = \frac{S1}{S2} * \frac{F2}{F1}[/tex]

S1 = 0.0277

f1 = 60 Hz

hence s2 = 0.018 therefore Ns2 = 1500

B) The speeds of operation at : 60 Hz, 45 Hz, 30 Hz, 15 Hz

for 60 Hz :

Ns = [tex]\frac{120f}{p}[/tex] = (120*60) / 4 = 1800 rpm

for 45 Hz:

Ns = 120(f) / p = (120*45) / 4 = 5400 /4 = 1350 rpm

for 30 Hz:

Ns = 120(f) / p = (120*30) / 4 = 3600 / 4 = 900 rpm

for 15 Hz:

Ns = 120(f) / p = (120*15) / 4 = 1800 / 4 = 450 rpm

Answer:

hello your question is incomplete here is the complete question

. “ABC construction Inc.” company becomes the lowest in the bed process to get a $21 million construction project for “Northern Inc.”. Now “ABC construction Inc.” planning to make a formal contract agreement with the “Northern Inc.”. What are the main elements of this agreement to consider it as a legal contract?

Answer : elements of the agreement

Explanation:

Offer : an offer is the beginning element for any valid agreement to be started or reached between two or more bodies. ABC construction would have to make an offer first for the agreement to be valid

Acceptance: This is part where by the company "Northern Inc" after receiving the offer from ABC construction Inc would have to consent to the approval of the offer made.

capacity : This the element of the agreement that helps to ensure that both parties have the legal and financial backings to embark on the contract agreement .

certainty : This element ensures that both parties understands the terms and conditions attached to the agreement and this to ensure that there are no bogus conditions

Consideration : This is a very vital element because the both parties have to give something in return while going into a valid agreement

Intention to create legal relation : Legal relations are applied to contract agreements whereby both parties want the contract agreement to b legally enforced and this is important in order to prevent contract breach by any party involved in the agreement

Answer:

A. Deter casual trespassers

Explanation:

In safeguarding valued property, fences are used to serve as a psychological and physical barrier to intending intruders. Depending on its height, the fence could serve several purposes. A fence that is three to four foot high would only serve to deter casual trespassers. This is because any intruder would make an effort to scale a four-foot-high fence.

However, when an intruder is determined he would make an effort to scale a fence that is even up to eight-foot-high. So to provide optimum security, a very high fence supported with barb wire and a closed-circuit television camera would help to ensure that the environment is monitored against any determined intruder.

Answer:

The discharge temperature is 259.82 K

Explanation:

In this question, we are concerned with calculating the discharge temperature

Please check attachment for complete solution

Answer:

T = 858.25 s

Explanation:

Given data:

Reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3,  c 550 J/kg K, k 48 W/m K),

initial uniform temperature ( Ti ) = 200 c

Final temperature = 550 c

convection coefficient  = 250 w/m^2 k

products combustion temp = 800 c

calculate how long the plate should be left in the furnace ( to attain 550 c )

first calculate/determine the Fourier series Number ( Fo )

[tex]\frac{T_{0}-T_{x} }{T_{1}-T_{x} } = C_{1} e^{(-0.4888^{2}*Fo )}[/tex]

= 0.4167 = [tex]1.0396e^{-0.4888*Fo}[/tex]

therefore Fo =  3.8264

Now determine how long the plate should be left in the furnace

Fo = [tex](\frac{k}{pc_{p} } ) ( \frac{t}{(L/2)^2} )[/tex]

k = 48

p = 7830

L = 0.1

Input the values into the relation and make t subject of the formula

hence t = 858.25 s

Answer:

At temperature is and relative humidity is 86% therefore,  the humidity ratio is 0.0223 and the specific volume is 14.289

At temperature is and Relative humidity is 40% therefore, the humidity ratio is  0.0066 and the specific volume is 13.535.

To calculate the mass of air can be calculated as follows:

Now , we going to calculate the volume,

The time which is required to fill the cistern can be calculated as follows:

Now, putting the value in above formula we get,

Therefore, the hours required to fill the cistern is 4.65 hours.

Explanation:

Answer:

max shear = R = V = 15 kN

Explanation:

given:

load = 10 kn/m

span = 3m

max shear = R = V = wL / 2

max shear = R = V = (10 * 3) / 2

max shear = R = V = 15 kN

Answer: 255

255 turns are required to create 25 ohms of  secondary impedance.

Explanation:

Given that,

Number of turns in primary wire N₁ = 900

impedance on Primary wire Z₁ = 400 ohms

Number of turns in Secondary wire N₂ = ?

impedance on Secondary wire Z₂ = 25 ohms

we know that, the relationship between turn and impedance is

Zp / Zs = ( Np / Ns )²

(Primary impedance / secondary impedance) = Number of turns in primary wire / Number of turns in secondary wire)²

there fore

Z₁ / Z₂ = ( N₁ / N₂ )²

Now we substitute

( 400 / 25 ) = ( 900 / N₂ )²

400 / 25 = 900² / N₂²

we cross multiple to get our N₂

400 × N₂² =  900² × 25

N₂² = ( 900² × 25 ) / 400

N₂² = ( 810000 × 25 ) / 400

N₂² = 20250000 / 400

N₂² = 50625

N₂ = √50625

N₂ = 225

Therefore 255 turns are required to create 25 ohms of  secondary impedance.

Answer:

[tex]\t(x)_{max} =\dfrac{p\times L}{2\times \pi}[/tex]

Explanation:

Given that

Shear modulus= G

Sectional area = A

Torsional load,

[tex]t(x) = p sin( \frac{2\pi}{ L} x)[/tex]

For the maximum value of internal torque

[tex]\dfrac{dt(x)}{dx}=0[/tex]

Therefore

[tex]\dfrac{dt(x)}{dx} = p cos( \frac{2\pi}{ L} x)\times \dfrac{2\pi}{L}\\ p cos( \frac{2\pi}{ L} x)\times \dfrac{2\pi}{L}=0\\cos( \frac{2\pi}{ L} x)=0\\ \dfrac{2\pi}{ L} x=\dfrac{\pi}{2}\\\\x=\dfrac{L}{4}[/tex]

Thus the maximum internal torque will be at x= 0.25 L

[tex]t(x)_{max} = \int_{0}^{0.25L}p sin( \frac{2\pi}{ L} x)dx\\t(x)_{max} =\left [p\times \dfrac{-cos( \frac{2\pi}{ L} x)}{\frac{2\pi}{ L}} \right ]_0^{0.25L}\\t(x)_{max} =\dfrac{p\times L}{2\times \pi}[/tex]

Answer:

Following are the answer to this question:

Explanation:

1)

Following the four operations in the airlines:

Landside operations:

In Airlines, the airports are divided into areas on the countryside and on the airside, in which landside region is available to the public, although strictly controlled access to the airside zone. Its area covers all areas of the airport across the aircraft, including parts of the buildings which can only be reached by customers and employees.

Airside Operations:

It's also committed to ensuring which air operations military exercises Ballarat airfields are safe and secure. It includes the provision of parking and flight escort services to itinerant and automates. Organizing operational response to incidents, accidents, or emergencies at the airport.

Billing and invoicing Operations:

This requires several steps, each of which must be performed with absolute accuracy to ensure that perhaps the airport operator is adequately paid for supplying passengers with all the services and infrastructure. After this, the receipts want to be produced and sent to customers on the airline.

Information management:

In this, it collects all the data about the customers and employees and it also helps in finding new routes and after collecting the data it processes on them.  

2)

It involves many activities in the airline, including dispatch, flight preparation, flight watch, weather information source, activities control, ground-to-air communications, and staff coordination, scheduling, and maintenance planning. Computing and expert programs are constantly being used to handle unpredictable activities.

Answer: 0.002174

Explanation:

Given that the

Yield strength rho = 450 MPa

Length = 2 m

Elastic modulus E= 207 GPa

According to Hook's law, if the elastic limits is not reached, the elastic modulus is the ratio of elastic strength to the elastic strain ə

E = rho/ə

Make ə the subject of formula

ə = rho/ E

ə = (450 × 10^6) / (207 × 10^9)

ə = 2.174 × 10^-3

Therefore, the engineering strain which depends on engineering stress and elastic modulus is 2.174 × 10^-3

Elastic Strain has no S.I Units.

Answer:

Hello the table which is part of the question is missing and below are the table values

For a 5V belt the available diameters are : 5.5, 5.8, 5.9, 6.2, 6.3, 6.6, 12.5, 13.9, 15.5, 16.1, 18.5, 20.1

Answers:

belt size = 140 in with diameter of 20.1n

actual speed of belt = 288.49 in/s

actual center distance = 49.345 in

Explanation:

Given data :

Electric motor (driver sheave) speed (w1) = 950 rpm

Driven sheave speed (w2) = 250 rpm

pick D1 ( diameter of driver sheave)  = 5.8 in  ( from table )

To select an appropriate belt size we apply the equation for the velocity ratio to get the diameter first

VR = [tex]\frac{w1}{w2}[/tex] = 950 / 250

also since the speed of  belt would be constant then ;

Vb = w1r1 = w2r2 ------- equation 1

r = d/2

substituting the value of r into equation 1

equation 2 becomes : [tex]\frac{w1}{w2} = \frac{d2}{d1}[/tex]    = VR

Appropriate belt size ( d2) can be calculated as

d2 = [tex]\frac{w1d1}{w2}[/tex] = [tex]\frac{950 * 5.8}{250}[/tex] = 22.04

From the given table the appropriate belt size would be : 20.1 because it is the closest to the calculated value

next we have to determine the belt length /size

[tex]L = 2C + \frac{\pi }{2} ( d1+d2) + \frac{(d2-d1)^2}{4C}[/tex]

inputting  all the values into the above equation including the value of C as calculated below

L ≈ 140 in

Calculating the center distance

we use this equation to get the ideal center distance

[tex]d2< C_{ideal} < 3( d1 +d2)[/tex]

22.04 < c < 3 ( 5.8 + 20.1 )

22.04 < c < 77.7

the center distance is between 22.04 and 77.7  but taking an average value

ideal center distance would be ≈ 48 in

To calculate the actual center distance we use

[tex]C = \frac{B+\sqrt{B^2 - 32(d2-d1)^2} }{16}[/tex] -------- equation 3

B = [tex]4L -2\pi (d2 + d1 )[/tex]

inputting all the values into (B)

B = 140(4) - 2[tex]\pi[/tex]( 20.01 + 5.8 )

B ≈ 399.15 in

inputting all the values gotten Back to equation 3 to get the actual center distance

C = 49.345 in ( actual center distance )

Calculating the actual belt speed

w1 = 950 rpm = 99.48 rad/s

belt speed ( Vb) = w1r1 = w1 * [tex]\frac{d1}{2}[/tex]

                           = 99.48 * 5.8 / 2 = 288.49 in/s

Answer:

P=11 kW

Explanation:

Given that

Number of poles= 8

I.E.C. 180L motor frame

From data book , for 8 poles motor at 50 Hz

Speed = 730 rpm

Power factor = 0.75

Efficiency at 100 % load= 89.3 %

Efficiency at 50 % load= 89.1 %

Output power = 11 kW

Therefore the rated output power of 8 poles motor will be 11 kW. Thus the answer will be 11 kW.

P=11 kW

Answer:

from the below explanation...  we can say that, in the series circuit, flowing current remains the same at each part of the circuit. While in parallel circuits, the voltage across two endpoints of the branches is the same as the supplied voltage.

Explanation:

1.

The components in a series circuit are arranged in a single path from one end of supply to another end. However, the multiple components in a parallel circuit are arranged in multiple paths wrt the two end terminals of the battery.

2.

In a series circuit, a common current flows through all the components of the circuit. While in a parallel circuit, a different amount of current flows through each parallel branch of the circuit.

3.

In the series circuit, different voltage exists across each component in the circuit. Whereas in the parallel circuit, the same voltage exists across the multiple components in the circuit.

4.

A fault in one of the components of the series circuit causes hindrance in the operation of a complete circuit. As against fault in a single component in a parallel network do not hinder the functioning of another part of the circuit.

5.

The detection of a fault in case of a series circuit is difficult, but it is quite easy in parallel circuits.

6.

The equivalent resistance in case of a series circuit is always more than the highest value of resistance in the series connection. While the equivalent resistance in the parallel circuit is always less than any of the individual resistances in parallel combination.

Answer:

1) There are  three (3) PWM generator blocks in LM3S1968 and they are

2)  Two (2) independent  PWM outputs can be generated on an LM3S1968

3) Applications for PWM

4) NVIC in a timer stand for ; Nested Vectored Interrupt Controller

its significance is that it is used to handle  and give priorities to exception and Interrupts

5) The counter/timer derive its time period from  counting the output pulses for one cycle which is the duration over which gate is open

Explanation:

1) There are  three (3) PWM generator blocks in LM3S1968 and they are

2)  Two (2) independent  PWM outputs can be generated on an LM3S1968

3) Applications for PWM

4) NVIC in a timer stand for ; Nested Vectored Interrupt Controller

its significance is that it is used to handle  and give priorities to exception and Interrupts

5) The counter/timer derive its time period from  counting the output pulses for one cycle which is the duration over which gate is open

6) THE WAVEFORM DIAGRAMS IS ATTACHED BELOW

it can be seen  that 50 % rises and goes down at half interval. 75 % goes down at half more of 50% and 25% goes down at half less of 50%

Answer:

There are six conditions

1. Poles should contain some residual flux.

2. Field and armature winding must be correctly connected so that initial mmm adds residual flux.

3. Resistance of field winding must be less than critical resistance.

4. Speed of prime mover of generator must be above critical speed.

5. Generator must be on load.

6. Brushes must have proper contact with commutators.

Explanation:

Answer:

A)  222.58 kJ / kg

B)  0.8897 M^3/ kg

c)  0.7737 m^3/kg

D)  746.542 k

E)  536.017 kj/kg

efficiency = 58% ( approximately )

Explanation:

Given Data :

Gas constant (R) =  0.287 kJ/ kg.K

T1 = 310 k

P1 ( Kpa ) = 100

r = 11.5 ( compression ratio )

rp = 1.95 ( pressure ratio )

A ) specific internal energy at state 1

 = Cv*T1 =  0.718 * 310 = 222.58 kJ / kg

B) Relative specific volume at state 1

= P1*V1 = R*T1 ( ideal gas equation )

V1 = R*T1 / P1 = (0.287* 10^3*310 ) / 100 * 10^3

V1 = 88.97 / 100 = 0.8897 M^3/ kg

C ) relative specific volume at state 2

Applying  r ( compression ratio) = V1 / V2

11.5 = 0.8897 / V2

V2 = 0.8897 / 11.5 = 0.7737 m^3/kg

D) The temperature (k) at state 2

since the process is an Isentropic process we will apply the p-v-t relation

[tex]\frac{T1}{T2} = (\frac{V1}{V2}^{n-1} ) = (\frac{P2}{P1} )^{\frac{n-1}{n} }[/tex]

hence T2 = [tex]9^{1.4-1} * 310[/tex] = 2.4082 * 310 = 746.542 k

e) specific internal energy at state 2

= Cv*T2 = 0.718  * 746.542 = 536.017 kj/kg

efficiency = output /input = 390.3511 / 667.5448 ≈ 58%

attached is a free hand diagram of an Otto cycle is attached below

Answer:

[tex]\mathbf{F_D \approx 1.071 \ lbf}[/tex]

Explanation:

Given that:

The height of a  triangular stabilizing fin on its stern is 1 ft tall

and it length is 2 ft long.

Temperature = 60 °F

The objective is to determine the drag on the fin when the submarine is traveling at a speed of 2.5 ft/s.

From these information given; we can have a diagrammatic representation describing how the  triangular stabilizing fin looks like as we resolve them into horizontal and vertical component.

The diagram can be found in the attached file below.

If we recall ,we know that;

Kinematic viscosity v = [tex]1.2075 \times 10^{-5} \ ft^2/s[/tex]

the density of water ρ = 62.36 lb /ft³

[tex]Re_{max} = \dfrac{Ux}{v}[/tex]

[tex]Re_{max} = \dfrac{2.5 \ ft/s \times 2 \ ft }{1.2075 \times 10 ^{-5} \ ft^2/s}[/tex]

[tex]Re_{max} = 414078.6749[/tex]

[tex]Re_{max} = 4.14 \times 10^5[/tex] which is less than < 5.0 × 10⁵

Now; For laminar flow;  the drag on  the fin when the submarine is traveling at 2.5 ft/s can be determined by using the expression:

[tex]dF_D = (\dfrac{0.664 \times \rho \times U^2 (2-x) dy}{\sqrt{Re_x}})^2[/tex]

where;

[tex](2-x) dy[/tex] = strip area

[tex]Re_x = \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}[/tex]

Therefore;

[tex]dF_D = (\dfrac{0.664 \times 62.36 \times 2.5^2 (2-x) dy}{\sqrt{ \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}}})[/tex]

[tex]dF_D = 1.136 \times(2-x)^{1/2} \ dy[/tex]

Let note that y = 0.5x from what we have in the diagram,

so , x = y/0.5

By applying the rule of integration on both sides, we have:

[tex]\int\limits \ dF_D = \int\limits^1_0 \ 1.136 \times(2-\dfrac{y}{0.5})^{1/2} \ dy[/tex]

[tex]\int\limits \ dF_D = \int\limits^1_0 \ 1.136 \times(2-2y)^{1/2} \ dy[/tex]

Let U = (2-2y)

-2dy = du

dy = -du/2

[tex]F_D = \int\limits^0_2 \ 1.136 \times(U)^{1/2} \ \dfrac{du}{-2}[/tex]

[tex]F_D = - \dfrac{1.136}{2} \int\limits^0_2 \ U^{1/2} \ du[/tex]

[tex]F_D = -0.568 [ \dfrac{\frac{1}{2}U^{ \frac{1}{2}+1 } }{\frac{1}{2}+1}]^0__2[/tex]

[tex]F_D = -0.568 [ \dfrac{2}{3}U^{\frac{3}{2} } ] ^0__2[/tex]

[tex]F_D = -0.568 [0 - \dfrac{2}{3}(2)^{\frac{3}{2} } ][/tex]

[tex]F_D = -0.568 [- \dfrac{2}{3} (2.828427125)} ][/tex]

[tex]F_D = 1.071031071 \ lbf[/tex]

[tex]\mathbf{F_D \approx 1.071 \ lbf}[/tex]

Answer:

1. Slope = 53.3 x 10⁻⁶

2. Deflection = -0.00016m

Explanation:

given:

let L = 4 m (span of cantilever beam)

let w = 300 N/m (distributed load)

let EI =60 MNm² (flexural stiffness)

                 dy      w * L³        300 x 4³

1. slope = ------- = --------- =  ------------------- =  53.3 x 10⁻⁶

                  dx        6EI           6 x 60x10⁶

                                  wL⁴               300 x 4⁴

2. Deflection = y = - ----------- =  - ------------------ =   -0.00016m

                                    8EI                8 x 60x10⁶

therefore the deflection is 0.16mm downwards.

Answer:

(a). -1.76i + 0.76j ft/s^2.

(b). -10.42i + 5.7j ft/s^2.

Explanation:

So, we are given the following data or parameters in the question/problem above;

=> "The fire truck is moving forward at a speed = 35 mi /hr and is decelerating at the rate = 10 ft/sec2."

=> "At the instant considered the angle = 30° and is increasing at the constant rate = 10 deg /sec. Also at this instant the extension b of the ladder = 5 ft, with b ˙ = 2 ft/sec and b ¨ =−1 ft/sec2"

STEP ONE: Convert the angular speed to rad/s and speed to ft/s.

Angular speed => π/18 rad/s

Speed => 35 × 1.46667 = 52.33 ft/s

STEP TWO: determine the coordinate form of the acceleration of the truck.

= - 10(cos 30°)i - sin(30°)j [π/18 × K] × 25i[π/18 × K] × 2i × [π/9 × K] - 1i.

= -10.42i + 5.7j ft/s^2.

PLEASE NOTE: The step TWO above is the solution to the second part (b) of this question which is the with respect to the ground.

STEP THREE:

This, from step two above;

( -10.42i + 5.7j ) - ( acceleration of the truck).

= - 10.42i + 5.7j - - 10(cos 30°)i - sin(30°)j = -1.76i + 0.76j ft/s^2.

Answer:

From the calculation, we can see that the invention's COP of 2.67 does not exceed the maximum theoretical COP of 14.65. Hence his claim is valid and could be possible.

Explanation:

Heat generated Q = 200 kW

power input W = 75 kW

Temperature of heated region [tex]T_{h}[/tex] = 293 K

Temperature of heat source [tex]T_{c}[/tex] = 273 K

For this engine,

coefficient of performance COP = Q/W  = 200/75 = 2.67

The maximum theoretical COP obtainable for a heat pump is given as

COP = [tex]\frac{T_{h} }{T_{h} - T_{c} }[/tex] =  [tex]\frac{293 }{293 - 273 }[/tex] = 14.65

From the calculation, we can see that the invention's COP of 2.67 does not exceed the maximum theoretical COP of 14.65. Hence his claim is valid and could be possible.

Given:

We have given two statements.

Statement 1: Proper footwear may include both leather and steel-toed shoes.

Statement 2:  Leather-soled shoes provide slip resistance.

Find:

Which statement is true.

Solution:

A slip-resistant outsole is smoother and more slip-resistant than other outsole formulations when exposed to water and oil. A smoother outsole in rubber ensures a slip-resistant shoe can handle a slippery floor more effectively.

Slip resistant shoes have an interlocked tread pattern that does not close the water in, enabling the slip resistant sole to touch the floor to provide better slip resistance.

Leather-soled shoes don't provide slop resistance.

Therefore, both the Technicians are wrong.

From the statements made by both technician A and Technician B, we can say that; both technicians are wrong.

We are given the statements made by both technicians;

Technician A: Proper footwear may include both leather and steel-toed shoes.

Technician B: Leather-soled shoes provide slip resistance.

Now, they are talking about safety shoes to be worn in workshops.

A shoe that is Slip resistant will have rubber soles and tread patterns that can help to have better grip of wet or greasy floors.

This is the type of shoe that should be worn by technicians in the workshop.

Thus, Technician A is wrong because proper footwear does not include leather shoes.

Similarly, technician B is also wrong because leather shoes are not safety shoes.

Read more about slip resistant shoes at; https://brainly.com/question/17411739

Answer:

a) 42 mm

b) 144.4 MPa

Explanation:

Load P = 200 kN = 200 x 10^3 N

Torque T = 1.5 kN-m = 1.5 x 10^3 N-m

maximum shear stress τ = 100 Mpa = 100 x 10^6 Pa

diameter of shaft d = ?

From T = τ * [tex]\frac{\pi }{16}[/tex] * [tex]d^{3}[/tex]

substituting values, we have

1.5 x 10^3 = 100 x 10^6 x [tex]\frac{3.142 }{16}[/tex] x [tex]d^{3}[/tex]

[tex]d^{3}[/tex] = 7.638 x 10^-5

d = [tex]\sqrt[3]{7.638 * 10^-5}[/tex] = 0.042 m = 42 mm

b) Normal stress = P/A

where A is the area

A = [tex]\frac{\pi d^{2} }{4}[/tex] = [tex]\frac{3.142*0.042^{2} }{4}[/tex] = 1.385 x 10^-3

Normal stress = (200 x 10^3)/(1.385 x 10^-3) = 144.4 x 10^6 Pa = 144.4 MPa

Answer:

μb = 0.096

μc  = 0.073

Explanation:

member AB:

-800( 4/3 ) + Nb (2) = 0

Nb (2) = 3200/3

Nb = 533.3N

Post BC:

summation of force along the y axis=0

Nc + Nb + 150(3/5 ) -50(9.81)=0

Nc + 533.3 + 150(3/5 ) -50(9.81)=0

Nc = 933.83 N

Also (-4/5)(150)(3) + Fb(0.7)= 0

Fb = (4/5)(150)(3)/0.7 = 51.429 N

Likewise alog the x axis,

4/5(150) - Fc -Fb = 0

4/5(150) - Fc -51.429 = 0

Fc = 4/5(150)  -51.429 =68.571 N

μb = Fb/Nb = 51.429/533.3  = 0.096

μc = Fc/Nc = 68.571 / 933.83 = 0.073

Answer:

It is difficult to introduce products modification

Answer:

a) The spring constant is 50 lb/ft

b) The man is 26.3 ft close to the ground.

Explanation:

Height of tower is 130 ft

Specification calls for a cable of length 85 ft

the maximum this length stretches is 100 ft when subjected to a load of 750 lb

The extension of the cable is calculated from the formula from Hooke's law

F = kx

where F is the load or force on the cable

k is the spring constant of the cable

x is the extension on the cable

a) The extension on the cable is

x = 100 ft - 85 ft = 15 ft

substituting into the formula above, we'll have

750 = k*15

k = 750/15 = 50 lb/ft

b) for a 185 lb man, jumping down will give an extension gotten as

F = kx

185 = 50*x

x = 185/50 = 3.7 ft

The total length of the cable will be extended to 100 ft + 3.7 ft = 103.7 ft

closeness to the ground = 130 ft - 103.7 ft = 26.3 ft

Answer:

Engineering drawing abbreviations and symbols are used to communicate and detail the characteristics of an engineering drawing.

There are many abbreviations common to the vocabulary of people who work with engineering drawings in the manufacture and inspection of parts and assemblies.

Technical standards exist to provide glossaries of abbreviations, acronyms, and symbols that may be found on engineering drawings. Many corporations have such standards, which define some terms and symbols specific to them; on the national and international level, like BS8110 or Eurocode 2 as an example.

Explanation: