Air at 27°C and a velocity of 5 m/s passes over the small region As (20 mm × 20 mm) on a large surface, which is maintained at Ts = 127°C. For these conditions, 0.5 W is removed from the surface As. To increase the heat removal rate, a stainless steel (AISI 304) pin fin of diameter 5 mm is affixed to As, which is assumed to remain at Ts = 127°C. (a) Determine the maximum possible heat removal rate through the fin. (Hint: look back to Chapter 3 to see which fin case produces the maximum heat transfer.) (b) What fin length would provide a close approximation to the heat rate found in part (a)? (Hint: refer to Example 3.9.) (c) Determine the fin effectiveness, εf. (d) What is the percentage increase in the heat rate from As due to installation of the fin? In other words, what is the percentage increase in q obtained by adding the fin as compared to not having the fin?

Answer:

See Explaination

Explanation:

# copy the function you have this is just for my convenniece

def finalGrade(scoresList):

weights = [0.05, 0.05, 0.40, 0.50]

grade = 0

for i in range(len(scoresList)):

grade += float(scoresList[i]) * weights[i]

return grade

import csv

def main():

with open('something.csv', newline='') as csvfile:

spamreader = csv.reader(csvfile, delimiter=',', quotechar='|')

file = open('something.txt')

for row in spamreader:

student = file.readline().strip()

scores = row

print(student, finalGrade(scores))

if __name__ == "__main__":

main()

Answer:

a) y (n + 1) = 1.005 y(n) + 5U n

y (n + 1) - 1.005 y(n) = 5U (n)

b) Z^-1(Z(y0)=y(n) = [1010(1.005)^n - 1000(1)^n] U(n)

c) h(n) = (1.005)^n U(n - 1) + 10(1.005)^n U(n)

Explanation:

Her bank account can be modeled as:

y (n + 1) = y (n) + 0.5% y(n) + $5

y (n + 1) = 1.005 y(n) + 5U n

Given that y (0) = $10

y (n + 1) - 1.005 y(n) = 5U (n)

Apply Z transform on both sides

= ZY ((Z) - Z(y0) - 1.005) Z = 5 U (Z)

U(Z) = Z {U(n)} = Z/ Z - 1

Y(Z) [Z- 1.005] = Z y(0) + 5Z/ Z - 1

= 10Z/ Z - 1.005 + 5Z/(Z - 1) (Z - 1.005)

Y(Z) = 10Z/ Z - 1.005 + 1000Z/ Z - 1.005 + 1000Z/ Z - 1

= 1010Z/Z- 1.005 - 1000Z/Z-1

Apply inverse Z transform

Z^-1(Z(y0)) = y(n) = [1010(1.005)^n - 1000(1)^n] U(n)

Impulse response in output when input f(n) = S(n)

That is,

y(n + 1)= 1. 005y (n) + 8n

y(n + 1) - 1.005y (n) = 8n

Apply Z transform

ZY (Z) - Z(y0) - 1.005y(Z) = 1

HZ (Z - 1.005) = 1 + 10Z [Therefore y(Z) = H(Z)]

H(Z) = 1/ Z - 1.005 + 10Z/Z - 1. 005

Apply inverse laplace transform

= h(n) = (1.005)^n U(n - 1) + 10(1.005)^n U(n)

Answer:

Modulus of Elasticity = 4.350×10⁶ lb/in²

q'n = 60.33 lb/in²

Side friction = 6.372 lb/in²

Explanation:

See workings in picture below.

Answer:

The power input, in kW is -86.396 kW

Explanation:

Given;

initial pressure, P₁ = 1.05 bar

final pressure, P₂ = 12 bar

initial temperature, T₁ = 300 K

final temperature, T₂ = 400 K

Heat transfer, Q = 6.5 kW

volumetric flow rate, V = 39 m³/min = 0.65 m³/s

mass of air, m = 28.97 kg/mol

gas constant, R = 8.314 kJ/mol.k

R' = R/m

R' = 8.314 /28.97 = 0.28699 kJ/kg.K

Step 1:

Determine the specific volume:

p₁v₁ = RT₁

[tex]v_1 = \frac{R'T_1}{p_1} = \frac{(0.28699.\frac{kJ}{kg.K} )(300 k)}{(1.05 bar *\ \frac{10^5 N/m^2}{1 bar} *\frac{1kJ}{1000N.m} )} \\\\v_1 = 0.81997 \ m^3/kg[/tex]

Step 2:

determine the mass flow rate; m' = V / v₁

mass flow rate, m' = 0.65 / 0.81997

mass flow rate, m' = 0.7927 kg/s

Step 3:

using steam table, we determine enthalpy change;

h₁ at T₁ = 300.19 kJ /kg

h₂ at T₂ = 400.98 kJ/kg

Δh = h₂ - h₁

Δh = 400.98 - 300.19

Δh = 100.79 kJ/kg

step 4:

determine work input;

W = Q - mΔh

Where;

Q is heat transfer = - 6.5 kW, because heat is lost to surrounding

W = (-6.5) - (0.7927 x 100.79)

W = -6.5 -79.896

W = -86.396 kW

Therefore, the power input, in kW is -86.396 kW

Answer:

highest possible ending pressure for this process is 8.0954 bar

Explanation:

We can say that Heat transfer is any or all of several kinds of phenomena, considered as mechanisms, that convey energy and entropy from one location to another. The specific mechanisms are usually referred to as convection, thermal radiation, and conduction.

Please see attachment for the solution.

Answer:

Please see the attached file for the complete answer.

Explanation:

Answer:

(a) The velocity of the exhaust gases. is 832.7 m/s

(b) The rate of fuel consumption is 0.6243 kg/s

Explanation:

For the given turbojet engine operating on an ideal cycle, the pressure ,temperature, velocity, and specific enthalpy of air at [tex]i^{th}[/tex] state are [tex]P_i[/tex] , [tex]T_i[/tex] , [tex]V_i[/tex] , and [tex]h_i[/tex] , respectively.

Use "ideal-gas specific heats of various common gases" to find the properties of air at room temperature.

Specific heat at constant pressure, [tex]c_p[/tex] = 1.005 kJ/kg.K

Specific heat ratio, k = 1.4

Answer:

determine the pipe diameter.  = 0.41ft

Explanation:

check the attached file for answer explanation

Answer: you need to do it on your own first

Explanation:

Answer:

decreased

Explanation:

when impaired you react slower then you would sober.

Alcohol adversely affects the complex reaction time, considerably decreasing the individual’s ability to respond swiftly and adequately in emergencies or unexpected situations. This impairment is attributed to alcohol's impact on the brain causing slow information processing, poor motor control, and a decrease in focus.

When a person's complex reaction time is compromised by alcohol, their ability to respond to unexpected situations or emergencies is greatly diminished. Alcohol's impact on the brain leads to slower processing of information, reduced concentration, and poorer motor control. As a result, they may not react as quickly or efficiently as they would if they were sober to changes in their environment. For example, if a situation arises that requires quick decision-making, such as stopping abruptly while driving to avoid a pedestrian, an intoxicated individual may not respond in time, leading to catastrophic outcomes.

Learn more about Alcohol impact on reaction time here:

https://brainly.com/question/1002220

#SPJ2

Answer:

1. Parg = 89.954 kw

  Pmax = 179.908 kw

2. Parg = 29.984 kw

  Pmax = 59.96 kw

3. Energy = 0.15 kWh

Explanation:

See the attached file for the calculation

Answer:

1) P = 81.74 kW

2) As seen on the pic.

3) E = 0.1362 kWh

Explanation:

1) Given

m = 1364 kg (mass of the vehicle)

vi = 0 mph = 0 m/s  (initial speed)

vf = 60 mph = (60 mph)(1609 m/ 1 mi)(1 h/ 3600 s) = 26.8167 m/s  (final speed)

t = 6 s

We get the acceleration as follows

a = (vf - vi)/t  ⇒   a = (26.8167 m/s - 0 m/s)/ 6s

⇒   a = 4.469 m/s²

then the Force is

F = m*a ⇒   F = 1364 kg*4.469 m/s²

⇒   F = 6096.32 N

Then we get the average power as follows

P = F*v(avg) = F*(vi + vf)/2

⇒   P = 6096.32 N*(0 m/s + 26.8167 m/s)/2

⇒   P = 81741.59 W = 81.74 kW

Knowing that that aerodynamic, rolling, and hill‐climbing force counts for an extra 10% of the needed acceleration force, we use the following formula to find the peak power

Pmax = (1 + 0.1)*F*vf  = 1.1*F*vf

⇒   Pmax = 1.1*6096.32 N*26.8167 m/s

⇒   Pmax = 179831.503 W = 179.83 kW

2) The pic 1 shows the average and peak power (kW) vs. time duration from 2 to 15 seconds.

In the first chart we use the equation

Pinst = F*v  where F is constant and v is the instantaneous speed, and Pavg is the mean of the values.

In the second chart, we use the equation

Ppeak = 1.1*F*v  where F is constant and v is the instantaneous speed.

3) For 0 s ≤ t ≤ 6 s

We can use the equation

E = ΔK = Kf - Ki = 0.5*m*(vf² - vi²)

⇒   E = 0.5*1364 kg*((26.8167 m/s)² - (0 m/s)²)

⇒   E = 490449.123 J = (490449.123 J)(1 kWh/3.6*10⁶J)

⇒   E = 0.1362 kWh

Answer:

[tex]\omega_{out} = 0.867\,\frac{kg\,H_{2}O}{kg\,DA}[/tex]

Explanation:

The final humidity ratio is computed by the Principle of Mass Conservation:

Dry Air

[tex]\dot m_{in} = \dot m_{out}[/tex]

Moist

[tex]\dot m_{in} \cdot \omega_{in} + \dot m_{w} = \dot m_{out}\cdot \omega_{out}[/tex]

Then, the final humidity ratio is:

[tex]\omega_{out} = \frac{\dot m_{in}\cdot \omega_{in}+\dot m_{w}}{\dot m_{out}}[/tex]

[tex]\omega_{out} = \omega_{in} + \frac{\dot m_{w}}{\dot m_{out}}[/tex]

[tex]\omega_{out} = 0.6\,\frac{kg\,H_{2}O}{kg\,DA} + \frac{0.4\,\frac{kg\,H_{2}O}{s} }{1.5\,\frac{kg\,DA}{s} }[/tex]

[tex]\omega_{out} = 0.867\,\frac{kg\,H_{2}O}{kg\,DA}[/tex]

The humidity ratio at the exit is approximately 0.87 kg(vapor)/kg(dry air) when moist air mixes with water vapor.

To solve this problem, we need to apply the conservation of mass principle for both the dry air and the water vapor.

Let's denote:

- [tex]\( w_{in} \)[/tex] as the initial humidity ratio of the moist air stream entering the air conditioner.

- [tex]\( \dot{m}_{dry} \)[/tex]  as the mass flow rate of the dry air stream.

- [tex]\( w_{water} \)[/tex]  as the humidity ratio of the separate stream of water vapor.

- [tex]\( \dot{m}_{water} \)[/tex] as the mass flow rate of the separate stream of water vapor.

- [tex]\( w_{out} \)[/tex] as the humidity ratio of the mixture at the exit.

According to the conservation of mass principle:

For dry air:

[tex]\[ \dot{m}_{dry} \times w_{in} = \dot{m}_{dry} \times w_{out} \][/tex]

For water vapor:

[tex]\[ \dot{m}_{water} \times w_{water} = \dot{m}_{dry} \times (w_{out} - w_{in}) \][/tex]

Given values:

- [tex]\( w_{in} = 0.6 \)[/tex] kg(vapor)/kg(dry air)

- [tex]\( \dot{m}_{dry} = 1.5 \)[/tex] kg/s

- [tex]\( w_{water} = 1 \) kg(vapor)/kg(dry air)[/tex] (since water vapor is pure, its humidity ratio is 1)

- [tex]\( \dot{m}_{water} = 0.4 \) kg/s[/tex]

Now, we can solve for [tex]\( w_{out} \)[/tex]:

From the first equation:

[tex]\[ w_{out} = \frac{\dot{m}_{dry} \times w_{in}}{\dot{m}_{dry}} = w_{in} = 0.6 \, \text{kg(vapor)/kg(dry air)} \][/tex]

Substitute this value into the second equation:

[tex]\[ \dot{m}_{water} \times w_{water} = \dot{m}_{dry} \times (w_{out} - w_{in}) \][/tex]

[tex]\[ 0.4 \times 1 = 1.5 \times (w_{out} - 0.6) \][/tex]

Now, solve for [tex]\( w_{out} \)[/tex]:

[tex]\[ 0.4 = 1.5 \times (w_{out} - 0.6) \][/tex]

[tex]\[ 0.4 = 1.5w_{out} - 0.9 \][/tex]

[tex]\[ 1.5w_{out} = 0.4 + 0.9 \][/tex]

[tex]\[ 1.5w_{out} = 1.3 \][/tex]

[tex]\[ w_{out} = \frac{1.3}{1.5} \][/tex]

[tex]\[ w_{out} \approx 0.87 \, \text{kg(vapor)/kg(dry air)} \][/tex]

So, the humidity ratio at the exit will be approximately 0.87 kg(vapor)/kg(dry air).

Answer:

[tex]\eta_{th} = 30\,\%[/tex], [tex]\eta_{th,max} = 40\,\%[/tex], [tex]\Delta S = \frac{1}{3}\,\frac{kJ}{K}[/tex], The cycle is irreversible.

Explanation:

The real cycle efficiency is:

[tex]\eta_{th} = \frac{1000\,kJ-700\,kJ}{1000\,kJ} \times 100\,\%[/tex]

[tex]\eta_{th} = 30\,\%[/tex]

The theoretical cycle efficiency is:

[tex]\eta_{th,max} = \frac{500\,K-300\,K}{500\,K} \times 100\,\%[/tex]

[tex]\eta_{th,max} = 40\,\%[/tex]

The reversible and real versions of the power cycle are described by the Clausius Inequalty:

Reversible Unit

[tex]\frac{1000\,kJ - 600kJ}{300\,K}= 0[/tex]

Real Unit

[tex]\Delta S = \frac{1000\,kJ-600\,kJ}{300\,K} -\frac{1000\,kJ-700\,kJ}{300\,K}[/tex]

[tex]\Delta S = \frac{1}{3}\,\frac{kJ}{K}[/tex]

The cycle is irreversible.

The cycle efficiency using clausius inequality is;

σ_cycle = 0.333 kJ/kg and is internally irreversible

η = 1 - Q_c/Q_h

Thus;

Q_c = (1 - η)Q_h

σ_cycle = -∫(dQ/dt)ₐ

Thus; σ_cycle = -[(Q_h/T_h) - (Q_c/T_c)]

We are given;

Q_h = 1000 kJ

T_h = 500 k

T_c = 300 k

Q_c = 700 kJ

σ_cycle = -[(1000/500) - (700/300)]

σ_cycle = -(2 - 2.333)

σ_cycle = 0.333 kJ/kg

Since σ_cycle > 0, then the cycle is internally irreversible

Read more about cycle efficiency at; https://brainly.com/question/16014998

Answer:

hL = 0.9627 m

Explanation:

Given

Q = 0.040 m³/s (constant value)

D₁ = 15 cm = 0.15 m  ⇒  R₁ = D₁/2 = 0.15 m/2 = 0.075 m

D₂ = 8 cm = 0.08 m  ⇒  R₂ = D₂/2 = 0.08 m/2 = 0.04 m

P₁ = 480 kPa = 480*10³Pa

P₂ = 440 kPa = 440*10³Pa

α = 1.05

ρ = 1000 Kg/m³

g = 9.81 m/s²

h₁ = h₂

hL = ?  (the irreversible head loss in the reducer)

Using the formula Q = v*A   ⇒  v = Q/A

we can find the velocities v₁ and v₂ as follows

v₁ = Q/A₁ = Q/(π*R₁²) = (0.040 m³/s)/(π*(0.075 m)²) = 2.2635 m/s

v₂ = Q/A₂ = Q/(π*R₂²) = (0.040 m³/s)/(π*(0.04 m)²) = 7.9577 m/s

Then we apply the Bernoulli law (for an incompressible flow)

(P₂/(ρ*g)) + (α*v₂²/(2*g)) + h₂ = (P₁/(ρ*g)) + (α*v₁²/(2*g)) + h₁ - hL

Since h₁ = h₂ we obtain

(P₂/(ρ*g)) + (α*v₂²/(2*g)) = (P₁/(ρ*g)) + (α*v₁²/(2*g)) - hL

⇒  hL = ((P₁-P₂)/(ρ*g)) + (α/(2*g))*(v₁²-v₂²)

⇒  hL = ((480*10³Pa-440*10³Pa)/(1000 Kg/m³*9.81 m/s²)) + (1.05/(2*9.81 m/s²))*((2.2635 m/s)²-(7.9577 m/s)²)

⇒  hL = 0.9627 m

Answer:

Minimum power output required = 1.1977 hp

Explanation:

Given Data:

Temperature outside = 100°F.

House temperature =  70°F

Rate of heat gain(Qw) =  800 Btu/min

Generation rate within(Ql) = 100 Btu/min.

Converting the outside temperature 100°F from fahrenheit to ranking, we have;

1°F = 460R

Therefore,

100°F = 460+100

 To     = 560 R

Converting the house temperature 70°F from fahrenheit to ranking, we have;

1°F = 460R

Therefore,

70°F = 460+70

    Th      = 530 R

Consider the equation for coefficient of performance (COP) of refrigerator in terms of temperature;

COP =Th/(To-Th)

        = 530/(560-530)

        = 530/30

        = 17.66

Consider the equation for coefficient of performance (COP) of refrigerator;

COP = Desired output/required input

         =  Q/Wnet

          = Ql + Qw/ Wnet

Substituting into the formula, we have;

17.667 = (100 + 800)/Wnet

17.667 = 900/Wnet

Wnet = 900/17.667

         = 50.94 Btu/min.

Converting from Btu/min. to hp, we have;

1 hp = 42.53 Btu/min.

Therefore,

50.94 Btu/min =  50.94 / 42.53

                         = 1.1977 hp =

Therefore, minimum power output required = 1.1977 hp

Given the following data:

We would convert the value of the temperatures in Fahrenheit to Rankine.

Note: 1°F = 460R

Conversion:

To calculate the minimum power input that is required for this air- conditioning system:

In Science, the coefficient of performance (COP) is a mathematical expression that is used to show the relationship between the power output of an air-conditioning system and the power input of its compressor.

Mathematically, the coefficient of performance (COP) is given by the formula:

[tex]COP =\frac{T_h}{T_o-T_h}[/tex]

Substituting the given parameters into the formula, we have;

[tex]COP =\frac{530}{560-530}\\ \\ COP =\frac{530}{30}[/tex]

COP = 17.66

For the power input:

[tex]COP = \frac{E_o}{E_i} \\ \\ COP = \frac{Q_l + Q_w}{Q_{net}}\\ \\ 17.67 = \frac{100+800}{Q_{net}}\\ \\ Q_{net}=\frac{900}{17.67} \\ \\ Q_{net}=50.93\;Btu/min[/tex]

Conversion:

1 hp = 42.53 Btu/min.

X hp = 50.93 Btu/min.

Cross-multiplying, we have:

[tex]X=\frac{50.93}{42.53} [/tex]

X = 1.1975 hp

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Answer:

(a) Fuel formula = C₁₀H₂₀

(b)The reactor represent higher heating value (HHV) of the fuel because the water in the product is liquid.

(c) Lower heating value = 44.36 mj/kg

(d) - 6669.8 KJ/Kmol

Explanation:

See the attached files for the calculation

Answer:

Please see the attached Picture for the complete answer.

Explanation:

Answer:

Explanation:

Given that,

Mass of boron fiber in unidirectional orientation

Mb = 5kg = 5000g

Mass of aluminum fiber in unidirectional orientation

Ma = 8kg = 8000g

A. Density of the composite

Applying rule of mixing

ρc = 1•ρ1 + 2•ρ2

Where

ρc = density of composite

1 = Volume fraction of Boron

ρ1 = density composite of Boron

2 = Volume fraction of Aluminum

ρ2 = density composite of Aluminum

ρ1 = 2.36 g/cm³ constant

ρ2 = 2.7 g/cm³ constant

To Calculate fractional volume of Boron

1 = Vb / ( Vb + Va)

Vb = Volume of boron

Va = Volume of aluminium

Also

To Calculate fraction volume of aluminum

2= Va / ( Vb + Va)

So, we need to get Va and Vb

From density formula

density = mass / Volume

ρ1 = Mb / Vb

Vb = Mb / ρ1

Vb = 5000 / 2.36

Vb = 2118.64 cm³

Also ρ2 = Ma / Va

Va = Ma / ρ2

Va = 8000 / 2.7

Va = 2962.96 cm³

So,

1 = Vb / ( Vb + Va)

1 = 2118.64 / ( 2118.64 + 2962.96)

1 = 0.417

Also,

2= Va / ( Vb + Va)

2 = 2962.96 / ( 2118.64 + 2962.96)

2 = 0.583

Then, we have all the data needed

ρc = 1•ρ1 + 2•ρ2

ρc = 0.417 × 2.36 + 0.583 × 2.7

ρc = 2.56 g/cm³

The density of the composite is 2.56g/cm³

B. Modulus of elasticity parallel to the fibers

Modulus of elasticity is defined at the ratio of shear stress to shear strain

The relation for modulus of elasticity is given as

Ec = = 1•Eb+ 2•Ea

Ea = Elasticity of aluminium

Eb = Elasticity of Boron

Ec = Modulus of elasticity parallel to the fiber

Where modulus of elastic of aluminum is

Ea = 69 × 10³ MPa

Modulus of elastic of boron is

Eb = 450 × 10³ Mpa

Then,

Ec = = 1•Eb+ 2•Ea

Ec = 0.417 × 450 × 10³ + 0.583 × 69 × 10³

Ec = 227.877 × 10³ MPa

Ec ≈ 228 × 10³ MPa

The Modulus of elasticity parallel to the fiber is 227.877 × 10³MPa

OR Ec = 227.877 GPa

Ec ≈ 228GPa

C. modulus of elasticity perpendicular to the fibers?

The relation of modulus of elasticity perpendicular to the fibers is

1 / Ec = 1 / Eb+ 2 / Ea

1 / Ec = 0.417 / 450 × 10³ + 0.583 / 69 × 10³

1 / Ec = 9.267 × 10^-7 + 8.449 ×10^-6

1 / Ec = 9.376 × 10^-6

Taking reciprocal

Ec = 106.66 × 10^3 Mpa

Ec ≈ 107 × 10^3 MPa

Note that the unit of Modulus has been in MPa,

Answer:

addi x31, x0, 4

addi x30, x0, 2

Explanation:

Recursion in computer sciencs is defined as a method of solving a problem in which the solution to the problem depends on solutions to smaller cases of the same problem. Such problems can generally be solved by iteration, but this needs to identify and index the smaller cases at time of programming.

addi x31, x0, 4

addi x30, x0, 2

addi x2, x0, 1600 // initialize the stack to 1600, x2= stackpointer

ecall x5, x0, 5 // read the input to x5

jal x1, rec_func

ecall x0, x10, 2 // print the result now

beq x0, x0, end

rec_func:

addi x2, x2, -8 // make room in stack

sd x1, 0(x2) // store pointer and result in stack

bge x5, x31, true // if i > 3, then go to true branch

ld x1, 0(x2)

addi x10, x0, 1 // if i <= 3, then return 1

addi x2, x2, 8 // reset stack point

jalr x0, 0(x1)

true:

addi x5, x5, -2 // compute i-2

jal x1, rec_func // call recursive func for i-2

ld x1, 0(x2) // load the return address

addi x2, x2, 8 // reset stack point

mul x10, x10, x30 // multiply by 2

addi x10, x10, 1 // add 1

jalr x0, 0(x1) // return

end:

Answer:

Please see the attached file for the complete answer.

Explanation:

Answer:

The expected settlement for the pile group using the given information is 19.92mm or 0.79 inch

Explanation:

In this question, we are asked to calculate the expected settlement for the pole group given some information.

Please check attachment for complete solution and step by step explanation

Answer:

The flux into the tissue at x=0 and t=5 s is 0.4476 uM/cm²s

Explanation:

Flux = [tex]\frac{Quantity}{Area * Time}[/tex]

given:

Area = 4.2 cm²

Time = 5 sec

Quantity ( concentration) = 9.4 uM

∴ Flux = Quantity / (Area × Time)

Flux = 9.4 uM / (4.2 cm² × 5 s)

Flux = 9.4 uM / 21 cm²s

Flux = 0.4476 uM/cm²s

Answer:

Explanation:

"Simple Multiples Logic Circuit Development"

From the attached file below;

The first diagram illustrate the truth table. The input consist of binary coded decimal with  its 16  possible output. As stated in the input, assuming we have 0010 which the value is 2, thus ; it is said to be divisible by 2 , then we enter the output of x  to be 1 and the remaining as zeros

The second diagram talks about the output x; the representation of x is 1 there and it was added up. The use case employed is  [tex]x + \bar {x} =1[/tex] which in turn yield output [tex]\bar {A3}[/tex]

The second to the last diagram represents output Y and Z.

As output Y can be further disintegrated ; we insert XNOR gate for the expressions. The expression is  [tex]a.b+(\bar{a}. \bar{b})[/tex]

The last diagram is the logic diagram.

Answer:

(c) Generator, 95.5 %

Explanation:

given data

voltage va = 120V

current Ia = - 5.5 A

Tmech =  5.5Nm

ns = 1200 RPM

solution

first we get here electric input power that is express as

electric input power = va × Ia    ......1

put here value and we get

electric input power = 120 × -5-5

electric input power -660 W

here negative mean it generate power

and here

Pin ( mech) will be

Pin ( mech) = Tl × ω   .........2

Pin ( mech) = 5.5 × [tex]\frac{2\pi N}{60}[/tex]    

Pin ( mech) =  5.5 × [tex]\frac{2\pi 1200}{60}[/tex]  

Pin ( mech) = 691.150 W

and

efficiency will be here as

efficiency = [tex]\frac{660}{691.150}[/tex]    

efficiency = 95.5 %

so correct option is (c) Generator, 95.5 %

The largest load P that can be applied to the steel bar, ensuring a factor of safety of 2.0 with respect to the weld material's tensile yield strength, is 32.5 kN. This is calculated based on the allowable tensile strength of the weld after applying the factor of safety and the cross-sectional area of the steel bar.

The student's question relates to the structural engineering concept of material strength and loading. The maximum load P that can be safely applied to a welded steel bar can be determined by considering the tensile and shear yield strengths of the two materials involved, i.e., the weld material and the bar material. Using the given factor of safety of 2.0, the tensile strength of the weld and the bar, and the cross section of the bar, we can calculate the allowable tensile stress and then the maximum load P by dividing the allowable tensile stress by the area of the cross section.

To begin with, we find the allowable tensile strength by dividing the weld material's yield strength (which is the weaker material concerning tensile strength) by the factor of safety:

Next, we need to calculate the area of the cross-section of the bar:

Now, we convert the units of the allowable tensile strength to N/m² (because 1 MPa = 1 x 10⁶ N/m²) and calculate the maximum tensile load (Ptensile) that the weld can sustain:

Thus, the largest load P that can be applied to the steel bar to satisfy the safety criteria is 32.5 kN.

Answer:

1st part: Section W18X76  is adequate

2nd part: Section W21X62 is adequate

Explanation:

See the attached file for the calculation

Answer:

h_f = 15 ft, so option A is correct

Explanation:

The formula for head loss is given by;

h_f = [10.44•L•Q^(1.85)]/(C^(1.85))•D^(4.8655))

Where;

h_f is head loss due to friction in ft

L is length of pipe in ft

Q is flow rate of water in gpm

C is hazen Williams constant

D is diameter of pipe in inches

We are given;

L = 1,800 ft

Q = 600 gpm

C = 120

D = 8 inches

So, plugging in these values into the equation, we have;

h_f = [10.44*1800*600^(1.85)]/(120^(1.85))*8^(4.8655))

h_f = 14.896 ft.

So, h_f is approximately 15 ft

Answer:

The total compressor work is 234.8 kJ/kg for a isentropic compression

Explanation:

Please look at the solution in the attached Word file

The total compressor work per unit of mass flow in a two-stage compressor with intercooling can be calculated by summing the compressor work in each stage and the work done in the intercooler.

The total compressor work per unit of mass flow in a two-stage compressor with intercooling can be calculated by summing the compressor work in each stage and the work done in the intercooler.

In the first stage, the air is compressed from 100 kPa to 300 kPa. The work done in this stage can be calculated using the isentropic compression process. In the second stage, the air is further compressed from 300 kPa to 1000 kPa. The work done in this stage can also be calculated using the isentropic compression process.

To calculate the total compressor work per unit of mass flow, you need to sum the work done in each stage and the work done in the intercooler.

Answer:

Explanation:

Find attach the solution

Answer:

a. BandStop filter

Explanation:

A band-stop filter or band-rejection filter is a filter that eliminates at its output all the signals that have a frequency between a lower cutoff frequency and a higher cutoff frequency. They can be implemented in various ways.  One of them is to implement a notch filter, which is characterized by rejecting a certain frequency that is interfering with a circuit. The transfer function of this filter is given by:

[tex]H(s)=\frac{s^2+\omega_o^2}{s^2+\omega_cs+\omega_o^2} \\\\Where:\\\\\omega_o=Central\hspace{3} rejected\hspace{3} frequency\\\omega_c=Width\hspace{3} of\hspace{3} the\hspace{3} rejected\hspace{3} band[/tex]

I attached you a graph in which you can see how the filter works.

Answer:

Rate of heat transfer from plate

6796.16 W

Corresponding rate of change of plate temperature

-2634 degrees.Celcius/sec

Explanation:

In this question, we are asked to calculate the rate of heat transfer and the corresponding rate of change of the plate temperature.

Please check attachment for complete solution and step by step explanation