Ethanol is produced in a one-liter batch fermenter by Saccharomyces cerevisiae (yeast). To begin the batch process, glucose and ammonia are added to the reactor, which has been inoculated by the yeast. The reaction produces glycerol and cell mass, in addition to ethanol, carbon dioxide, and water according to the (unbalanced) reaction below. Note that that the mass ratio of glycerol produced to glucose consumed is 0.2556. Also note that the mass ratio of H.O produced to NH, consumed is 1.058 Glucose + Ammonia → Glycerol + Ethanol + Yeast Cell Mass + Carbon Dioxide + Water a CH (8) + b NH(g)}p CH 0 (1) + (CHO) +ỊCH K04 Nụ (8) + $ 0 g) + t H (1) Balance the equation and fill out the table below. (1 point cach, 6 total) a- 1 b = 115 g of NH, are added to the reactor. At the end of the batch, 640 g of ethanol has been produced. How much glucose was added (assume complete conversion of the limiting reactant). mlar B 16 points) BONUS QUESTION USING CHAPTER 4 CONCEPTS: The standard heat of combustion (AHC) for the yeast is-22.1 kJ/mol. In order to maintain a constant temperature, how much heat must be added or removed from the reactor? Both the sign and magnitude of your answer will be graded J 3 points)

Answer:

(a) The volume rate of flow per meter width = 5.6*10⁻³ m²/s

(b) The shear stress  acting on the bottom plate = 157.5 N/m²

(c) The velocity along the centerline of the channel = 0.93 m/s

Explanation:

(a)

Calculating the distance of plate from centre line using the formula;

h = d/2

where h = distance of plate

d = diameter of flow = 9 mm

Substituting, we have;

h = 9/2

  = 4.5 mm = 4.5*10^-3 m

Calculating the volume flow rate using the formula;

Q = (2h³/3μ)* (Δp/L)

Where;

Q = volume flow rate

h = distance of plate = 4.5*10^-3 m

μ = dynamic viscosity = 0.38 N.s/m²

(Δp/L) = Pressure drop per unit length = 35 kPa/m = 35000 Pa

Substituting into the equation, we have;

Q = (2*0.0045³/3*0.38) *(35000)

    = (1.8225*10⁻⁷/1.14) * (35000)

    = 1.60*10⁻⁷ * 35000

   = 5.6*10⁻³ m²/s

Therefore, the volume flow rate = 5.6*10⁻³ m³/s

(b) Calculating the shear stress acting at the bottom plate using the formula;

τ  = h*(Δp/L)

    = 0.0045* 35000

    = 157.5 N/m²

(c) Calculating the velocity along the centre of the channel using the formula;

u(max) = h²/2μ)* (Δp/L)

   = (0.0045²/2*0.38) * 35000

   =2.664*10⁻⁵ *35000

   = 0.93 m/s

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

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:

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) the cutting time to complete the facing operation = 11.667mins

b) the cutting speeds and metal removal rates at the beginning= 12.89in³/min and end of the cut. = 8.143in³/min

Explanation:

check attached files below for answer.

Answer:

Explanation:

Please go through the attached file first before you you come back to the explanation page.

Description of code used:

G00: Rapid positioning

G01: Linear interpolation

G02: Circular interpolation (Clockwise)

G20: Program in inch

G90: Absolute programing

M02: Program end

M03: Spindle start

M05: Spindle stop

T04: Tool no. 04

F: Feed rate

S: Spindle speed

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:

Please see the attached Picture 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:

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

For beam design, the cross section dimensions of the beam are determined A. Based on allowable normal stress, but the allowable shear stress should always be checked to be sure it is not exceeded.

In beam design, the cross-sectional dimensions of the beam are primarily determined based on the allowable normal stress, which ensures that the beam can support the applied loads without exceeding the material's strength limit in tension or compression.

However, while designing the beam, it is crucial to also check the allowable shear stress to ensure that it is not exceeded. Shear stress can occur in beams due to transverse loads and can lead to failure if not adequately considered.

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:

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#SPJ2

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:

G(s)=1000/(s+100)(s+10)

Using the inverse Laplace transformation

Magnitude of this function=10

Phase shift=10,20,700

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:

(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 %

Answer:

1st part: Section W18X76  is adequate

2nd part: Section W21X62 is adequate

Explanation:

See the attached file for the calculation

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:

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:

[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

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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:

A) β_max = 20.64

B) TH = 68.25°C

C) TC = 54.27°C

Explanation:

A) We are given;

TH = 16°C = 16 + 273K = 289K

TC = 2°C = 2 + 273K = 275K

Formula for maximum cycle coefficient of performance is given as;

β_max = TH/(TH - TC)

β_max = 289/(288 - 275)

β_max = 20.64

B) We are given;

Heat rejected to system at hot reservoir; Q_H = 10.5 KW

Heat provided to system at cold reservoir; Q_C = 8.75 KW

Cold reservoir temperature; TC = 0°C = 0 + 273K = 273K

Formula for actual cycle COP is given as;

β_actual = Q_C/W_cycle

Where W_cycle is the work done and is given by;

W_cycle = Q_H - Q_C

W_cycle = 10.5 - 8.75 = 1.75 KW

Thus,

β_actual = 8.75/1.75

β_actual = 5

Actual cycle COP is defined as;

β_actual = TH/(TH - TC)

And we are looking for TH.

Thus,

TH = TC/(1 - (1/β_actual))

TH = 273/(1 - 1/5)

TH = 273/(4/5)

TH = 341.25K = 341.25 - 273°C = 68.25°C

C) We are given;

TH = 27°C = 27 + 273 = 300°C

β_max = 12

Thus, from,

β_max = TH/(TH - TC)

TC = TH(1 - (1/β_max))

TC = 300/(1 - 1/12)

TC = 327.27K = 327.27 - 273 °C = 54.27°C

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

Calculating the diameter of a thermocouple junction based on given properties and conditions for temperature measurement.

The diameter of the thermocouple junction can be determined using the given properties and conditions. To achieve a response time of 5 seconds for 42 percent of the initial temperature difference, calculations involving

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:

The discharge [tex]Q = \frac{2}{3} Cd*L\sqrt2g*H^{\frac{3}{2}}[/tex] or other than [tex]H = [\frac{3Q}{2CdL\sqrt2g}]^\frac{2}{3}[/tex]

Explanation:

see picture

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:

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:

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