Diesel-fueled generators are frequently used as backup electrical power sources for homes and hospitals. Consider a Diesel powered generator with an efficiency of 39 percent for an engine speed of idle to about 1,800 rpm (revolutions per minute). If Diesel fuel has a chemical formula of C12H23.
a. Determine the chemical reaction for 1 kmol of Diesel fuel burning with the stoichiometric amount of air.
b. For each kg of Diesel fuel burned, how much CO2 is generated, in kg?
c. Find the higher heating value (HHV) for C12H23 at 25°C, 1 atm.
d. Calculate the amount of Diesel fuel, in kg and in gallons, required to produce a power output of 18 kW to a home for a period of 8 h.
e. Comment on your results.

Answer:

The force on the strut is 909000 lbm ft/s²

Explanation:

please look at the solution in the attached Word file

Answer:

a. The fuel formula is C10H20 (dec-1-ene)

b. The heat given-off to the cooling system of the reactor represents the Higher Heating Value (HHV) of the gaseous hydrocarbon fuel

c. LHV = 44.36 KJ/kg

d. ΔH (formation) = -6669.6 KJ/mol

Explanation:

Molecular weight = 140, Molar H/C ratio = 2.0

a) To determine the fuel formula, we represent the information we have as equations

Molecular weight of C = 12, Molecular weight of H = 1

CxHy = 140g

12 · x + 1 · y = 140      ·········Eqn 1

y = 2x                        ·········Eqn 2

Substitute y = 2x into Eqn 1, we have:

12 (x) + 1 (2x) = 140

12x + 2x = 140

14x = 140

x = 10

Substitute x into Eqns 2, we have:

y = 2 x 10 = 20

∴ the fuel formula is C10H20 (dec-1-ene)

b) The heat given-off to the cooling system of the reactor represents the Higher Heating Value (HHV) of the gaseous hydrocarbon fuel . HHV refers to the amount of heat given-off in the combustion of a specified amount of fuel which was initially at 25°C and cooled back to a temperature of 25°C.

c) We will use LHV formula given as:

LHV = HHV – [2.441 (0.09H + M) ÷ 1000]

where:

LHV = lower heating value of fuel in MJ/kg

HHV = higher heating value of fuel in MJ/kg

M = Weight % of moisture in fuel

H = Weight % of hydrogen in fuel

HHV = 47.5 MJ/kg, M = 0, H = mass of hydrogen ÷ total mass of fuel

⇒ H = 20 ÷ 140 = 0.143 * 100 = 14.3 kg per 100 kg, Latent heat of steam (at 25°C) = 2441 KJ/kg

LHV = 47.5 - [2441 * (0.09 * 14.3 + 0) ÷ 1000]

LHV = 47.5 - 3.14 = 44.36

LHV = 44.36 KJ/kg

d) To get the enthalpy of the combustion of the fuel, we have:

fuel + air - - - - > carbon dioxide + water

C10H20 + 15 O2 - - - - > 10 CO2 + 10 H20

Enthalpy (formation) = Enthalpy (product) - Enthalpy (reactant)

ΔH (CO2) = -393.5 kJ/mol, ΔH (H2O) = -285.8 kJ/mol, ΔH (C10h20) = - 123.4KJ/mol

ΔH (formation) = 10 * (-393.5) + 10 * (-285.8) - (-123.4)

ΔH (formation) = -3935 - 2858 + 123.4

ΔH (formation) = -6669.6 KJ/mol

Answer:

(a) The mass glow rate of water is 0.0032kg/s

(b) The heat removal from air is 14.17KJ/sec

Explanation:

In this question, we are asked to use the formula to calculate the mass flow rate and the heat removal from water based on the data in the question.

To answer this question, we take values of absolute humidity and enthalpy values from table from corresponding temperature

Please check attachment for complete solution and step by step explanation

Answer:

472,826 lb/hr

Explanation:

As per the given data:

Solids in feed= 2%

Solids in concentrate= 25%

HTA1 = HTA2 = 1000 ft^2

U1 = 500 Btu/ h ft^2 F & U2 = 700 Btu/ h ft^2 F

Overall material balance: Feed= Distillate + concentrate ----> Eq-1

Component balance: Feed * 0.02 = Distillate * 0 + concentrate * 0.25

Feed = 12.5 * concentrate ---> Eq-2

Boiling point rise = negligible, so solution & solvent vapor temperature will be same.

Assumed that the 1st effect is operating under atmospheric pressure (Boiling point - 212F).

As per the data:

Latent heat 212F = 300 Btu/lb

Latent heat 100F = 320 Btu/lb

As per material balance:

Vapor flowrate * latent heat = Overall HT coefficient * HTA * DT

1st effect: M-1= (500 * 1000 * (326-212)) / 300 = 190,000 lb/hr

2nd effect: M-2= (700*1000 * (212-100)) / 320 = 245,000 lb/hr

Distillate = M-1 + M-2 = 190,000+245,000 = 435,000 lb/hr

Substituting the above in Eq-1

Feed = 435,000 + concentrate

Substitute Eq-2 in the above

12.5 * concentrate = 435,000 + concentrate

Concentrate, L1 = 435,000/11.5 = 37826 lb /hr

Feed, F = 435,000 + 37826 = 472,826 lb/hr

1st effect operating pressure is not given, That may be the reason we are not getting the given answer. But procedure is right.

[ Find the figure in the attachment ]

The elevation difference between the two reservoirs is 10.28 ft

Given information:

- Length of the 3-in. cast iron pipe = 150 ft

- Number of flanged elbows = 4

- Entrance condition: well-rounded

- Exit condition: sharp-edged

- Fully open gate valve

- Flow rate = 75 gal/min

- Water temperature = 50 °F

Step 1: Convert the flow rate from gal/min to ft³/s.

Flow rate in ft³/s = (75 gal/min) × (1 ft³/7.48 gal) × (1 min/60 s) = 0.167 ft³/s

Step 2: Calculate the velocity of flow.

Velocity of flow, V = Flow rate / Cross-sectional area of the pipe

Cross-sectional area of the 3-in. cast iron pipe = (π/4) × (3/12)² ft² = 0.0491 ft²

Velocity of flow, V = 0.167 ft³/s / 0.0491 ft² = 3.4 ft/s

Step 3: Calculate the Reynolds number to determine the flow regime.

Reynolds number, Re = (ρVD) / μ

Where,

ρ = density of water at 50 °F = 1.94 slugs/ft³ (from tables)

V = velocity of flow = 3.4 ft/s

D = diameter of the pipe = 3/12 ft

μ = dynamic viscosity of water at 50 °F = 2.73 × 10⁻⁵ lb.s/ft² (from tables)

Re = (1.94 slugs/ft³) × (3.4 ft/s) × (3/12 ft) / (2.73 × 10⁻⁵ lb.s/ft²) = 1.9 × 10⁵

Since Re > 4000, the flow is turbulent.

Step 4: Calculate the head loss due to friction in the straight pipe.

Head loss due to friction, hf = f × (L/D) × (V²/2g)

Where,

f = friction factor (depends on Reynolds number and pipe roughness)

L = length of the pipe = 150 ft

D = diameter of the pipe = 3/12 ft

V = velocity of flow = 3.4 ft/s

g = acceleration due to gravity = 32.2 ft/s²

From the Moody diagram or friction factor tables for cast iron pipe, assuming a relative roughness of 0.00085, and Re = 1.9 × 10⁵, the friction factor, f = 0.021.

Substituting the values, hf = 0.021 × (150 × 12/3) × [(3.4)²/(2 × 32.2)] = 8.4 ft

Step 5: Calculate the head loss due to fittings (elbows, entrance, exit, and valve).

Head loss due to fittings, hf_fittings = Σ(K × V²/2g)

Where K is the loss coefficient for each fitting.

For 4 flanged elbows, K = 0.3 (for each elbow)

For a well-rounded entrance, K = 0.03

For sharp-edged exit, K = 1.0

For a fully open gate valve, K = 0.2

Substituting the values, hf_fittings = [4 × 0.3 + 0.03 + 1.0 + 0.2] × [(3.4)²/(2 × 32.2)] = 1.7 ft

Step 6: Calculate the velocity head.

Velocity head, hv = V²/2g = (3.4)²/(2 × 32.2) = 0.18 ft

Step 7: Calculate the total head loss.

Total head loss, H = hf + hf_fittings + hv

H = 8.4 ft + 1.7 ft + 0.18 ft = 10.28 ft

Therefore, the elevation difference between the two reservoirs is 10.28 ft for the given piping system and a flow rate of 75 gal/min of water at 50 °F.

Answer:

6 m²

Explanation:

application of fluid pressure according to  Pascal's principle for the two pistons is given as:

[tex]P_1=P_2[/tex]

Where P₁ is the pressure at the input and P₂ is the pressure at the output.

But P₁ = F₁ / A₁ and P₂ = F₂ / A₂

Where F₁ and F₂ are the forces applied at the input and output respectively and A₁ and A₂ are the area of  the input pipe and output pipe respectively

Since, [tex]P_1=P_2[/tex]

[tex]\frac{F_1}{A_1} =\frac{F_2}{A_2}\\[/tex]

But A₁ = 0.2 m², F₁ = 250 N, F₂ = 7500 N. Substituting values to get:

[tex]\frac{F_1}{A_1} =\frac{F_2}{A_2}\\\frac{250}{0.2}=\frac{7500}{A_2}\\ A_2=\frac{7500*0.2}{250} = 6m^2[/tex]

Therefore, the area of the pipe below the load is 6 m²

Answer:

(a) Frequency induced voltage of motor is 120 Hz

(b) The speed at which the machine will produce an average torque is 240π rad/s. Maximum Torque is 1.0 Nm

Explanation:

Explanation:

a. The output of an ideal full wave rectifier is zero volts only when the input is zero volts.

True

The output of an ideal full wave rectifier is zero volts only when the input is zero since 1st diode is forward biased during one half cycle of the input and 2nd diode is forward biased during the other half cycle of the input therefore, it fully utilizes both the input cycles so the output voltage is only zero when the input is zero.

b. Half-wave rectifier circuits need a minimum of 4 diodes to operate.

False

A Half-wave rectifier circuits need a minimum of 1 diode to operate, whereas a full-wave bridge rectifier need minimum of 4 diodes to operate.

c. In an ideal full wave bridge rectifier, half of the diodes are in the ON state and half of the diodes are in the OFF state at any given time the input voltage is not zero.

True

A full-wave bridge rectifier consists of 4 diodes, where 2 diodes are functional in half of the cycle(so the other 2 are off) and other 2 diodes are functional in the other half cycle( so the other 2 are off).

d. The output of an ideal half wave rectifier is zero volts only when the input is zero volts.

False

The output of an ideal half wave rectifier is zero during half of the cycle when the diode is reversed biased and doesn't conduct even though input voltage is not zero volts at this point.

e. A turn-on voltage of a diode (y,) greater than zero can cause the output of a full wave rectifier to be zero volts even when the input is not zero volts.

False

A turn-on voltage of a diode (y,) greater than zero cannot cause the output of a full wave rectifier to be zero rather there will be a little voltage drop across the output of full wave rectifier due to this turn-on voltage of diode which is usually 0.7 volts for silicon based diodes.

f. An advantage of the half wave rectifier is that is can use a smoothing capacitor, while a full wave rectifier cannot.

False

Smoothing capacitor can be used in both half wave rectifier as well as full wave rectifier.

Answer:

P = 98052.64 Pa or 98.05 kPa

Explanation:

Using Bernoulli's equation;

P+rho*v^2/2+ rho*g*z = constant

at high end;

dia= 1.2 m

flow, Q = 5400 L/min = 0.09 m^3/s

therefore, velocity at the high end, v1 = Q/A =0.09/(pi()*(1.2/2)^2) = 0.08 m/s

pressure, P = 68.67 kPa

Solving for elevation, z

assume lower end is reference line. then higher end will be 'x' m high wrt to lower end.

x= 300*sin(tan^-1(1/100)) = 3 m

that means higher end will be 3 m above with respect to lower end

Similarly for lower end;

dia= 0.6 m

flow, Q = 5400 L/min = 0.09 m^3/s

therefore, velocity at the high end, v2 = Q/A =0.09/(pi()*(0.6/2)^2) = 0.318 m/s

assume pressure, P

z=0

put all values in the formula, we get;

68.67*10^3+1000*0.08^2/2+ 1000*9.81*3 =P+1000*0.318^2/2+ 0

solving this, we get;

P = 98052.64 Pa or 98.05 kPa

According to Bernoulli's principle, a lower pressure than expected at the

lower end because the velocity of the fluid is higher.

The given parameter are;

Length of the pipe, L = 300 m

Slope of the pipe = 1 in 100

Diameter at the high end, d₁ = 1.2 m

Diameter at the low end, d₂ = 0.6 m

The volume flowrate, Q = 5,400 L/min

Pressure at the high end, P = 68.67 kPa

Pressure at the low end

The elevation of the pipe, z₁ = [tex]300 \, m \times \frac{1}{100} = 3 \, m[/tex]  

z₂ = 0

The continuity equation is given as follows;

Q = A₁·v₁ = A₂·v₂

[tex]\sqrt[n]{x} \displaystyle Q = 5,400 \, L/min = 5,400 \, \frac{L}{min} \times \frac{1 \, m^3}{1,000 \, L} \times \frac{1 \, min}{60 \, seconds} = \mathbf{ 0.09 \, m^3/s}[/tex]

Therefore;

[tex]\displaystyle 0.09 = \frac{\pi}{4} \times 1.2^2 \times v_1 = \mathbf{\frac{\pi}{4} \times 0.6^2 \times v_2}[/tex]

[tex]\displaystyle v_1 = \frac{0.09}{\frac{\pi}{4} \times 1.2^2 } \approx 0.0796[/tex]

[tex]\displaystyle v_2 =\frac{0.09}{\frac{\pi}{4} \times 0.6^2 } \approx \mathbf{0.318}[/tex]

[tex]\displaystyle \frac{p_1}{\rho \cdot g} + \frac{v_1^2}{2 \cdot g} + z_1 = \mathbf{ \frac{p_2}{\rho \cdot g} + \frac{v_2^2}{2 \cdpt g} + z_2}[/tex]

Therefore;

[tex]\displaystyle p_2 = \left(\frac{p_1}{\rho \cdot g} + \frac{v_1^2}{2 \cdot g} + z_1 - \left( \frac{v_2^2}{2 \cdot g} + z_2 \right)\right) \times \rho \cdot g = p_1 + \frac{\rho}{2} \cdot \left(v_1^2} -v_2^2 \right)+ \rho \cdot g \left(z_1 - z_2\right)[/tex]

The density of water, ρ = 997 kg/m³

Which gives;

Learn more about Bernoulli's principle here:

https://brainly.com/question/6207420

Answer:

Here, The CML is used for efficient portfolios whereas the SML applies to all portfolios or securities

Considering the seperation theorem, The separation theorem states that the investment decision is sepaarte from the financing decision.

This implies that The SML can be used to analyze the relationship between risk and requried return for all assets.

Under the separation theorem investors should Hold the same portfolio of risky assets and the same expected return but at different levels of risk

Explanation:

See answer

Answer:

Copy MATLAB code to plot the magnitude of magnetic field strength with respect to z on the axis of solenoid:

z=-20:0.01:20;

H=120.*(((20-(2.*z))./sqrt((20-(2.*z)).^2+100))+((20+(2.*z))./sqrt((20+(2.*z)).^2+100)));

plot(z,H)

title('plot of |H| vs z along the axis of solenoid')

ylabel('Magnitude of magnetic field 'H")

xlabel('position on axis of solenoid 'z")

Explanation:

full explanation is attached as picture and the resultant plot also.

Answer:

Explanation:

Given that,

Omega is a function of the following

ω = f(b, h, v, ρ, k)

Where, all unit have a dimension of

ω = T^-1

b = L

h = L

V = LT^-1

ρ = FL^-4T²

k = FL

Then,

From the pie theorem

The required pi term is 6—3 = 3 terms,

So, we use V, p and b as a repeating term.

For first pi

π1 = ω•b^a•v^b•ρ^c.

Since

ω= T^-1, b = L, v = LT^-1 and

ρ= FL^-4T²

Since π is dimensionless then,

π = F^0•L^0•T^0

(T^-1)•(L^a)•(LT^-1)^b•(FL^-4T²)^c = F^0•L^0•T^0

Rearranging

F^0•L^0•T^0 = F^c•T^(2c-b-1)• L^(a+b-4c)

Comparing coefficient

c = 0

2c - b - 1 = 0

b = 2c - 1 = 0 - 1 = -1

a + b - 4c = 0

a = 4c - b = 0 - -1 = 0+1

a = 1

Then, a = 1, b = -1 and c = 0

So, π1 = ω•b^a•v^b•ρ^c.

π1 = ω•b^1•v^-1•ρ^0

π1 = ω•b / v

Check dimensions

ωb/v = (T^-1)L / LT^-1 = L^0•T^0 = 1

Then, π1 is dimensionless

For second pi

π2 = h•b^a•v^b•ρ^c.

Since

h = L, b = L, v = LT^-1 and ρ= FL^-4T²

Since π is dimensionless then,

π = F^0•L^0•T^0

(L)•(L^a)•(LT^-1)^b•(FL^-4T²)^c = F^0•L^0•T^0

Rearranging

F^0•L^0•T^0 = F^c•T^(2c-b)• L^(1+a+b-4c)

Comparing coefficient

c = 0

2c - b = 0

b = 2c = 0

1 + a + b - 4c = 0

a = 4c - b - 1 = 0 -0 - 1  = -1

a = -1

Then, a = -1, b = 0 and c = 0

So, π2 = h•b^a•v^b•ρ^c.

π2 = h•b^-1•v^0•ρ^0

π2 = h / b

Check dimensions

h / b = L / L = 1

Then, π2 is dimensionless

For third pi

π3 = k•b^a•v^b•ρ^c.

Since

k= FL, b = L, v = LT^-1 and ρ=FL^-4T²

Since π is dimensionless then,

π = F^0•L^0•T^0

(FL)•(L^a)•(LT^-1)^b•(FL^-4T²)^c = F^0•L^0•T^0

Rearranging

F^0•L^0•T^0 = F^(c+1)•T^(2c-b)• L^(1+a+b-4c)

Comparing coefficient

c + 1= 0

Then, c = -1

2c - b = 0

b = 2c = -2

1 + a + b - 4c = 0

a = 4c - b - 1 = -4 +2 - 1  = -3

a = -3

Then, a = -3, b = -2 and c = -1

So, π3 = k•b^a•v^b•ρ^c.

π3 = k•b^-3•v^-2•ρ^-1

Therefore,

π3 = k / b³•v²•ρ

Let check for dimension

π3 = FL / (L³• L²T^-2 • FL^-4T²)

π3 = FL / (L^(3+2-4) • T^(-2+2) •F)

π3 = FL / (L• T^(0) •F)

π3 = FL / LF = 1

π3 is also dimensions less

So.

I. There are three none dimensional pi

II. The none dimensional group are

π1 = ω•b / v

π2 = h / b

π3 = k / b³•v²•ρ

III. Reynolds Number. The Reynolds number is the ratio of inertial forces to viscous forces and it is dimensionless

So, the π3 can be considered as a Reynolds number

Answer:

The reactances vary with frequency, with large XL at high frequencies and large Xc at low frequencies, as we have seen in three previous examples. At some intermediate frequency fo, the reactances will be the same and will cancel, giving Z = R; this is a minimum value for impedance and a maximum value for Irms results. We can get an expression for fo by taking

XL=Xc

Substituting the definitions of XL and XC,

2[tex]\pi[/tex]foL=1/2[tex]\pi[/tex]foC

Solving this expression for fo yields

fo=1/2[tex]\pi[/tex][tex]\sqrt{LC}[/tex]

where fo is the resonant frequency of an RLC series circuit. This is also the natural frequency at which the circuit would oscillate if it were not driven by the voltage source. In fo, the effects of the inductor and capacitor are canceled, so that Z = R and Irms is a maximum.

Explanation:

Resonance in AC circuits is analogous to mechanical resonance, where resonance is defined as a forced oscillation, in this case, forced by the voltage source, at the natural frequency of the system. The receiver on a radio is an RLC circuit that oscillates best at its {f} 0. A variable capacitor is often used to adjust fo to receive a desired frequency and reject others is a graph of current versus frequency, illustrating a resonant peak at Irms at fo. The two arcs are for two dissimilar circuits, which vary only in the amount of resistance in them. The peak is lower and wider for the highest resistance circuit. Thus, the circuit of higher resistance does not resonate as strongly and would not be as selective in a radio receiver, for example.

A current versus frequency graph for two RLC series circuits that differ only in the amount of resistance. Both have resonance at fo, but for the highest resistance it is lower and wider. The conductive AC voltage source has a fixed amplitude Vo.

It will take approximately 0.0238 s to cut the hole.

To calculate the time required to cut the hole using Electrochemical Machining (ECM), we can use the following formula:

[tex]\text { Time }=\frac{\text { Volume of Material to be Removed }}{\text { Material Removal Rate (MRR) }}[/tex]

First, let's calculate the volume of material to be removed:

Volume of Material=Frontal Area × Thickness of Plate

Given:

[tex]Frontal Area =245 \mathrm{~mm}^2\\Thickness of Plate $=12 \mathrm{~mm}$[/tex]

Volume of Material = [tex]245 \mathrm{~mm}^2 \times 12 \mathrm{~mm}=2940 \mathrm{~mm}^3[/tex]

Next, we need to calculate the Material Removal Rate (MRR). MRR is usually given in units of volume removed per unit time per unit current density. Let's assume it is given as [tex]X \mathrm{~mm}^3 / \mathrm{min} / \mathrm{A} / \mathrm{mm}^2 .[/tex] Given the current density [tex]\text { Current Density }=\frac{\text { Applied Current }}{\text { Frontal Area }}[/tex], we can calculate the MRR as:

MRR=X×Current Density

Given:

Applied Current = 1200 A

Efficiency = 95%

Frontal Area = 245 [tex]mm^2[/tex]

[tex]\begin{aligned}& \text { Current Density }=\frac{\text { Applied Current }}{\text { Frontal Area }}=\frac{1200 \mathrm{~A}}{245 \mathrm{~mm}^2} \\& \text { Current Density } \approx 4.897 \mathrm{~A} / \mathrm{mm}^2\end{aligned}[/tex]

Let's assume X=0.2[tex]mm^3[/tex] /s/A/[tex]mm^2[/tex](hypothetical value).

[tex]\mathrm{MRR}=0.2 \times 4.897 \approx 0.979 \mathrm{~mm}^3 / \mathrm{s}[/tex]

To convert mm³/s to g/s, we need to know the density of the material. Let's assume the density of low alloy steel is approximately [tex]7.85 \mathrm{~g} / \mathrm{cm}^3[/tex], which is [tex]7.85 \times 10^{-6} \mathrm{~g} / \mathrm{mm}^3 .[/tex]

Now, we can calculate the MRR in grams per second:

[tex]MRR in $\mathrm{g} / \mathrm{s}=\mathrm{MRR} \times$ Density of material\\MRR in $\mathrm{g} / \mathrm{s}=0.979 \mathrm{~mm}^3 / \mathrm{s} \times 7.85 \times 10^{-6} \mathrm{~g} / \mathrm{mm}^3$\\MRR in $\mathrm{g} / \mathrm{s} \approx 0.00000768515 \mathrm{~g} / \mathrm{s}$[/tex]

Now, we can use these values to calculate the time required:

[tex]\text { Time }=\frac{2940 }{0.00000768515} \\& \text { Time } \approx 0.022594341[/tex]

However, we need to consider the efficiency. Since the efficiency is 95%, the actual time required will be:

[tex]\begin{aligned}& \text { Actual Time }=\frac{\text { Time }}{\text { Efficiency }} \\& \text { Actual Time }=\frac{0.022594341}{0.95}\end{aligned}[/tex]

≈ 0.0238 s

Answer:

a) check the attached files below

b)  What power will be developed by a geometrically and dynamically similar prototype, of diameter 15 m, in winds of 35 m/s at 500 m standard altitude from sea level = 2184.56KW

c) What is the appropriate rotation rate = 122.5 rpm

Explanation:

please kindly check the attachment below.

1. Proton assignments for each isomer: p-bromoaniline: Hd signal around 7.5 ppm (doublet, 2H), Hb signal around 6.7 ppm (doublet, 2H) o-bromoaniline: Ha and Hc signals overlapping around 7.6-7.65 ppm (multiplets, 4H total) o,p-dibromoaniline: Hb signal around 6.6 ppm (doublet, 2H), Hd signal around 7.55 ppm (doublet, 1H) o,o,p-tribromoaniline: No clear signals observed, likely not formed.

2. The isomer not observed to be formed is o,o,p-tribromoaniline. A plausible experimental explanation is that the highly brominated product is less favored due to steric hindrance and deactivation of the ring towards further electrophilic substitution.

3. Diagnostic protons for product ratio calculations: p-bromoaniline: Hb around 6.7 ppm o-bromoaniline: Ha/Hc around 7.6-7.65 ppm o,p-dibromoaniline: Hd around 7.55 ppm o,o,p-tribromoaniline: N/A (not formed)

4. Calculating product ratios using diagnostic proton integrations: p-bromoaniline: Integration of Hb signal ≈ 2.0, relative ratio = 2.0/2.0 = 100% o-bromoaniline: Integration of Ha/Hc signals ≈ 2.8, relative ratio = 2.8/4.8 ≈ 58% o,p-dibromoaniline: Integration of Hd signal ≈ 1.0, relative ratio = 1.0/3.0 ≈ 33% o,o,p-tribromoaniline: 0% (not formed)

5. Theoretical reasons for relative product abundance: i. p-bromoaniline: Dominant product due to the directing effect of the amino group favoring para substitution. ii. o-bromoaniline: Formed in moderate amounts due to some ortho substitution, but less favored than para. iii. o,p-dibromoaniline: Formed in smaller amounts due to the decreased reactivity of the ring after the first substitution. iv. o,o,p-tribromoaniline: Not formed, likely due to high steric hindrance and deactivation of the ring towards further substitution

The complete question:

The chemist’s crude product mixture was submitted for NMR analysis to give Spectrum C. Using the expected splitting patterns and chemical shift changes you predicted in part a, assign each proton for each of the isomers above (see structures on Spectrum C) by labeling each signal on the spectrum with the appropriate letter. Keep the integration ratios in mind when assigning each proton.

2. One of the four potential products above was NOT observed to be formed. Which one? Propose a very brief experimental explanation for the absence of this product.

3. For each of the three products that are observed in Spectrum C, there is a diagnostic proton that can be used to accurately calculate the product ratios. (Remember that a good diagnostic proton has a “clean” signal that has minimal overlap with other signals- see the “Using Integrations to Calculate Compound Ratios” page in the Assignment Information folder on Top Hat.) What diagnostic proton can be used to calculate product ratios for each isomer? If the isomer was not formed, write “N/A.”

4. Using the diagnostic protons you listed in part c, calculate the percent of each isomer. Show calculations for each isomer. If an isomer was not formed, write “0%.”

5. Provide a theoretical reason for the relative abundance of the following products

i. p-bromoaniline

ii. o-bromoaniline

iii. o,p-dibromoaniline

iv. o,o,p-tribromoaniline

Spectrum C bromination of aniline, crude solvent: CDCl Notes 1. Spectrum C is the zoomed-in aromatic region (6.3-7.9 ppm); amine protons are not shown 2. There are 2 overlapping signals from relevant protons at 6.60-6.65 ppm. You should assign both NH2 NH2 NH2 NH2 Br Br Chloroform H Sovent Но Hb He Br Hd ignore (possible oxidation product) 7.8 7.7 7.6 7.5 7.4 7.3 7.2 7.1 7.0 6.9 6.8 6.7 6.6 6.5 6.4 ppm p-bromoaniline o-bromoaniline o,p- dibromoaniline o,o,p- tribromoaniline %p-bromoaniline %o-bromoaniline % o,p- dibromoaniline % o,o,p- tribromoaniline. show all working. solve in hum, and form

The image of the elevation and wall section is missing, so i have attached them.

Answer:

Number of concrete blocks = 1020 blocks

Number of lintel blocks = 240

Number of plain blocks = 780

Explanation:

First of all, we'll find the net area of the wall as follows:

The height of the wall is 17' - 4", so we need to convert it to ft, thus, h = 17' + (4/12)' = 17.33'

So, Gross wall area = 80′ × 17.333' = 1,387 ft²

From the image, there are 4 doors, thus, Area of doors = 4 × 10′ × 12′ = 480 ft²

Thus, Net area = Gross wall area - area of doors

Net area = 1,387 – 480 = 907 ft²

We are told that the block is 8 inches length by 16inches width. Thus, converting to ft; (8/12)ft by (16/12)ft.

So area of one block = (8/12) x (16/12) = 128/144 ft² = 0.88889 ft²

So, number of blocks per ft² = 1/0.88889 = 1.125 blocks

So, there are 1.125 blocks per ft²

Thus, Number of concrete blocks = 907 ft² × 1.125 blocks per ft² = 1,020 blocks

From the image attached, there are five rows of lintel blocks, located at 4′, 8′, 12′, 16′, and 17′-4″. The bottom three rows pass through the doors and are only a total of (80′ – (4 × 10′) ) = 40′ long .

The top two rows are 80′ long. Thus, there is 40 feet (4 × 10′) of lintel block above the doors.

Lintel blocks = (3 × 40′) + (2 × 80′) + 40′ = 320 ft

Number of Lintel blocks = 320′ × 12 in per ft / 16″ = 240 blocks

Number of Plain blocks = 1,020 blocks – 240 blocks = 780 blocks

Without specific wall dimensions provided, it is not possible to calculate the exact number of plain and lintel concrete blocks required for the construction of the wall. However, the general procedure involves calculating the wall's total area, subtracting the area of doors or windows, and then dividing by the size of one block, while also considering the requirements for lintel blocks above openings and where supports are needed.

Determining the Quantity of Concrete Blocks for a Wall

To determine the number of 8-inch-high by 8-inch-wide by 16-inch-long concrete blocks required to complete a wall, one must calculate the total volume of the wall and then divide by the volume of a single block. Unfortunately, the dimensions of the wall are not provided in the question, so we cannot calculate the exact number of blocks needed. As for the lintel blocks, which are used above the doors and wherever the #4 horizontal bars are located, the number would depend on the linear footage that needs to be covered by the lintels and the length of each lintel block.

Without specific wall dimensions, we must refer to the typical process where one would calculate the wall's total surface area, subtract the area occupied by doors or windows, and divide by the area covered by one block. Then, identify the areas requiring lintel blocks and count the number of lintels needed based on their standard lengths.

The plain blocks and lintel blocks calculation requires detailed measurements of the wall and openings. Since the details provided are related to the overhead doors with known dimensions, one would need to consider the door's dimensions when calculating the plane wall areas and the lintel requirements above them.

Question: The program was not attached to your question. Find attached of the program and the answer.

Answer:

See the explanation for the answer.

Explanation:

#include <iostream>

using namespace std;

int main()

{

   cout << "How many students will you enter? ";

   int n;

   cin>>n;

   string *name = new string[n];

   double *gpa = new double[n];

   for(int i=0;i<n;i++){

       cout<<"Enter student"<<(i+1)<<"'s name: ";

       cin>>name[i];

       cout<<"Enter student"<<(i+1)<<"'s gpa: ";

       cin>>gpa[i];

   }

   cout<<"The list students"<<endl;

   cout<<"Name             GPA"<<endl;

   cout<<"----------------------"<<endl;

   for(int i=0;i<n;i++){

       cout<<name[i]<<"              "<<gpa[i]<<endl;

   }

   return 0;

}

OUTPUT : See the attached file.

Answer:

q' = 5826 W/m

Explanation:

Given:-

- The length of the rectangular fin, L = 0.15 m

- The surface temperature of fin, Ts = 250°C

- The free stream velocity of air, U = 80 km/h

- The temperature of air, Ta = 27°C

- Parallel flow over both surface of the fin, assuming turbulent conditions through out.

Find:-

What is the rate of heat removal per unit width of the fin?

Solution:-

- Assume steady state conditions, Negligible radiation and flow conditions to be turbulent.

- From Table A-4, evaluate air properties (T = 412 K, P = 1 atm ):

    Dynamic viscosity , v = 27.85 * 10^-6 m^2/s  

    Thermal conductivity, k = 0.0346 W / m.K

    Prandlt number Pr = 0.69

- Compute the Nusselt Number (Nu) for the - turbulent conditions - the appropriate relation is as follows:

                          [tex]Nu = 0.037*Re_L^\frac{4}{5} * Pr^\frac{1}{3}[/tex]

Where,    Re_L: The average Reynolds number for the entire length of fin:

                          [tex]Re_L = \frac{U*L}{v} \\\\Re_L = \frac{80*\frac{1000}{3600} * 0.15}{27.85*10^-^6} \\\\Re_L = 119688.80909[/tex]

Therefore,

                         [tex]Nu = 0.037*(119688.80909)^\frac{4}{5} * 0.69^\frac{1}{3}\\\\Nu = 378[/tex]

- The convection coefficient (h) can now be determined from:

                          [tex]h = \frac{k*Nu}{L} \\\\h = \frac{0.0346*378}{0.15} \\\\h = 87 \frac{W}{m^2K}[/tex]

- The rate of heat loss q' per unit width can be determined from convection heat transfer relation, Remember to multiply by (x2) because the flow of air persists on both side of the fin:

                          [tex]q' = 2*[h*L*(T_s - T_a)]\\\\q' = 2*[87*0.15*(250 - 27)]\\\\q' = 5826\frac{W}{m}[/tex]

- The rate of heat loss per unit width from the rectangular fin is q' = 5826 W/m

- The heat loss per unit width (q') due to radiation:

                  [tex]q' = 2*a*T_s^4*L[/tex]

Where, a: Stefan boltzman constant = 5.67*10^-8

                  [tex]q' = 2*5.67*10^-^8*(523)^4*0.15\\\\q' = 1273 \frac{W}{m}[/tex]

- We see that radiation loss is not negligible, it account for 20% of the heat loss due to convection. Since the emissivity (e) of the fin has not been given. So, in the context of the given data this value is omitted from calculations.  

Answer:

As Ray L zapp is thinking about new strategies to test his new hash Table,hence therefore the best testing technique is  Stubbing& mocking .

The question is an engineering problem that involves calculating the average temperature of a truck's refrigeration compartment's outer surface. A precise answer would require additional thermal information and properties, which are not provided in the question.

The question is related to the field of thermodynamics and heat transfer, specific to engineering. It requires determining the average temperature of the outer surface of the refrigeration compartment of a truck. However, to provide a precise answer, additional data would be required, including the thermal properties of the truck's material, the heat transfer coefficient, and the difference between the interior and exterior temperatures. Typically, such a problem would involve calculations using the concepts of convection and possibly conduction, considering the truck as a heat exchanger with heat being removed by the refrigeration system and added from the external environment. Without specific values for heat transfer coefficients and material properties, a calculation cannot be accurately made.

Answer:

[tex]\dot W_{out} = 399.47\,kW[/tex]

Explanation:

The turbine is modelled after the First Law of Thermodynamics:

[tex]-\dot Q_{out} -\dot W_{out} + \dot m\cdot (h_{in}-h_{out}) = 0[/tex]

The work done by the turbine is:

[tex]\dot W_{out} = \dot m \cdot (h_{in}-h_{out})-\dot Q_{out}[/tex]

The properties of the water are obtained from property tables:

Inlet (Superheated Steam)

[tex]P = 10\,MPa[/tex]

[tex]T = 520\,^{\textdegree}C[/tex]

[tex]h = 3425.9\,\frac{kJ}{kg}[/tex]

Outlet (Superheated Steam)

[tex]P = 1\,MPa[/tex]

[tex]T = 280\,^{\textdegree}C[/tex]

[tex]h = 3008.2\,\frac{kJ}{kg}[/tex]

The work output is:

[tex]\dot W_{out} = \left(1.1\,\frac{kg}{s}\right)\cdot \left(3425.9\,\frac{kJ}{kg} -3008.2\,\frac{kJ}{kg}\right) - 60\,kW[/tex]

[tex]\dot W_{out} = 399.47\,kW[/tex]

Answer:

 IA = 80/3 kgm^2

Explanation:

Given:-

- The mass of rod AB, m1 = 20 kg

- The length of rod AB, L1 = 2m

- The mass of rod CD, m2 = 10 kg

- The length of rod CD, L2 = 1m

Find:-

What is the moment of inertia about A for member AB?

Solution:-

- The moment of inertia About point "O" the center of rod AB is given as:

                               IG = 1/12*m1*L^2

- To shift the axis of moment of inertia for any object at a distance "d" from the center of mass of that particular object we apply the parallel axis theorem. The new moment of inertia about any arbitrary point, which in our case A end of rod AB is:

                              IA = IG + m1*d^2

- Where the distance "d" from center of rod AB to its ends is 1/2*L1 = 1 m.

So the moment of inertia for rod AB at point A would be:

                              IA = 1/12*m1*L^2 + m1*0.5*L1^2

                              IA = 1/3 * m1*L1^2

                              IA = 1/3*20*2^2

                             IA = 80/3 kgm^2

Answer:

[tex]D_c(s) = 1.8(\frac{s+1.11}{s+0.2})[/tex]

Explanation:

Code attached with picture.

Answer:

Dude just like forget the highway and drive on the interstate bruh

Explanation:

Answer:

(a) Calculate the rod base temperature (°C). = 299.86°C

(b) Determine the rod length (mm) for the case where the ratio of the heat transfer from a finite length fin to the heat transfer from a very long fin under the same conditions is 99 percent.  = 0.4325m

Explanation:

see attached file below

Answer:

So these are the RTL representation:

MAR<----X

MBR<-----AC

M[MAR]<------MBR

Control signal sequence are:

P3T0:MAR<----X

P2P3 P4T1:MBR<-----AC

P0P1 P3T2:M[MAR]<------MBR

Explanation:

STORE X instruction is used for storing the value of AC to the memory address pointed by X. This operation can be done by using the Register Transfers at System Level and this can be represented by using a notation called Register Transfer Language RTL. Let us see what are the register transfer operations happening at the system level.

1. First of all the address X has to be tranfered on to the Memory Address Register MAR.  

MAR<----X

2. Next we have to tranfer the contents of AC into the Memory Buffer Register MBR

MBR<-----AC

3. Store the MBR into memory where MAR points to.

M[MAR]<------MBR

So these are the RTL representation:

MAR<----X

MBR<-----AC

M[MAR]<------MBR

Control signal sequence are:

P3T0:MAR<----X

P2P3 P4T1:MBR<-----AC

P0P1 P3T2:M[MAR]<------MBR

Answer:

a) 7.5A

b) 370V

Explanation:

Note that the Line current is the current through any one line between a three-phase source and load.

From the given question, The source line current =7.5A(approximately)

While the Source line voltage = 370V.

Kindly go through the attached file for a step by step detail of how I arrived at these answers.

The source line current in the three-phase system is 2.08 A and the source line voltage is approximately 371 V.

To calculate the source line current and the source line voltage in a three-phase system, we can use the formula:

I = P / (sqrt(3) * V * PF)

Where I is the source line current, P is the power delivered to the load, V is the phase voltage, and PF is the power factor.

Given that the power delivered is 4.8 kVA, the phase voltage is 214 V, and the power factor is 0.9 lagging, we can substitute these values into the formula and calculate:

I = 4.8 kVA / (sqrt(3) * 214 V * 0.9)

Calculating this gives us a source line current of 2.08 A.

To find the source line voltage, we can use the formula:

Vline = sqrt(3) * Vphase

Substituting the phase voltage of 214 V into the formula gives us a source line voltage of approximately 371 V.

Complete Question

The diagram for this question is shown on the first uploaded image  

Answer:

The largest offset that can be used is  [tex]h = 0.455 \ in[/tex]

Explanation:

From the question we are told that

   The diameter of the metal tube is  [tex]d_m = 0.75 \ in[/tex]

    The thickness of the wall is  [tex]D = 0.08 \ in[/tex]

Generally the inner diameter is mathematically evaluated as

           [tex]d_i = d_m -2D[/tex]

               [tex]= 0.75 - 2(0.08)[/tex]

               [tex]= 0.59 \ in[/tex]

Generally the tube's cross-sectional area can be evaluated as

         [tex]a = \frac{\pi}{4} (d_m^2 - d_i^2)[/tex]

             [tex]= \frac{\pi}{4} (0.75^2 - 0.59^2)[/tex]

              [tex]= 0.1684 \ in^2[/tex]

Generally the maximum stress of the metal is mathematically evaluated as

          [tex]\sigma = \frac{P}{A}[/tex]

              [tex]\sigma = \frac{P}{ 0.1684}[/tex]

The diagram showing when the stress is been applied is shown on the second uploaded image  

        Since the internal forces in the cross section are the same with the force P and the bending couple M then  

           [tex]M = P * h[/tex]

Where h is the offset

       The maximum stress becomes

                 [tex]\sigma_n = \frac{P}{A} + \frac{M r_m }{I}[/tex]

Where [tex]r_m[/tex] is the radius of the outer diameter  which is evaluated as

                 [tex]r_m = \frac{0.75}{2}[/tex]

                 [tex]r_m = 0.375 \ in[/tex]

and I is the moment of inertia which is evaluated as

           [tex]I = \frac{\pi}{64} (d_m^4 - d_i^4 )[/tex]

              [tex]= \frac{\pi}{64}(0.75^4 - 0.59^4)[/tex]

              [tex]= 0.009583 \ in^4[/tex]

So the maximum stress becomes  

                [tex]\sigma' = \frac{P}{0.1684} + \frac{Phr}{0.009583}[/tex]

Now the question made us to understand that the maximum stress when the offset was introduced must not exceed the 4 times the original stress

So

        [tex]\sigma ' = 4 \sigma[/tex]

=>    [tex]\frac{P}{0.1684} + \frac{Phr_m }{0.009583} = 4 [\frac{P}{0.1684} ][/tex]

  The P would cancel out  

            [tex]\frac{1}{0.1684} + \frac{h(0.375)}{0.009583} = \frac{4}{0.1684}[/tex]

           [tex]5.94 + 39.13h = 23.753[/tex]

             [tex]39.13h = 17. 813[/tex]

                      [tex]h = 0.455 \ in[/tex]

Answer:

a. 5m

b. r = 0.16 e^-80.5◦

c. Zpn = (115.7 + j27.4) ohms

d. Vi = 2.2e^-j22.56◦ volts

e. Vi(t) = 2.2 cos (8π × 107t − 22.56◦ ) Volts

Explanation:

In this question, we are tasked with calculating a series of terms.

Please check attachment for complete solution and step by step explanation