Products exit a combustor at a rate of 100 kg/sec, and the air-fuel ratio is 9. Determine the air flow rate. a. 9 kg/sec b. 90 kg/sec c. 100 kg/sec d. 10 kg/sec

Answer: 98.5% of pearlite was formed during the equilibrium cooling

Explanation:

First we calculate the fraction of pro-eutectoid phase which forms for equilibrium cooling of the 1085 steel from 1000°C at room temperature;

we know that in 1085 steel, last two digits denotes the carbon percentage

so 1085 steel contains 0.85% carbon.

Now from the diagram, carbon percentage is greater than the eutectoid com[psition

i.e 0.85 > 0.76

it is a hyper eutectoid steel

so

fraction of pro eutectoid phase W_Fe₃C = (0.85 - 0.76) / ( 6.7 - 0.76)

= 0.09 / 5.94 = 0.015 = 1.5%

Now, the amount of pearlite formed during the equilibrium cooling of the 1055 steel from 1000°C to room temperature will be;

pearlite (C') = (1 - W_Fe₃C)

= 1 - 0.015

= 0.985 = 98.5%

Therefore 98.5% of pearlite was formed during the equilibrium cooling

Answer:

406.140 KHz

Explanation:

Given data:

Rsig = 100 kΩ

Rin = 100kΩ

Cgs = 1 pF,

Cgd = 0.2 pF,  and   etc.

Determine the expected 3-dB cutoff frequency

first find the CM miller capacitance

CM = ( 1 + gm*ro || RL )( Cgd )

     = ( 1 + 5*10^-3 * 25 || 20 ) ( 0.2 )

     = ( 11.311 ) pF

now we apply open time constant method to determine the cutoff frequency

Th = 1 / Fh

hence : Fh = 1 / Th = [tex]\frac{1}{(Rsig +Rin) (Cm + Cgs )}[/tex]

                               = [tex]\frac{1}{( 200*10^3 ) ( 12.311 * 10^{-12} )}[/tex] =  406.140 KHz

Answer:Decay rate constant,k  = 0.00376/hr

Explanation:

IsT Order  Rate of reaction is given as

In At/ Ao = -Kt

where [A]t is the final concentration at time  t  and  [A]o  is the inital concentration at time 0, and  k  is the first-order rate constant.

Initial concentration = 80 mg/L

Final concentration = 50 mg/L

Velocity = 40 m/hr

Distance= 5000 m

Time taken = Distance / Time

              5000m / 40m/hr = 125 hr

In At/ Ao = -Kt

In 50/80 = -Kt

-0.47 = -kt

- K= -0.47 / 125

k = 0.00376

Decay rate constant,k  = 0.00376/hr

Answer:Total Voltage = 14V

Explanation: it is possible that a circuit  can contain more than one source of electromotive force which can cause flow of current in the same or opposite direction . When the  connection to  voltage sources  allows for current  from the voltage sources to flow in  same direction,it is termed  Series aiding  Thus, the  Total/effective voltage in a series aiding circuit is  computed as the sum of series aiding voltages .

 Here we have the series aiding voltages to be 5V and 9V ,

therefore,

Total Voltage = 5V + 9V

= 14V

Answer:

what do you mean by that ? snap points ?

Answer:

Irreversibility = 5.361 kW

Explanation:

From the given information:

By applying ideal gas equation at entry:

PV =  mRT

600 × 0.3 = m × 0.287 × 300      (where R = 0.287 kJ/kg)

180 = m × 86.1

m = 180/86.1

m = 2.0905 kg/min

At the hot end, using the same ideal gas equation:

PV = mRT

100 × V = 1.4905 × 0.287 × 325

V = 139.026/100

V = 1.3903 m³/ min

This implies that: The total entropy change = Entropy of the universe

So,

[tex]m\bigg [ c_p \ In \dfrac{T_2}{T_o}-R \ In \dfrac{P_2}{P_o} \bigg] + m_2 \bigg [ c_p \ In \dfrac{T_2}{T_o}- R \ In\dfrac{P_2}{P_o} \bigg][/tex]

[tex]= 0.6\bigg [ 1.004 \ In \dfrac{245}{300}-0.287 \ In \dfrac{100}{600} \bigg] +1.4905\bigg [1.004 \ In \dfrac{325}{300}- 0.287\ In\dfrac{100}{600} \bigg][/tex]

= 0.6[-0.2033 + 0.5142] + 1.4905 [0.08036 + 0.5142]

= 1.0727 kJ/min.K

= 0.01787 kw/K

Irreversibility = [tex]T_o [ \Delta S][/tex]

Irreversibility = 300 × 0.01787

Irreversibility = 5.361 kW

Answer:

155 KJ

Explanation:

The total enthalpy is given by

ΔH=ΔU + PV

Where;

ΔH = enthalpy

ΔU = internal energy = 25000 kJ/kg/ 5 kg = 5000 KJ

P = 150 kPa = 150,000 Pa

V =  1 m3

ΔH=  5000 + (150,000 * 1)

ΔH=  155 KJ

Answer:

There are several similarities between the nucleus and a liquid droplet.

Explanation:

A droplet of liquid simply is is very small or tiny drop of liquid. It is also considered as a tiny column of liquid that is surrounded by surfaces that have zero shear stress.

A nucleus on the other hand is an assembly between protons and neutrons. The latter is electrically charged whilst the former is positively charged. The number of protons present in an element is very crucial to the qualities of an element.

The main similarities between a nucleus and a liquid droplet are:

1. a nucleus consists of a large amount of neutrons and protons in the same volume as would a liquid which contains large numbers of molecules in the same volume;

2. both the nucleus and the droplet are similar for their homogeneity in electric charge and density;

3. the molecules exert the same amount for forces towards one another as would the nuclear forces in the nucleons.

4. both of them cannot be compressed

5. both molecules and nucleus are can be subject to nuclear fission which simply mean the breaking apart into smaller units (in the case of the nucleus) or the breaking apart into smaller droplets in the case of the liquid molecule.

6. There are two types of phenomena which occurs in both the liquid droplet and the nucleus which are similar to one another. They are:

Evaporation (in the case of the liquid molecule) and reaction emission (in the case of the nucleus). In evaporation, particles are lost, in Atomic transmutation, particles are lost as well.

B)  the forces which determine the stability of droplets are surface tension and gravitation. The smaller the area, the stronger the surface tension available to keep the drops from going out of shape.

Cheers

Answer:

Super critical

1.2 m

Explanation:

Q = Flow rate = [tex]12\ \text{m}^3/\text{s}[/tex]

w = Width = 3 m

d = Depth = 90 cm = 0.9 m

A = Area = wd

v = Velocity

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

[tex]Q=Av\\\Rightarrow v=\dfrac{Q}{wd}\\\Rightarrow v=\dfrac{12}{3\times 0.9}\\\Rightarrow v=4.44\ \text{m/s}[/tex]

Froude number is given by

[tex]Fr=\dfrac{v}{\sqrt{gd}}\\\Rightarrow Fr=\dfrac{4.44}{\sqrt{9.81\times 0.9}}\\\Rightarrow F_r=1.5[/tex]

Since [tex]F_r>1[/tex] the flow is super critical.

Flow is critical when [tex]Fr=1[/tex]

Depth is given by

[tex]d=(\dfrac{Q^2}{gw^2})^{\dfrac{1}{3}}\\\Rightarrow d=(\dfrac{12^2}{9.81\times 3^2})^{\dfrac{1}{3}}\\\Rightarrow d=1.2\ \text{m}[/tex]

The depth of the channel will be 1.2 m for critical flow.