Steam enters an adiabatic nozzle at 1 MPa, 250°C, and 30 m/s. At one point in the nozzle the enthalpy dropped 40 kJ/kg from its inlet value. Determine velocity at that point. (A) 31 m/s (B) 110 m/s (C) 250 m/s (D) 280 m/s

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.

Answer: its A

Explanation:

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:

Following are the solution to this question:

Explanation:

The number of vacancies by the cubic meter is determined.  

[tex]N_V =N exp(\frac{Q_v}{kT})[/tex]

      [tex]= \frac{N_A \rho}{A} exp (\frac{Q_v}{kT})[/tex]

      [tex]= \frac{6.022 \times 10^{23} \times 6.10}{43.41} \exp(\frac{-0.95}{8.62\times 10^{-5} \times (783+273)})\\\\= \frac{36.7342 \times 10^{23}}{43.41} \exp(\frac{-0.95}{0.0313626})\\\\= 0.846215158 \times 10^{23} \exp(-30.290856)\\\\[/tex]

      [tex]=1.57 \times 10^{25} \ cm^{-3}[/tex]

Answer: the air flow rate a is 90 kg/sec; Option b) 90 kg/sec is the correct answer

Explanation:

Given that;

product of combustor flow rate m = 100 kg/s

air-fuel = 9

Airflow rate = ?

⇒We know that in the combustor, air fuel are mixed and then ignited,

⇒air fuel products are exited at the combustor

let air and fuel be a and b respectively

⇒ a + b = 100 kg/sec ----- let this be equation 1

now

⇒ air / fuel = 9

a / b = 9

a = 9b -----------let this be equation 2

now input a = 9b in equation 1

9b + b = 100 kg/sec

10b = 100 kg/sec

b = 10 kg/sec

we know that

a = 9b

so a = 9 × 10 = 90 kg/sec

Therefore the air flow rate a is 90 kg/sec

Answer: The 8-core machine saves  87.5% of the dynamic power.

Explanation:

Let Fold = f , Vold = V , Cold = Capacitance

so

Old Dynamic power = Cold × (Vold × Vold) × f

therefore for the 8-core machine

 Fnew / Fold = 1/4

Fnew = Fold/4

we were told that Voltage decreases proportional to frequency,

so

Vnew / Vold = 1/4

Vnew = V / 4

So New Capacitance will be;

Cnew = Cold

Thus, New Dynamic power = 8 × Cnew × ( Vnew × Vnew ) ×  Fnew

= 8 × Cold × (Vold × Vold/16) × ( f/4 )

=  8 × ( Cold ) × ( Vold × Vold ) × ( f ) / 64

= (Old Dynamic Power) / 8

therefore

Old Dynamic Power / New Dynamic Power = 8

Thus, Percentage of power saved will be;

Percentage power saved = 100 × ( Old Dynamic Power - New Dynamic Power ) / Old Dynamic Power

=   100 × (8-1) / 8

= 87.5 %

Therefore The 8-core machine saves  87.5% of the dynamic power.

Answer:

The appropriate solution will be "6.4 cm".

Explanation:

The given values are:

Length,

l = 0.2 m

Thermal conductivity,

K₁ = 1.82 W/m-K

K₂ = 0.095 W/m-K

Temperature,

T = 950 K

T = 300 K

Heat transfer rate,

Q = 830 W/m²

Now,

⇒  [tex]Q = \frac{\Delta T}{\frac{L_1}{K_1 A} +\frac{L_2}{K_2 A} }=\frac{A \Delta T}{\frac{L_1}{K_1 } +\frac{L_2}{K_2 } }[/tex]

⇒ [tex]\frac{Q}{A} =\frac{\Delta T}{\frac{L_1}{K_1} +\frac{L_2}{K_2} }[/tex]

On substituting the above given values in the equation, we get

⇒ [tex]830=\frac{(980-300)}{\frac{0.2}{1.82} +\frac{x}{0.095} }[/tex]

On applying cross-multiplication, we get

⇒ [tex]\frac{0.2}{1.82} +\frac{x}{0.095} =\frac{950-300}{830}[/tex]

⇒ [tex]\frac{0.2}{1.82} +\frac{x}{0.095} =\frac{650}{830}[/tex]

⇒                [tex]x =0.639 \ m[/tex]

⇒                [tex]x=6.345 \ i.e., 6.4 \ m[/tex]  

Solution :

Given :

radius of perigee, [tex]$r_p$[/tex] = 10,000 km

radius of apogee, [tex]$r_a$[/tex] = 100,000 km

a). Eccentricity of the orbit

  [tex]$e=\frac{|r_p-r_a|}{r_p+r_a}$[/tex]

[tex]$e=\frac{|10,000-100,000|}{10,000+100,000}$[/tex]

[tex]$e=\frac{9}{11}$[/tex]

or e = 0.818

b). Semi major axis of the orbit

  [tex]$a=\frac{r_p+r_a}{2}$[/tex]

  [tex]$a=\frac{10,000+100,000}{2}$[/tex]

     = 55,000 km

c). period of orbit

  [tex]$T=\frac{2\pi}{\sqrt{\mu}}\times a^{3/2}$[/tex]

Replacing μ with [tex]$398600 \ km^3/s^2$[/tex]

[tex]$T=\frac{2\pi}{\sqrt{398600}}\times (55,000)^{3/2}$[/tex]

[tex]$T=128304.04 \ s \left(\frac{1 \ hr}{3600 \ s}\right)$[/tex]

T = 35.64 hr

d). Specific energy of the orbit

[tex]$\varepsilon = -\frac{\mu}{2a}$[/tex]

[tex]$\varepsilon = -\frac{398600}{2 \times 55000}$[/tex]

[tex]$\varepsilon = -3.62 \ km^2/s^2$[/tex]

e). the equation of the distance to the focus

[tex]$\theta = \cos^{-1}\left(\frac{a(1-e^2)}{r}-\frac{1}{e}\right)$[/tex]

[tex]$\theta = \cos^{-1}\left(\frac{55000(1-(0.818)^2)}{(1000+6378)}-\frac{11}{9}\right)$[/tex]

[tex]$\theta = \cos^{-1}\left(\frac{55000(0.33)}{(7378)}-\frac{11}{9}\right)$[/tex]

[tex]$\theta = \cos^{-1}\left(2.4-1.2\right)$[/tex]

[tex]$\theta = \cos^{-1}\left(1.2\right)$[/tex]

θ = 1.002°

f).Calculating the angular momentum

[tex]$r_p=\frac{h^2}{\mu(1+e)}$[/tex]

or [tex]$h=\sqrt{r_p \mu(1+e)}$[/tex]

Now calculate the radial velocity

[tex]$v_r=\frac{\mu}{h} e \sin \theta$[/tex]

Substituting for h,

[tex]$v_r=\frac{\mu}{h}e \sin \theta$[/tex]

[tex]$v_r=\frac{e\mu \sin \theta}{\sqrt{r_p \mu(1+e)}}$[/tex]

[tex]$v_r=\frac{\frac{9}{11}\sqrt{398600} \sin 20}{\sqrt{10,000 (1+0.818)}}$[/tex]

[tex]$v_r= 1.30 \ km/s$[/tex]

Now calculating the azimuthal velocity

[tex]$v_{\perp}=\frac{\mu}{h}(1+e \cos \theta)$[/tex]

[tex]$v_{\perp}=\frac{\mu (1+e \cos \theta)}{\sqrt{r_p \mu(1+e)}}$[/tex]

[tex]$v_{\perp}=\frac{\sqrt{398600} (1+0.818 \cos 20)}{\sqrt{10000(1+0.818)}}$[/tex]

[tex]$v_{\perp}=7.58 \ km/s$[/tex]

g). Velocity at perigee

[tex]$v_p=\frac{h}{r_p}$[/tex]

[tex]$v_p=\frac{\sqrt{r_p \mu (1+e)}}{r_p}$[/tex]

[tex]$v_p=\frac{\sqrt{10000 (398600) (1+0.818)}}{10000}$[/tex]

[tex]$v_p=8.52 \ km/s$[/tex]

Now calculate the velocity of the apogee

[tex]$v_a=\frac{h}{r_a}$[/tex]

[tex]$v_a=\frac{\sqrt{r_p \mu (1+e)}}{r_a}$[/tex]

[tex]$v_p=\frac{\sqrt{10000 (398600) (1+0.818)}}{100000}$[/tex]

[tex]$v_a= 0.85 \ km/s$[/tex]

Answer:

The answer is "[tex]\bold{W_{in} = 645.434573 \ Btu}[/tex]".

Explanation:

Its air enthalpy is obtained only at a given temperature by A-17E.  

The solution from the carbon cycle is acquired:

[tex]\to \Delta U= W_{in}-m_{out}h_{out}=0\\\\\to W_{in} = (m_1 - m_2)h[/tex]

           [tex]=\frac{Vh}{RT}(P_1- P_2)\\\\= \frac{40 \times 138.66}{0.3704\times 580}(50-25) Btu\\\\= \frac{5,546.4}{214.832}(25) Btu\\\\= 25.8173829(25) Btu\\\\ =645.434573 Btu[/tex]

[tex]W_{in} = 645.434573 \ Btu[/tex]

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:

The answer is "Option A".

Explanation:

Series:

[tex]9, 94, 916, 964, 9256, ........[/tex]

Solving the above series:

[tex]\to 9\\ \to 9(4) =94\\\to 9 (4^2) = 9(16) =916\\\to 9 (4^3) = 9(64) =964\\\to 9 (4^4) = 9(256) =9256\\\to 9 (4^5) = 9(1024) =91024\\\to 9 (4^6) = 9(4096) =94096\\[/tex]

So, the series is:  [tex]9, 94, 916, 964, 9256, 91024, 94096, .................[/tex]

Answer:

a)  ( ∝ ) = 69.6548

b) supply power factor = 0.6709

  displacement power factor = 0.8208

Explanation:

Given data:

Nominal voltage ( E ) = 200-V

resistance (r) = 10 milliohms

constant current ( I )  = 20 amps

Phase ; 3-phase

semi-converter  with 220-v ( line-to-line ) ,  220√2  ( phase voltage )

frequency ; 60 Hz

a) determine the firing angle of thyristors

Vo = E + I*r

    = 200 + 20*10*10^-3

    = 200.2 v

attached below is the remaining part of the solution

firing angle of thyristors for charging process ( ∝ ) = 69.6548

b) determine displacement power factor and supply power factor

attached below is the detailed solution

Displacement power factor ( Dpf ) = cos ( ∝ /2 ) = 0.8208

displacement power factor = g * Dpf

                                             = 0.81747 * 0.8208 = 0.6709

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:

Following are the solution to this question:

Explanation:

To copy 3.0 bits in 50 dollars or run at 50 dollars, it takes just 3.0 bits as well as other bits but masks, and 50 dollars.  

Instead of shifting the $ 50 by 3 bits to 6...3 bits of [tex]\$ \ 50, \$ \ 50,0*0000 0003,[/tex] This procedure instead took place at $53 and $50  

AND [tex]\$ \ 50,\$ \ 50,0*0000 000f[/tex], take 3..0  bits

SLL [tex]\$ \ 50, \$ \ 50,0*0000 0003,[/tex]Shifts the bits to 6..3

O R [tex]\$ \ 53,\$ \ 53,\$ \ 50 ,[/tex] coping to [tex]\$ \ 53[/tex]

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:

b

Explanation:

Answer:

divide then add XD my guy this is easy

Explanation:

The linear convolution is a mathematical operation that finds the output of the linear time-variant system that is given its impulse and linear time-invariant responses.

Learn more about the achieve the linear convolution of a 1.

brainly.com/question/24452045.

Answer:

the motor input power is 19.42 KW

the Reactive power is 17.65 KVAR

Current is 31.56 A

Explanation:

Given that;

V = 480V

h.p = 25 hp

p.f = 0.74 lagging

n_motor = 96%

so output = 25hp

and we know that;

1hp = 746 watt

watt = hp × 1hp

so output in watt = 25 × 746 = 18650 Watt = 18.65 KW

n_motor = (output / input) × 100

96 = 1865 / Input

96Input = 1865

Input = 1865 / 96

Input = 19.42 KW

Therefore the motor input power is 19.42 KW

P = √( 3 × V × I × cos∅)

19.42 = √( 3 ×480 × I × 0.74)

I = 31.56 A

Therefore Current is 31.56 A

Q = √( 3 × V × I × sin∅)

we know that

cos∅ = 0.74

so ∅ = cos⁻¹(0.74) = 42.26

so we substitute

Q = √( 3 × 480 × 31.56 × sin(42.26))

 = 17.65 KVAR

Therefore the Reactive power is 17.65 KVAR

Explanation:

D. B. C. A. E. Is this a good idea

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:

what do you mean by that ? snap points ?

Answer:

C/C++

Explanation:

C/C++

Answer:

the answer would be d its d

Answer:

Pretty sure the answer is "C"

Explanation:

"The media doesn't affect me at all because I'm smart enough to know the difference between right and wrong."

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:

In Crystalline metals or materials, plasticity is examined from the perspective of the motion of linear defects or dislocations within the polymer chains.

Explanation:

When a temperature range below and near the glass transition temperature is reached, there is warping or contortion of structureless or malformed polymers. This warping happens as the polymer chains move over one another.

Unlike elasticity when requires or enables an object to resume its original dimensions, ductility is the quality of an element or material to change form albeit permanently.

Cheers

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:

hmmm.........

Explanation:

Answer:people tend to do this when they are in a different environment they lose something or just have something going on in their life

Explanation: