Ammonia enters the expansion valve of a refrigeration system at a pressure of 10 bar and a temperature of 24 C and exits at 1 bar. If the refrigerant undergoes a throttling process, what is the quality of the refrigerant exiting the expansion valve.

Answer:

a.  164 °F b. 91.11 °C c. 1439.54 kJ

Explanation:

a. [1 pts] How many degrees Fahrenheit (°F) must you raise the temperature?

Since the starting temperature is 48°F and the final temperature which water boils is 212°F, the number of degrees Fahrenheit we would need to raise the temperature is the difference between the final temperature and the initial temperature.

So, Δ°F = 212 °F - 48 °F = 164 °F

b. [2 pts] How many degrees Celsius (°C) must you raise the temperature?

To find the degree change in Celsius, we convert the initial and final temperature to Celsius.

°C = 5(°F - 32)/9

So, 48 °F in Celsius is

°C₁ = 5(48 - 32)/9

°C₁ = 5(16)/9

°C₁ = 80/9

°C₁ = 8.89 °C

Also, 212 °F in Celsius is

°C₂ = 5(212 - 32)/9

°C₂ = 5(180)/9

°C₂ = 5(20)

°C₂ = 100 °C

So, the number of degrees in Celsius you must raise the temperature is the temperature difference between the final and initial temperatures in Celsius.

So, Δ°C = °C₂ - °C₁ = 100 °C - 8.89 °C = 91.11 °C

c. [2 pts] How much energy is required to heat the four quarts of water from

48°F to 212°F (boiling)?

Since we require 15.8 kJ for every degree Celsius of temperature increase of the four quarts of water, that is 15.8 kJ/°C and it rises by 91.11 °C, then the amount of energy Q required is Q = amount of heat per temperature rise × temperature rise =  15.8 kJ/°C × 91.11 °C = 1439.54 kJ

Answer:

first mark me as a brainleast

Answer:

(a) the reversible power output of turbine is 5810 kw

(b) The second-law efficiency of he turbine = 86.05%

Explanation:

In state 1: the steam has a pressure of 6 MPa and 600°C. Obtain the enthalpy and entropy at this state.

h1 = 3658 kJ/kg s1=7.167 kJ/kgK

In state 2: the steam has a pressure of 50 kPa and 100°C. Obtain the enthalpy and entropy at this state

h2 = 2682kl/kg S2= 7.694 kJ/kg

Assuming that the energy balance equation given  

Wout=m [h1-h2+(v1²-v2²) /2]

Let

W =5 MW

V1= 80 m/s  V2= 140 m/s

h1 = 3658kJ/kg  h2 = 2682 kJ/kg

∴5 MW x1000 kW/ 1 MW =m [(3658-2682)+ ((80m/s)²-(140m/s)²)/2](1N /1kg m/ s²) *(1KJ/1000 Nm)

m = 5.158kg/s

Consider the energy balance equation given  

Wrev,out =Wout-mT0(s1-s2)

Substitute Wout =5 MW m = 5.158kg/s 7

s1=  7.167 kJ/kg-K            s2= 7.694kJ/kg-K and 25°C .

Wrev,out=(5 MW x 1000 kW /1 MW) -5.158x(273+25) Kx(7.167-7.694)

= 5810 kW

(a) Therefore, the reversible power output of turbine is 5810 kw.

The given values of quantities were substituted and the reversible power output are calculated.

(b) Calculating the second law efficiency of the turbine:  

η=Wout/W rev,out

Let Wout =  5 MW and Wrev,out = 5810 kW  

η=(5 MW x 1000 kW)/(1 MW *5810)  

η= 86.05%

Answer:

53 sec / cycle

Explanation:

Displayed red time = 35 seconds

Start up lost time = 2 seconds

Yellow time = 4 seconds

Red time = 1 second

Total lost time L = 2n + r

L = lost time

n = number of phase

R = red time

35+2+4+4*1

= 45

L = 2x4+45

= 53 sec/cycle

The total lost time is typically calculated as 53 seconds per cycle

Answer:

a)  Vr = - a^2/r cosθ  + aß / r

    Vθ = 1/r [ -a^2/r * sinθ ]

b) attached below

Explanation:

potential function

Ø= a^2 /r  cosØ + aßlnr ----- ( 1 )

a = radius ,  ß = constant

a) Expressions for Vr and Vθ

Vr =  dØ / dr  ----- ( 2 )

hence expression : Vr = - a^2/r cosθ  + aß / r

Vθ = 1/r dØ / dθ ------ ( 3 )

back to equation 1

dØ / dr = - a^2/r sinθ + 0  --- ( 4 )

Resolving equations 3 and 4

Vθ = 1/r [ -a^2/r * sinθ ]

b) expression for stream function

attached below

Answer:

1.c

2.c

3.c

Explanation:

James is a  pilot, whistle. He is wearing a flight suit. Paris is the palace where does he fly to. Hence, option C, C, and C are correct.

When flying an aircraft, such as a military aircraft, a glider, or a helicopter, one must wear a full-body suit called a flight suit. These outfits are typically meant to keep the user warm and are also functional (they have many of pockets) (including fire ). In most cases, it looks like a jumpsuit.

The G suit, sometimes known as a "anti-G suit," is a one-piece jumpsuit that shields a pilot from the pressure of G forces pressing down on him and causing discomfort or unconsciousness.

The traditional attire for pilots of military and commercial aircraft, helicopters, and even gliders is flight suits or flyers coveralls. In areas where there is a risk of fire, ground personnel—including aircrews—often wear flight suits as well.

Thus, option C, C, and C are correct.

For more information about point of a flight suit, click here:

https://brainly.com/question/12302183

#SPJ2

Answer:

The right answer is "16.5 hrs".

Explanation:

Given values are:

[tex]x_1=2.6 \ mm[/tex]

[tex]t_1=3 \ hrs[/tex]

[tex]x_2=6.1 \ mm[/tex]

As we know,

⇒ [tex]\frac{x^2}{Dt}=constant[/tex]

or,

⇒ [tex]\frac{x_1^2}{t_1} =\frac{x_2^2}{t_2}[/tex]

⇒ [tex]t_2=(\frac{x_2}{x_1})^2\times t_1[/tex]

By putting the values, we get

       [tex]=(\frac{6.1}{2.6} )^2\times 3[/tex]

       [tex]=5.5\times 3[/tex]

       [tex]=16.5 \ hrs[/tex]      

Answer:

there are generally two types of toilet bowl types- round and elongated.

Explanation:

ask your dad please and use your brain

Answer: False

Explanation:

Reinforced concrete is simply a combination of the traditional cement concrete with the steel bars which are the reinforcements.

Reinforced concrete is utilized for construction purpose mostly on a large scale. The reinforced concrete was invented by French gardener Joseph in 1849 therefore, it has always been available and appreciated by architects before the 19th century.

Answer:

a)  Δh = 2 cm,  b) Δh = 0.4 cm

Explanation:

Let's start by using Bernoulli's equation for the Pitot tube, we define two points 1 for the small entry point and point 2 for the larger diameter entry point.

            P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Point 1 is called the stagnation point where the fluid velocity is reduced to zero (v₁ = 0), in general pitot tubes are used  in such a way that the height of point 2 of is the same of point 1

           y₁ = y₂

subtitute

           P₁ = P₂ + ½ ρ v₂²

           P₁ -P₂ = ½ ρ v²

where ρ is the density of fluid  

now we measure the pressure on the included beforehand as a pair of communicating tubes filled with mercury, we set our reference system at the point of the mercury bottom surface

           ΔP =ρ_{Hg} g h - ρ g h

           ΔP =  (ρ_{Hg} - ρ) g h

as the static pressure we can equalize the equations

          ΔP = P₁ - P₂

         (ρ_{Hg} - ρ) g h = ½ ρ v²

         v = [tex]\sqrt{\frac{2 (\rho_{Hg} - \rho) g}{\rho } } \ \sqrt{h}[/tex]

in this expression the densities are constant

        v = A  √h

       A =[tex]\sqrt{\frac{2(\rho_{Hg} - \rho ) g}{\rho } }[/tex]

They indicate the density of mercury rhohg = 13600 kg / m³, the density of dry air at 20ºC is rho air = 1.29 kg/m³

we look for the constant

        A = [tex]\sqrt{\frac{2( 13600 - 1.29) \ 9.8}{1.29} }[/tex]

        A = 454.55

we substitute

       v = 454.55 √h

to calculate the uncertainty or error of the velocity

         h = [tex]\frac{1}{454.55^2} \ v^2[/tex]

       Δh = [tex]\frac{dh}{dv}[/tex]   Δv

       [tex]\frac{\Delta h}{h } = 2 \ \frac{\Delta v}{v}[/tex]

Suppose we have a height reading of h = 20 cm = 0.20 m

a) uncertainty 2.5 m / s ( 0.05)

        [tex]\frac{\delta v}{v} = 0.05[/tex]

       [tex]\frac{\Delta h}{h}[/tex] = 2 0.05  

       Δh = 0.1 h

       Δh = 0.1  20 cm

       Δh = 2 cm

b) uncertainty 0.5 m / s ( Δv/v= 0.01)

        [tex]\frac{\Delta h}{h}[/tex] =  2 0.01

        Δh = 0.02 h

        Δh = 0.02 20

        Δh = 0.1 20 cm

        Δh = 0.4 cm = 4 mm

Answer:

The heat input from the combustion phase is 2000 watts.

Explanation:

The energy efficiency of the heat engine ([tex]\eta[/tex]), no unit, is defined by this formula:

[tex]\eta = \frac{\dot W}{\dot Q}[/tex] (1)

Where:

[tex]\dot Q[/tex] - Heat input, in watts.

[tex]\dot W[/tex] - Power output, in watts.

If we know that [tex]\dot W = 600\,W[/tex] and [tex]\eta = 0.3[/tex], then the heat input from the combustion phase is:

[tex]\eta = \frac{\dot W}{\dot Q}[/tex]

[tex]\dot Q = \frac{\dot W}{\eta}[/tex]

[tex]\dot Q = \frac{600\,W}{0.3}[/tex]

[tex]\dot Q = 2000\,W[/tex]

The heat input from the combustion phase is 2000 watts.

Answer:

8.6 ft³/s

Explanation:

The force due to the water jet F = mv where m = mass flow rate = ρQ where ρ = density of water = 62.4 lbm/ft³ and Q = volume flow rate. v = velocity of water jet = 30 ft/s

So, F = mv

F = ρQv

making Q subject of the formula, we have

Q = F/ρv

Since F = force due to water jet = force needed to hold the plate against the water stream = 500 lbf = 500 × 1 lbf = 500 × 32.2 lbmft/s² = 16100 lbmft/s²

Since

Q = F/ρv

Substituting the values of the variables into the equation for Q, we have

Q = F/ρv

Q = 16100 lbmft/s²/(62.4 lbm/ft³ × 30 ft/s)

Q = 16100 lbmft/s²/1872 lbm/ft²s

Q = 8.6 ft³/s

So, the volume flow rate is 8.6 ft³/s.

Answer:

(i) 0

(iv) 8a+6b+c-70

Explanation:

Hope this helps you

Answer:

The explanation according to the given query is summarized in the explanation segment below.

Explanation:

Answer:

Explanation:

#include<iostream>

using namespace std;

int main()

{

int n1,n2;

cout<<"Enter a number:"<<endl; //Entering first number

cin>>n1;

cout<<"Enter a number:"<<endl; //Entering second number

cin>>n2;

if(n1%2==0 && n1%2==0) //Checking whether the two number are even or not

{

if(n1>n2)

{

cout<<n1<<" is the larger number"<<endl;

}

else if(n1==n2)

{

cout<<"Numbers are equal"<<endl;

}

else

{

cout<<n2<<" is the larger number"<<endl;

}

}

else

{

cout<<"The number are not even"<<endl;

}

}

1.5 nautical miles per 1,000 feet

Solution :

Given :

[tex]$H(S) =\frac{2S-2}{S^2+\left(\frac{10}{3}\right) S+1}$[/tex]

Transfer function, [tex]$H(S) =\frac{Y(S)}{K(S)}= \frac{2S-2}{S^2+\left(\frac{10}{3}\right) S+1}$[/tex]

[tex]$Y(S) \left(S^2+\frac{10}{3}S+1\right) = (2S-2) \times (S)$[/tex]

[tex]$S^2Y(S) + \frac{10}{3}(SY(S)) + Y(S) = 2(S \times (S)) - 2 \times (S)$[/tex]

Apply Inverse Laplace Transforms,

[tex]$\frac{d^2y(t)}{dt^2} + \frac{10}{3} \frac{dy(t)}{dt} + y(t)=2 \frac{dx(t)}{dt} - 2x(t)$[/tex]

The above equation represents the differential equation of transfer function.

Given : [tex]$x(t)=e^{-t} u(t) \Rightarrow X(S) = \frac{1}{S+1}$[/tex]

We have : [tex]$H(S) =\frac{Y(S)}{K(S)}= \frac{2S-2}{S^2+\left(\frac{10}{3}\right) S+1}$[/tex]

[tex]$Y(S) = X(S) \times \frac{6S-6}{3S^2+10 S + 3} = \frac{6S-6}{(S+1)(3S+1)(S+3)}$[/tex]

[tex]$Y(S) = \frac{A}{S+1}+\frac{B}{3S+1} + \frac{C}{S+3}[/tex]

[tex]$A = Lt_{S \to -1} (S+1)Y(S)=\frac{6S-6}{(3S+1)(S+3)} = \frac{-6-6}{(-3+1)(-1+3)} = 3$[/tex]

[tex]$B = Lt_{S \to -1/3} (3S+1)Y(S)=\frac{6S-6}{(S+1)(S+3)} = \frac{-6/3-6}{(1/3+1)(-1/3+3)} = \frac{-9}{2}$[/tex]

[tex]$C = Lt_{S \to -3} (S+3)Y(S)=\frac{6S-6}{(S+1)(3S+1)} = \frac{-18-6}{(-3+1)(-9+1)} = \frac{-3}{2}$[/tex]

So,

[tex]$Y(S) = \frac{3}{S+1} - \frac{9/2}{3S+1} - \frac{3/2}{S+3}$[/tex]

        [tex]$=\frac{3}{S+1} - \frac{3/2}{S+1/3} - \frac{3/2}{S+3}$[/tex]

Applying Inverse Laplace Transform,

[tex]$y(t) = 3e^{-t}u(t)-\frac{3}{2}e^{-t/3}u(t) - \frac{3}{2}e^{-3t} u(t)$[/tex]

       [tex]$=\frac{-3}{2}e^{-\frac{1}{3}t}u(t) + \frac{3}{1}e^{-t}u(t)-\frac{3}{2}e^{-3t} u(t)$[/tex]

where, a = 2

            b = 1

            c= 2

Answer: the attached picture is the answer.

Explanation:

Assuming:

the switch position connect to 1, hence 4.5V exist at across lamp1

the switch position connects to 2 hence 4.5 V exist across lamp 1 and lamp 2

the switch position connects to 3, hence, 4.5 V exist across lamp 1, lamp 2 and lamp 3.

Answer:

D. Turn on the light source and the spectrophotometer.

Explanation:

A spectrophotometer is a machine used to measure the presence of any light-absorbing particle in a solution as well as its concentration. To prepare a spectrophotometer for use, the first step is to turn on the spectrophotometer and allow it to warm up for at least 15 minutes. After this is done, the next step will be to ensure that the samples and blank are ready. Next, an appropriate wavelength is set for the solute being determined. Finally, the absorbance is measured of both the blank and samples.

Answer:

1709.07 ft^3/s

Explanation:

Annual peak streamflow = Log10(Q [ft^3/s] )

mean = 1.835

standard deviation = 0.65

Probability of levee been overtopped in the next 15 years = 1/5

Determine the design flow ins ft^3/s

P₁₅ = 1 - ( q )^15 = 1 - ( 1 - 1/T )^15 = 0.2

                         ∴  T = 67.72 years

Q₁₅ = 1 - 0.2 = 0.8

Applying Lognormal distribution : Zt = mean + ( K₂ * std ) --- ( 1 )

K₂ = 2.054 + ( 67.72 - 50 ) / ( 100 - 50 ) * ( 2.326 - 2.054 )

    = 2.1504

back to equation 1

Zt = 1.835 + ( 2.1504 * 0.65 )  = 3.23276

hence:

Log₁₀ ( Qt(ft^3/s) ) = Zt  = 3.23276

hence ; Qt = 10^3.23276

                  = 1709.07 ft^3/s

Answer:

To become ASE certified, you must pass an ASE test and have relevant hands-on work experience. The amount of work experience required can vary by test, and is specified in detail here. ASE recommends submitting the form after you've registered to take an ASE certification test.

Good luck!

Explanation:

Answer:  One theme in White Fang is adapting in order to survive. White Fang finally submits to Gray Beaver. He also copes with fighting other dogs. White Fang changes his behaviors so that he can live.

Explanation: its the sample response

Answer:

The physical topology addresses how devices are connected, while a logical topology addresses how devices actually communicate to one another ( C )

Explanation:

Network physical is simply the process/method of connecting the Network using cables while Logical topology is the general architecture of the communication mechanism in the network for all nodes.

Hence The correct statement is the physical topology addresses how devices are connected, while a logical topology addresses how devices actually communicate to one another

Answer:

≈ 0.516 m^3/hr

Explanation:

Inlet of compressor = 200 SCMH

sheer standard conditions = 1 atm and 288.5 K

For oxygen :

critical pressure(Pc) = 49.8 atm

critical temperature Tc = 154.6 K

hence at compressor inlet

Tr = T / Tc = 288.5/154.6 = 1.866

Pr = P / Pc = 1 / 49.8 = 0.0204

Z1 ( from compressibility chart ) = 0.98

at compressor outlet

P2 = 500 bar = 500*0.9869 = 493.45 atm  , T2 = 360 k

hence : Pr = P / Pc = 493.45 / 49.8 = 9.91

            Tr = T / Tc = 360 / 154.6 = 2.33

Z2 ( from compressibility chart ) ≈ 1

V2( volumetric flow rate ) = V1*(P₁Z₂T₂) / (P₂Z₁T₁)

                                         = 200 ( 1 * 1* 360) / (493.45 *0.98*288.5)

                                         = 0.516 m^3/hr

Answer: Attached below is the missing detail and Mohr's circle.

i) б1 =  9.6 Ksi

б2 = -10.7 ksi

ii) 10.2 Ksi

iii)  -0.51Ksi

Explanation:

First step :

direct compressive stress on shaft

бd = P / π/4 * d^2

      = -20 / 0.785 * 5^2  = -1.09 Ksi

shear stress at the outer surface due to torsion

ζ = 16*T / πd^3

  = (16 * 250 ) / π * 5^3  = 010.19 Ksi

Calculate the Principal stress, maximum in-plane shear stress and average normal stress

Using Mohr's circle ( attached below )

i) principal stresses:

б1 = 4.8 cm * 2 = 9.6 Ksi

б2 = -5.35 cm * 2 = -10.7 ksi

ii) maximum in-plane shear stress

ζ  = radius of Mohr's circle

   = 5.1 cm = 10.2 Ksi   ( Given that ; 1 cm = 2Ksi )

iii) average normal stress

 = 9.6 + ( - 10.7 ) / 2

  = -0.51Ksi

Answer:

The pseudocode is as follows:

Input SalesAmount

Input CommissionRate

CommissionEarned= SalesAmount * (CommissionRate/100)

Print CommissionEarned

Explanation:

This gets input for SalesAmount

Input SalesAmount

This gets input for CommissionRate

Input CommissionRate

This calculates the CommissionEarned

CommissionEarned= SalesAmount * (CommissionRate/100)

This prints the calculated CommissionEarned

Print CommissionEarned

Answer:

The Space Needle is a cut away with minimal residual deflection due to load transfer.

Answer:

Veracity, Value and Visualization are not only the characteristics of Big Data but are also the characteristics of relational databases. Veracity of data is issue with smallest data stores this is the reason that it is important in relation...

Answer:

can you please give us the option for this question