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:
[tex]h_{1} = h_2} = 293.45 KJ/kg[/tex].
The quality of the refrigerant exiting the expansion valve is
[tex]x_{2}=0.193596[/tex].
Explanation:
Fluid given Ammonia.
Inlet 1:-
Temperature [tex]T_{1}[/tex] = [tex]24^{o} C[/tex].
Pressure [tex]P_{1}[/tex] = 10 bar.
Exit 2:-
Pressure [tex]P_{2}[/tex] = 1 bar.
Solution:-
HELP PLEASE!! ASAP!!!!
can some answer this 2 questions please as paragraph i want it nowww it is graded what action should be taken to make it safe ? also the first question
Actions violated:
Long hair isn't tied upThe girl isn't wearing a lab coatThe girl isn't wearing safety gogglesExtra: There doesn't seem to be an emergency fire blanket in the safeActions to be taken:
Make sure the girl wears a lab coat or kick her outMake sure the girl wears safety goggles or kick her outMake sure her hair is tied up or kick her outEdit: Use these to write your paragraph.
Do you know who Candice is
Answer: Can these nuts fit in your mouth?
Explanation:
im just here for the points >:)
The system is initially moving with the cable taut, the 15-kg block moving down the rough incline with a speed of 0.080 m/s, and the spring stretched 39 mm. By the method of this article, (a) determine the velocity v of the block after it has traveled 99 mm, and (b) calculate the distance d traveled by the block before it comes to rest.
Solution :
The spring is expanded by 2 times of the block when it moves down an inclined by x times.
Here, [tex]$x_1$[/tex] = 39 mm
[tex]x_2[/tex] = 225 mm
a). From the work energy principal,
Work forces = kinetic energy
[tex]$(mg \sin 50^\circ)\times \frac{99}{1000}-(\mu_k mg \cos 50^\circ) \times \frac{99}{1000} -\frac{1}{2}k(0.225^2 - 0.039^2)=\frac{1}{2}m(V^2_2-0.08^2)$[/tex]
[tex]$(112.6 \times 0.099)-(14.17 \times 0.099)-4.91= 7.5(V^2_2-0.08^2)$[/tex]
[tex]$9.75= 7.5(V^2_2-0.08^2)$[/tex]
[tex]$1.3= V^2_2-0.08^2$[/tex]
[tex]$V_2=1.14\ m/s$[/tex]
b). calculating the distance travelled by the block before it comes to rest.
Substitute the value of [tex]V_2[/tex] in (1),
[tex]$-(\mu_kmg \cos 50^\circ)x + (mg \sin 50^\circ)x-\frac{1}{2}k\left( ( 2x+0.039)^2 - 0.039^2\right)= -\frac{1}{2}m(0.08)^2$[/tex]
[tex]$-14.17x+112.6x - 100(4x^2+0.156x)=-0.048$[/tex]
[tex]$98.43x - 100(4x^2+0.156x)+0.048=0$[/tex]
[tex]$98.43x - 400x^2-15.6x+0.048=0$[/tex]
[tex]$82.83x - 400x^2+0.048=0$[/tex]
[tex]$ 400x^2- 82.83x-0.048=0$[/tex]
x = 0.20 m
If you are driving down to the sleep downgrade and you have reached the speed of 40 mph , you would apply the setrvice break until your speed dropped to below_____mph.
Answer:
35 miles
Explanation:
When you are driving a truck that has an air brake system you have to keep in mind that when driving down a steep downgrade, the truck will automatically accelerate due to the inclination of the road, so in order to keep the speed to a controllable situation, you need to activate the service brake until you've reached the 35 miles per hour mark.
An ideal neon sign transformer provides 9130 V at 51.0 mA with an input voltage of 240 V. Calculate the transformer's input power and current.
Answer:
Input power = 465.63 W
current = 1.94 A
Explanation:
we have the following data to answer this question
V = 9130
i = 0.051
the input power = VI
I = 51.0 mA = 0.051
= 9130 * 0.051
= 465.63 watts
the current = 465.63/240
= 1.94A
therefore the input power is 465.63 wwatts
while the current is 1.94A
the input power is the same thing as the output power.
(50 POINTS) How many people use pipes in the world? How do you know this?
Answer:
7.9 billion people
Explanation:
what are qualifications that are required to be an architect
Answer:
Bachelor's degree
In order to become a licensed architect in the US and the District of Columbia, applicants are required to complete a professional degree in architecture, gain on-the-job experience through a paid internship.
Blocks A and B each have a mass m. Determine the largest horizontal force P which can be applied to B so that A will not move relative to B. All surfaces are smooth.
Answer:
The answer is "15 N".
Explanation:
Please find the complete question in the attached file.
In frame B:
For just slipping:
[tex]\to \frac{P}{2} \cos \theta =mg \sin \theta\\\\\to P=2 mg \tan \theta \\\\[/tex]
[tex]=2 \times 1 \times g \times \tan 37^{\circ}\\\\ =2 \times 10 \times \frac{3}{4}\\\\ =15 \ N[/tex]
A designer needs to select the material for a plate under tensile stress. Assuming that the applied tensile force is 13,000 lb and the area under the stress is 4 square inches, determine which material should be selected to assure safety. Assume safety factor is 2. Material A: Ultimate Tensile stress is 8000 lb/in2Material B: Ultimate Tensile stress is 5500 lb/in2
If there is a discrepancy between Chick-fil-A food safety requirements and local Health Department
regulations, what should Team Member do?
The following should be done by the team member:
It is important to follow both Chick-fil-A food safety requirements and local Health Department regulations. In the case when there is a discrepancy between the two, always follow the more stringent requirement. Any other appearance or grooming issue not covered in these materials may be addressed at the discretion of the Operator.
Learn more: brainly.com/question/17429689
Activity 1. Fill the blank with the correct answer. Write your answer on the blank. 1. ___________________ is a regular pattern of dots displayed on the screen which acts as a visual aid and also used to define the extent of your drawing. 2. Ortho is short for ___________________, which means either vertical or horizontal. 3. Tangent is a point where two _______________________ meet at just a single point. 4. If you want to create a new drawing, simply press ___________________ for the short cut key. 5. There are _______________Osnap that can help you performs your task easier.
Answer:
1. Drawing grid.
2. Orthogonal.
3. Geometries.
4. CTRL+N.
5. Thirteen (13).
Explanation:
CAD is an acronym for computer aided design and it is typically used for designing the graphical representation of a building plan. An example of a computer aided design (CAD) software is AutoCAD.
Some of the features of an AutoCAD software are;
1. Drawing grid: is a regular pattern of dots displayed on the screen of an AutoCAD software, which acts as a visual aid and it's also used to define the extent of a drawing.
2. Ortho is short or an abbreviation for orthogonal, which means either vertical or horizontal.
3. Tangent is a point where two geometries meet at just a single point.
4. If you want to create a new drawing, simply press CTRL+N for the short cut key.
5. There are thirteen object snaps (Osnap) that can help you perform your task on AutoCAD easily. The 13 object snaps (Osnap) are; Endpoint, Midpoint, Apparent intersect, Intersection, Quadrant, Extension, Tangent, Center, Insert, Perpendicular, Node, Parallel, and Nearest.
Steam at 4 MPa and 350°C is expanded in an adiabatic turbine to 125kPa. What is the isentropic efficiency (percent) of this turbine if the steam is exhausted as a saturated vapor?
Answer:
[tex]\eta_{turbine} = 0.603 = 60.3\%[/tex]
Explanation:
First, we will find actual properties at given inlet and outlet states by the use of steam tables:
AT INLET:
At 4MPa and 350°C, from the superheated table:
h₁ = 3093.3 KJ/kg
s₁ = 6.5843 KJ/kg.K
AT OUTLET:
At P₂ = 125 KPa and steam is saturated in vapor state:
h₂ = [tex]h_{g\ at\ 125KPa}[/tex] = 2684.9 KJ/kg
Now, for the isentropic enthalpy, we have:
P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K
Since s₂ is less than [tex]s_g[/tex] and greater than [tex]s_f[/tex] at 125 KPa. Therefore, the steam is in a saturated mixture state. So:
[tex]x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88[/tex]
Now, we will find [tex]h_{2s}[/tex](enthalpy at the outlet for the isentropic process):
[tex]h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg[/tex]
Now, the isentropic efficiency of the turbine can be given as follows:
[tex]\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%[/tex]
In an international film festival, a penal of 11 judges is formed to judge the best film. At
last two films FA and FB were considered to be the best where the opinion of judges got
divided. Six judges where in favor of FA whereas five in favor of FB. A random sample
of five judges was drawn from the panel. Find the probability that out of five judges,
three are in favor of film FA.Enunciate demerits of classical probability.
Answer:
International Film Festival
Judging the best best film:
a. The probability that out of five judges (random sample), three are in favor of film FA is:
= 33%.
b. The demerits of classical probability are:
1. Classical probability can only be used with events that have definite numbers of possible outcomes.
2. Classical probability can only handle events where each outcome is equally likely.
3. Classical probability is based on the assumption of linear relationship (which is not always true in real life) between the latent variable and observed scores.
Explanation:
a) Number of judges = 11
Number of judges in favor of FA film = 6
Number of judges in favor of FB film = 5
Probability of judges in favor of FA film = 6/11
Probability of judges in favor of FB film = 5/11
Random sample of judges = 5
Probability that out of five judges, three are in favor of film FA = 3/5 * 6/11
= 18/55
= 33%
b) Classical probability is the simple probability showing that each event has equal chance of happening. It can be contrasted with empirical probability that is obtained from experiments.
Find the general solution of the given differential equation. Give the largest interval over which the general solution is defined. Determine whether there are any transient terms in the general solution.
x dx/dy−y=x^2sinx
Answer:
Interval: x∈ ( 0, ∞ )
There are no transient terms
Explanation:
x (dy/dx) – y= x^2sinx
Attached below is the detailed solution of the Given problem
There are no transient terms found in the general solution
Interval: x∈ ( 0, ∞ )
Represent each of the following units as a combination of primitive
dimensions where M=mass, L=length, T=time. As an example, miles per hour would
correspond to [L/T].
a. kilometer
b. quart
c. pascal
d. watt
e. newton
f. horsepower
Answer:
a. unit of length: [L]
b. unit of volume: [[tex]L^3[/tex]]
c. unit of pressure:[tex]P=\frac{F}{A} \equiv\frac{[MLT^{-2}]}{[L^2]}[/tex] [tex][ML^{-1}T^{-2}][/tex]
d. unit of power: [tex]N.m.s^{-1}\equiv [ML^2T^{-3}][/tex]
e. unit of force: [tex][kg.m/s^2]\equiv [MLT^{-2}][/tex]
f. unit of power: [tex]N.m.s^{-1}\equiv [ML^2T^{-3}][/tex]
Force: [tex]F=m.a=m.\frac{v}{t}=m.\frac{x}{t}\div t[/tex]
Power: [tex]P=\frac{W}{t}=\frac{F.x}{t}[/tex]
where:
F = force
A = area
W = work
t = time
a = acceleration
v = velocity
x = displacement
ow Pass Filter Design 0.0/5.0 points (graded) Determine the transfer function H(s) for a low pass filter with the following characteristics: a cutoff frequency of 100 kHz a stopband attenuation rate of 40 dB/decade. a nominal passband gain of 20 dB, which drops to 14 dB at the cutoff frequency Write the formula for H(s) that satisfies these requirements:
Answer:
H(s) = 20 / [ 1 + s / 10^5 ]^2
Explanation:
Given data:
cutoff frequency = 100 kHz
stopband attenuation rate = 40 dB/decade
nominal passband gain = 20 dB
new nominal passband gain at cutoff = 14 dB
Represent the transfer function H(s)
The attenuation rate show that there are two(2) poles
H(s) = k / [ 1 + s/Wc ]^2 ----- ( 1 )
where : Wc = 100 kHz = 10^5 Hz , K = 20 log k = 20 dB ∴ k = 20
Input values into equation 1
H(s) = 20 / [ 1 + s / 10^5 ]^2
Explain by Research how a basic generator works ? using diagram
Calculate density, specific weight and weight of one litter of petrol having specific gravity 0.7
Explanation:
mass=19kg
density=800kg/m³
volume=?
as we know that
density=mass/volume
density×volume=mass
volume=mass/density
putting the values
volume=19kg/800kg/m³
so volume=0.02375≈0.02m³
Set the leak rate to zero and choose a non-zero value for the proportional feedback gain.Restart the simulation and turn on the outflow valve.What happens to the liquid level in the tank?Repeat this process with higher and lower values for the proportional feedback gain.What happens when the proportional feedback gain is increased?What happens when it is decreased?Find the proportional gain that will reach steady state the quickest without oscillationin the state of the valve and restart the simulation.What is the system time constant, as determined from the tank level versus time plot.
Answer:
Explanation:
The proportional gain K is usually a fixed property of the controller . If proportional gain is increased , The sensitivity of the controller to error is increased but the stability is impaired. The system approaches the behaviour of on off controlled system and it response become oscillatory
On a two-way roadway with a center lane, drivers from either direction can _________ from the center lane.
Water steam enters a turbine at a temperature of 400 o C and a pressure of 3 MPa. Water saturated vapor exhausts from the turbine at a pressure of 125 kPa. At steady state, the work output of the turbine is 530 kJ/kg. The surrounding air is at 20 o C. Neglect the changes in kinetic energy and potential energy. Determine (20 points) (a) the heat transfer from the turbine to the surroundings per unit mass flow rate, (b) the entropy generation during this process.
Answer:
a) -505.229 kJ/Kg
b) -1.724 kJ/kg
Explanation:
T1 = 400°C
P1 = 3 MPa
P2 = 125 kPa
work output = 530 kJ/kg
surrounding temperature = 20°C = 293 k
A) Calculate heat transfer from Turbine to surroundings
Q = h2 + w - h1
h ( enthalpy )
h1 = 3231.229 kj/kg
enthalpy at P2
h2 = hg = 2676 kj/kg
back to equation 1
Q = 2676 + 50 - 3231.229 = -505.229 kJ/Kg ( i.e. heat is lost )
b) Entropy generation
entropy generation = Δs ( surrounding ) + Δs(system)
= - 505.229 / 293 + 0
= -1.724 kJ/kg
Hans, a scheduler at Mangel-Wurzel Transport, gets a call from a regular customer needing to move of rock and soil, which Hans knows from previous experience has an average density of . Hans has available a dump truck with a capacity of and a maximum safe load of .Calculate the number of trips the dump truck will have to make to haul the customer's load away.
Complete Question:
Hans, a scheduler at Mangel-Wurzel Transport, gets a call from a regular customer needing to move 19.8m³ of rock and soil, which Hans knows from previous experience has an average density of 650/kgm³. Hans has available a dump truck with a capacity of 4m³ and a maximum safe load of 3700kg. Required: calculate the number of trips the dump truck will have to make to haul the customer's load away.
Answer:
Mangel-Wurzel Transport
The number of trips that the dump truck will have to make to haul the customer's load away is:
= 5 trips.
Explanation:
a) Data and Calculations:
Volume of customer's load (rock and soil) = 19.8m³
Density of load = 650 kg/m³
Mass of load = Volume of load * Density of load
= 19.8m³ × 650 kg/m³
= 12,870 kg
The maximum safe load (mass) of the dump truck = 3,700 kg
Volume of the dump truck = 4m³
Assuming the truck is to carry 4m³ of the load.
The mass of load that the 4m³ capacity truck can carry = 4m³ × 650kg/m³
= 2,600kg
Quick Check:
Mass = 2,600kg < 3,700 kg, satisfying required conditions.
The number of trips that the truck would make to haul the customer's load away is, therefore, calculated as follows:
Number of trips = N
N = total volume of load/ volume per trip
N = 19.8/4
N = 4.95
N = 5 trips approx.
The following is a correlation for the average Nusselt number for natural convection over spherical surface. As can be seen in the above, the Nusselt number approaches 2 as Rayleigh number approaches zero. Prove that this situation corresponds to conduction heat transfer and in conduction heat transfer over sphere, the Nusselt number becomes 2. Hint: First step: Write an expression for heat transfer between two spherical shells that share the same center. Second step: Assume the outer spherical shell is infinitely large.
Answer:
Explanation:
[tex]r_2=[/tex]∞
[tex]q=4\pi kT_1(T_2-T_1)\\[/tex]
[tex]q=2\pi kD.[/tex]ΔT--------(1)
[tex]q=hA[/tex] ΔT[tex]=4\pi r_1^2(T_2_s-T_1_s)\\[/tex]
[tex]N_u=\frac{hD}{k} = 2+\frac{0.589 R_a^\frac{1}{4} }{[1+(\frac{0.046}{p_r}\frac{9}{16} )^\frac{4}{9} } ------(3)[/tex]
By equation (1) and (2)
[tex]2\pi kD.[/tex]ΔT=h.4[tex]\pi r_1^2[/tex]ΔT
[tex]2kD=hD^2\\\frac{hD}{k} =2\\N_u=\frac{hD}{k}=2\\[/tex]-------(4)
From equation (3) and (4)
So for sphere [tex]R_a[/tex]→0
Technician A says that a graphing multi-meter may be used to verify signals going to and from electrical and electronic components. Technician B says that digital storage oscilloscope may be used to verify signals going to and from electrical and electronic components. Who is correct
Answer:
Both are correct.
Explanation:
Graphing multi meter is used to verify signals that move from electrical components. Digital oscilloscope is an equipment which stores and analyzes input signals with digital technique.
1. A manufacturing cell with two workers is responsible for producing a small frying pan with a required takt time of 496 seconds. The material passes through two processes: a deep drawing process and a trimming process. The average cycle time for the deep drawing process is 450 seconds and average cycle time for trimming is 430 seconds. (2 pts.)
a. Does the work cell have adequate capacity to meet demand? Explain.
b. What is the required daily production capacity of the work cell (in number of frying pans per day)? Assume 480 minutes/workday of available time.
2. What is the total daily idle time for both workers in Problem 1? Report your answer in (a) seconds of idle time and (b) as a percentage of total working time for the cell. (2 pts.)
Answer:
Explanation:
[tex]496=\frac{480\times 60}{demand}[/tex]
demand per day = 58 pans
Due to availability of two workers we can have parallel we can have deep drawing and trimming operations simultaneously.
Hence the cycle time would be the greater time of the two operations.
cycle time = 450 seconds
[tex]\text{capacity of work cell}=\frac{\text{available working time}}{\text{cycle time}}[/tex]
[tex]\text{capacity of work cell}=\frac{480\times 60}{450}[/tex]
[tex]\text{capacity of work cell}=64 ~pans[/tex] (which is greater than the demand of 58 pans)
Therefore the work cell has sufficient capacity and time (496 sec.>cycle time 450 sec) to meet the demand.
b)
Required daily production is 58 pans
g Steel plates (AISI 1010) of 4 cm thickness initially at a uniform temperature of 500 deg C are cooled by air at 50 deg C with a convection coefficient of 30 W-m2-K-1. Estimate the time it will take for their midplane temperature to reach 100 deg C.
Solution :
Characteristic length = thickness / 2
[tex]$=\frac{0.04}{2}$[/tex]
= 0.02 m
Thermal conductivity for steel is 42.5 W/m.K
[tex]$\text{Biot number} = \frac{\text{convective heat transfer coefficient} \times \text{characteristic length}}{\text{thermal conductivity}}$[/tex]
[tex]$=\frac{30 \times 0.02}{42.5}$[/tex]
= 0.014
Since the Biot number is less than 0.01, the lumped system analysis is applicable.
[tex]$\frac{T-T_{\infty}}{T_0-T_{\infty}} = e^{-b\times t}$[/tex]
Where,
T = temperature after t time
[tex]$T_{\infty}$[/tex] = surrounding temperature
[tex]$T_0$[/tex] = initial temperature
[tex]$b=\frac{\text{heat transfer coefficient}}{\text{density} \times {\text{specific heat } \times \text{characteristic length }}}$[/tex]
t = time
We calculate B:
[tex]$b=\frac{30}{7833 \times 460 \times 0.02}$[/tex]
= 0.000416
Thus, [tex]$\frac{100-50}{500-50}=e^{-0.00416 \times t}$[/tex]
t = 5281.78 second
= 88.02 minutes
Thus the time taken for reaching 100 degree Celsius is 88.02 minutes.
How do your arm muscles work to lift a mug of coffee to your lips?
A.
The biceps muscle contracts while the triceps muscle relaxes. Only then can the forearm move up to lift the mug.
B.
The muscle tendons in your arms stretch to their limit as a result of the flexibility exercises that lift the mug.
C.
Your muscles rapidly increase the speed at which they twitch to trigger your arm movement.
D.
Your biceps and triceps muscles first relax, and then the arm muscle raises the mug.
Answer:
A. The biceps muscle contracts while the triceps muscle relaxes. Only then can the forearm move up to lift the mug.
Explanation:
Skeletal muscle is also known as voluntary muscle and it can be defined as a type of muscle connected with the skeleton by tendons, so as to form a mechanical system that enables the movement of the limbs and other body parts with respect to another.
Generally, skeletal muscles are only found in vertebrates.
The muscles in the arm work to lift a mug of coffee to your lips when the biceps muscle contracts while the triceps muscle relaxes.
Basically, the only way the forearm move up to lift a mug for example, is through the contraction of the biceps muscle and the relaxation of the triceps muscle.
This ultimately implies that, the elbow joint is straightened or bent due to the actions of the biceps muscle and triceps muscle against each other.
To bend the elbow and raise the forearm, the biceps muscle contracts and the triceps muscle relaxes.To straighten the elbow and lower the forearm, the biceps muscle relaxes and the triceps muscle contracts.Answer:A
Explanation:
got it right on plato
Determine the transfer function H(s) for a high pass filter with the following characteristics:
1. a cutoff frequency of 100 kHz
2. a stopband attenuation rate of 40 dB/decade
3. a nominal passband gain of 20 dB, which drops to 14 dB at the cutoff frequency
Write the formula for H(s) that satisfies these requirements.
Answer:
H(s) = 10 / [ 1 + s / (200*10^3π ) ]^2
Explanation:
Characteristics of the high pass filter
Cutoff frequency = 100 kHz
stopband attenuation rate = 40 dB/decade
nominal passband gain = 20dB = 20logK = 20
Formula for H(s) satisfying the requirements above
given that the stopband attenuation = 40 dB/decade the formula for H(s) that will satisfy the requirements is a second order filter
H(s) = K / ( 1 + s/Wo ) ^2 ----- ( 1 )
Wo = 2πf = 2π ( 100 * 10^3 ) = 200 * 10^3 π
K = 10
back to equation ( 1 )
H(s) = 10 / [ 1 + s / (200*10^3π ) ]^2
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Add a pair of radio buttons to your form, each nested in its own label element.
One should have the option of car and the other should have the option of bike.
Both should share the name attribute of “vehicle” to create a radio group
Make sure the radio buttons are nested with the form
Make sure that the name attributes appear after the type
Answer:
The code is as follows:
<form name = "myForm">
<div>
<input type="radio" name="vehicle" value="D0" id="D0"/>
<label for="D0">Car</label>
</div>
<div>
<input type="radio" name="vehicle" value="D1" id="D1"/>
<label for="D1">Bike</label>
</div>
</form>
Explanation:
This defines the first button
<input type="radio" name="vehicle" value="D0" id="D0"/>
<label for="D0">Car</label>
This defines the second button
<input type="radio" name="vehicle" value="D1" id="D1"/>
<label for="D1">Bike</label>
The code is self-explanatory, as it follows all the required details in the question