After adding refrigerant to a nonpressurized container used with a system-dependent recovery
device, the pressure of the container will:
Select one:
A. Increase to the saturation point.
B. Remain at 0 psig.
C. Increase in proportion to the amount of refrigerant added.
D. Decrease in proportion to the refrigerant added.
There are different aspect to system-dependent recovery equipment. The pressure of the container will Increase to the saturation point.
System-dependent is known to be a passive recovery system. It is known as the recovery of refrigerant from a system by using the refrigeration system's internal pressure and/or compressor so as to simulate the recovery process.
System-dependent equipment is not used used with appliances that has more than 15 pounds of refrigerant.
When refrigerant pressure increases, saturation temperature is known to also increases and as pressure decreases, the saturation temperature will also decrease. Saturation is known to have a pressure-temperature relationships for all refrigerants.
Learn more about saturation from
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Explain the advantages of using register indirect addressing mode over direct addressing mode with an 8051 assembly code. Also provide an example where it is inefficient to code using register indirect addressing mode.
Derive the expression for electrical-loading nonlinearity error (percentage) in a rotatory potentiometer in terms of the angular displacement, maximum displacement (stroke), potentiometer element resistance, and load resistance. Plot the percentage error as a function of the fractional displacement for the three cases: RL/RC = 0.1, 1.0, and 10.0
Answer:
The plot for percentage error as a function of fractional displacement ( [tex]\frac{R_{L} }{R_{C} }[/tex]) for the values of 0.1,1.0,10.0 is shown in image attached below.
Explanation:
Electrical loading non linearity error (percentage) is shown below.
[tex]E=\frac{(\frac{v_{o} }{v_{r} }-\frac{Q}{Q_{max} } )}{\frac{Q}{Q_{max} } }[/tex]×[tex]100[/tex]
where Q= displacement of the slider arm
[tex]Q_{max}=[/tex] maximum displacement of a stroke
[tex]\frac{v_{o} }{ v_{r} } =[/tex][tex]\frac{(\frac{Q}{Q_{max} }(\frac{R_{L} }{R_{C} } ) )}{(\frac{R_{L} }{R_{C} } ) +(\frac{Q}{Q_{max} })-(\frac{Q}{Q_{max} })^{2} }[/tex]
here [tex]R_{L}=load resistance[/tex]
[tex]R_{C}=[/tex]total resistance of potentiometer.
Now the nonlinearity error in percentage is
[tex]E=\frac{(\frac{(\frac{Q}{Q_{max} }(\frac{R_{L} }{R_{C} } ) )}{(\frac{R_{L} }{R_{C} } ) +(\frac{Q}{Q_{max} })-(\frac{Q}{Q_{max} })^{2} }-\frac{Q}{Q_{max} } )}{\frac{Q}{Q_{max} } }[/tex]×[tex]100[/tex]
The following attached file shows nonlinear error in percentage as a function of [tex]\frac{R_{L} }{R_{C} }[/tex] displacement with given values 0.1, 1.0, 10.0. The plot is drawn using MATLAB.
The MATLAB code is given below.
clear all ;
clc ;
ratio=0.1 ;
i=0 ;
for zratio=0:0.01:1 ;
i=i+1 ;
tratioa (1,i)=zratio ;
E1(1,i)=((((zratio*ratio)/(ratio+zratio-zratio^2))-zratio)/zrtio)*100 ;
end
ratio=1.0 :
i=0 ;
for zratio=0:0.01:1 ;
i=i+1 ;
tratiob (1,i)=zratio ;
E2(1,i)=((((zratio*ratio)/(ratio+zratio-zratio^2))-zratio)/zratio)*100 ;
end
ratio=10.0 :
i=0 ;
for zratio=0:0.01:1 ;
i=i+1 ;
tratioc (1,i)=zratio ;
E3(1,i)=((((zratio*ratio)/(ratio+zratio-zratio^2))-zratio)/zrtio)*100 ;
end
k=plot(tratioa,E1,tratiob,E2,tratioc,E3)
grid
title({non linear error in % as a function of R_L/R_C})
k(1). line width = 2;
k(1).marker='*'
k(1).color='red'
k(2).linewidth=1;
k(2).marker='d';
k(2).color='m';
k(3).linewidth=0.5;
k(3).marker='h';
k(3).color='b'
legend ('location', 'south east')
legend('R_L/R_C=0.1','R_L/R_C=1.0','R_L/R_C=10.0')
The current flowing into the collector lead of a certain bipolar junction transistor (BJT) is measured to be 1 nA. If no charge was transferred in or out of the collector lead prior to t = 0, and the current flows for 1 min. calculate the total charge which crosses into the collector.
Answer:
the total charge which crosses into the collector is 60 nC
Explanation:
Given the data in the question;
current flowing into the collector lead of the bipolar junction transistor (BJT); i = 1 nA = 10⁻⁹ A
no charge was transferred in or out of the collector lead prior to t = 0
the current flow time t = 1 min = 60 sec
Now we write the relation between current, charge, and time;
i = dq / dt
where i is current, q is charge and t is time. { d refers to change }
Now,
[tex]q=\int\limits^t_{t=0} {i(t)} \, dt[/tex]
[tex]q=\int\limits^{t=60}_{t=0} { (10^{-9}) } \, dt[/tex]
[tex]q = ( 10^{-9}) (t)_0^{60[/tex]
[tex]q = ( 10^{-9}) ( 60 - 0 )[/tex]
q = 60 × 10⁻⁹ C
q = 60 nC
Therefore, the total charge which crosses into the collector is 60 nC
Hot combustion gases, modeled as air behaving as an ideal gas, enter a steady-state turbine at 145 psi, 2700oR, and exit at 29 psi and 1974oR. The power output is 74.26 hp, and heat transfer from the turbine to the surroundings occurs at a rate of 14 BTU/s.
Required:
Determine the power output of the turbine, in hp.
Hot combustion gases, modeled as air behaving as an ideal gas, enter a steady-state turbine at 145 psi, 2700oR, and exit at 29 psi and 1974oR. The power output is 74.26 hp, and heat transfer from the turbine to the surroundings occurs at a rate of 14 BTU/s.
Required:
a)Determine the power output of the turbine, in hp.
b) The Flow rate
Answer:
a) [tex]w=74.26hp[/tex]
b) [tex]m=0.22[/tex]
Explanation:
From the question we are told that:
Initial Pressure [tex]p_1= 145 psi[/tex]
Initial Temperature [tex]T_1 =2700oR=>2240.33^oF[/tex]
Final Pressure [tex]p_2= 29 psi[/tex]
Final Temperature [tex]t_2=1974oR=>1514.33^oF[/tex]
Output Power [tex]w=74.26hp[/tex]
Heat transfer Rate [tex]Q=14BTU/s[/tex]
Generally the equation for Steady flow energy is mathematically given by
[tex]Q-w=m(h_2-h_1)[/tex]
Where
[tex]m=Flow\ rate[/tex]
From Steam table
[tex]h_1=704btu/ib (at\ p_1= 145\ psi,\ T_1 =2700oR=>2240.33^oF )[/tex]
[tex]h_2=401btu/ib (at\ p_2= 29psi\ t_2=1974oR=>1514.33^oF )[/tex]
Therefore
[tex]-14-74.26=m(401-704)[/tex]
[tex]m=\frac{-14-74.26}{(401-704)}[/tex]
[tex]m=0.22[/tex]
A wastewater treatment plant treats 20 MGD of wastewater containing 950 mg/L of suspended solids in a primary clarifier that has a 20% suspended solids removal efficiency. The rate of sludge collection, the flow rate out the bottom of the clarifier, is 0.08 MGD.
a) What is the solids concentration (in mg/L) in the sludge leaving the clarifier?
b) What mass of solids (in kg/y) is removed annually by the primary clarifier?
Answer:
a) 47500 mg/L
b) 5250366.444 kg/year
Explanation:
Given data:
suspended solids removal efficiency = 20%
Flowrate in the primary clarifier ( Q ) = 20 MGD ( change to Liters/day
Q = 20* 10^6 * 3.785412 Liters /day
settled concentration ( St ) = 950mg/L * 0.2 = 190 mg/L
amount of settled solid = Q * St
= ( 20* 10^6 * 3.785412 ) * 190 = 14384.5656 kg/day
∴ Amount going into sludge with a flowrate of 0.08 MGD = 14384.5656 kg/day
a) concentration of solid in sludge ( leaving the clarifier )
= amount of settled solid / flow rate out of the clarifier in liters/day
= 14384.5656 / ( 0.08 * 10^6 * 3.785412 )
= 0.0475 kg/L
= 47500 mg/L
b) Determine mass of solids that is removed annually
= 14384.5656 kg/day * 365 days
= 5250366.444 kg/year
When you double the velocity, you _____ the kinetic energy.
Answer:
you quadruple the kinetic energy
Consider the same piping system. this time, the same pipe is buried underground. assuming that there is a constant heat flux of 100w/m^2 from the outer surfance of the pipe to the soil determine the exit temperature of the water.
a. 129.1
b. 111.1
c. 82.1
d. 68.1
Complete Question
Complete Question is attached below
Answer:
Option A
Explanation:
From the question we are told that:
inner Diameter of pipe [tex]d_i=100^c[/tex]
Thickness [tex]t=50mm[/tex]
Outer diameter of pipe [tex]d_o=1.1m[/tex]
Length [tex]l=5m[/tex]
Temperature [tex]T_i=130^oC[/tex]
Generally the equation for Heat Balance is mathematically given by
[tex]q*\pi d_oL=mC_p(T_i-T_o)[/tex]
Therefore
[tex]T_o=T_i+\frac{q*\pi d_oL}{mC_p}[/tex]
[tex]T_o=130+\frac{100*3.142 *1.1*5}{0.5*4000}[/tex]
[tex]T_o=129.136^oC[/tex]
Therefore the exit temperature of the water.is [tex]T_o=129.136^oC[/tex]
Option A
A 6-hp pump is used to raise water to an elevation of 12 meters. If the mechanical efficiency of the pump is 85 percent, determine the maximum volume flow rate of water in gallons per hour.
Answer:
29481.60 gallons/hour
Explanation:
Pump ( in ) = 6 - hp
Elevation ( h ) = 12 meters
mechanical efficiency of pump ( η pump ) = 85% = 0.85
Calculate for the maximum volume flow rate of water
1 - hp = 745.7 w
η pump = mgh / W * in
mgh = η pump * W * in
i.e. pVgh = η pump * W * in
∴ V = η pump * W * in / pgh
= 0.82 * ( 745.7 * 6 ) / 1000 * 9.81 * 12
= 3668.844 / 117720
= 0.031 m^3 /s = 29481.601 gallons/hour
A spring having a stiffness k is compressed a distance δ. The stored energy in the spring is used to drive a machine which requires power P. Determine how long the spring can supply energy at the required rate.
Units Used: kN = 103 N
Given: k = 5 kN/m; δ = 400 mm; P = 90 W
Answer:
30w
Explanation:
You are hired as the investigators to identify the root cause and describe what should have occurred based on the following information. The mass of jet fuel required to travel from Toronto to Edmonton is 22,300 kg. The fuel gage correctly indicated that the plane already had 7,682 L of jet fuel in the tank. The specific gravity of the jet fuel is 0.803. Using this information, the crew added 4,916 L of fuel and took off, only to run out of fuel and crash a short while later. Use your knowledge of dimensions and units to work out what went wrong.
Answer:
The mass of fuel added, which is 10,166.2 kg is less than 22,300 kg which is the mass of fuel required to travel from Toronto to Edmonton, the plane therefore crashed.
Explanation:
Since density ρ = m/v where m = mass of fuel and v = volume of fuel, we need to find the mass of each volume of fuel.
So, m = ρv now ρ = specific gravity × density of water = 0.803 × 1000 kg/m³ = 803 kg/m³.
To find the mass of the 7,682 L of fuel, its volume is 7,682 dm³ = 7,682 dm³ × 1 m³/1000 dm³ = 7.682 m³.
It's mass, m = 803 kg/m³ × 7.682 m³ = 6168.646 kg
To find the mass of the extra 4,916 L of fuel added, we have
m' = ρv' where v' = 4,916 L = 4,916 dm³ = 4916 dm³ × 1 m³/1000 dm³ = 4.916 m³
m' = 803 kg/m³ × 4.916 m³ = 3947.548 kg
So, the total mass of the fuel is m" = m + m' = 6168.646 kg + 3947.548 kg = 10116.194 kg ≅ 10,166.2 kg
Since this mass of fuel added, which is 10,166.2 kg is less than 22,300 kg which is the mass of fuel required to travel from Toronto to Edmonton, the plane therefore crashed.
A steel plate of width 120mm and thickness of 20mm is bent into a circular arc radius of 10. You are required to calculate the maximum stress induced and the bending moment which will give the maximum stress. You are given that E=2*10^5
Answer:
Hence the magnitude of the pure moment m will be [tex]2\times 10^5.[/tex]
Explanation:
Width of steel fleet = 120 mm The thickness of steel fleet = 10 mm Let the circle of radius = 10 mNow,
We know that,
[tex]\frac{M}{I} = \frac{E}{R}[/tex]
Thus, [tex]M =\frac{EI}{R}[/tex]
Here
R = 10000 mm
[tex]I=\frac{1}{12}\times 120\times 10^{3}\\= 10^{4} mm^{4}[/tex]
[tex]E=2\times 10^{5}n/mm^{2}\\\\E=2\times 10^{5}n/mm^{2}\\\\M={(2\times 10^{5}\times 10^{4})/{10000}}\\\\M=2\times 10^{5}[/tex]
Hence, the magnitude of the pure moment m will be [tex]2\times 10^5.[/tex]
Greek engineers had the unenviable task of moving largecolumns from the quarries to the city.One engineer,Chersiphron, tried several different techniques to do this. Onemethodwas to cut pivot holes into the ends of the stone andthen use oxen to pull the column. The 5-ft diameter columnweighs 14,000 lbs, andthe team of oxen generates a constantpull force of 2,000 lbs on the center of the cylinder,G. Knowingthat the column starts from rest and rolls without slipping,determine:(a) the velocity of its center,G,after it has moved7ft,and (b) the minimum static coefficient of friction that will keepit from slipping.
Answer:
yea
Explanation:
okkudjeheud email and delete the electronic version of this communication
The term _______________refers to the science of using fluids to perform work.
Answer:
Hi, there your answer is hydraulics
Explanation:
Is it permissible to install recessed fixture directly against wood ceiling joists? Explain why or why not.
Answer:
No
Explanation:
It is not permissible to install recessed fixture directly against wood ceiling joists when the label on the fixture does not indicate that it can be used for Insulation in contact with the recessed fixture .
Because The trim finishing and points of support of openings in finished surfaces are not allowed to be in contact with the combustible materials
Find at the terminals of the circuit
lời mở đầu cho môn lí thuyết tài chính tiền tệ
Answer:
No puedo entenderte solo en español
Every time I take a photo, that photo has to be stored in a file somewhere within "My Files" correct?
How would I be able to take that photo out of the file it was stored?
Cut that photo by
1. Left click your mouse on the photo
2. Click cut
Then enter the file where you want to transfer and press
1. ctrl+v
Answer:
you can go to your file and then select the phpto and hold on a little bit and choose the delete option
Which of the following is not a correct statement about a probability select one
It must have a value between 0 and 1
It can be reported as a decimal or a fraction
A value near means that the event is not likely to occur/happens
It is the collection of several experiments
Answer:
but how does that make sense
Answer:
option b is the wrong statement
Step-by-step explanation:
it can be reported as decimal and fraction. probability must have a value between 0 and 1.probabilty zero means it is an impossible event, which is not likely to occur or happen.
4 points
Para transportar una Unidad
Evaporadora Horizontal que mide 5
pies de largo, 3 pies de ancho y 4
pies de alto se necesita un volumen
de: *
23 pies cúbicos
12 pies cúbicos
60 pies cúbicos
O
63 pies cúbicos
Planos de construcción
Answer:
4 points
Para transportar una Unidad
Evaporadora Horizontal que mide 5
pies de largo, 3 pies de ancho y 4
pies de alto se necesita un volumen
de: *
23 pies cúbicos
12 pies cúbicos
60 pies cúbicos
O
63 pies cúbicos
Planos de construcciónExplanation:
The answer to the question is what the
A 2*8 inches wood member has a length =11 ft it's density is =30 Lb/ft^3 find the actual weight of this member in Lb
Answer:
359.046 Ib
Explanation:
density = m / v ---- ( 1 )
m = mass
v = volume = l * b * h
Given data :
density = 30 Ib / ft^3
l = 2 inches = 0.1667 ft
b = 8 inches = 0.6667 ft
h = 11 ft
v ( volume ) = 0.1667 * 0.6667 * 11 = 1.22 ft^3
back to equation 1
m = 30 * 1.22 = 36.6 Ib
Actual weight ( F ) = m * g
= 36.6 * 9.81 = 359.046 Ib.
The inputs of two registers R0 and R1 are controlled by a 2-to-1 multiplexer. The multiplexer select line and the register load enable inputs are controlled by inputs C0 and C1. Only one of the control inputs may be equal to 1 at a time. The required transfers are:
Answer: Hello your question is incomplete attached below is the complete question
answer :
Attached below
Explanation:
Given data:
The inputs of two registers are controlled by a 2-to-1 multiplexer.
The multiplexer select line and the register load enable inputs are controlled by inputs Co, C1, and C2.
Using the required transfers in the question to complete the detailed logic diagrams ( attached below )
A project has an initial cost of $100,000 and uniform annual benefits of $12,500. At the end of its 8-year useful life, its salvage value is $30,000. At a 10% interest rate, the net present worth of the project is approximately:__________
1- $-19,318
2- $0
3- $+30,000
4- $+100,000
Answer:
[tex]X=-\$19318[/tex]
Explanation:
From the question we are told that:
Initial cost [tex]P= $100,000[/tex]
Annual benefits [tex]A= $12,500[/tex]
Salvage value [tex]S= $30,000[/tex]
Interest rate [tex]I=10\%=>0.10[/tex]
Time [tex]t=8years[/tex]
Generally the equation for Net Project worth X is mathematically given by
[tex]X=-P+A+S[/tex]
Where
Present worth of Annual benefits A is
[tex]A'=A(P/a,0.10,8)[/tex]
[tex]A'=12500*5.3349[/tex]
[tex]A'=\$66686.25[/tex]
Present worth of Salvage Price S is
[tex]S'=S(P/a,0.10,8)[/tex]
[tex]S'=30000*0.46651[/tex]
[tex]S'=\$13995.3[/tex]
Therefore
[tex]X=-P+A'+S'[/tex]
[tex]X=-100000+66686.25+\$13995.3[/tex]
[tex]X=-\$19318[/tex]
3. Which of these instruments is used to measure wind speed? A. anemometer C. wind sock B. thermometer D. wind vane It is an instrument that can show both the wind speed and direction. A. Anemometer C. Wind sock B. thermometer D. wind vane 4. 5. At what time does air temperature changes? A. from time to time C. in the evening only B. in the afternoon only D. in the morning only
Answer:
wind vane if it can be used to show wind speed and the other is a
Explanation:
please mark 5 star if im right and brainly when ya can
Consider a single crystal of nickel oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a (111) plane and in a [101]direction and is initiated at an applied tensile stress of 13.9 MPa (2020 psi), compute the critical resolved shear stress
Answer:
[tex]\mathbf{\tau_c =5.675 \ MPa}[/tex]
Explanation:
Given that:
The direction of the applied tensile stress =[001]
direction of the slip plane = [[tex]\bar 1[/tex]01]
normal to the slip plane = [111]
Now, the first thing to do is to calculate the angle between the tensile stress and the slip by using the formula:
[tex]cos \lambda = \Big [\dfrac{d_1d_2+e_1e_2+f_1f_2}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_2^2+e_2^2+f_2^2) }} \Big][/tex]
where;
[tex][d_1\ e_1 \ f_1][/tex] = directional indices for tensile stress
[tex][d_2 \ e_2 \ f_2][/tex] = slip direction
replacing their values;
i.e [tex]d_1[/tex] = 0 ,[tex]e_1[/tex] = 0 [tex]f_1[/tex] = 1 & [tex]d_2[/tex] = -1 , [tex]e_2[/tex] = 0 , [tex]f_2[/tex] = 1
[tex]cos \lambda = \Big [\dfrac{(0\times -1)+(0\times 0) + (1\times 1) }{\sqrt{(0^2+0^2+1^2)+((-1)^2+0^2+1^2) }} \Big][/tex]
[tex]cos \ \lambda = \dfrac{1}{\sqrt{2}}[/tex]
Also, to find the angle [tex]\phi[/tex] between the stress [001] & normal slip plane [111]
Then;
[tex]cos \ \phi = \Big [\dfrac{d_1d_3+e_1e_3+f_1f_3}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_3^2+e_3^2+f_3^2) }} \Big][/tex]
replacing their values;
i.e [tex]d_1[/tex] = 0 ,[tex]e_1[/tex] = 0 [tex]f_1[/tex] = 1 & [tex]d_3[/tex] = 1 , [tex]e_3[/tex] = 1 , [tex]f_3[/tex] = 1
[tex]cos \ \phi= \Big [ \dfrac{ (0 \times 1)+(0 \times 1)+(1 \times 1)} {\sqrt {(0^2+0^2+1^2)+(1^2+1^2 +1^2)} } \Big][/tex]
[tex]cos \phi= \dfrac{1} {\sqrt{3} }[/tex]
However, the critical resolved SS(shear stress) [tex]\mathbf{\tau_c}[/tex] can be computed using the formula:
[tex]\tau_c = (\sigma )(cos \phi )(cos \lambda)[/tex]
where;
applied tensile stress [tex]\sigma =[/tex] 13.9 MPa
∴
[tex]\tau_c =13.9\times ( \dfrac{1}{\sqrt{2}} )( \dfrac{1}{\sqrt{3}})[/tex]
[tex]\mathbf{\tau_c =5.675 \ MPa}[/tex]
Question 3
Marked out of 1.00
Question text
When you sell a car, what do you need to do to tell the state that you are not liable for that car anymore, including parking tickets?
Select one:
a. Call the DMV and tell them you sold the car
b. On the "Title" of the car, fill out the "Release of Liability" form and mail it in as instructed on the form.
c. The person you are selling the car to will take care of it.
d. There is nothing to do; if you do not own the car anymore, it is not your responsibility.
Answer:
D
Explanation:
Your answer is D.There is nothing to do; if you do not own the car anymore, it is not your responsibility
A column carries 5400 pounds of load and is supported on a spread footing. The footing rests on coarse sand. Design the smallest square footing (to the next 3-inch increment) that will safely carry the column load. The footing will be 1 ft 9 in. deep and will be constructed of cast-in-place concrete.
Answer:
Following are the responses to the given question:
Explanation:
The following true stresses produce the corresponding true plastic strains for a brass alloy: True Stress (psi) True Strain 49300 0.11 61300 0.21 What true stress is necessary to produce a true plastic strain of 0.25
Answer:
64640.92 psi
Explanation:
True stress ( psi ) True strain
49300 0.11
61300 0.21
Determine the true stress necessary to produce a true plastic strain of 0.25
бT1 = 49300
бT2 = 61300
бT3 = ?
∈T1 = 0.11
∈T2 = 0.21
∈T3 = 0.25
note : бTi = k ∈Ti^h
∴ 49300 = k ( 0.11 )^h ----- ( 1 )
61300 = k ( 0.21)^h ------ ( 2 )
solving equations 1 and 2 simultaneously
49300/61300 = ( 0.11 / 0.21 )^h
0.804 = (0.52 )^h
next step : apply logarithm
log ( 0.804 ) = log(0.52)^h
h = log 0.804 / log (0.52)
= 0.33
back to equation 1
49300 = k ( 0.11 )^0.33
k = 49300 / (0.11)^0.33
= 102138
therefore бT3 = K (0.25)^h
= 102138 ( 0.25 )^ 0.33
= 64640.92
La fuerza necesaria para girar los birlos de una llanta es de 40 N, si se tiene una cruceta de 0.5m. ¿Cuál es el torque del birlo?
To Find :
The torque of the stud.
Solution :
We know, torque can be calculated by force and distance between the center and force applied.
So,
[tex]\tau = F\times R\\\\\tau = 40 \times 0.5 \ N .\ m\\\\\tau = 20 \ N .\ m[/tex]
Therefore, the torque on the stud is 20 N m .