An n- channel enhancement- mode MOSFET with 50 nm thick HfO2 high- k gate dielectric (Pr = 25) has a flat band voltage of 0.5 V, and substrate doping of 1018 cm-3. The intrinsic carrier concentration is 1011 cm-3, effective electron channel mobility is 250 cm2/Vs, and Pr = 15. What is the drive current for a 50 om wide and 2 om long device at VG = 3 V and VD = 0.05 V? What is the saturation current at this gate bias?
Answers
Answer:
Find the complete solution in the given attachments
Related Questions
Water vapor at 100 psi, 500 F and a velocity of 100 ft./sec enters a nozzle operating at steady sate and expands adiabatically to the exit, where the pressure is 40 psia. If the isentropic nozzle efficiency is 95%, determine for the nozzle,
(a) the exit velocity of the steam in ft./sec, and
(b) the amount of entropy produced in BTU/ lbm R.
Answers
Answer:
a)exit velocity of the steam, V2 = 2016.8 ft/s
b) the amount of entropy produced is 0.006 Btu/Ibm.R
Explanation:
Given:
P1 = 100 psi
V1 = 100 ft./sec
T1 = 500f
P2 = 40 psi
n = 95% = 0.95
a) for nozzle:
Let's apply steady gas equation.
[tex] h_1 + \frac{(v_1) ^2}{2} = h_2 + \frac{(v_2)^2}{2} [/tex]
h1 and h2 = inlet and exit enthalpy respectively.
At T1 = 500f and P1 = 100 psi,
h1 = 1278.8 Btu/Ibm
s1 = 1.708 Btu/Ibm.R
At P2 = 40psi and s1 = 1.708 Btu/Ibm.R
1193.5 Btu/Ibm
Let's find the actual h2 using the formula :
[tex] n = \frac{h_1 - h_2*}{h_1 - h_2} [/tex]
[tex] n = \frac{1278.8 - h_2*}{1278.8 - 1193.5} [/tex]
solving for h2, we have
[tex] h_2 = 1197.77 Btu/Ibm [/tex]
Take Btu/Ibm = 25037 ft²/s²
Using the first equation, exit velocity of the steam =
[tex] (1278.8 * 25037) + \frac{(100)^2}{2}= (1197.77*25037)+ \frac{(V_2)^2}{2}[/tex]
Solving for V2, we have
V2 = 2016.8 ft/s
b) The amount of entropy produced in BTU/ lbm R will be calculated using :
Δs = s2 - s1
Where s1 = 1.708 Btu/Ibm.R
At h2 = 1197.77 Btu/Ibm and P2 =40 psi,
S2 = 1.714 Btu/Ibm.R
Therefore, amount of entropy produced will be:
Δs = 1.714Btu/Ibm.R - 1.708Btu/Ibm.R
= 0.006 Btu/Ibm.R
Derive the stress strain relationship for each model A and model B. (5 points) b) Is model B equivalent to model A (same stress strain general equation)? If your answer is yes, express E1, E2, η1, and η2 in terms of E 0 1 , E 0 2 , η 0 1 , and η 0 2 . (5 points) c) If applicable, repeat the derivation in part (b) and express E 0 1 , E 0 2 , η 0 1 , and η 0 2 in terms of E1, E2, η1, and η2. (5 points) d) Derive the expression for stress relaxation for both these models and compare the expressions. How are they similar? How are they different? (15 points)
Answers
Answer:
See attached images
Consider schedules S3, S4, and S5 below. Determine whether each schedule is
strict, cascadeless, recoverable, or nonrecoverable. (Determine the strictest
recoverability condition that each schedule satisfies.)
S3: r1 (X); r2 (Z); r1 (Z); r3 (X); r3 (Y); w1 (X); c1; w3 (Y); c3; r2 (Y); w2 (Z);
w2 (Y); c2;
S4: r1 (X); r2 (Z); r1 (Z); r3 (X); r3 (Y); w1 (X); w3 (Y); r2 (Y); w2 (Z); w2 (Y); c1;
c2; c3;
S5: r1 (X); r2 (Z); r3 (X); r1 (Z); r2 (Y); r3 (Y); w1 (X); c1; w2 (Z); w3 (Y); w2 (Y);
c3; c2;
Answers
Answer:
A schedule is strict if it satisfies the following conditions:
Tj reads a data item X after Ti has written to X and Ti is terminated means aborted or committed.
Tj writes a data item X after Ti has written to X and Ti is terminated means aborted or committed.
Explanation:
See attached image
Refrigerant 22 flows in a theoretical single-stage compression refrigeration cycle with a mass flow rate of 0.05 kg/s. The condensing temperature is 48 oc, and the evaporating temperature is —16 oc. If the power input to the cycle is 2.5 kW, determine: (a) the work done by the compressor in kJ/kg, (b) the heat rejected from the condenser in kJ/kg, (c) the heat absorbed by the evaporator in kJ/kg, (d) the coefficient of performance, and (e) the refrigerating efficiency.
Answers
Answer:
a. The work done by the compressor is 447.81 Kj/kg
b. The heat rejected from the condenser in kJ/kg is 187.3 kJ/kg
c. The heat absorbed by the evaporator in kJ/kg is 397.81 Kj/Kg
d. The coefficient of performance is 2.746
e. The refrigerating efficiency is 71.14%
Explanation:
According to the given data we would need first the conversion of temperaturte from C to K as follows:
Temperature at evaporator inlet= Te=-16+273=257 K
Temperatue at condenser exit=Te=48+273=321 K
Enthalpy at evaporator inlet of Te -16=i3=397.81 Kj/Kg
Enthalpy at evaporator exit of Te 48=i1=260.51 Kj/Kg
b. To calculate the the heat rejected from the condenser in kJ/kg we would need to calculate the Enthalpy at the compressor exit by using the compressor equation as follows:
w=i4-i3
W/M=i4-i3
i4=W/M + i3
i4=2.5/0.05 + 397.81
i4=447.81 Kj/kg
a. Enthalpy at the compressor exit=447.81 Kj/kg
Therefore, the heat rejected from the condenser in kJ/kg=i4-i1
the heat rejected from the condenser in kJ/kg=447.81-260.51
the heat rejected from the condenser in kJ/kg=187.3 kJ/kg
c. Temperature at evaporator inlet= Te=-16+273=257 K
The heat absorbed by the evaporator in kJ/kg is Enthalpy at evaporator inlet of Te -16=i3=397.81 Kj/Kg
d. To calculate the coefficient of performance we use the following formula:
coefficient of performance=Refrigerating effect/Energy input
coefficient of performance=137.3/50
coefficient of performance=2.746
the coefficient of performance is 2.746
e. The refrigerating efficiency = COP/COPc
COPc=Te/(Tc-Te)
COPc=255/(321-255)
COPc=3.86
refrigerating efficiency=2.746/3.86
refrigerating efficiency=0.7114=71.14%
The journals in a high speed oil engine are 80 mm in diameter. and 40 mm long. The radial clearance is 0.060mm. Each supports a load of 9 kN when the shaft is rotating at 3600 rpm. The bearing is lubricated with SAE 40 oil supplied at atmospheric pressure and average operating temperature is about 65oC. Using Raimondi- Boyd charts analyze the bearing under steady state operation.
Answers
Answer: S = 0.284
Explanation:
Data: d = 80 mm;
l =40 mm;
c = 0.06 mm;
F = 9kN;
n = 3600rpm = 60 rps
SAE 40 oil
T= 65°C
Therefore:
p = F / ld
= 9 x1000 /40 x 80
= 2.813 MPa
μ = 30 cp at 65°C for SAE 40 oil
S = r^2 x μ x n / c^2 x p
S= ( 40 )^2 x 30*10^-3 x 60 / (0.06)^2 x 2.813*10^6
S = 2880 / 10,126.6
S = 0.284
l/d = ½,
h o /c = 0.38
ε = e /c = 0.62
h o = 0.38 x C
= 0.382 x 0.06
=0.023mm
= 23µm
e = 0.62 x C
= 0.62 x 0.06
= 0.037 mm
Viscosity temperature curves of SAE graded oils
(r /c) f = 7.5,
S = 0.284
l /d = ½
f = 7.5 x (c / r)
= 7.5x (0.06/40)
= 0.0113
Logical variables: Running late? Complete the tunction RunningLate such that the logical variable on Time is true if no Traffic is true and gasEmpty is false. Ex >> n°Traffic = true ; >> gasEmpty-false; >onTime - Runninglate(noTraffic,gasEmpty) logical Your Function B Save C Reset EE MATLAB Documentation 1 function onTime - RunningLate (noTraffic,gasEmpty) 2 % complete the logical expression to the right of . using the variables n°Traffic and gasEmpty 4 onTime - 6 end Code to call your function C Reset 1 noTraffic true; gasEmpty true; 2 onTime-Runninglate(noTraffic, gasEmpty)
Answers
The function RunningLate returns true for the variable onTime when noTraffic is true and gasEmpty is false, which is achieved by the logical expression noTraffic && ~gasEmpty in MATLAB.
Explanation:The task is to complete the function RunningLate in MATLAB, which returns a logical variable onTime that is true if noTraffic is true and gasEmpty is false. The finished MATLAB code inside the function should be:
onTime = noTraffic && ~gasEmpty;This code uses the logical AND operator (&&) to check that there is no traffic, and the NOT operator (~) to check that the gas tank is not empty.
A phase angle in the frequency domain corresponds to
a. the initial slope of the sinusoidal signal.
b. the angle at which measurements should be taken for the most accurate results.
c. a delay or advance in time as compared to a pure cosine wave.
d. the angle of the approach vector of the voltage.
Answers
Answer:
c. a delay or advance in time as compared to a pure cosine wave.
Explanation:
Electrical phase is measured in degrees, with 360° corresponding to a complete cycle. A sinusoidal voltage is proportional to the cosine or sine of the phase. Phase difference , also called phase angle , in degrees is conventionally defined as a number greater than -180, and less than or equal to +180.
The phase angle corresponds to delay or advance in time as compared to a pure cosine wave.
3-For the problems in this exercise, assume that there are no pipeline stalls and that the breakdown of executed instructions is as follows: a. In what fraction of all cycles is the data memory used b. In what fraction of all cycles is the input of the sign-extend circuit needed? c. What is this circuit doing in cycles in which its input is not needed?
Answers
Answer:
(a) 35% (b) 80% (c) Control signal is sent to a resource for activation of it's usage.
The operations are performed even in areas where it's not needed, it is ignored, because it is not used in cycles.
Explanation:
Solution
The computation of fraction is defined below:
(a) The data memory used in lw and sw instructions
Now,
The fraction value is = sw + lw
=10 + 25
= 35
therefore the fraction value is 35%
(b) The needed sign is extended for all other instructions other than ADD
which is shown below
The Fraction value = addi + beq + lw + sw
=20 +25+25+ 10 =80 %
The fraction value here is 80%
(c) Now, a control signal is been sent to resource for activation of it's usage.
The operations are carried out even in case where it's not needed, it is ignored, because it is not used in cycles.
Note: The complete question to this exercise is attached below.
The fraction of all cycles that is the data memory used is 35%.
The fraction of all cycles that is the input of the sign-extend circuit needed is 80%.The circuit is known to be sending a resource for activation of it's usage and it is also done in areas of the circuit where it's not needed and so it is left that way.What is a circuit ?A circuit is known to be a kind of closed loop via which electricity often pass through.
To solve for (a) which is the data memory that has been used in lw and sw instructions, you then add the value together:
That is: fraction value is = sw + lw
=10 + 25
So question A fraction value is 35%
To solve for me the required sign-extend circuit is depicted as:
The Fraction value = addi + beq + lw + sw
=20 +25+25+ 10 =80 %
The question (b) fraction value here is 80%
So therefore, the fraction value are 35 percent and 80 percent respectively
Learn more about circuit from
https://brainly.com/question/2969220
A specific internal combustion engine has a displacement volume VD of 5.6 liters. The processes within each cylinder of the engine are modeled as a cold air-standard Diesel cycle with a cutoff ratio rc = 2.5. The pressure, temperature, and volume of the air at the beginning of compression are p1 = 95 kPa, T1 = 26◦C, and V1 = 6.0 liters. Use values of cv = 0.72 kJ/kg·K and γ = 1.38 for air. Determine: (a) the net work per cycle, in kJ. (b) the thermal efficiency η
Answers
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
Derive an expression for the axial thrust exerted by a propeller if the thrust depends only on forward speed, angular speed, size, and viscosity and density ofthe fluid. How would the expression change if gravity were a relevant variable in the case of a ship propeller?
Answers
Answer:
Find the given attachment
An air standard cycle with constant specific heats is executed in a closed pistoncylinder system and is composed of the following four processes: 1-2 Isentropic compression 2-3 Constant volume heat addition 3-4 Isentropic expansion with a volume ratio, r1=V4/V3 4-1 Constant pressure heat rejection with a volume ratio, r2=V4/V1 (a) Sketch the P-v and T-s diagrams for this cycle. (b)Find out T2/T1 as a function of k, r1, r2 only. (c) Find out T4/T1 as a function of k, r1, r2 only. (d)Find out T3/T4 as a function of k, r1, r2 only. (e) Find out T3/T2 as a function of k, r1, r2 only. (f) Obtain an expression for the back work ratio for a fixed minimum-tomaximum temperature ratio T1/T3. The expression should be of a function of T1/T3, k, r1, r2 only. (g)Obtain an expression for the cycle thermal efficiency as a function of k, r1, r2 only.
Answers
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to get the step by step explanation to the question above.
Two production methods are being compared. One manual and the other automated. The manual method produces 10 pc per hour and requires one worker at $ 15.00 per hour. Fixed cost of the manual method is $ 5,000 per year. The automated method produces 25 pc per hour, has a fixed cost of $ 55,000 per year, and a variable cost of $ 4.50 per hour. Determine the Break – Even Point (BEP) for the two methods; that is, determine the annual production quantity at which the two methods have the same annual cost. Ignore the cost of material used in the two methods
Answers
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
The break-even point is approximately 27,778 pieces per year for both manual and automated methods.
To determine the break-even point (BEP) for the two production methods, we need to equate the total annual costs of the manual method with the total annual costs of the automated method. Let's denote:
- [tex]\( Q \)[/tex] as the annual production quantity (in pieces)
- [tex]\( C_{\text{manual}} \)[/tex] as the total annual cost of the manual method
- [tex]\( C_{\text{automated}} \)[/tex] as the total annual cost of the automated method
For the manual method:
[tex]\[ C_{\text{manual}} = \text{Fixed cost} + (\text{Hourly wage} \times \text{Hours per year}) \][/tex]
[tex]\[ C_{\text{manual}} = 5000 + (15 \times Q/10 \times 24 \times 365) \][/tex]
For the automated method:
[tex]\[ C_{\text{automated}} = \text{Fixed cost} + (\text{Variable cost per hour} \times \text{Hours per year}) \][/tex]
[tex]\[ C_{\text{automated}} = 55000 + (4.50 \times Q/25 \times 24 \times 365) \][/tex]
To find the break-even point, we set [tex]\( C_{\text{manual}} = C_{\text{automated}} \) and solve for \( Q \):[/tex]
[tex]\[ 5000 + (15 \times Q/10 \times 24 \times 365) = 55000 + (4.50 \times Q/25 \times 24 \times 365) \][/tex]
Let's solve this equation for [tex]\( Q \)[/tex]:
[tex]\[ 5000 + 54 \times Q = 55000 + 52.20 \times Q \][/tex]
[tex]\[ 54 \times Q - 52.20 \times Q = 55000 - 5000 \][/tex]
[tex]\[ 1.80 \times Q = 50000 \][/tex]
[tex]\[ Q = \frac{50000}{1.80} \][/tex]
[tex]\[ Q \approx 27777.78 \][/tex]
So, the break-even point for the two methods is approximately 27778 pieces per year.
Water vapor at 5 bar, 320°C enters a turbine operating at steady state with a volumetric flow rate of 1.825 m3/s and expands adiabatically to an exit state of 1 bar, 200°C. Kinetic and potential energy effects are negligible. Determine for the turbine: (a) the power developed, in kW. (b) the rate of entropy production, in kW/K. (c) the percent isentropic turbine efficiency.
Answers
Answer: (a).power developed = 776.1 kW
(b). Rate of entropy production = 1.023 kW/K
(c). efficiency = 63%
Explanation:
Let us carry a step by step process to solve this problem;
from the question we have that
P₁ = 5 bar
T₁ = 320°C
where V₁ = 0.5416 m³/Kg, S₁ = 7.5308 KJ/Kg-K and R₁ = 0.3105.6 KJ/Kg
the volumetric flow rate is given as (φ) = 1.825 m³/s
Remember that φ = ṁ V
where ṁ is the mass flowrate, and V is the volume
ṁ = φ/V = 1.825/0.5416 = 3.37 Kg/s
Also given for the Exit state;
P₂ = 1 bar
T₂ = 200°C
where V₂ = 0.5416 m³/Kg, S₂ = 7.5308 KJ/Kg-K and R₂ = 0.3105.6 KJ/Kg
(a). we are asked to determine the power developed in the Kw.
using the Flow energy equation to turbine we have;
ṁ(R₁ + V₁²/2 + gZ₁) + φ = ṁ(R₂ + V₂²/2 + gZ₂₂) + ш
canceling out terms from both steps we have that
ш = 3.37 (3105-2815.3) = 776.1 kW
Therefore the Power output is 776.1 kW
(b). The rate of entropy production in Kw/K.
Rate(en) = ṁ (S₂-S₁) = 3.37 (7.8343 - 7.5308)
Rate(en) = 1.023 kW/K
(c). The percent isentropic turbine efficiency.
Πt = (R₁-R₂) / (h₁ - h₂s)
Πt = (3105.6 - 2875.3) / (3105.6 - 2740) = 63%
Πt = 63%
cheers i hope this helped!!!!!
Consider the following grooves, each of width W, that have been machined from a solid block of material. (a) For each case obtain an expression for the view factor of the groove with respect to the surroundings outside the groove. (b) For the V groove, obtain an expression for the view factor F12, where A1 and A2 are opposite surfaces. (c) If H
Answers
The heat transfer while the reference information pertains to thermal expansion and physics-related work. Thermal expansion affects the volume, cross-sectional area, and height of objects, and these changes can be calculated using the coefficient of thermal expansion, initial dimensions, and temperature change.
Explanation:The view factor calculations in heat transfer, specifically related to grooves machined from a solid block of material. While the initial question seems to relate to this topic, the provided reference information does not align with the question and seems to cover thermal expansion and work done by forces, which are different aspects of Physics.
However, to answer the student's question regarding thermal expansion, we can consider that when temperature changes, all dimensions of an object change. The volume change ΔV can be calculated using the formula ΔV = α·V·ΔT, where α is the coefficient of thermal expansion, V is the original volume, and ΔT is the temperature change. For block A with volume L·2L·L and block B with volume 2L·2L·2L, the change in volume will be proportional to each block's respective original volume.
The change in cross-sectional area, typically lw for block A and 2Lw for block B, and change in height h or 2L for each block, will be affected in a similar manner and can be calculated using α, the original dimensions, and the temperature change ΔT.
Insulated Gas Turbine Air enters an adiabatic gas turbine at 1050 K and 1 MPa and leaves at 400 kPa. Kinetic and potential energy changes can be ignored. Treat the air as an ideal gas with constant specific heats. Let k = 1.4 a. Determine the theoretical exit temperature - corresponding to part "b" below, the maximum theoretical power output. b. Determine the maximum theoretical work output for the gas turbine in kJ/kg. c. If the isentropic turbine efficiency is 0.8, what is the actual work output of the turbine in kJ/kg?
Answers
Answer:A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.
A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.
Explanation:A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.
A cylinder of length L would be made to carry a torque T with an angle of twist ɸ. There are two options considered: 1) hollowed cylinder with an inner radius that is equal to 0.9 of the outer radius, and 2) solid cylinder (with a different radius). If both options would be made from the same material and must have the same angle of twist ɸ under the torque T, find the ratio of the weight between the cylinder designed for option 1 and option 2.
Answers
Answer:
See explaination
Explanation:
To compare the hollow and solid cylinder we need ro use torsional formula.
And since same material and length are given.
For same torque and angle of twist there will be same polar moment of area of the section for both the cylinder.
Please kindly check attachment for further solution
g Two Standard 1/2" B18.8.2 dowel pins are to be installed in part B with an LN1 fit. The thickness of plate A is .750 +/- .005" The thickness of plate B is .750 +/- .005" The position tolerance of the clearance holes to one another is .014" The position tolerance of the precision holes to one another is .028" What is an appropriate MMC clearance hole diameter to allow the blocks to assemble?
Answers
Answer:
nmuda mudaf A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.
Explanation:
g A food department is kept at -12oC by a refrigerator in an environment at 30oC. The total heat gain to the food department is estimated to be 3300 kJ/h, which should be transferred out of the food department by the refrigerator. The heat rejection from the refrigerator to the environment is 4800 kJ/h. Determine the power input required by the refrigerator, in kW and the COP of the refrigerator. Is the refrigeration cycle reversible, irreversible, or impossible
Answers
Answer:
a) [tex]\dot W = 0.417\,kW[/tex], b) [tex]COP_{R} = 2.198[/tex], c) Irreversible.
Explanation:
a) The power input required by the refrigerator is:
[tex]\dot W = \dot Q_{H} - \dot Q_{L}[/tex]
[tex]\dot W = \left(4800\,\frac{kJ}{h} - 3300\,\frac{kJ}{h}\right)\cdot \left(\frac{1}{3600} \,\frac{h}{s} \right)[/tex]
[tex]\dot W = 0.417\,kW[/tex]
b) The Coefficient of Performance of the refrigerator is:
[tex]COP_{R} = \frac{\dot Q_{L}}{\dot W}[/tex]
[tex]COP_{R} = \frac{3300\,\frac{kJ}{h} }{(0.417\,kW)\cdot \left(3600\,\frac{s}{h} \right)}[/tex]
[tex]COP_{R} = 2.198[/tex]
c) The maximum ideal Coefficient of Performance of the refrigeration is given by the inverse Carnot's Cycle:
[tex]COP_{R,ideal} = \frac{T_{L}}{T_{H}-T_{L}}[/tex]
[tex]COP_{R,ideal} = \frac{261.15\,K}{303.15\,K - 261.15\,K}[/tex]
[tex]COP_{R,ideal} = 6.218[/tex]
The refrigeration cycle is irreversible, as [tex]COP_{R} < COP_{R,ideal}[/tex].
Diborane is used in silicon chip manufacture. One facility uses a 500-lb bottle. If the entire bottle is released continuously during a 20-min period, determine the location of the 5 mg/m3 ground-level isopleth. It is a clear, sunny day with a 5 mph wind. Assume that the release is at ground level. Assume now that the bottle ruptures and that the entire contents of diborane are released instantaneously. Determine, at 10 min after the release,
Answers
Answer:
We want to determine the location after 10mins
Explanation:
The release of diborane is continuous
It is release at a rate of 500lb per 20mins
Then, let find the rate in mg/s
1 lb = 453592.37 mg
So, the mass rate Q is
Q = 500lb / 20mins
Q = 500 × 453592.37 mg / 20 × 60sec
Q = 188,996.82 mg/s
Given that mass concentration of
m~ = 5mg/m³
Then,
Rate of volume is
V~ = Q / m~
V~ = 188,996.82 / 5
V~ = 37,799.364 m³/s
The wind speed is
V = 5mph
Let convert to m/s
1 mph = 0.447 m/s
Then, 5mph = 2.235 m/s
From Pasquill Gilford, the cloud atmosphere characteristic is class A
To know the location, we will divide the velocity by heat rate
X = V / Q
X = 2.235 / 188,996.82 mg/s
X = 2.235 / 0.188996kg/s
X = 11.83 m / kg
The location is 0.000001183 m per mg of diborane
Paint can shaker mechanisms are common in paint and hardware stores. While they do a good job of mixing the paint, they are also noisy and transmit their vibrations to the shelves and counters. A better design of the paint can shaker is possible using a balanced fourbar linkage. Design such a portable device to sit on the floor (not bolted down) and minimize the shaking forces and vibrations while still effectively mixing the paint.
Answers
Answer:
A good design for a portable device to mix paint minimizing the shaking forces and vibrations while still effectively mixing the paint. Is:
The best design is one with centripetal movement. Instead of vertical or horizontal movement. With a container and system of holding structures made of materials that could absorb the vibration effectively.
Explanation:
First of all centripetal movement would be friendlier to our objective as it would not shake the can or the machine itself with disruptive vibrations. Also, we would have to use materials with a good grade of force absorption to eradicate the transmission of the movement to the rest of the structure. Allowing the reduction of the shaking forces while maintaining it effective in the process of mixing.
Water at 15°C is to be heated to 65°C by passing it over a bundle of 7-m-long, 1-cm-diameter resistance heater rods maintained at 90°C. Water approaches the heater rod bundle in normal direction at a mean velocity of 0.8 m/s. The rods are arranged in-line with longitudinal and transverse pitches of SL = 4 cm and ST = 3 cm. Determine the number of tube rows NL in the flow direction needed to achieve the indicated temperature rise.
Answers
Answer:
NL = 207
Explanation:
Solution
Now,
The mean temperature is measured as:
Tm = (T₁ - T₀)/2
= (15 + 65)/2
= 40°C
So,
we find all the thermo-physical properties of water from the table, that is properties of saturated water at T =40°C
Thermo conductivity, k = 0.631 W/m . K
Specific heat Cp = 4179 J/kg . K
Density р = 992. 1 kg/m³
The dynamic viscosity, μ = 0.653 * 10 ^⁻3 kg/m *s
Prandtl number, Pr = 4.32
At T = 15°C
рi = 992.1 kg/m³
At T = 90°C
Prandtl number, Prs = 1.96
Thus,
The maximum flow of velocity is known from the equation stated as:
Vmax = ST/ST - D *V
Here,
ST is refereed to as the transverse pitch for inline arrangements of the rods
so,
Vmax = 3/3-1 * 0.8
= 1.2 m/s
Now
The Reynolds number is determined from the equation given below
ReD =ρVmax D/μ
= 922.1 * 1.2 *(1 *10^⁻²)/ 0.653 * 10^⁻³
= 18231.55
From the table, The Nusselt number correlations fro cross flow over the tube banks for inline arrangement over the range of ReD is shown as
1000 - 2 * 10⁵
Now, the Nusselt number is determined by
NuD = 0.27ReD ^0.63 Pr^ 0.36 (Pr/Prs)^0.25
= 0.27 * (18231.55)^0.63 (4.32)^0.36 * (4.32/1.96)^0.25
=269.32
Then,
The convective transfer of heat water coefficient is determined from the equation shown by Diametral Nusselt Number
NuD =hD/k
So,
we re-write and solve for h
h = NuD * k/D
=269.32 * 0.631/(1 * 10 ^⁻2)
=16993.9 W/m² .K
Now,
The heat transfer surface area for a tube in a row is NT = 1
As = NT NLπDL
= 1*NL* π * (1 * 10^⁻2) * 4
= 0.1257NL
The logarithmic mean temperature of water is represented as
ΔTlm = Te - Ti/ln (Ts - Ti/Ts- Te)
= 65- 15/ln (90 -15/ 90 -65) = 45.51°C
Thus,
The rate of the heat transfer is determined from the equation shown below,
Q =hAsΔTlm
=16993.9 *0.1257 * NL* 45.51.......equation (1)
The mas flow rate of water is determined by the equation below
m =ρiAcV
= ρi * (STL) * V
= 999.1 8 ( 3* 10^⁻2 * 4) * 0.8
= 95.91 kg/s
The rate of heat transfer of water is determined by the equation below
Q = mcp (Te- Ti)
= 95.91 * 4179 * (65-15)
=20041146.72 W..........(Equation 2)
Now,
The number of tube rows in the direction flow is determined by measuring both equations 1 and 2 as
97219.61 NL = 20041146.72
NL =206.14
NL = 207
Therefore, the number of tube rows NL in the flow direction needed to achieve the indicated temperature rise is NL = 207
=
Convert 580,000 kW to [GW] with correct number of significant digits.
Answers
Answer:
[tex]x = 0.58\,GW[/tex]
Explanation:
The conversion is:
[tex]x = (580000\,kW)\cdot \left(\frac{1\,GW}{1000000\,kW} \right)[/tex]
[tex]x = 0.58\,GW[/tex]
For the following circuit, V"#$=120∠30ºV.Redraw the circuit in your solution.a.(4) Calculate the total input impedance seen by the source. Express in rectangular form.b.(3) Calculate the input phasor current(express answer in polar form).c.(6) Using the voltage division, calculate the phasor voltages across each component. Express final answers in polar form.d.(6) Using current divider, calculate phasor currents through L1, and C1.Show all steps. Express final answers in polar form
Answers
Answer:
Check the explanation
Explanation:
Kindly check the attached images for the step by step explanation to the question
Consider casting a concrete driveway 40 feet long, 12 feet wide and 6 in. thick. The proportions of the mix by weight are given as 1:2:3 for cement, sand, and gravel with w/c of 0.50. The specific gravities of sand and gravel are 2.60 and 2.70 respectively. Entrained air content is 7.5%. How many pounds of cement, water, sand, and gravel are needed for the driveway?
Answers
Answer:
Weight of cement = 10968 lb
Weight of sand = 18105.9 lb
Weight of gravel = 28203.55 lb
Weight of water = 5484 lb
Explanation:
Given:
Entrained air = 7.5%
Length, L = 40 ft
Width,w = 12 ft
thickness,b= 6 inch, convert to ft = 6/12 = 0.5 ft
Specific gravity of sand = 2.60
Specific gravity of gravel = 2.70
The volume will be:
40 * 12 * 0.5 = 240 ft³
We need to find the dry volume of concrete.
Dry volume = wet volume * 1.54 (concrete)
Dry volume will be = 240 * 1.54 = 360ft³
Due to the 7% entarained air content, the required volume will be:
V = 360 * (1 - 0.07)
V = 334.8 ft³
At a ratio of 1:2:3 for cement, sand, and gravel respectively, we have:
Total of ratio = 1+2+3 = 6
Their respective volume will be =
Volume of cement = [tex] \frac{1}{6}*334.8 = 55.8 ft^3 [/tex]
Volume of sand = [tex] \frac{2}{6}*334.8 = 111.6 ft^3 [/tex]
Volume of gravel = [tex] \frac{3}{6}*334.8 = 167.4 ft^3 [/tex]
To find the pounds needed the driveway, we have:
Weight = volume *specific gravity * density of water
Specific gravity of cement = 3.15
Weight of cement =
55.8 * 3.15 * 62.4 = 10968 pounds
Weight of sand =
111.6 * 2.60 * 62.4 = 18105.9 lb
Weight of gravel =
167.4 * 2.7 * 62.4 = 28203.55 lb
Given water to cement ratio of 0.50
Weight of water = 0.5 of weight of cement
= 1/2 * 10968 = 5484 lb
Show the bias polarities and depletion regions of an npn BJT in the normal active, saturation, and cutoff modes of operation. Draw the three sketches one below the other to (qualitatively) reflect the depletion widths for these biases, and the relative emitter, base, and collector doping.
Consider a BJT with a base transport factor of 1.0 and an emitter injection efficiency of 0.5.Calculate roughly by what factor would doubling the base width of the BJT would increase, decrease, or leave unchanged the emitter injection efficiency and base transport factor?
Answers
Complete Question:
Show the bias polarities and depletion regions of an npn BJT in the normal active, saturation, and cutoff modes of operation. Draw the three sketches one below the other to (qualitatively) reflect the depletion widths for these biases, and the relative emitter, base, and collector doping.
Consider a BJT with a base transport factor of 1.0 and an emitter injection efficiency of 0.5.
Calculate roughly by what factor would doubling the base width of a BJT would increase, decrease, or leave unchanged the emitter injection efficiency and base transport factor? Repeat for the case of emitter doping increased 5 × =. Explain with key equations, and assume other BJT parameters remain unchanged!
Answer & Explanation:
[Find the attachments]
Step 1 :
Emitter and base, collector, and base are forward biased then BJT is in saturation region. Emitter and base is forward biased and base and collector in reverse biased then BJT is in active region.
Emitter and base, collector and base are reverse biased then BJT in cut off region.
Three sketches one below the other is shown in Figure 1.
[find the figure in attachment]
Step 2:
Value of base widths of saturation, active and cut off operated BJT are value of Base width of saturated region operated BJT is less than base width in active region operated BJT. Value of base width of active region operated BJT is less than base width in cut off region operated BJT.
Saturation region operated base width of BJT is < Active region operated base width of BJT is < Cut off region operated base width of BJT.
[For Steps 3 4 5 6 and 7 find attachments]
Air enters the first compressor stage of a cold air-standard Brayton cycle with regeneration and intercooling at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The overall compressor pressure ratio is 10, and the pressure ratios are the same across each compressor stage. The temperature at the inlet to the second compressor stage is 300 K. The temperature at the inlet to the turbine is 1400 K. The compressor stages and turbine each have isentropic efficiencies of 80% and the regenerator effectiveness is 80%.
For k = 1.4, calculate:
(A) the thermal efficiency of the cycle.
(B) the back work ratio.
(C) the net power developed, in kW.
(D) the rates of exergy destruction in each compressor stage and the turbine stage as well as the regenerator, in kW, for T0 = 300 K.
Answers
An air-conditioning system is to be filled from a rigid container that initially contains 5 kg of saturated liquid R-134a at 26 °C. The valve connecting this container to the air-conditioning system is now opened until the mass in the container is 0.5 kg, at which time the valve is closed. During this time, only liquid R-134a flows from the container. Presuming that the process is isothermal while the valve is open, determine the final quality of the R-134a in the container and the total heat transfer.
Answers
Answer:
x2 = 0.5056
Qin = 22.62Kj
Explanation:
An aluminum cylinder bar ( 70 GPa E m = ) is instrumented with strain gauges and is subject to a tensile force of 5 kN. The diameter of the bar is 10 cm. The Poisson’s ratio of the bar is 0.33. A Wheatstone bridge is constructed to measure the axial strain. Gauge 1 measures the axial strain and gauge 2 measures the lateral strain.
Answers
Find the complete solution in the given attachments.
Note: The complete Question is attached in the first attachment as the provided question was incomplete
(a) Determine the temperature of the insulated walls. (b) Determine the net radiation heat rate from surface 2 per unit conduit length. 13.48 A long conduit is constructed with diffuse, gray walls 0.5 m wide. The top and bottom of the conduit are insulated. The emissivities of the walls are ε 1 = 0.45, ε 2 = 0.65, and ε 3 = 0.15, respectively, while the temperatures of walls 1 and 2 are 500 K and 700 K, respectively.
Answers
Answer:
Explanation:
the solution to the problem is given in the pictures attached. (b) is answered first then (a). I hope the explanation helps you.Thank you
A very long pipe of 0.05 m (r1) radius and 0.03 m thickness (r2 - r1) is buried at a depth of 2m (z) to transport liquid nitrogen. The pipe wall has a very low thermal conductivity of 0.0035 Wm- 1K-1 and receives a cooling power of 10 W/m to keep the liquid nitrogen at 77 K. If the thermal conductivity of the ground is 1 Wm-1K-1, what is the surface temperature of the ground
Answers
Answer:
See explaination
Explanation:
thermal conductivity is A measure of the ability of a material to transfer heat. Given two surfaces on either side of the material with a temperature difference between them, the thermal conductivity is the heat energy transferred per unit time and per unit surface area, divided by the temperature difference.
Please kindly check attachment for the step by step solution of the given problem.