Answer:
$14253
Explanation:
Write a program that read two integers and display their MOD,VID and their floating-point division in both settings x/y and y/x
e.g 5/3 and 3/5
Answer:
#!/usr/bin/env python
def calculate(x, y):
return {
"MOD": x % y,
"DIV": x/y, # you mean div instead of “VID”, right?
"floating-point division": float(x)/y,
}
def calculateInBothSettings(x, y):
return {
"x/y": calculate(x, y),
"y/x": calculate(y, x),
}
if __name__ == "__main__":
x = int(input("x: "))
y = int(input("y: "))
print(calculateInBothSettings(x, y))
Explanation:
I wrote a python script. Example output:
x: 2
y: 3
{'x/y': {'MOD': 2, 'DIV': 0.6666666666666666, 'floating-point division': 0.6666666666666666}, 'y/x': {'MOD': 1, 'DIV': 1.5, 'floating-point division': 1.5}}
Yeah order for a firm voltage dividers to operate properly The load resistance value should be at least Times greater than resistance value of the voltage divider bleeder resistor
Answer:
A voltage divider is a simple series resistor circuit. It's output voltage is a fixed fraction of its input voltage. The divide-down ratio is determined by two resistors.
A program consists of 100,000 instructions as follows:
Instruction Type Instruction Count Cycles per Instruction
Integer arithmetic 45,000 4
Data transfer 32,000 6
Floating point arithmetic 15,000 10
Control transfer 8,000 3
Determine:
a. the program execution time
b. the effective CPI for the machine
c. MIPS rate for the following processors
Answer:
Execution time is 13.65ms
CPI is 5.46
MIPS is 73.3\ MIPs
Explanation:
Given:
[tex]s = 40MHz[/tex] --- processor speed [Missing from the question]
[tex]\begin{array}{ccc}{Instruction\ Type} & {Instruction\ Count} & {Cycles\ per\ Instruction} & {Integer\ Arithmetic} & {45000} & {4} \ \\ {Data\ Transfer} & {32000} & {6} & {Floating\ point\ arithmetic} & {15000} & {10} & {Control\ Transfer} &{8000} & {3} \ \end{array}[/tex]
Solving (a): The program execution time
First, we solve for (b)
Solving (b): The effective CPI
This is calculated as:
[tex]CPI = \frac{\sum IC * CI}{\sum IC}[/tex]
Where: IC = Instruction Count and CI = Cycles per Instruction
So, the equation becomes:
[tex]CPI = \frac{45000 *4 + 32000 * 6 + 15000 * 10 + 8000 * 3}{45000+32000+15000+8000}[/tex]
[tex]CPI = \frac{546000}{100000}[/tex]
[tex]CPI = 5.46[/tex]
Solving (c): MIPS
This is calculated as:
[tex]MIPS = Speed * \frac{1}{CPI} * \frac{1}{\sum IC}[/tex]
[tex]MIPS = 40 * \frac{1}{5.46} * \frac{1}{100000}[/tex]
[tex]MIPS = \frac{40 * 1 * 1}{5.46*100000}[/tex]
[tex]MIPS = \frac{40}{546000}[/tex]
[tex]MIPS = 0.00007326007[/tex]
Convert to MIPs
[tex]MIPS = 73.3\ MIPs[/tex]
Solving (a): Execution Time
This is calculated as:
[tex]Time = Instructions * CPI * \frac{1}{Speed}[/tex]
[tex]Time = 100000 * 5.46* \frac{1}{40M}[/tex]
[tex]Time = \frac{546000}{40M}[/tex]
[tex]Time = \frac{546000}{40*1000000}[/tex]
[tex]Time = \frac{546000}{40000000}[/tex]
[tex]Time = 0.01365s[/tex]
[tex]Time = 13.65ms[/tex]
Execution time is 13.65ms
The dry weather average flow rate for a river is 8.7 m3/s. During dry weather flow, the average COD concentration in the river is 32 mg/L. An industrial source continuously discharges 18,000 m3/d of wastewater contains an average 342 mg/L COD concentration into the river. What is the COD mass loading in the river upstream of the industrial source discharge
Answer:
6156 kg /day
Explanation:
Determine the COD mass loading in the river upstream of the industrial source discharge
Given data:
Flow rate of river = 8.7 m^3/s
Average COD concentration in river = 32 mg/L
Industrial source continuous discharge ( Qw )= 18,000 m^3/d
Yw = 342 mg/l
since :
1 m^3 = 1000 liters
Qw = 18 * 10^6 liters = ( 18 million per day )
Hence the COD mass loading
= Yw * Qw
= 342 * 18 liters
= 6156 kg /day
1. A drilling operation is to be performed with a 10 mm diameter twist drill in a steel workpart. The hole is a blind hole at a depth of 60 mm and the point angle is 118. The cutting speed is 30 m/min and the feed is 0.25 mm/rev. Determine (a) the cutting time to complete the drilling operation, and (b) metal removal rate during the operation, after the drill bit reaches full diameter.
Answer:
The answer is below
Explanation:
v = velocity = 30 m/min = 30 * 10³ mm/min, D = diameter = 10 mm, f = feed = 0.25 mm/rev, point angle = 118, cutting time = Tm, d = depth = 60 mm
[tex]a)\\N=\frac{v}{\pi D}=\frac{30*10^3}{\pi * 10}=954.9\ rev/min\\\\f_r=Nf =954.9(0.25)=238\ mm/min\\\\A=0.5Dtan(90-\frac{point\ angle}{2} )=0.5*10*tan(90-\frac{118}{2} )=3\ mm\\\\T_m=\frac{(d+A)}{f_r} =\frac{60+3}{238}=0.265 \ s\\\\b)\\metal\ removal\ rate(R_{MR})=0.25\pi D^2f_r\\\\R_{MR}=0.25\pi (10)^2(238)=18692\ mm^3/min[/tex]
Truckco manufactures two types of trucks: 1 and 2. Each truck must go through the painting shop and assembly shop. If the painting shop were completely devoted to painting Type 1 trucks, then 800 per day could be painted; if the painting shop were completely devoted to painting Type 2 trucks, then 700 per day could be painted. If the assembly shop were completely devoted to assembling truck 1 engines, then 1,500 per day could be assembled; if the assembly shop were completely devoted to assembling truck 2 engines, then 1,200 per day could be assembled. Each Type 1 truck contributes $300 to profit; each Type 2 truck contributes $500. How much capacity in percent does a single truck of each type uses at each shop?
Water is pumped from a lake to a storage tank 18 m above at a rate of 70 L/s while consuming 20.4 kW of electric power. Disregard any frictional losses in the pipes and any changes in kinetic energy, determine (a) the overall efficiency of the pump-motor unit (5-point), and (b) the pressure difference between the inlet and the exit of the pump (5-point).
Trichloroethylene (TCE) is a common groundwater contaminant. Based on an adult ingesting the water under residential exposure parameters, which of the following presents the greatest cancer risk:
a. to drink unchlorinated groundwater with 10 ppb of TCE
b. to switch to a surface water supply that, as a result of chlorination, has a chloroform concentration of 50 ppb
Answer:
To drink unchlorinated groundwater with 10 ppb of TCE ( A )
Explanation:
The option that presents the greatest cancer risk for ingesting water
contaminated with Trichloroethylene under the residential exposure parameters is to drink unchlorinated groundwater with 10 ppb of TCE
This is because suitable water for drinking has chloroform concentration that ranges from 4 to 44 ppb but drinking under groundwater with ppb value above 4 ppb will have a more severe damage to the body
For an Na+—Cl- ion pair, attractive and repulsive energies EA and ER, respectively, depend on the distance between the ions r, according to EA = -1.436/r ER =(7.32 *10-6 )/r8 For these expressions, energies are expressed in electron volts per Na+—Cl- pair, and r is the distance in nanometers. The net energy EN is just the sum of the preceding two expressions
Answer:
Explanation:
[tex]\text{The curve of the plot}[/tex] [tex]\mathbf{E_A,E_R, \ and \ E_N}[/tex] [tex]\text{can be seen in the attached diagram below}[/tex]
[tex]\text{From the plot}[/tex], [tex]\mathbf{r_o = 0.2 4nm \ and \ E_o =-5.3 eV}[/tex]
[tex]\mathbf{We \ knew \ that: E_N = E_A + E_R}[/tex]
[tex]\mathtt{GIven \ E_A = \dfrac{-1.436}{r}\ \ \ , E_R = \dfrac{7.32 \times 10^{-6}}{r^n} \ \ and \ \ n=8 }[/tex]
[tex]\mathtt{Then; E_N = -\dfrac{-1.436}{r}+ \dfrac{7.32\times 10^{-6}}{r^8}}[/tex]
[tex]\mathtt{Also; r_o = \Big( \dfrac{A}{nB} \Big)^{\dfrac{1}{1-n}}} \\ \\ \mathtt{ r_o = \Big( \dfrac{1.986}{8 \times 7.32\times 10^{-6}} \Big)^{\dfrac{1}{1-8}}} \\ \\ \mathbf{r_o = 0.236 nm}[/tex]
[tex]E_o = \dfrac{-1.436}{\Big[\dfrac{1.436}{8(732\times 10^{-6})}\Big]^{\dfrac{1}{1-8}}} + \dfrac{7.32 \times 10^{-6}}{\Big[ \dfrac{1.436}{8\times7.32 \times 10^{-6} } \Big]^{\dfrac{8}{1-8}}}[/tex]
[tex]\mathbf{E_o = -5.32 \ eV}[/tex]
2 A SQUARE GIVEN
LENGTH OF DIAGONAL = 70 mm
Answer:
Area of square = 2,450 mm²
Explanation:
Given:
Length of diagonal = 70 mm
Find:
Area of square
Computation:
Area of square = diagonal² / 2
Area of square = 70² / 2
Area of square = 4900 / 2
Area of square = 2,450 mm²
A company purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate of the device is 3%. 1) The inspector randomly picks 20 items from a shipment. What is the probability that there will be at least one defective item among these 20? 2) Suppose that the company receives 10 shipments in a month and the inspector randomly tests 20 devices per shipment. What is the probability that there will be exactly 3 Shipments each containing at least one defective device among the 20 that are selected and tested from the shipment?
Answer:
1) The probability of at least 1 defective is approximately 45.621%
2) The probability that there will be exactly 3 shipments each containing at least 1 defective device among the 20 devices that are tested from the shipment is approximately 16.0212%
Explanation:
The given parameters are;
The defective rate of the device = 3%
Therefore, the probability that a selected device will be defective, p = 3/100
The probability of at least one defective item in 20 items inspected is given by binomial theorem as follows;
The probability that a device is mot defective, q = 1 - p = 1 - 3/100 = 97/100 = 0.97
The probability of 0 defective in 20 = ₂₀C₀(0.03)⁰·(0.97)²⁰ ≈ 0.543794342927
The probability of at least 1 = 1 - The probability of 0 defective in 20
∴ The probability of at least 1 = 1 - 0.543794342927 = 0.45621
The probability of at least 1 defective ≈ 0.45621 = 45.621%
2) The probability of at least 1 defective in a shipment, p ≈ 0.45621
Therefore, the probability of not exactly 1 defective = q = 1 - p
∴ q ≈ 1 - 0.45621 = 0.54379
The probability of exactly 3 shipment with at least 1 defective, P(Exactly 3 with at least 1) is given as follows;
P(Exactly 3 with at least 1) = ₁₀C₃(0.45621)³(0.54379)⁷ ≈ 0.160212
Therefore, the probability that there will be exactly 3 shipments each containing at least 1 defective device among the 20 devices that are tested from the shipment is 16.0212%
Engine oil flows at a rate of 1 kg/s through a 5-mmdiameter straight tube. The oil has an inlet temperature of 45°C and it is desired to heat the oil to a mean temperature of 80°C at the exit of the tube. The surface of the tube is maintained at 150°C. Determine the required length of the tube. Hint: Calculate the Reynolds numbers at the entrance and
Answer:
length of tube = 4.12 m
Explanation:
Given data:
flow rate of engine oil = 1 kg/s
diameter of tube = 5-mm
inlet temperature of oil = 45°C
exit temperature of oil = 80°C
surface temperature of tube = 150°C
Determine the required length of the tube
attached below is a detailed solution to the given problem
length of tube = 4.12 m
A demand factor of _____ percent applies to a multifamily dwelling with ten units if the optional calculation method is used.
A laissez-faire leadership style works best for what type of group?
Disciplined
Disorganized
Uncooperative
Intelligent
Answer:
a
Explanation:
A 3-phase , 1MVA, 13.8kV/4160V, 60 Hz, transformer with Y-Delta winding connection is supplying a3-phase, 0.75 p.u. load on the 4160V side of the transformer. The load has leading power factor of 0.9. It issupplied by 1 p.u. voltage on the 13.8kV side. The transformer per unit impedance is j0.12 referred to thesecondary side.
a. Find the load impedance.
b. Find the input current on the primary side in real units.
c. Find the input power factor
Answer:
a) 23.89 < -25.84 Ω
b) 31.38 < 25.84 A
c) 0.9323 leading
Explanation:
A) Calculate the load Impedance
current on load side = 0.75 p.u
power factor angle = 25.84
[tex]I_{load}[/tex] = 0.75 < 25.84°
attached below is the remaining part of the solution
B) Find the input current on the primary side in real units
load current in primary = 31.38 < 25.84 A
C) find the input power factor
power factor = 0.9323 leading
attached below is the detailed solution
A signal is assumed to be bandlimited to kHz. It is desired to filter this signal with an ideal bandpass filter that will pass the frequencies between kHz and kHz by a system for processing analog signals composed of a digital filter with frequency response sandwiched between an ideal A/D and an ideal D/A, both operating at sampling interval . 1. Determine the Nyquist sampling frequency, (in kHz), for the input signal. 2. Find the largest sampling period (in s) for which the overall system comprising A/D, digital filter and D/A realize the desired band pass filter.
Answer:
Hello your question is poorly written attached below is the complete question
answer :
1) 60 kHz
2) Tmax = ( 1 / 34000 ) secs
Explanation:
1) Determine the Nyquist sampling frequency, (in kHz), for the input signal.
F(s) = 2 * Fmax
Fmax = 30 kHz ( since Xa(t) is band limited to 30 kHz )
∴ Nyquist sampling frequency ( F(s) ) = 2 * 30 = 60 kHz
2) Determine the largest sampling period (in s) .
Nyquist sampling period = 1 / Fs = ( 1 / 60000 ) s
but there is some aliasing of the input signal ( minimum aliasing frequency > cutoff frequency of filter ) hence we will use the relationship below
= 2π - 2π * T * 30kHz ≥ 2π * T * 4kHz
∴ T ≤ [tex]\frac{1}{34kHz}[/tex]
largest sampling period ( Tmax ) = ( 1 / 34000 ) secs
how do we succeed in mechanical engineering?
6. Find the heat flow in 24 hours through a refrigerator door 30.0" x 58.0" insulated with cellulose fiber 2.0" thick. The temperature inside the refrigerator is 38°F. Room temperature is 72°F. [answer in BTUs]
Answer:
The heat flow in 24 hours through the refrigerator door is approximately 1,608.57 BTU
Explanation:
The given parameters are;
The duration of the heat transfer, t = 24 hours = 86,400 seconds
The area of the refrigerator door, A = 30.0" × 58.0" = 1,740 in.² = 1.122578 m²
The material of the insulator in the door = Cellulose fiber
The thickness of the insulator in the door, d = 2.0" = 0.0508 m
The temperature inside the fridge = 38° F = 276.4833 K
The temperature of the room = 78°F = 298.7056 K
The thermal conductivity of cellulose fiber = 0.040 W/(m·K)
By Fourier's law, the heat flow through a by conduction material is given by the following formula;
[tex]\dfrac{Q}{t} = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d}[/tex]
[tex]Q = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d} \times t[/tex]
Therefore, we have;
[tex]Q = \dfrac{0.04 \times 1.122578 \times (298.7056 - 276.4833 ) }{0.0508} \times 86,400 =1,697,131.73522[/tex]
The heat flow in 24 hours through the refrigerator door, Q = 1,697,131.73522 J = 1,608.5705140685 BTU
Some of our modern kitchen cookware is made of ceramic materials. (a) List at least three important characteristics required of a material to be used for this application. (b) Make a comparison of three ceramic materials as to their relative properties and, in addition, to cost. (c) On the basis of this comparison, select the material most suitable for the cookware.
Answer:
A)
It should be Non- toxic
It should possess high Thermal conductivity
It should have the Required Thermal diffusivity
B)
stoneware : This material has good thermal diffusivity and it is quite affordable and it is used in making pizza stonesporcelain: mostly used for mugs and it is non-toxic Pyrex : posses good thermal conductivity used in ovenC) All the materials are suitable because they serve different purposes when making modern kitchen cookware
Explanation:
A) characteristics required of a ceramic material to be used as a kitchen cookware
It should be Non- toxicIt should possess high Thermal conductivityIt should have the Required Thermal diffusivityB) comparison of three ceramic materials as to their relative properties
stoneware : This material has good thermal diffusivity and it is quite affordable and it is used in making pizza stonesporcelain: mostly used for mugs and it is non-toxic Pyrex : posses good thermal conductivity used in ovensC) material most suitable for the cookware.
All the materials are suitable because they serve different purposes when making modern kitchen cookware
It describes the physical and social elements common to this work. Note that common contexts are listed toward the top, and less common contexts are listed toward the bottom. According to O*NET, what are common work contexts for Reporters and Correspondents? Check all that apply.
Answer:
Acef
Explanation:
Edginuity 2021
Answer:
2,3,4,5
Explanation:
guy above me is wrong
What is the importance of ethics in emerging technologies?
Explanation:
the Ethics of emerging Technology can only make use of speculative data about future products,uses and impacts.
Question 1: What is the power observed in the energy analyzer when the rated voltage(U1) is applied to the primary of the transformer, and there is no load at the secondary?
Question 2: Find the transformation ratio of the transformer using the values U1,U2 recorded in the experiment.
Question 3: Sketch the no-load operation graph of the transformer using the values U1, I2 and the values read in the energy analyzer.
Question 4: How can we find the number of turns of transformer?
Question 5: Explain the operation principle of the transformer.
Question 6: State your final observations about the experiment.
Answer:
preguntas a parte o no???????
Two fluids, A and B exchange heat in a counter – current heat exchanger. Fluid A enters at 4200C and has a mass flow rate of 1 kg/s. Fluid B enters at 200C and also has a mass flow rate of 1 kg/s, Effectiveness of heat exchanger is 75%. Determine the heat transfer rate and exit temperature of fluid B.
Answer:
Your question has some missing information below is the missing information
Given that ( specific heat of fluid A = 1 kJ/kg K and specific heat of fluid B = 4 kJ/kg k )
answer : 300 kW , 95°c
Explanation:
Given data:
Fluid A ;
Temperature of Fluid ( Th1 ) = 420° C
mass flow rate (mh) = 1 kg/s
Fluid B :
Temperature ( Tc1) = 20° C
mass flow rate ( mc ) = 1 kg/s
effectiveness of heat exchanger = 75% = 0.75
Determine the heat transfer rate and exit temperature of fluid B
Cph = 1000 J/kgk
Cpc = 4000 J/Kgk
Given that the exit temperatures of both fluids are not given we will apply the NTU will be used to determine the heat transfer rate and exit temperature of fluid B
exit temp of fluid B = 95°C
heat transfer = 300 kW
attached below is a the detailed solution
Answer the question on the image and a brianiest will be given to the person that provided the right answer to it.
Answer:
(a) The distance up the slope the wagon moves before coming to rest is approximately 21.74 m
(b) The distance the wagon comes to rest from the starting point is approximately 12.06 m
(c) The value of 'U' at which the wagon should be propelled if it is to come finally to rest at its starting point is approximately 3.214 m/s (the difference in value can come from calculating processes)
Explanation:
The wagon motion parameters are;
The mass of the wagon, m = 7,200 kg
The initial velocity with which the wagon is projected along the horizontal rail, v = U
The length of the horizontal portion of the rail = 100 m
The angle of inclination of the inclined portion of the rail, θ = sin⁻¹(0.01)
The exerted frictional resistance to motion of the rail, [tex]F_f[/tex] = 140 N
∴ θ = sin⁻¹(0.01)
The work done by the frictional force on the horizontal portion of the rail = 140 N × 100 m = 14,000 J
(a) If U = 3 m/s, we have;
Kinetic energy = 1/2·m·v²
The initial kinetic energy of the wagon, K.E. is given with the known parameters as follows;
K.E. = 1/2 × 7,200 kg × (3 m/s)² = 32,400 J
The energy, E, required to move a distance, 'd', up the slope is given as follows;
E = [tex]F_f[/tex] × d + m·g·h
Where;
[tex]F_f[/tex] = The friction force = 140 N
m = The mass of the wagon = 7,200 kg
g = The acceleration due to gravity ≈ 9.81 m/s²
h = The height reached = d × sin(θ) = d × 0.01
Therefore;
E = 140 N × d₁ + 7,200 kg × 9.81 m/s² × d₁ × 0.01 = 846.32 N × d
The energy, [tex]E_{NET \ horizontal}[/tex], remaining from the horizontal portion of the rail is given as follows;
[tex]E_{NET \ horizontal}[/tex] = Initial kinetic energy of the wagon - Work done on frictional resistance on the horizontal portion of the rail
∴ [tex]E_{NET \ horizontal}[/tex] = 32,400 J - 14,000 J = 18,400 J
[tex]E_{NET \ horizontal}[/tex] = 18,400 J
Therefore, for the wagon with energy, [tex]E_{NET \ horizontal}[/tex] to move up the train, we get;
[tex]E_{NET \ horizontal}[/tex] = E
∴ 18,400 J = 846.32N × d
d₁ = 18,400 J/(846.36 N) ≈ 21.7401579 m
d₁ ≈ 21.74 m
The distance up the slope the wagon moves before coming to rest, d₁ ≈ 21.74 m
(b) Given that the initial velocity of the wagon, U = 3 m/s, the distance up the slope the wagon moves before coming to rest is given above as d₁ ≈ 21.74 m
The initial potential energy, PE, of the wagon while at the maximum height up the slope is given as follows;
P.E. = m·g·h = 7,200 kg × 9.81 m/s² × 21.74 × 0.01 m = 15,355.3968 J
The work done, 'W', on the frictional force on the return of the wagon is given as follows;
W = [tex]F_f[/tex] × d₂
Where d₂ = the distance moved by the wagon
By conservation of energy, we have;
P.E. = W
∴ 15,355.3968 = 140 × d₂
d₂ = 15,355.4/140 = 109.681405714
Therefore;
The distance the wagon moves from the maximum height, d₂ ≈ 109.68 m
The distance the wagon comes to rest from the starting point, d₃, is given as follows;
d₃ = Horizontal distance + d₁ - d₂
d₃ = 100 m + 21.74 m - 109.68 m ≈ 12.06 m
The distance the wagon comes to rest from the starting point, d₃ ≈ 12.06 m
(c) For the wagon to come finally to rest at it starting point, we have;
The initial kinetic energy = The total work done
1/2·m·v² = 2 × [tex]F_f[/tex] × d
∴ 1/2 × 7,200 × U² = 2 × 140 × d₄
d₄ = 100 + (1/2·m·U² - 140×100)
(1/2·m·U² - 140×100)/(m·g) = h = d₁ × 0.01
∴ d₁ = (1/2·m·U² - 140×100)/(m·g×0.01)
d₄ = 100 + d₁
∴ d₄ = 100 + (1/2·m·U² - 140×100)/(m·g×0.01)
∴ 1/2 × 7,200 × U² = 2 × 140 × (100 + (1/2 × 7,200 × U² - 140×100)/(7,200 × 9.81 ×0.01))
3,600·U² = 280·(100 + (3,600·U² - 14,000)/706.32)
= 28000 + 280×3,600·U²/706.32 - 280 × 14,000/706.32
= 28000 - 280 × 14,000/706.32 + 1427.11518858·U²
3,600·U² - 1427.11518858·U² = 28000 - 280 × 14,000/706.32
U²·(3,600 - 1427.11518858) = (28000 - 280 × 14,000/706.32)
U² = (28000 - 280 × 14,000/706.32)/(3,600 - 1427.11518858) = 10.3319363649
U = √(10.3319363649) = 3.21433295801
The value of 'U' at which the wagon should be propelled if it is to come finally to rest at its starting point is U ≈ 3.214 m/s
Percentage error = (3.214-3.115)/3.214 × 100 ≈ 3.1% < 5% (Acceptable)
The difference in value can come from difference in calculating methods
Thermal energy storage systems commonly involve a packed bed of solid spheres, through which a hot gas flows if the system is being charged, or a cold gas if it is being discharged. In a charging process, heat transfer from the hot gas increases thermal energy stored within the colder spheres; during discharge, the stored energy decreases as heat is transferred from the warmer spheres to the cooler gas. Consider a packed bed of 75-mm-diameter aluminum spheres (p = 2,700 kg/m^3; c = 950 J/kg*K; k = 240 W/m*K) and a charging process for which gas enters the storage unit at a temperature of 300 degrees C. The initial temperature of the spheres is Ti = 25 degrees C and the convection heat transfer coefficient is h = 75 W/m^2*K.
a. How long does it take a sphere near the inlet of the system to accumulate 90% of the maximum possible thermal energy? What is the corresponding temperature at the center of the sphere?
b. Is there any advantage to using copper (p = 8,900 kg/m^3; c = 380 J/kg*K; k = 390W/m*K) instead of aluminum?
c. Consider the same packed bed operating conditions, but with Pyrex (p = 2,200 kg/m^3; c = 840 J/kg*K; k = 1.4 W/m*K) used instead of aluminum. How long does it take a sphere near the inlet of the system to accumulate 90% of the maximum possible thermal energy? What is the corresponding temperature at the center of the sphere?
Answer:
A) i) 984.32 sec
ii) 272.497° C
B) It has an advantage
C) attached below
Explanation:
Given data :
P = 2700 Kg/m^3
c = 950 J/kg*k
k = 240 W/m*K
Temp at which gas enters the storage unit = 300° C
Ti ( initial temp of sphere ) = 25°C
convection heat transfer coefficient ( h ) = 75 W/m^2*k
A) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere
First step determine the Biot Number
characteristic length( Lc ) = ro / 3 = 0.0375 / 3 = 0.0125
Biot number ( Bi ) = hLc / k = (75)*(0.0125) / 40 = 3.906*10^-3
Given that the value of the Biot number is less than 0.01 we will apply the lumped capacitance method
attached below is a detailed solution of the given problem
B) The physical properties are copper
Pcu = 8900kg/m^3)
Cp.cu = 380 J/kg.k
It has an advantage over Aluminum
C) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere
Given that:
P = 2200 Kg/m^3
c = 840 J/kg*k
k = 1.4 W/m*K
A thin silicon chip and an 8-mm-thick aluminum substrate are separated by a 0.02-mm-thick epoxy joint. The chip and substrate are each 10 mm on a side, and their exposed surfaces are cooled by air, which is at a temperature of 25 C and provides a convection coefficient of 100 W/m2 K. If the chip dissipates 104 W/m2 under normal conditions, will it operate below a maximum allowable temperature of 85 C
Answer:
The chip will operate below a maximum allowable temperature of 85°C
Explanation:
Given data:
8-mm-thick aluminum
0.02 mm-thick epoxy joint
chip and substrate = 10 mm on a side
temperature = 25°C
attached below is a detailed solution
Tc = 75.3 ° c which is less than 85°c . hence the chip will operate below a maximum allowable temperature of 85°C
What is your favorite Electronic company? (E.g. Windows, Apple, Samsung...)
Answer:
apple
Explanation:
is the answer toWhat is your favorite Electronic company? (E.g. Windows, Apple, Samsung...)
In addition to being good problem solvers, which of the following do engineers need to be?
O wealthy
O rigid
O respected
O practical
A fill covering a wide area is to be placed at the surface of this profile. The fill has a total unit weight of 20 kN/m^3 and is 3 m thick. Assume that the data for the sample at 7.0 m are representative of the entire clay profile. Also assume that the clay is heavily over consolidated and that the danse sands at the surface of the profile are so stiff that they do not contribute to the settlement. Find the settlement of the surface due to compression of the clay layer
Answer:
hello your question lacks some information attached below is the complete question with the required information
answer : 81.63 mm
Explanation:
settlement of the surface due to compression of the clay ( new consolidated )
= 81.63 mm
attached below is a detailed solution to the given problem
Past evidence shows that when a customer complains of an out-of-orderphone there is an 8% chance that the problem is with the inside wiring. During a 1-month period,100 complains are lodged. Assume that there have been no wide-scale problems that could beexpected to affect many phones at once, and that, for this reason, these failures are consideredto be independent.
Required:
a. Find the expected number of failures due to a problem with the inside wiring.
b. Find the probability that at least 10 failures are due to a problem with the inside wiring.
c. Would it be unusual if at most 5 were due to problems with the inside wiring? Explain, based on the probability of this occurring.
Answer:
a. The expected number of failures due to a problem with inside wiring is 8 failures
b. The probability that at least 10 failures are due to inside wiring is approximately 0.176
c. It will be not unusual
Explanation:
The probability that the problem of an out of order is the inside wiring, P(x) = 8%
The number of complaints in a month period, x = 100
a. The expected number of failures due to a problem with inside wiring, E(x) = x·P(x)
∴ E(x) = 100 × 8% = 8
The expected number of failures due to a problem with inside wiring, E(x) = 8 failures
b. The probability that at least 10, P₁₀, failures are due to inside wiring is given as follows;
The probability of success, P = 0.08, therefore, the probability of failure, q = 1 - 0.08 = 0.92
P = [tex]_nC_r[/tex]·[tex]P^r[/tex]·[tex]q^{n-r}[/tex]
P₀ = ₁₀₀C₀·(0.08)⁰·(0.92)¹⁰⁰ = 0.000239211874657
P₁ = ₁₀₀C₁·(0.08)¹·(0.92)⁹⁹ = 0.00208010325
P₂ = ₁₀₀C₂·(0.08)²·(0.92)⁹⁸ = 0.00895348793
P₃ = ₁₀₀C₃·(0.08)³·(0.92)⁹⁷ = 0.02543309616
P₄ = ₁₀₀C₄·(0.08)⁴·(0.92)⁹⁶ = 0.0536306593
P₅ = ₁₀₀C₅·(0.08)⁵·(0.92)⁹⁵ = 0.0895398833653
P₆ = ₁₀₀C₆·(0.08)⁶·(0.92)⁹⁴ = 0.123279549561
P₇ = ₁₀₀C₃·(0.08)⁷·(0.92)⁹³ = 0.143953759735
P₈ = ₁₀₀C₈·(0.08)⁸·(0.92)⁹² = 0.145518474516
P₉ = ₁₀₀C₉·(0.08)⁹·(0.92)⁹¹ = 0.129349755125
P₁₀ = ₁₀₀C₁₀·(0.08)¹⁰·(0.92)⁹⁰ = 0.10235502362
∴ P₀ + P₁ + P₂ + P₃ + P₄ + P₅ + P₆ + P₇ + P₈ + P₉ + P₁₀ = 0.82433004464
The probability of at least 10 failures are due problem with the inside wiring = 1 - (P₀ + P₁ + P₂ + P₃ + P₄ + P₅ + P₆ + P₇ + P₈ + P₉ + P₁₀) = 1 - 0.82433004464 = 0.175666995536
The probability of at least 10 failures are due problem with the inside wiring = 0.175666995536 ≈ 0.176
c. The probability of at most 5 failures are due problem with the inside wiring = P₀ + P₁ + P₂ + P₃ + P₄ + P₅ = 0.179876441908
Therefore, given that probability of at most 5 failures > The probability of 8 failures it will be not unusual since the cause of failure is more (92%) due to other causes which are more likely and therefore increase in the probability that there are fewer failures due inside wiring