Answer:
2727000 Liters
Explanation:
Determine the daily methane production in liters
For waste cooking oil ;
Delivered amount in liters = 757.082 liters per week
daily = 757.082 / 7 = 108.154 L/day
waste vs concentration = 25,000 mg/L
Its volatile solid = 25000 * 10^-3 (g/l) * 108.154 L/day
= 2703.864 g/day
hence methane production in liters
= 600 L/g * 2703.864 g/day
= 1622318.57 L
For cow manure
Delivered amount in liters = 500 liters / day
Vs concentration = 5000 mg/l
Hence amount of Vs = 5000 * 10^-3 * 500 = 2500 ( g/day)
methane production in liters
= 150 * 2500 = 375000 liters
For glass clippings
Delivered amount in liters per day = 1135.62 / 7 = 162.23 L/day
amount of Vs = 15000 * 10^-3 (g/L) * 162.23 L/day = 2438.45 ( g/day )
hence the methane production in liters
= 300 * 2433.45 = 730035 L
Hence ∑ daily methane production of each waste
= 1622318.57 + 375000 + 730035 = 2727000 Liters
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]
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
A laissez-faire leadership style works best for what type of group?
Disciplined
Disorganized
Uncooperative
Intelligent
Answer:
a
Explanation:
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
In addition to being good problem solvers, which of the following do engineers need to be?
O wealthy
O rigid
O respected
O practical
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
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
how do we succeed in mechanical engineering?
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.
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
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
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
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???????
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...)
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?
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 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