It is desired to filter a cell broth at a rate of 3000 liters/h on a rotary vacuum filter at a vacuum pressure of 70kPa. The cycle time for the drum will be 60 s and the cake formation time (filtering time) will be 20 s. The broth to be filtered has a viscosity of 2.0 cP and a cake solids (dry basis) per volume of filtrate of 10 g/liter. From laboratory test, the specific cake resistance has been determined to be 9 x 1010 cm/g. Determine the area of the filter that is required.
Answer:
0.5667 m^2
Explanation:
Given data :
Flow rate = 3000 liters/h
vacuum pressure = 70 kPa
cycle time for drum = 60 s
cake formation time ( filtering time ) = 20 s
Broth viscosity = 2.0 cP
Cake solids per volume = 10 g/liter
specific cake resistance = 9 * 10^10 cm/g
Calculate the area of filter required
Area of filter required = 0.5667 m^2
Attached below is the detailed solution
A rectangular channel 3.0 m wide has a flow rate of 5.0 m3/s with a normal depth of 0.50 m. The flow then encounters a dam that rises 0.25 m above the channel bottom. Will a hydraulic jump occur?
Answer:
The hydraulic will jump since the flow is subcritical ( i.e. Y2 > Yc )
Explanation:
width of channel = 3.0 m
Flow rate = 5 m^3/s
Normal depth = 0.50 m
Flow encounters a dam rise of 0.25 m
To know if the hydraulic jump will occur
we will Determine the new normal depth
Y2 = 3.77m
Yc ( critical depth )= 0.66m
Attached below is the detailed solution
Q2 [45 marks] Consider Ibra region where the installed solar panels cost on average 2 OMR /W.
[10 marks] What is the cost to install a 5kW PV system for a residence?
[10 marks] If the solar irradiance in Ibra is on average 800W/m2 and the installed panels have efficiency of 18%. How many panels are required if the panel’s area is 2m2?
[15 marks] Assume Ibra has an average of 10 day-hours, dusty environment which causes the efficiency of the solar system to drop by 10% on average, and 30 cloudy days/year which cause the efficiency of the solar panels drops by 50%. If electrical power cost per kWh is 0.05 OMR determine the break-even time for the 5kW PV system.
[10 marks] If the system to be off-grid, what would be the backup time if three 12-V batteries were selected each with a capacity of 200Ah. Assume that you can discharge the batteries up to 80% of their capacities.
Answer:
so hard it is
Explanation:
I don't know about this
please mark as brainleast
byýyy
A center-point bending test was performed on a 2 in. x d in. wood lumber according to ASTM D198 procedure with a span of 4 ft and the 4 in. side is positioned vertically. If the maximum load was 240 kips and the modulus of rupture was 940.3 ksi, what is the value of d
Answer:
3.03 INCHES
Explanation:
According to ASTM D198 ;
Modulus of rupture = ( M / I ) * y ----- ( 1 )
M ( bending moment ) = R * length of span / 2
= (120 * 10^3 ) * 48 / 2 = 288 * 10^4 Ib-in
I ( moment of inertia ) = bd^3 / 12
= ( 2 )*( d )^3 / 12 = 2d^3 / 12
b = 2 in , d = ?
length of span = 4 * 12 = 48 inches
R = P / 2 = 240 * 10^3 / 2 = 120 * 10^3 Ib
y ( centroid distance ) = d / 2 inches
back to equation ( 1 )
( M / I ) * y
940.3 ksi = ( 288 * 10^4 / 2d^3 / 12 ) * d / 2
= ( 288 * 10^4 * 12 ) / 2d^3 ) * d / 2
940300 = 34560000* d / 4d^3
4d^3 ( 940300 ) = 34560000 d ( divide both sides with d )
4d^2 = 34560000 / 940300
d^2 = 9.188 ∴ Value of d ≈ 3.03 in
An engineer sets up an experiment to determine the coefficient of static friction "us" for an unknown material. She cuts the material in to a disc and places a test mass on top, L = .75m from the center, and she proceeds to spin the disc with an angular acceleration of theta-double-dot = 40t rad/s^2 counterclockwise. The engineer notes that the test mass slips at t=.2s. What is "us" (friction)?
Answer:
HF I am writing this email with your company is it a try and make them easier for us and we will need a new job and then email
In a ground-water basin of 12 square miles, there are two aquifers: an upper unconfined aquifer 500 ft in thickness and a lower confined aquifer with an available hydraulic head drop of 150 ft. Hydraulic tests have determined that the specific yield of the upper unit is 0.12 and the storativity of the lower unit is 4x10-4. What is the amount of recoverable ground water in the basin
Answer:
0.1365 m^3
Explanation:
thickness of upper aquifer = 500 ft
lower aquifer head drop = 150 ft
area of ground water basin = 12 m^2
specific yield of upper unit = 0.12
Storativity of lower unit = 4 * 10^-4
determine the amount of recoverable ground water
first step : calculate volume of unconfined aquifer
= 12 * 500/5280 = 1.1364 miles^3
The recoverable volume of water from unconfined aquifer
= 1.1364 * 0.12 = 0.1364 miles^3
next : calculate volume of confined aquifer
= 12 * 150/5250 = 0.341 miles^3
The recoverable volume of water from confined aquifer
= 0.341 * ( 4 * 10^-4 ) = 1.364 * 10^-4 miles^3
Hence the amount of recoverable ground water in the basin
= ∑ recoverable ground water from both aquifer
= 0.1365 m^3
The blue and white “Alfa” flag signifies what?
A) A tow boat is towing another vessel to safety
B) A ski boat that has a participant in the water
C) A dive boat that is restricted in its ability to maneuver
D) A Coast Guard boat that is on patrol
The blue and white "Alfa" flag signifies: C. a dive boat that is restricted in its ability to maneuver.
What is a dive boat?A dive boat can be defined as a type of boat that is commonly used by scuba divers or recreational divers to reach a deep dive site which cannot be conveniently accomplished by swimming from the bank or sea shore.
As a safety precaution, the blue and white "Alfa" flag is designed and developed to signify a dive boat that is restricted in its ability to maneuver.
Read more on dive boat here: https://brainly.com/question/23462572
#SPJ2
The volume fraction of particles in a WC-particle Cu-matrix CERMET is 0.84 Calculate the minimum expected elastic modulus, in GPa, of this particle reinforced composite. The moduli of the two components are 682 GPa and 110 GPa, respectively. Use what you know about these composite phases to discern which modulus is which.
Answer:
the minimum expected elastic modulus is 372.27 Gpa
Explanation:
First we put down the data in the given question;
Volume fraction [tex]V_f[/tex] = 0.84
Volume fraction of matrix material [tex]V_m[/tex] = 1 - 0.84 = 0.16
Elastic module of particle [tex]E_f[/tex] = 682 GPa
Elastic module of matrix material [tex]E_m[/tex] = 110 GPa
Now, the minimum expected elastic modulus will be;
[tex]E_{CT[/tex] = ([tex]E_f[/tex] × [tex]E_m[/tex] ) / ( [tex]E_f[/tex][tex]V_m[/tex] + [tex]E_m[/tex] [tex]V_f[/tex] )
so we substitute in our values
[tex]E_{CT[/tex] = (682 × 110 ) / ( [ 682 × 0.16 ] + [ 110 × 0.84] )
[tex]E_{CT[/tex] = ( 75,020 ) / ( 109.12 + 92.4 )
[tex]E_{CT[/tex] = 75,020 / 201.52
[tex]E_{CT[/tex] = 372.27 Gpa
Therefore, the minimum expected elastic modulus is 372.27 Gpa
An asphalt concrete mixture includes 94% aggregate by weight. The specific gravities of aggregate and asphalt are 2.65 and 1.0, respectively. The unit weight of water is 62.4 lb/ft3. The total weight of the mix is 147 Ib and the total volume is 1 ft3. Ignoring absorption, what is the percent voids in the total mix
Answer:
percent voids = 2.36%
Explanation:
Theoretical specific gravity ( Gt ) = ( w1 + wL ) / (( w1/G1) + (wL/GL ) )
= ( 94 + 6 ) / ((94/2.65) + ( 6 / 1.0 ))
∴ Gt = 2.411
next determine mass specific gravity = Ym / 1000
Bulk density = mass / volume = 147 Ib / 1ft^3 = 147 Ib/ft^3
given that: 1 Ib/ft^3 = 16.018 kg/m63
147 Ib/ft^3 = 2354.646 kg/m^3 ( Ym )
hence mass specific gravity = 2354.646 kg/m^3 / 1000
Gm = 2.354
finally determine percent void
= ( Gt - Gm ) / ( Gt ) * 100
= ( 2.411 - 2.354 ) / ( 2.411 ) * 100
= 2.36 %
3. When mixing repair adhesive, how do you know when the material is ready?
A) O The mix is uniform in color
B) O The mix has set for 2 minutes
C)The mix has no lumps
D)The mix turns blue
Answer:
O The mix is uniform in color
You do a simple experiment at home with the plastic body of a syringe, where you close the exit with one thumb, and push in the plunger with your other thumb. You are able to compress the air inside the syringe from 5ml to 1ml. Assume that you hold the plunger for long enough such that the temperature equalizes to ambient conditions.
Required:
a. Given that the circular plunger's diameter is 1.5cm, how much force is being exerted to hold the plunger in the compressed state?
b. Given that the opening of the syringe has a diameter of 2mm, how much force is exerted on the thumb used to trap the air from escaping?
Answer:
a. 89.5 N b. 1.59 N
Explanation:
a. Given that the circular plunger's diameter is 1.5cm, how much force is being exerted to hold the plunger in the compressed state?
Using Boyle's law, we find the final pressure at the compressed state given that the initial pressure is atmospheric pressure
So, P₁V₁ = P₂V₂ where P₁ = initial atmospheric pressure in syringe = 1 atm = 1.013 × 10⁵ N/m²,V₁ = initial volume of syringe = 5 ml, P₂ = final pressure in syringe at compression and V₂ = final volume of syringe = 1 ml
So, making P₂ subject of the formula, we have
P₂ = P₁V₁/V₂
Substituting the values of the variables into the equation, we have
P₂ = P₁V₁/V₂
P₂ = 1.013 × 10⁵ N/m² × 5 ml/1 ml
P₂ = 1.013 × 10⁵ N/m² × 5
P₂ = 5.065 × 10⁵ N/m²
Since pressure, P = F/A where F = force and A = cross-sectional area of syringe = πd²/4 where d = diameter of syringe = 1.5 cm = 1.5 × 10⁻² m.
So, F = PA
F = P₂πd²/4
substituting the values of the variables into the equation, we have
F = P₂πd²/4
F = 5.065 × 10⁵ N/m²π(1.5 × 10⁻² m)²/4
F = 5.065 × 10⁵ N/m²π(2.25 × 10⁻⁴ m²)/4
F = 35.8 × 10/4 N
F = 8.95 × 10
F = 89.5 N
b. Given that the opening of the syringe has a diameter of 2mm, how much force is exerted on the thumb used to trap the air from escaping?
Since the pressure in the syringe after compression is constant, we have
P₂ = F₁/A₁ where F₁ = force exerted on thumb and A₁ = cross-sectional area of opening of syringe = πd₁²/4 where d = diameter of opening of syringe = 2 mm = 2 × 10⁻³ m.
So, F₁ = P₂A₁
F = P₂πd₁²/4
substituting the values of the variables into the equation, we have
F = P₂πd²/4
F = 5.065 × 10⁵ N/m²π(2 × 10⁻³ m)²/4
F = 5.065 × 10⁵ N/m²π(4 × 10⁻⁶ m²)/4
F = 15.91 × 10⁻¹
F = 1.591 N
F ≅ 1.59 N
Consider a turbofan engine installed on an aircraft flying at an altitude of 5500m. The CPR is 12 and the inlet diameter of this engine is 2.0m The bypass ratio of this engine 8. The bypass ratio (BPR) of a turbofan engine is the ratio between the mass flow rate of the bypass stream to the mass flow rate entering the core. The inlet temperature is 253K and the outlet temperature is 233K. Determine the thrust of this engine in order to fly at the velocity of 250 m/s. Assume cold air approach. The engine is ideal.
Answer:
The thrust of the engine calculated using the cold air is 34227.35 N
Explanation:
For the turbofan engine, firstly the overall mass flow rate is considered. The mass flow rate is given as
[tex]\dot{m}=\rho AV_a[/tex]
Here
ρ is the density which is given as [tex]\dfrac{P}{RT}[/tex]P is the pressure of air at 5500 m from the ISA whose value is 50506.80 PaR is the gas constant whose value is 286.9 J/kg.KT is the temperature of the inlet which is given as 253 KA is the cross-sectional area of the inlet which is given by using the diameter of 2.0 mV_a is the velocity of the aircraft which is given as 250 m/sSo the equation becomes
[tex]\dot{m}=\rho AV_a\\\dot{m}=\dfrac{P}{RT} AV_a\\\dot{m}=\dfrac{50506.80}{286.9\times 253} \times (\dfrac{\pi}{4}\times 2^2)\times 250\\\dot{m}=546.4981\ kgs^{-1}[/tex]
Now in order to find the flow from the fan, the Bypass ratio is used.
[tex]\dot{m}_f=\dfrac{BPR}{BPR+1}\times \dot{m}[/tex]
Here BPR is given as 8 so the equation becomes
[tex]\dot{m}_f=\dfrac{BPR}{BPR+1}\times \dot{m}\\\dot{m}_f=\dfrac{8}{8+1}\times 546.50\\\dot{m}_f=485.77\ kgs^{-1}[/tex]
Now the exit velocity is calculated using the total energy balance which is given as below:
[tex]h_4+\dfrac{1}{2}V_a^2=h_5+\dfrac{1}{2}V_e^2[/tex]
Here
h_4 and h_5 are the enthalpies at point 4 and 5 which could be rewritten as [tex]c_pT_4[/tex] and [tex]c_pT_5[/tex] respectively.The value of T_4 is the inlet temperature which is 253 KThe value of T_5 is the outlet temperature which is 233KThe value of c_p is constant which is 1005 J/kgKV_a is the inlet velocity which is 250 m/sV_e is the outlet velocity that is to be calculated.So the equation becomes
[tex]h_4+\dfrac{1}{2}V_a^2=h_5+\dfrac{1}{2}V_e^2\\c_pT_4+\dfrac{1}{2}V_a^2=c_pT_5+\dfrac{1}{2}V_e^2[/tex]
Rearranging the equation gives
[tex]\dfrac{1}{2}V_e^2=c_pT_4-c_pT_5+\dfrac{1}{2}V_a^2\\\dfrac{1}{2}V_e^2=c_p(T_4-T_5)+\dfrac{1}{2}V_a^2\\V_e^2=2c_p(T_4-T_5)+V_a^2\\V_e=\sqrt{2c_p(T_4-T_5)+V_a^2}\\V_e=\sqrt{2\times 1005\times (253-233)+(250)^2}\\V_e=320.46 m/s[/tex]
Now using the cold air approach, the thrust is given as follows
[tex]T=\dot{m}_f(V_e-V_a)\\T=485.77\times (320.46-250)\\T=34227.35\ N[/tex]
So the thrust of the engine calculated using the cold air is 34227.35 N
forty gal/min of a hydrocarbon fuel having a spesific gravity of 0.91 flow into a tank truck with load limit of 40,000 lb of fuel. How long will it takee to fill the tank in the truck?
Answer: 131.75minutes
Explanation:
First if all, we've to find the density of liquid which will be:
= Specific gravity × Density to pure water
= 0.91 × 8.34lb/gallon
= 7.59lb/gallon
Then, the volume that's required to fill the tank will be:
= Load limit/Density of fluid
= 40000/7.59
= 5270.1gallon
Now, the time taken will be:
= V/F
= 5270.1/40
= 131.75min
It'll take 131.75 minutes to fill the tank in the truck.
Air is compressed from 100 kPa, 300 K to 1200 kPa in a two-stage compressor with intercooling between the stages. The intercooler pressure is 400 kPa. The air is cooled back to 300 K in the intercooler before entering the second compressor stage. Each compressor stage is isentropic. For steady-state operation and negligible changes in kinetic and potential energy from inlet to exit, determine (a) the temperature at the exit of the seco
Answer:
410.7 K
Explanation:
Given data:
P1 ( initial air pressure ) = 100 kPa
Pi ( intercooler pressure ) = 400 kPa
P2( compressed air pressure ) = 1200 kPa
T1 = 300 k
Ti = 300 k
Calculate for the temperature at the exit of the second stage ( T2 )
we can calculate this using the relation below
T2 / Ti = ( P2/Pi )^(r-1/r)
∴ T2 = 300 ( 1200 / 400 )^( 1.4 - 1 / 1.4 )
= 300 ( 3 )^0.286
= 300 * 1.369 = 410.7 K
Air at 308C, 1 bar, 50% relative humidity enters an insulated chamber operating at steady state with a mass flow rate of 3 kg/min and mixes with a saturated moist air stream entering at 58C, 1 bar with a mass flow rate of 5 kg/min. A single mixed stream exits at 1 bar. Determine (a) the relative humidity and temperature, in 8C, of the exiting stream. (b) the rate of exergy destruction, in kW, for T0 5 208C. Neglect kinetic and potential energy effects.
Answer: the question of whether or is that a new place for a person to come in the morning and then the day that we have a good day at school or to come home with us to go to the church or we could meet up at my place in about a week or so to get the rest of the kids and I can go out to the school to go to the gym to go to the doctor to pick them out or not I have a good time to come over to get the stuff out of the car so I’m going out of the house to go to the store to pick it out I don’t have any money for that
Explanation:
Great amounts of electro-magnetic energy from our son and other bodies n space travel through space. Who's is a logical conclusion about these electro-magnetic waves?
Answer:
Logical conclusion : there are more electromagnetic waves than sunlight
Explanation:
The traveling of electromagnetic energy from the sun and other bodies through space leads to Electromagnetic radiation.
Hence the logical conclusion about Electromagnetic waves is that there are more electromagnetic waves than sunlight
While the travelling of electromagnetic waves through space is described as gliding through space
A cylindrical rod of copper (E = 110 GPa) having a yield strength of 240 MPa is to be subjected
to a load of 6660 N. If the length of the rod is 380 mm, what must be the diameter to allow an
elongation of 0.50 mm?
Answer:
"7.654 mm" is the correct solution.
Explanation:
According to the question,
[tex]E=110\times 10^3 \ N/mm^2[/tex][tex]\sigma_y = 240 \ mPa[/tex][tex]P = 6660 \ N[/tex][tex]L = 380 \ mm[/tex][tex]\delta = 0.5 \ mm[/tex]Now,
As we know,
The Elongation,
⇒ [tex]E=\frac{\sigma}{e}[/tex]
[tex]=\frac{\frac{P}{A} }{\frac{\delta}{L} }[/tex]
or,
⇒ [tex]\delta=\frac{PL}{AE}[/tex]
By substituting the values, we get
[tex]0.5=\frac{6660\times 380}{(\frac{\pi}{4}D^2)(110\times 10^3)}[/tex]
then,
⇒ [tex]D^2=58.587[/tex]
[tex]D=\sqrt{58.587}[/tex]
[tex]=7.654 \ mm[/tex]
A continuous and aligned fiber-reinforced composite is manufactured using 80 vol% aramid fiber (a kevlar-like compound) embedded nylon 6-6. A part for a high-performance aircraft utilizes this composite. If the part experiences 953 lb-f (pounds force) along the fiber alignment axis, what is the force conveyed by the fibers ?
Answer:
the force conveyed by the fibers is 947.93 lb-f
Explanation:
Given the data in the question;
V_f = 80% = 0.8
V_m = 1 - V_f = 1 - 0.8 = 0.2
Now,
length of fibre L_f = length of Nylon L_n
V_f = A_f × L_f = 0.8
V_m = A_n × L_n = 0.2
so
V_f/V_m = A_f/A_n = 0.8/0.2
A_f/A_n = 4
now, the strains in fibre is equal to strains in nylon
(P/AE)f = (P/AE)n
P_f/A_fE_f = P_n/A_nE_n
P_f = (A_f/A_n)(E_f/E_n)(P_n)
P_f = ( 4 )( 131 / 2.8 )(Pn)
P_f = 187.14Pn
and P_n = Pf / 187.14
Hence
given that P_total = 953 lb-f
P_f + P_n = 953
P_f + ( P_f / 187.14 ) = 953
P_f( 1 + ( 1 / 187.14 ) ) = 953
P_f( 1.00534359 = 953
P_f = 953 / 1.00534359
P_f = 947.93 lb-f
Therefore, the force conveyed by the fibers is 947.93 lb-f
Could anyone answer this, please? It's about solid mechanics. I will give you 100 points!!! It's due at midnight.
Answer:
sorry i don't know
Explanation:
rosbel or Janette lol baakkaaa
Answer:
t5g5gtttttttttttttttttttttttttttttttttttttttttttt
Explanation:gt555555555555555555555555555555555555555555555555
Answer:
dawbkjbjwwjhjfbfjewfaekfhawkjndwkja
Explanation: dum*as*