Answer:
The following statements are true:
A. For flows over a flat plate, in the laminar region, the heat transfer coefficient is decreasing in the flow direction
C. For flows over a flat plate, the transition from laminar to turbulence flow only happens for rough surface
E. In general, turbulence flows have a larger heat transfer coefficient compared to laminar flows 6.
Select ALL statements that are TRUE
B. In the hydrodynamic fully developed region, the mean velocity of the flow becomes constant
D. For internal flows, if Pr>1, the flows become hydrodynamically fully developed before becoming thermally fully developed
Explanation:
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
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
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.
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:
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
6 A series of vane shear tests has been performed in a stratum of inorganic clay that has a plasticity index of 50. The vane had a diameter of 50 mm and a height of 100 mm. The test results were as follows: Depth (m) Torque at Failure (N-m) 3.4 12.7 4.1 18.1 5.0 15.8 6.6 20.1 Compute the undrained shear strength, su, for each test, then combine this data to determine a single su value for this stratum.
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
An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standard power cycle for which the working fluid is evaporated, passed through a turbine, and subsequently condensed. The system is to be used in very special locations for which the oceanic water temperature near the surface is approximately 300 K, while the temperature at reasonable depths is approximately 280 K. The warmer water is used as a heat source to evaporate the working fluid, while the colder water is used as a heat sink for condensation of the fluid. Consider a power plant that is to generate 2 MW of electricity at an efficiency (electric power output per heat input) of 3%. The evaporator is a heat exchanger consisting of a single shell with many tubes executing two passes. If the working fluid is evaporated at its phase change temperature of 290 K, with ocean water entering at 300 K and leaving at 292 K.
Required:
a. What is the heat exchanger area required for the evaporator?
b. What flovw rate must be maintained for the water passing through the evaporator?
Answer:
a) the heat exchanger area required for the evaporator is 11178.236 m²
b) the required flow rate is 1993630.38 kg/s
Explanation:
Given the data in the question;
Water temperature near the surface = 300 K
temperature at reasonable depths ( cold ) = 280 K
power plant output W' = 2 MW
efficiency η = 3% = 0.03
we know that; efficiency η = W'[tex]_{power-out[/tex] / Q[tex]_{supplied[/tex]
we substitute
0.03 = 2 / Q[tex]_{supplied[/tex]
Q[tex]_{supplied[/tex] = 2 / 0.03
Q[tex]_{supplied[/tex] = 66.667 MW = 66.667 × 10⁶ Watt
T[tex]h_{in[/tex] = 300 K T[tex]h_{out[/tex] = 292 K
T[tex]c_{in[/tex] = 290 K T[tex]c_{out[/tex] = 290 K
Now, Heat transfer in evaporator;
Q = UA( LMTD )
so
LMTD = (ΔT₁ - ΔT₂) / ln( ΔT₁ / ΔT₂ )
first we get ΔT₁ and ΔT₂
ΔT₁ = T[tex]h_{in[/tex] - T[tex]c_{out[/tex] = 300 - 290 = 10 K
ΔT₂ = T[tex]h_{out[/tex] - T[tex]c_{in[/tex] = 292 - 290 = 2 K
so we substitute into our equation;
LMTD = (10 - 2) / ln( 10 / 2 )
LMTD = 8 / ln( 5 )
LMTD = 8 / 1.6094379
LMTD = 4.97
a) Heat transfer Area will be;
Q[tex]_H[/tex] = UA( LMTD )
we substitute
66.667 × 10⁶ = 1200 × A × 4.97
66.667 × 10⁶ = 5964 × A
A = (66.667 × 10⁶) / 5964
A = 11178.236 m²
Therefore, the heat exchanger area required for the evaporator is 11178.236 m²
b) Flow rate
we know that;
Q[tex]_H[/tex] = m'C[tex]_P[/tex]( [tex]T_{in[/tex] - [tex]T_{out[/tex] )
specific heat capacity of water Cp = 4.18 (kJ/kg∙°C)
we substitute
66.667 × 10⁶ = m' × 4.18 × ( 300 - 292 )
66.667 × 10⁶ = m' × 33.44
m' = ( 66.667 × 10⁶ ) / 33.44
m' = 1993630.38 kg/s
Therefore, the required flow rate is 1993630.38 kg/s
Determine the mean effective pressure of an ideal Otto Cycle that uses air as the working fluid; its state at the beginning of the compression is 14 psia and 60oF; its temperature at the end of the combustion is 1500oF; and its compression ratio is 9. Use constant specific heats at room temperature.
Answer:
31.282 psia
Explanation:
Calculate the mean effective pressure of an ideal Otto cycle
Given data:
compression ratio = 9
inlet pressure ( P1 ) = 14 psia
Initial temperature ( T1 ) = 60°F = 520 R
Final/maximum temp ( T3 ) = 1500°F = 1960 R
using the constant specific heats at room temperature
attached below is the detailed solution
The mean effective pressure of the ideal Otto Cycle that uses air as the working fluid is; 31.28 psia
We are given;
Initial pressure; P₁ = 14 psia
Initial temperature; T₁ = 60°F = 520 R
Maximum temperature; T₃ = 1500°F = 1960 R
Compression ratio = 9
From tables, the specific heat capacities and constants are;
c_v = 0.171 btu/lbm.R
c_p = 0.24 btu/lbm.R
R = 0.3704 btu/lbm.R
ratio of specific heats; k = 1.4
Let us first calculate the initial volume from the formula;
V₁ = RT₁/P₁
V₁ = (0.3704 * 520)/14
V₁ = 13.7577 ft³/lb.m
Compression ratio is; V₁/V₂ = 9Thus;
V₂ = V₁/9
V₂ = 13.7577/9
V₂ = 1.5286 ft³/lb.m
To get the temperature T₂, we will use the formula;T₂ = T₁(V₁/V₂)^(k - 1)
T₂ = 520(9)^(1.4 - 1)
T₂ = 1252.277 R
Similarly;
T₄ = T₃/9
T₄ = 1960/(9^(1.4 - 1))
T₄ = 813.87 R
Formula for heat entering and heat exiting are;Q_in = c_v(T₃ - T₂)
Q_in = 0.171(1960 - 1252.277)
Q_in = 121.02 btu/lbm
Q_out = c_v(T₄ - T₁)
Q_out = 0.171(813.87 - 520)
Q_out = 50.251 btu/lbm
Net work done is given by;W_net = Q_in - Q_out
W_net = 121.02 - 50.251
W_net = 70.769 btu/lbm
Formula for the mean effective pressure is;
mean effective pressure = W_net/(V₁ - V₂)
mean effective pressure = 70.769/(13.7577 - 1.5286)
mean effective pressure = 5.7869 btu.ft³
Converting to psia gives;
Mean effective pressure = 31.28 psia
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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 liquid propellant engine has the following characteristics: chamber pressure of 7 MPa, constant ratio of specific heats of 1.3, and a characteristic velocity of 1600 m/s. The nozzle has the following characteristics: throat area of 0.010 m2 and an expansion ratio of 10. Calculate the following:
1. Thrust coefficient at sea level
2. Specific impulse at sea level
3. Altitude at optimal expansion
4. Thrust coefficient at optimal expansion
5. Mass flux through the throat
Answer:
1.55260 N.s3370 m1.643.75 kg/sExplanation:
1) Thrust coefficient at sea level.
Cfsl = TSL / Pca
TSL = Mp * Vc + ( Pc - Pa )Ac
Mp = mass flux = 43.75 kg/s
∴ Cfsl = Mp Vc / Pca + ( Pc - Pa )/Pc * ( Ac / A* )
= 1.6 - 0.04923 = 1.55
2) Specific impulse at sea
Isp = Vc / g = 2549.75 / 9.81
= 260 N.s
3) Altitude at optimal expansion
H = 3370 m
4) thrust coefficient at optimal expansion
CF = 1.6
attached below is the detailed solution
5) Mass flux through the throat
Mass flux = P1 * At / Cc
= ( 7*10^6 * 0.01 ) / 1600
= 43.75 kg/s
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
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
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
An ideal Diesel Cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the maximum air temperature and the rate of heat addition to this cycle when it produces 200 hp of power; the cycle is repeated 1200 times per minute; and the state of the air at the beginning of the compression is 95 kPa and 17oC. Use constant specific heats at room temperature.
Answer:
A) Q'_in = 304.66 hp
B) T_max = 1109.29 °C
Explanation:
We are given;
compression ratio; r = 18
cutoff ratio; r_c = 1.5
Power; W' = 200 hp
Temperature; T1 = 17°C = 290 K
A) To find the rate of heat addition, we will use the derived formula;
Q'_in = (W' × r^(k - 1) × k(r_c - 1))/(r^(k - 1) × k(r_c - 1)) - ((r_c)^(k) - 1)
Where k is a constant = 1.4
Thus;
Q'_in = (200 × 18^(1.4 - 1) × 1.4(1.5 - 1))/((18^(1.4 - 1) × 1.4(1.5 - 1)) - ((1.5^(1.4)) - 1)
Q'_in = 304.66 hp
B) To find the maximum temperature, we will use the Formula;
T_max = T1 × r^(k - 1) × r_c
T_max = 290 × 18^(1.4 - 1) × 1.5
T_max = 1382.29 K
Converting to °C gives;
T_max = 1109.29 °C
Explain when it is appropriate to use Tier II?
????? what do u mean tier 2
superposition theorem
Answer:
mark me brainliest
Explanation:
Superposition theorem is one of those strokes of genius that takes a complex subject and simplifies it in a way that makes perfect sense. A theorem like Millman’s certainly works well, but it is not quite obvious why it works so well. Superposition, on the other hand, is obvious.
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]
An aircraft engine operates on a simple ideal Brayton Cycle with a pressure ratio of 10. Heat is added to the cycle at a rate of 500 kW; air passes through the engine at a rate of 1 kg/s; and the air at the beginning of the compression is at 70 kPa and 0oC. Determine the power produced by this engine and its thermal efficiency. Use constant specific heats at room temperature.
Answer: look at the screenshot
The power and thermal efficiency of this engine is equal to 241 Kilowatts and 48.2% respectively.
How to calculate the power and thermal efficiency?
First of all, we would determine the thermal efficiency of this engine by applying the following formula:
[tex]\eta= 1-\frac{1}{r_p^{k-1/k}} \\\\\eta= 1-\frac{1}{10^{1.4-1/1.4}} \\\\\eta= 1-\frac{1}{10^{0.4/1.4}}\\\\\eta= 1-\frac{1}{10^{0.4/1.4}}\\\\\eta= 1-\frac{1}{10^{0.2857}}\\\\\eta = 0.482[/tex]
Thermal efficiency = 48.2%.
Now, we can determine net power output as follows:
[tex]W_{out}=nq_{in}\\\\W_{out}= 0.482 \times 500\\\\W_{out}=241\;kW.[/tex]
Power = 241 Kilowatts.
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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
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:
In a certain balanced three phase system each line current is a 5a and each line voltage is 220v . What is the approximate real power if the power factor is 0.7
Answer:
1,334
Explanation:
In a certain balanced three phase system each line current is a 5a and each line voltage is 220v . 3.87 kilowatts is the approximate real power if the power factor is 0.7.
The idea of real power in a balanced three-phase system is critical in electrical power systems. It denotes the actual power transferred and used by the system, which is usually measured in kilowatts (kW) or megawatts (MW). Understanding real power is critical for evaluating electrical system efficiency and performance. Real power is the component of power in a balanced three-phase system that does useful work, such as operating motors, generating heat, or powering appliances.
Real Power (P) = √3 × Line Current (I) × Line Voltage (V) × Power Factor (PF)
P = √3 × 5 A × 220 V × 0.7
P = 3.87 kilowatts (kW)
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The acronym ABS stands for _____.
A. Antilock Braking System
B. Antilock Buoyancy System
C. Automatic Braking System
D. American Braking System
The acronym ABS stands for Antilock Braking System. The correct option is A.
Thus, ABS, as it is frequently abbreviated, stands for "Anti-lock Braking System."
Especially in emergency situations or on slick conditions, an anti-lock braking system is a safety device in cars that helps prevent the wheels from locking up when braking.
ABS lowers the possibility of sliding or losing control of the vehicle by enabling the driver to maintain steering control and maximize braking efficiency.
Thus, The acronym ABS stands for Antilock Braking System. The correct option is A.
Learn more about ABS, refer to the link:
https://brainly.com/question/34459929
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Draw a sketch of the following situations identifying the system or control volume, and the boundary of the system or the control surface:
(a) The combustion gases in a cylinder during the power stroke.
(b) the combustion gases in a cylinder during the exhaust stroke.
(c) a balloon exhausting air.
(d) an automobile tire being heated while driving.
(e) a pressure cooker during operation.
Answer:
Attached below
Explanation:
a) combustion gases in a cylinder during the power stroke is A closed system
b) Combustion gases in a cylinder during the exhaust stroke is an Open system
c) A balloon exhausting air is an Open system
d) It is a system
e) This is a control volume
Attached below is the sketch of the following situations
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
describe five tools used in suspension system service and repair
Answer:
Explanation:
You'll still need wrenches, sockets and screwdrivers, but there are other things that it may be necessary to buy to complete the work.
Spring Compressor. One part of suspension repair is replacing coil springs.
Hydraulic Puller. -Hydraulic pullers are used to remove shaft-fitted parts (bearings or couplings). Pullers use a controlled hydraulic force in an effective way and can quickly separate (especially compared to the manual alternative) the parts.
CV Boot Tool. he CV Boot is a ribbed, rubber flexible boot that keeps water and dirt out of the joint and the special grease inside the joint.
Torque Wrench. .A torque wrench is a tool used to apply a specific torque to a fastener such as a nut, bolt, or lag screw. ... A torque wrench is used where the tightness of screws and bolts is crucial.
Ball Joint Separator. This tool is used to separate the ball joint from the spindle support arm. It works on many domestic and import front wheel drive vehicles and is adjustable up to 2" for different size ball joints.
Strut Nuts.
Tie Rod Puller.
etc....