Solution:
Given :
D = 6.35 cm
[tex]$\bar f = 0.005$[/tex]
[tex]$P_s = 280 \ kPa$[/tex]
[tex]$T_s= 825 K[/tex]
a). From fanno flow table (γ = 1.4)
At [tex]$M_1 = 0.12$[/tex] , [tex]$\left(\frac{4 \bar f L_{max}}{D}\right)_1 = 45.408$[/tex]
At [tex]$M_2 = 1$[/tex] , [tex]$\left(\frac{4 \bar f L_{max}}{D}\right)_2 = 0$[/tex]
∴ [tex]$\left(\frac{4 \bar f L_{max}}{D}\right)_1 - \left(\frac{4 \bar f L_{max}}{D}\right)_2 = \frac{4 \bar f L}{D}$[/tex]
[tex]$45.408 - 0 = \frac{4 \times 0.005 \times L}{0.0635}$[/tex]
[tex]$45.408 = \frac{4 \times 0.005 \times L}{0.0635}$[/tex]
L = 144.17 m
b). If L = 25 m
[tex]$\frac{4 \bar f L}{D}=\frac{4 \times 0.005 \times 25}{0.0635} = 7.874$[/tex]
From fanno flow , (γ = 1.4)
[tex]$At, M_2 = 0.26 , \frac{4 \bar f L}{D} = 7.874$[/tex]
[tex]$\frac{P_s}{P_1}=\frac{T_s}{T_1}^{\frac{\gamma}{\gamma - 1}} = (1+\frac{\gamma-1}{2}M_1^2)^{\frac{\gamma}{\gamma - 1}}$[/tex]
[tex]$\frac{280}{P_1}=\frac{825}{T_1}^{\frac{1.4}{1.4 - 1}} = (1+\frac{1.4-1}{2}(0.12)^2)^{\frac{1.4}{1.4 - 1}}$[/tex]
[tex]$\frac{280}{P_1}=\left(\frac{825}{T_1}\right)^{3.5} =1.0101$[/tex]
[tex]$P_1 = 277.2 \ kPa$[/tex]
[tex]$T_1=822.63 \ K$[/tex]
[tex]$\frac{T_2}{T_1}=\frac{1+\frac{\gamma -1}{2}M_1^2}{1+\frac{\gamma -1}{2}M_2^2}$[/tex]
[tex]$\frac{T_2}{822.63}=\frac{1+\frac{1.4 -1}{2}(0.12)^2}{1+\frac{1.4 -1}{2}(0.26)^2}$[/tex]
[tex]$\frac{T_2}{822.63}=\frac{1.00288}{1.01352}$[/tex]
[tex]$T_2=814 \ K$[/tex]
[tex]$\frac{P_2}{P_1}=\frac{M_1}{M_2}\left(\frac{1+\frac{\gamma -1}{2}M_1^2}{1+\frac{\gamma -1}{2}M_2^2}\right)^{1/2}$[/tex]
[tex]$\frac{P_2}{P_1}=\frac{0.12}{0.26}\left(\frac{1+\frac{1.4 -1}{2}(0.12)^2}{1+\frac{1.4 -1}{2}(0.26)^2}\right)^{1/2}$[/tex]
[tex]$\frac{P_2}{277.2}=0.459$[/tex]
[tex]$P_2=127.27 \ kPa$[/tex]
A single phase inductive load draws 10 MW at 0.6 power factor lagging. Draw the power triangle and determine the reactive power of a capacitor to be connected in parallel with the load to raise the power factor to 0.85.
Answer: attached below is the power triangles
7.13589 MVAR
Explanation:
Power ( P1 ) = 10 MW
power factor ( cos ∅ ) = 0.6 lagging
New power factor = 0.85
Calculate the reactive power of a capacitor to be connected in parallel
Cos ∅ = 0.6
therefore ∅ = 53.13°
S = P1 / cos ∅ = 16.67 MVA
Q1 = S ( sin ∅ ) = 13.33 MVAR ( reactive power before capacitor was connected in parallel )
note : the connection of a capacitor in parallel will cause a change in power factor and reactive power while the active power will be unchanged i.e. p1 = p2
cos ∅2 = 0.85 ( new power factor )
hence ∅2 = 31.78°
Qsh ( reactive power when power factor is raised to 0.85 )
= P1 ( tan∅1 - tan∅2 )
= 10 ( 1.333 - 0.6197 )
= 7.13589 MVAR
Which of the following is an example of someone who claims that the media has a shooting blanks effect?
A. "Along with parents, peers, and teachers, the media socializes children about how boys and girls are supposed to behave."
B. "My kid saw a cigarette ad in a magazine and now he's smoking. It's the magazine's fault!"
C. "The media doesn't affect me at all because I'm smart enough to know the difference between right and wrong."
D. "There is no definitive evidence that the media affects our behavior"
Answer:
the answer would be d its d
Answer:
Pretty sure the answer is "C"
Explanation:
"The media doesn't affect me at all because I'm smart enough to know the difference between right and wrong."
For a 3-Phase, Wye connected system the Line to Line Voltage measures 12,470 Volts, the Phase current measures 120 Amps.
Required:
a. What will the Line-to-Neutral/Phase voltage be?
b. What will the Line current be?
Answer:
A. 7199.55 volts
B. 120A
Explanation:
In this question we have the
line voltage = VLL = 12470volts
Phase current = Iph = 120 amps
A.)
We are to calculate the line-to-neutral/phase voltage here
VLL = √3VL-N
VL-N = VLL/√3
VL-N = 12470/√3
This gives a line to neutral phase/voltage of 7199.55 volts.
B.
We are to calculate the line current here:
In this connection, the line current and the phase current are equal
ILL = Iph = 120A
please help me make a lesson plan. the topic is Zigzag line. and heres the format.
A. Objective
B. Subject matter
C. Learning activities.
D. Assessment.
E. Reinforcement
Explanation:
D. B. C. A. E. Is this a good idea
Calculate the Lee for the same wing if we increase the span to 0.245 m. By increasing the span we also increase the glider weight to 0.0523
Answer:
0.21
Explanation:
This would have been a fairly easy one, except for that the first part of the question is missing, and as such, I'd assume a value.
We need to use chord, so, I'm assuming the length of the chord to be 0.045 m
The Area is given by the formula
Area = span * chord
Area = 0.245 * 0.045
Area = 0.011 m²
This area gotten, is what we then divide the glider weight by to get our answer.
Lee = area / weight
Lee = 0.011 / 0.0523
Lee = 0.21
Therefore, using the values of the chord I'd assumed, the Lee of the same wing is 0.21
Answer: 0.2108
Explanation:got it correct
Series aiding is a term sometimes used to describe voltage sources of the same polarity in series. If a 5 V and a 9 V source are connected in this manner, what is the total voltage?
Answer:Total Voltage = 14V
Explanation: it is possible that a circuit can contain more than one source of electromotive force which can cause flow of current in the same or opposite direction . When the connection to voltage sources allows for current from the voltage sources to flow in same direction,it is termed Series aiding Thus, the Total/effective voltage in a series aiding circuit is computed as the sum of series aiding voltages .
Here we have the series aiding voltages to be 5V and 9V ,
therefore,
Total Voltage = 5V + 9V
= 14V
A 13.7g sample of a compound exerts a pressure of 2.01atm in a 0.750L flask at 399K. What is the molar mass of the compound?a. 318 g/mol
b. 204 g/mol
c. 175 g/mol
d. 298 g/mol
Answer: Option D) 298 g/mol is the correct answer
Explanation:
Given that;
Mass of sample m = 13.7 g
pressure P = 2.01 atm
Volume V = 0.750 L
Temperature T = 399 K
Now taking a look at the ideal gas equation
PV = nRT
we solve for n
n = PV/RT
now we substitute
n = (2.01 atm x 0.750 L) / (0.0821 L-atm/mol-K x 399 K )
= 1.5075 / 32.7579
= 0.04601 mol
we know that
molar mass of the compound = mass / moles
so
Molar Mass = 13.7 g / 0.04601 mol
= 297.7 g/mol ≈ 298 g/mol
Therefore Option D) 298 g/mol is the correct answer
Products exit a combustor at a rate of 100 kg/sec, and the air-fuel ratio is 9. Determine the air flow rate. a. 9 kg/sec b. 90 kg/sec c. 100 kg/sec d. 10 kg/sec
Answer: the air flow rate a is 90 kg/sec; Option b) 90 kg/sec is the correct answer
Explanation:
Given that;
product of combustor flow rate m = 100 kg/s
air-fuel = 9
Airflow rate = ?
⇒We know that in the combustor, air fuel are mixed and then ignited,
⇒air fuel products are exited at the combustor
let air and fuel be a and b respectively
⇒ a + b = 100 kg/sec ----- let this be equation 1
now
⇒ air / fuel = 9
a / b = 9
a = 9b -----------let this be equation 2
now input a = 9b in equation 1
9b + b = 100 kg/sec
10b = 100 kg/sec
b = 10 kg/sec
we know that
a = 9b
so a = 9 × 10 = 90 kg/sec
Therefore the air flow rate a is 90 kg/sec
please what is dif
ference between building technology and building engineering.
Answer:
Building technology is building technology such as coding an app or a website
Building engineering is making computers or cars or phones
Explanation:
Solar energy stored in large bodies of water, called solar ponds, is being used to generate electricity. If such a solar power plant has an efficiency of 4.5 percent and a net power output of 150 kW, determine the average value of the required solar energy collection rate, in Btu/h.
Answer: 1.137*10^7 Btu/h.
Explanation:
Given data:
Efficiency of the plant = 4.5percent
Net power output of the plant = 150kw
Solution:
The required collection rate
QH = W/n
= 150/0.045 * 0.94782/ 1 /60 */60 Btu/h.
= 3333.333 *3412.152Btu/h.
= 11373840 Btu/h
= 1.137*10^7 Btu/h.
Compute the solution to x + 2x + 2x = 0 for Xo = 0 mm, vo = 1 mm/s and write down the closed-form expression for the response.
Answer:
β = [tex]\frac{c}{\sqrt{km} }[/tex] = 0.7071 ≈ 1 ( damping condition )
closed-form expression for the response is attached below
Explanation:
Given : x + 2x + 2x = 0 for Xo = 0 mm and Vo = 1 mm/s
computing a solution :
M = 1,
c = 2,
k = 2,
Wn = [tex]\sqrt{\frac{k}{m} }[/tex] = [tex]\sqrt{2}[/tex]
next we determine the damping condition using the damping formula
β = [tex]\frac{c}{\sqrt{km} }[/tex] = 0.7071 ≈ 1
from the condition above it can be said that the damping condition indicates underdamping
attached below is the closed form expression for the response
People tend to self-disclose to others that are in age, social status, religion, and personality.
Answer:people tend to do this when they are in a different environment they lose something or just have something going on in their life
Explanation:
the reaction of 4A+3B→2C+D is studied. Unknown masses of the reactants were mixed . After a reaction time of 1 hour the analysis of the mixture showed 2 kmol, 1 kmol of B and 4 kmol of C. product D was present in the mixture but could not be analysed. what is the mole fraction of D in the mixture?
Answer: the mole fraction of D in the mixture is 0.2222
Explanation:
Given that;
mixture analysis shows 2 kmol A, 1 kmol B, 4 kmol C and some unknown kmol of D was present.
4A+3B→2C+D
As from reaction stoichiometry, for every 2 kmol of C produced, kmol of D produced = 1 kmol
so, for 4 kmol C, kmol of D produced = 4/2 × 1 kmol = 2 kmol
Now our mixture has 2 kmol A, 1 kmol B, 4 kmol C and also 2 kmol of D
so, total moles in mixture, we have (2 + 1 + 4 + 2) kmol = 9 kmol
mole fraction of D in mixture will be;
( Kmol of D) / (total moles in mixture) = 2 / 9 = 0.2222
Therefore the mole fraction of D in the mixture is 0.2222
If ice homogeneously nucleates at -44.6°C, calculate the critical radius given values of -3.1 × 10^8 J/m3 and 25 × 10^-3 J/m^2, respectively, for the latent heat of fusion and the surface free energy.
Answer:the critical radius for the homogeneous nucleating ice is 0.986 nm
Explanation:
Using the formulae below to calculate the critical radius for homogeneous nucleation,
We Have that
Critical radius ( r *) = [2γ Tm/ΔHf]{ 1/ Tm- T]
where γ = surface free energy =25 × 10^-3 J/m^2
Tm= solidification temperature at equilibrium =273K
Hf= latent heat of fusion = -3.1 × 10^8 J/m3
Temperature , T = -44.6°C
Critical radius ( r *) = (2 X 25 × 10^-3 J/m^2 x 273)/ (-3.1 × 10^8 J/m3 ) X ( 1/ 273 - ( -44.6 +273)
4.40x 10^-8X ( 1/273-228.4)
4.40x 10^-8 X 1/44.6
4.40x 10^-8 x 0.0224
9.86x 10 ^-10m =0.986 x 10 ^-9m = 0.986nm
How would you expect an increase in the austenite grain size to affect the hardenability of a steel alloy? Why?
Answer:
The hardenability increases with increasing austenite grain size, because the grain boundary area is decreasing. This means that the sites for the nucleation of ferrite and pearlite are being reduced in number, with the result that these transformations are slowed down, and the hardenability is therefore increased.
Which of these words was first used during the 1970s economic crisis?
influx
stagflation
deficit
programs
Answer:
stagflation
Explanation:
it was used in the article
A word which was first used during the 1970s economic crisis is stagflation.
The economic crisis of the 1970s.In the 1970s, an energy crisis took place in the United States of America due to the oil embargo that was imposed on it by OPEC. This oil embargo was imposed on the United States of America by the Organization of Petroleum Exporting Countries (OPEC) in 1973 because of its role in the Arab-Israeli War.
Consequently, the economy of the United States of America experienced stagflation in the following ways:
Slow economic growth.Relatively high unemployment.Read more on stagflation here: https://brainly.com/question/25505087
what substance does light travel through before putting water in the cup
Steam enters an adiabatic nozzle at 1 MPa, 250°C, and 30 m/s. At one point in the nozzle the enthalpy dropped 40 kJ/kg from its inlet value. Determine velocity at that point. (A) 31 m/s (B) 110 m/s (C) 250 m/s (D) 280 m/s
Answer:
284.4 m/s
Explanation:
At the inlet of the nozzle P =1 atm.
Temperature T = 250° C
Velocity of the steam at the inlet V_1 = 30 m/s
Change in enthalpy Δh = 40 KJ/kg
let V_2 be the final velocity
then
[tex]V_2 =\sqrt{2\Delta h+V_1^2} \\=\sqrt{2\times40+30^2}\\= 284.4 m/s[/tex]
Explain combined normal and shear stresses with sketch. Write the general expression for (a) Normal and shear stresses on inclined plane (b) principal and maximum shear stresses and identify all the terms in the expression including units.
Answer:
a) Normal stress :
бn =[ ( бx + бy ) / 2 + ( бx - бy ) / 2 ] cos2∅ + Txysin2∅
shear stress
Tn = ( - бx - бy ) / 2 sin2∅ + Txy cos2∅
b) principal stress :
б1 = ( бx + бy ) / 2 - [tex]\sqrt{}[/tex]( ( бx - бy ) / 2 )^2 + T^2xy
maximum shear stress:
Tmax = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy
Explanation:
Combined normal stress and shear stress sketches attached below
The terms in the sketch are :
бx = tensile stress in x direction
бy = tensile stress in y direction
Txy = y component of shear stress acting on the perpendicular plane to x axis
бn = Normal stress acting on the inclined plane EF
Tn = shear stress acting on the inclined plane EF
A) Normal and shear stresses on inclined plane
Normal stress :
бn =[ ( бx + бy ) / 2 + ( бx - бy ) / 2 ] cos2∅ + Txysin2∅
shear stress
Tn = ( - бx - бy ) / 2 sin2∅ + Txy cos2∅
B) principal and maximum shear stresses
principal stress :
б1 = ( бx + бy ) / 2 - [tex]\sqrt{}[/tex]( ( бx - бy ) / 2 )^2 + T^2xy
maximum shear stress:
Tmax = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy
A fluid has a mass of 5 kg and occupies a volume of 1 m3 at a pressure of 150 kPa. If the internal energy is 25000 kJ/kg, what is the total enthalpy?
Answer:
155 KJ
Explanation:
The total enthalpy is given by
ΔH=ΔU + PV
Where;
ΔH = enthalpy
ΔU = internal energy = 25000 kJ/kg/ 5 kg = 5000 KJ
P = 150 kPa = 150,000 Pa
V = 1 m3
ΔH= 5000 + (150,000 * 1)
ΔH= 155 KJ
A three-phase motor rated 25 hp, 480 V, operates with a power factor of 0.74 lagging and supplies the rated load. The motor efficiency is 96%. Calculate the motor input power, reactive power and current.
Answer:
the motor input power is 19.42 KW
the Reactive power is 17.65 KVAR
Current is 31.56 A
Explanation:
Given that;
V = 480V
h.p = 25 hp
p.f = 0.74 lagging
n_motor = 96%
so output = 25hp
and we know that;
1hp = 746 watt
watt = hp × 1hp
so output in watt = 25 × 746 = 18650 Watt = 18.65 KW
n_motor = (output / input) × 100
96 = 1865 / Input
96Input = 1865
Input = 1865 / 96
Input = 19.42 KW
Therefore the motor input power is 19.42 KW
P = √( 3 × V × I × cos∅)
19.42 = √( 3 ×480 × I × 0.74)
I = 31.56 A
Therefore Current is 31.56 A
Q = √( 3 × V × I × sin∅)
we know that
cos∅ = 0.74
so ∅ = cos⁻¹(0.74) = 42.26
so we substitute
Q = √( 3 × 480 × 31.56 × sin(42.26))
= 17.65 KVAR
Therefore the Reactive power is 17.65 KVAR
A magnesium–lead alloy of mass 7.5 kg consists of a solid α phase that has a composition just slightly below the solubility limit at 300°C.
(a) What mass of lead is in the alloy?
(b) If the alloy is heated to 400°C, how much more lead may be dissolved in the αα phase without exceeding the solubility limit of this phase?
Answer:
(a)This portion of the problem asks that we calculate, for a Pb-Mg alloy, the mass of lead in 7.5 kg of thesolidphase at 300C just below the solubility limit.From Figure 9.20, the solubility limit for thephase at
Explanation:
A production line manufactures 10-liter gasoline cans with a volume tolerance of up to 5%. The probability that any one is out of tolerance is 0.03. If five cans are selected at random. a) What is the probability that they are all out of tolerance? b) What is the probability that exactly two are out of tolerance?
Answer:
In the case of the production Line, we know that,
No of gasoline cans = 5
probability that 1st can is out of tolerance = 0.03
probability that 2nd can is out of tolerance = 0.03
.
.
probability that the 5th can is out of tolerance = 0.03
Therefore,
probability of 1st can out of tolerance + probability of 1st can not out of tolerance = 1
Probability of 1st can not out of tolerance = 1 -- 0.03 = 0.97
probability of 2nd can not out of tolerance = 0.97
.
.
probability of 5th can not out of tolerance = 0.97
Question A:
Probability that they are all out of tolerance
= P(1st can out of tolerance) * P(2nd can out of tolerance) * P(3rd can out of tolerance) * P(4th can out of tolerance) * P(5th can out of tolerance)
= (0.03 ) * (0.03) * (0.03) * (0.03) * (0.03) = 2.43 E⁻⁸ (2.43 ˣ 10⁻⁸)
Question B:
Probability that exactly two are out of tolerance
= P(1st can is out of tolerance) * P(2nd can is out of tolerance) * P(3rd can is not out of tolerance) * P(4th can is not out of tolerance) * P(5th can is not out of tolerance)
= (0.03) * (0.03) * (0.97) * (0.97) * (0.97) = 0.0008214057
Explanation:
Indicate similarities between a nucleus and a liquid droplet; why small droplets are stable and very big droplets are not?
Answer:
There are several similarities between the nucleus and a liquid droplet.
Explanation:
A droplet of liquid simply is is very small or tiny drop of liquid. It is also considered as a tiny column of liquid that is surrounded by surfaces that have zero shear stress.
A nucleus on the other hand is an assembly between protons and neutrons. The latter is electrically charged whilst the former is positively charged. The number of protons present in an element is very crucial to the qualities of an element.
The main similarities between a nucleus and a liquid droplet are:
1. a nucleus consists of a large amount of neutrons and protons in the same volume as would a liquid which contains large numbers of molecules in the same volume;
2. both the nucleus and the droplet are similar for their homogeneity in electric charge and density;
3. the molecules exert the same amount for forces towards one another as would the nuclear forces in the nucleons.
4. both of them cannot be compressed
5. both molecules and nucleus are can be subject to nuclear fission which simply mean the breaking apart into smaller units (in the case of the nucleus) or the breaking apart into smaller droplets in the case of the liquid molecule.
6. There are two types of phenomena which occurs in both the liquid droplet and the nucleus which are similar to one another. They are:
Evaporation (in the case of the liquid molecule) and reaction emission (in the case of the nucleus). In evaporation, particles are lost, in Atomic transmutation, particles are lost as well.
B) the forces which determine the stability of droplets are surface tension and gravitation. The smaller the area, the stronger the surface tension available to keep the drops from going out of shape.
Cheers
For proper function hydraulics systems need a reservoir of which of the following?
A.) Compressible fluid
B.) Non-compressible fluid C.) Non-compressible air
Calculate the molar volume of ammonia at 92oC and 310 bar. What phase is the ammonia in?
Answer:
The ammonia is still in the gas phase
Explanation:
Given that 1 bar is approximately = 1 atm
From;
PV=nRT
P= 310 atm
V= the unknown
n= 1
R = 0.082atm LK-1mol-1
T = 92oC + 273 = 365 K
V= nRT/P
V= 1 * 0.082 * 365/310
V = 0.0965 L = 96.5 mL
molar mass of NH3 = 17 g/mol
Molar density of NH3 = 17g/ 96.5mL = 0.176g/mL
The ammonia is still in the gas phase
Laminar flow normally persists on a smooth flat plate until a critical Reynolds number value is reached. However, the flow can be tripped to a turbulent state by adding roughness to the leading edge of the plate. For a particular situation, experimental results show that the local heat transfer coefficients for laminar and turbulent conditions are
h_lam(x)= 1.74 W/m^1.5. Kx^-0.5
h_turb(x)= 3.98 W/m^1.8 Kx^-0.2
Calculate the average heat transfer coefficients for laminar and turbulent conditions for plates of length L = 0.1 m and 1 m.
Answer:
At L = 0.1 m
h⁻_lam = 11.004K W/m^1.5
h⁻_turb = 7.8848K W/m^1.8
At L = 1 m
h⁻_lam = 3.48K W/m^1.5
h⁻_turb = 4.975K W/m^1.8
Explanation:
Given that;
h_lam(x)= 1.74 W/m^1.5. Kx^-0.5
h_turb(x)= 3.98 W/m^1.8 Kx^-0.2
conditions for plates of length L = 0.1 m and 1 m
Now
Average heat transfer coefficient is expressed as;
h⁻ = 1/L ₀∫^L hxdx
so for Laminar flow
h_lam(x)= 1.74 . Kx^-0.5 W/m^1.5
from the expression
h⁻_lam = 1/L ₀∫^L 1.74 . Kx^-0.5 dx
= 1.74k / L { [x^(-0.5+1)] / [-0.5 + 1 ]}₀^L
= 1.74k/L = [ (x^0.5)/0.5)]⁰^L
= 1.74K × L^0.5 / L × 0.5
h⁻_lam= 3.48KL^-0.5
For turbulent flow
h_turb(x)= 3.98. Kx^-0.2 W/m^1.8
form the expression
1/L ₀∫^L 3.98 . Kx^-0.2 dx
= 3.98k / L { [x^(-0.2+1)] / [-0.2 + 1 ]}₀^L
= (3.98K/L) × (L^0.8 / 0.8)
h⁻_turb = 4.975KL^-0.2
Now at L = 0.1 m
h⁻_lam = 3.48KL^-0.5 = 3.48K(0.1)^-0.5 W/m^1.5
h⁻_lam = 11.004K W/m^1.5
h⁻_turb = 4.975KL^-0.2 = 4.975K(0.1)^-0.2
h⁻_turb = 7.8848K W/m^1.8
At L = 1 m
h⁻_lam = 3.48KL^-0.5 = 3.48K(1)^-0.5 W/m^1.5
h⁻_lam = 3.48K W/m^1.5
h⁻_turb = 4.975KL^-0.2 = 4.975K(1)^-0.2
h⁻_turb = 4.975K W/m^1.8
Therefore
At L = 0.1 m
h⁻_lam = 11.004K W/m^1.5
h⁻_turb = 7.8848K W/m^1.8
At L = 1 m
h⁻_lam = 3.48K W/m^1.5
h⁻_turb = 4.975K W/m^1.8
Calculate the LER for the rectangular wing from the previous question if the weight of the glider is 0.0500 Newton’s.
Answer:
0.2
Explanation:
Since the span and chord of the rectangular wing is missing, due to it being from the other question, permit me to improvise, or assume them. While you go ahead and substitute the ones from your question to it, as it's both basically the same method.
Let the span of the rectangular wing be 0.225 m
Let the chord of the rectangular wing be 0.045 m.
Then, the area of any rectangular chord is
A = chord * span
A = 0.045 * 0.225
A = 0.010 m²
And using the weight of the glider given to us from the question, we can find the LER for the wing.
LER = Area / weight.
LER = 0.010 / 0.05
LER = 0.2.
Therefore, using the values of the rectangular wing I adopted, and the weight of the glider given, we can see that the LER of the glider is 0.2
Please mark brainliest...
Answer: 0.2025
Explanation: I got it correct
Air enters an adiabatic turbine at 900 K and 1000 kPa. The air exits at 400 K and 100 kPa with a velocity of 30 m/s. Kinetic and potential energy changes are negligible. If the power delivered by the turbine is 1000 kW.
Required:
a. Find the mass flow rate.
b. Find the diameter of the duct at the exit.
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It is important to keeo a copy of your written plan and safety record s off-site. True or false
Answer:
The answer for the question is true
Explanation:
If you get a virus or get hacked you will still have it saved