A jet aircraft is in level flight at an altitude of 30,000 ft with an airspeed of 500 ft/s. The aircraft has a gross weight of 19,815 lb, a wingspan of 53.3 ft, and an average chord length of 6 ft. The Oswald efficiency factor is 0.81 and the zero-lift drag coefficient is equal to 0.02. The jet has two turbofan engines, each producing a maximum thrust of 3,650 lb at sea level.
Required:
a. Create a plot of the drag polar for this aircraft for CL from 0 to 5. Plot CL on the vertical axis, CD on the horizontal axis, and do not include negative CL values.
b. What is the total drag coefficient at the flight condition described above?
c. What is the required thrust for level flight at this altitude in lb?
d. If the pilot runs the engines at maximum thrust, what is the instantaneous rate of climb at this altitude and velocity?
Answer:
a) attached below
b) 0.0337
c) 2730.206 Ib
d) 2320.338 ft/min
Explanation:
a) Plot of the drag polar for this aircraft
first we will calculate :
Wing area (s) = Wing span (b) * Average chord length(c)
= 53.3 * 6 = 319.8 ft^2
Aspect ratio = b^2 / s = 8.883
K = 1 / [tex]\pi[/tex]eAR = 1 /
Drag polar ( Cd ) = 0.02 + 0.044 C^2L
attached below is a plot of the drag polar
Attached below is the detailed solution of the remaining part of the question
7.13 An intersection approach has a saturation flow rate of 1500 veh/h, and vehicles arrive at the approach at the rate of 800 veh/h. The approach is controlled by a pretimed signal with a cycle length of 60 seconds and D/D/1 queuing holds. Local standards dictate that signals should be set such that all approach queues dissipate 10 seconds before the end of the effective green portion of the cycle. Assuming that approach capacity exceeds arrivals, determine the maximum length
Answer:
23.34 seconds
Explanation:
Flow rate = 1500
Arrival = 800 vehicle per hour
Cycle c = 60 seconds
Dissipation time = 10 seconds
Arrival time = 800/3600 = 0.2222
Rate of departure = 1500/3600 = 0.4167
Traffic density p = 0.2222/0.4167 = 0.5332
Real time = r
r + to + 10 = c
to = c-r-10 ----1
t0 = p*r/1-p ----2
Equate both 1 and 2
C-r-10 = p*r/1-p
60-r-10 = 0.5332r/1-0.5332
50-r = 0.5332r/0.4668
50-r = 1.1422r
50 = 1.1422r + r
50 = 2.1422r
r = 50/2.1422
r = 23.34 seconds
Design a filter that has infinite DC gain, a gain of one from 1Hz to 100 Hz and filters (1storder) any signals above 100 Hz.a) Sketch the bode plotb) Sketch the s-planec) write the transfer function of the filterd) write the differential equatione) write out the unforced transient responsef) write out the frequency response
Answer:
Attached below are the sketches
answer :
c) G(s) = 100 / ( s + 100 )
d) y'(t) + 100Y(s) = 100 X(s)
e) g(t) = e^-100t u(t)
Explanation:
a) Sketch the bode plot
The filter here is a low pass filter
b) Sketch the s-plane
attached below. pole ( s ) is at 100
c) write the transfer function of the filter
Transfer function ; G(s) = 100 / ( s + 100 )
d) write the differential equation
Y(s) / X(s) = 100 / s + 100
Y(s) [ s + 100 ] = 100 X(s)
= sY(s) + 100Y = 100 X(s)
∴ differential equation = y'(t) + 100Y(s) = 100 X(s)
e) write out the unforced transient response
g(t) = e^-100t u(t)
f) write out the frequency response
attached below
Hot engine oil with heat capacity rate of 4440 w/k (product of mass flow rate and specific heat) and an inlet temperature of 150°c flows through a double pipe heat exchanger. the double pipe heat exchanger is constructed of a 1.5-m-long copper pipe (k = 250 w/m·k) with an inner tube of inside diameter 2 cm and outside tube diameter of 2.25 cm. the inner diameter of the outer tube of the double pipe heat exchanger is 6 cm. oil flowing at a rate of 2 kg/s through inner tube exits the heat exchanger at a temperature of 50°c. the cold fluid, i. e., water enters the heat exchanger at 20°c and exits at 70°c. assuming the fouling factor on the oil side and water side to be 0.00015 m2 ·k/w and 0.0001 m2 ·k/w, respectively, determine the overall heat transfer coefficient on inner and outer surface of the copper tube.
Explanation:
fluid, i. e., water enters the heat exchanger at 20°c and exits at 70°c. assuming the fouling factor on the oil side and water side to be 0.00015 m2 ·k/w and 0.0001 m2 ·k/w, respectively, determine the overall heat transfer coefficient on inner and outer surface of the copper tube.xgjicbb .follow me
8. Which of these plastics is a themoplastic - melts or softens with heat.
Acrylic
Bakelite
Polyester resin
Melamine
Epoxy Resin
Could someone please help me, this is very urgent, I will pay an extra bonus if done correct.
Answer:
Acryclic
Explanation:
Hope it helps you!
A baker deck decorates 42 cupcakes and 30 minute she decorated cupcakes at a constant rate how many cups cage the baker decorate pre-minute
Answer: 1.4 cupcakes per minute.
Explanation:
Here's the correct question:
A baker decorates 42 cupcakes in 30 minutes. How many cupcakes can the baker decorate per minute?
Since the baker decorates 42 cupcakes in 30 minutes, the number of cupcakes that the baker decorates for each minutes since it's at a constant rate will be the total number of cupcakes made divided by the number of minutes. This will be:
= 42cupcakes / 30minutes
= 1.4 cupcakes per minute.
Therefore, the number of cupcakes that the baker can decorate per minute is 1.4cupcakes/minute
A transformer is to be used to provide power for a computer disk drive that needs 6.4 V (rms) instead of the 120 V (rms) from the wall outlet. The number of turns in the primary is 300, and it delivers 500 mA (the secondary current) at an output voltage of 6.4 V (rms). (a) Should the transformer have more turns in the secondary compared to the primary, or fewer turns
Answer:
The secondary coil should have fewer turns compared to the primary coil.
Explanation:
[tex]N_p[/tex] = Number of turns in primary coil = 300
[tex]N_s[/tex] = Number of turns in secondary
[tex]V_p[/tex] = Voltage in primary coil = 120 V
[tex]V_s[/tex] = Voltage in secondary coil = 6.4 V
We have the relation
[tex]\dfrac{N_p}{N_s}=\dfrac{V_p}{V_s}\\\Rightarrow N_s=\dfrac{N_p}{V_p}\times V_s\\\Rightarrow N_s=\dfrac{300}{120}\times 6.4\\\Rightarrow N_s=16[/tex]
The secondary coil has 16 turns which is less than the turns in the primary coil.
There are gauges available for checking panel flushness. TRUE OR FALSE?
Answer: true
Explanation:
because flushness means your dead so you are dead now
multimeter and the LCD is showing Hz. What's she measuring?
A. Resistance
B. Amplitude
C. Voltage
D. Frequency
Answer:
i think its d frequency
Explanation:
hz on a multimeter means frequency setting
Q.13 In order to produce maximum starting torque in a split-phase motor, how many degrees out of phase should the start- and run-winding currents be with each other?
Select one:
A. 180°
B. 0°
C. 120°
D. 90°
Answer:
D. 90 degrees.
Explanation:
Torque is a rotational force which moves an object in other direction. There should be 90 degrees out of phase to start, run winding currents with each other. Torque is produced by the rotational motion of an object. The angle of the object must be 90 degrees set in order to create torque.