Answer:
- the mass flow rate of air is 7.53 kg/s
- the exit area is 0.108 m²
Explanation:
Given the data in the question;
lets take a look at the steady state energy equation;
m" = W"[tex]_{cv[/tex] / [ (h₁ - h₂ ) -[tex]\frac{V_2^2}{2}[/tex] ]
Now at;
T₁ = 900K, h₁ = 932.93 k³/kg
T₂ = 500 K, h₂ = 503.02 k³/kg
so we substitute, in our given values
m" = [ 3200 kW × [tex]\frac{1\frac{k^3}{s} }{1kW}[/tex] ] / [ (932.93 - 503.02 )k³/kg -[tex]\frac{100^2\frac{m^2}{s^2} }{2}[/tex]|[tex]\frac{ln}{kg\frac{m}{s^2} }[/tex]||[tex]\frac{1kJ}{10^3N-m}[/tex]| ]
m" = 7.53 kg/s
Therefore, the mass flow rate of air is 7.53 kg/s
now, Exit area A₂ = v₂m" / V₂
we know that; pv = RT
so
A₂ = RT₂m" / P₂V₂
so we substitute
A₂ = {[ [tex](\frac{8.314}{28.97}\frac{k^3}{kg.K})[/tex]×[tex]500 K[/tex]×[tex](7.54 kg/s)[/tex] ] / [(1 bar)(100 m/s )]} |[tex]\frac{1 bar}{10N/m^2}[/tex]||[tex]10^3N.m/1k^3[/tex]
A₂ = 0.108 m²
Therefore, the exit area is 0.108 m²
You would use a _____________ gauge to check the pressure in each tire. A technician should compare the tire-pressure reading with the tire pressure specified on the side of the driver's _____________ . When servicing tires follow the _____________ procedure outlined in the owner’s manual or online service information. Correct tire-inflation pressure is printed on a placard on the _____________. Why are tires rotated? Which tire rotation method is most often recommended? _____________ An easy way to remember effective tire rotation is, “Drive wheels straight, _____________ the nondrive wheels.” A _____________ is used to ensure that the proper torque is completed on a
Answer:
the first one is tire pressure gauge :)
Explanation:
In this lab, we assumed that the flip-flops did not contribute to the timing constraints of the circuit. Unfortunately, this is not the case. As you saw when you simulated the D flip-flop, the sampling action does not happen instantaneously. In fact, a flip-flop will become unstable if the inputs do not remain stable for a certain amount of time prior to the rising-edge event (setup time) and a certain amount of time after the rising-edge event (hold time). Assume a setup and hold time of 2ns and 1ns, respectively. What would the theoretical maximum clock rate for the synchronous adder be in this scenario
is a baby duck swimming in a lake a learned behavior
Answer:
True because some ducks can't swim but have to learn it
Name some technical skills that are suitable for school leavers .
Answer:
Welding, carpentry, masonry, construction worker, barber
Explanation:
Water vapor at 5 bar, 3208C enters a turbine operating at steady state with a volumetric flow rate of 0.65 m3/s and expands adiabatically to an exit state of 1 bar, 1608C. Kinetic and potential energy effects are negligible. Determine for the turbine (a) the power developed, in kW, (b) the rate of entropy production, in kW/K, and (c) the isentropic turbine efficiency.
Answer:
A) 371.28 kW
b) 0.1547 Kw/K
c) 85%
Explanation:
pressure (p1) = 5 bar
exit pressure ( p2 ) = 1 bar
Initial Temperature ( T1 ) = 320°C
Final temp ( T2 ) = 160°C
Volume ( V ) = 0.65 m^3/s
A) Calculate power developed ( kW )
P = m( h1 - h2 ) = 1.2 ( 3105.6 - 2796.2 ) = 371.28 kW
B) Calculate the rate of entropy production
Δs = m ( S2 - S1 ) = 1.2 ( 7.6597 - 7.5308 ) = 0.1547 Kw/K
c) Calculate the isentropic turbine efficiency
For an isentropic condition : S2s = S1
therefore at state , value of h2 at isentropic condition
attached below is the remaining part of the solution
Note : values of [ h1, h2, s1, s2 , v1 and m ] are gotten from the steam tables at state 1 and state 2
1. A wastewater treatment plant (WWTP) releases effluent into a stream with mean depth 2 m and mean velocity 0.75 m/s. The BOD concentration at the WWTP is 15 mg/L, and the oxygen deficit is negligible. The deoxygenation rate in the stream is 0.8 d-1 and the reaeration rate is 1.2 d-1. a) Calculate the BOD concentration and DO deficit at a point 20 km downstream from the WWTP. (10 pts) b) What assumptions are inherent in these predictions (give at least two)
Answer:
A) BOD = 6.51 mg/l , DO = 2.46 mg/l
B) BOD of stream is negligible and DO of stream is at saturation level
Explanation:
Mean depth = 2 m
Mean velocity = 0.75 m/s
Bod concentration at WWTP = 15 mg/L
deoxygenation rate = 0.8 d-1
reaeration rate = 1.2 d-l
a) Calculate the BOD concentration and DO deficit
at 20 km
tc = (20 * 10^3) / (0.75 * 3600 * 24 )
= 0.309 days
[tex]BOD_{t}[/tex] = lo ( 1 - 10^- 0.8 * 0.309 )
= 15 ( 1 - 10^ - 0.2472 )
= 15 ( 0.434 ) = 6.51 mg/l
DO = ( Kd * lo / Kr ) * 10^ -Kd*tc
= ( 0.8 * 6.51 / 1.2 ) * 10 ^ - 0.8 * 0.309
= 4.34 * 10^-0.2472 = 2.46 mg/l
B) The assumptions are : BOD of stream is negligible and DO of stream is at saturation level
Sarah is a site investigator for a large construction firm. She is considering Miguel, a former geology student with experience as an intern at an architecture firm, for an assistant site investigation position. Which of the following is most relevant to her decision?
Answer: A. whether his geology studies exposed him to principles of geotechnical engineering
Explanation:
The options include:
a. whether his geology studies exposed him to principles of geotechnical engineering
b. the size of the geology program he attended
c. the size of the architecture firm
d. whether the architecture firm was intending to offer Miguel a full-time position
Since Miguel, is a former geology student with experience as an intern at an architecture firm, and Sarah is considering him for an assistant site investigation position, the option that will be relevant for her to make a decision is to know whether his geology studies exposed him to principles of geotechnical engineering.
Geotechnical engineering, is a branch of engineering that makes use of principles of rock mechanics to solve engineering challenges. Since Sarah needs him for an assistant site investigation position, he'll need to investigate souls, rocks and evaluate them.
Identify the true statements about the lumped system analysis.
A. The entire body temperature remains essentially uniform at all times during a heat transfer process.
B. The temperature of lumped system bodies can be taken to be a function of time only.
C. The Biot number is less than or equal to 0.1.
D. The Biot number is greater than or equal to 1.
Also discuss how bandwidth is affected by increasing the number of signal elements.
What 2 things can you never eat for breakfast?
i know the answer but lets see if you do
Answer:
you can't eat your school computer or a pencil.
Can someone tell me what car, year, and model this is please
Answer:
Explanation:
2019 nissan altima 2.5 SV
have a good day /night
may i please have a branlliest
Alicia had to get over her fear of heights in order to become comfortable maintaining the generators in wind turbines. professional certification exam verifies the knowledge of the testee, whereas the CMP verifies green activity of the certified professional. A professional certification exam verifies the current qualifications of the testee, whereas the CMP ensures the certified professional continues to exhibit those qualifications. A professional certification exam verifies the qualifications of the testee, whereas the CMP measures the actual green commitment of the certified professional. A professional certification exam verifies the current knowledge of the testee, whereas the CMP ensures the certified professional continues to add to that knowledge.
Answer:a
Explanation:
Where do you prefer to live?
USA UK Canada ????????
Answer for POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Test if a number grade is an A (greater than or equal to 90). If so, print "Great!". Hint: Grades may be decimals. Sample Run Enter a Number: 98.5 Sample Output Great!
Answer:
In Python:
grade = float(input("Enter a Number: "))
if grade >= 90:
print("Great!")
Explanation:
This prompts the user for grade
grade = float(input("Enter a Number: "))
This checks for input greater than or equal to 90
if grade >= 90:
If yes, this prints "Great"
print("Great!")
In low-speed external water flow over a bluff object, vortices are shed from the object. The vortex shedding produced by a particular object is to be studied in a water tunnel at a 1/4 scale model (model 1/4 the size of the prototype). After a dimensional study, it is found that the Pi terms of this phenomenon are the Reynolds number and the Strouhal number:Re = pVd/muSt = fd/Vwhere f is the frequency of the vortex shedding, V the velocity of the flow, d the characteristic length of the object, and p and mu the density and viscosity of the flow. 1. If the prototype speed is 7 m/s, determine the water velocity in the tunnel for the model tests. 2. If the model tests of part 1 produced a model shedding frequency of 200 Hz, determine the expected vortex shedding frequency on the prototype.
Answer:
1) the water velocity in the tunnel for the model tests is 28 m/s
2) the expected vortex shedding frequency on the prototype is 12.5 Hz
Explanation:
Given the data in the question;
1)
using the Reynolds number relation for prototype and model,
PpVpdp/mu(Prototype) = PmVmdm/mu(for model)
but we know that, density and viscosity in prototype and model will remain same so;
Vp × dp = Vm × dm
vm = Vp × dp/dm
we substitute
vm = 7 m/s × 4
vm = 28 m/s
Therefore, the water velocity in the tunnel for the model tests is 28 m/s
2)
we make use of the Strouhal number relation as given in the question;
fp × dp/Vp = fm × dm/Vm
frequency of the prototype will be;
fp = fm × dm/Vm / dp/Vp
we substitute
fp = 200 × 7 / ( 4 × 28 )
fp = 1400 / 112
fp = 12.5 Hz
Therefore, the expected vortex shedding frequency on the prototype is 12.5 Hz
Rainfall rates for successive 20-min period of a 140min storm are 1.5, 1.5, 6.0, 4.0, 1.0, 0.8, and 3.2 in/hr, totaling 6.0in. Determine the rainfall excess during successive 20-min periods by the NRCS method. The soil in the basin belongs to group A. It is an agriculture row crop land with contoured pattern in good hydrologic condition. The soil is in average condition before the storm (moisture condition II).
What can be used to measure the alcohol content in gasoline? A. Graduated cylinder B. Electronic tester C. Scan tool D. Either a graduated cylinder or an electronic tester
Answer:
GRADUATED CYLINDER
Explanation:
Suppose we are given three boxes, Box A contains 20 light bulbs, of which 10 are defective, Box B contains 15 light bulbs, of which 7 are defective and Box C contains 10 light bulbs, of which 5 are defective. We select a box at random and then draw a light bulb from that box at random. (a) What is the probability that the bulb is defective? (b) What is the probability that the bulb is good?
Answer:
0.49
0.51
Explanation:
Probability that bulb is defective :
Let :
b1 = box 1 ; b2 = box 2 ; b3 = box 3
d = defective
P(defective bulb) = (p(b1) * (d|b1)) + (p(b2) * p(d|b2)) + (p(b3) * p(d|b3))
P(defective bulb) = (1/3 * 10/20) + (1/3 * 7/15) + (1/3 * 5/10))
P(defective bulb) = 10/60 + 7/45 + 5/30
P(defective bulb) = 1/6 + 7/45 + 1/6 = 0.4888
= 0.49
P(bulb is good) = 1 - P(defective bulb) = 1 - 0.49 = 0.51
Air is a....
O Solid
O Liquid
O Gas
O Plasma
Answer:
Air is a gas
Explanation:
i think. beavuse it cant be a liqued or a solid. i dont think a plasma. i would answer gas
Factors such as brake shoe orientation, pin location, and direction of rotation determine whether a particular brake shoe is considered to be self-energizing or self-deenergizing. Sketch a brake drum where the left shoe is self-energizing and the right shoe is self-deenergizing. Which shoe will produce more braking torque, assuming equal actuation forces and identical brake shoes
Answer:
Self energizing brake shoe produces more Torque
Explanation:
Attached below is the sketches of the various cases
The cases are :
case 1 ; deenergizing
case 2 ; self energizing
case 3 ; produces high braking torque
For a brake to be self energizing when the Frictional torque and braking torque are in the same direction
summary from equation 3
when: b > uc the brake is controllable
b = uc It is self locking
b < uc The brake is uncontrollable
A low-altitude meteorological research balloon, temperature sensor, and radio transmitter together weigh 2.5 lb. When inflated with helium, the balloon is spherical with a diameter of 4 ft. The volume of the transmitter can be neglected when compared to the balloon's size. The balloon is released from ground level and quickly reaches its terminal ascent velocity. Neglecting variations in the atmosphere's density, how long does it take the balloon to reach an altitude of 1000 ft?
Answer:
12 mins
Explanation:
The summation of the forces in vertical direction
= Fb - Fd - w = 0 ∴ Fd = Fb - w ----- ( 1 )
Fb ( buoyant force ) = Pair * g * Vballoon ------- ( 2 )
Pair = air density , Vballoon = volume of balloon
Vballoon = [tex]\frac{\pi D^3}{6}[/tex] , where D = 4 ∴ Vballoon = 33.51 ft^3
g = 32.2 ft/s^2
From property tables
Pair = 2.33 * 10^-3 slug/ft^3
μ ( dynamic viscosity ) = 3.8 * 10^-7 slug/ft.s
Insert values into equation 2
Fb = ( 2.33 * 10^-3 ) * ( 32.2 ) *( 33.51 ) = 2.514 Ib
∴ Fd = 2.514 - 2.5 = 0.014 Ib ( equation 1 )
Assuming that flow is Laminar and RE < 1
Re = (Pair * vd) / μair -------- ( 3 )
where: Pair = 2.33 * 10^-3 slug/ft^3 , vd = ( 987 * 4 ) ft^2/s , μair = 3.8 * 10^-7 slug/ft.s
Insert values into equation 3
Re = 2.4 * 10^7 ( this means that the assumption above is wrong )
Hence we will use drag force law
Assume Cd = 0.5
Express Fd using the relation below
Fd = 1/2* Cd * Pair * AV^2
therefore V = 1.39 ft/s
Recalculate Reynold's number using v = 1.39 ft/s
Re = 34091
from the figure Cd ≈ 0.5 at Re = 34091
Finally calculate the rise time ( time taken to reach an altitude of 1000 ft )
t = h/v
= 1000 / 1.39 = 719 seconds ≈ 12 mins
FOR BRAINLIST HELP PLEASE IS A DCP
A- Causes of the 13t Amendment
B- Reasons for Women's Suffrage
C- Reasons for the Freedmen's Bureau
D- Causes of the Plantation System
Answer:
C
Explanation: Freedmens Bureau provided resources for southerners and newly freed slaves