In an axial flow compressor air enters the compressor at stagnation conditions of 1 bar and 290 K. Air enters with an absolute velocity of 145 m/s axially into the first stage of the compressor and axial velocity remains constant through the stage. The rotational speed is 5500 rpm and stagnation temperature rise is 22 K. The radius of rotor-blade has a hub to tip ratio of 0.5. The stage work done factor is 0.92, and the isentropic efficiency of the stage is 0.90. Assume for air Cp=1005 kJ/(kg·K) and γ= 1.4
Determine the followings. List your assumptions.
i. The tip radius and corresponding rotor angles at the tip, if the inlet Mach number for the relative velocity at the tip is limited to 0.96.
ii. The mass flow at compressor inlet.
iii. The stagnation pressure ratio of the stage and power required by the first stage.
iv. The rotor angles at the root section.
Answers
Answer:
i) r_t = 0.5101 m
ii) m' = 106.73 kg/s
iii) R_s = 1.26
P = 2359.8 kW
iv) β2 = 55.63°
Explanation:
We are given;
Stagnation pressure; T_01 = 290 K
Inlet velocity; C1 = 145 m/s
Cp for air = 1005 kJ/(kg·K)
Mach number; M = 0.96
Ratio of specific heats; γ = 1.4
Stagnation pressure; P_01 = 1 bar
rotational speed; N = 5500 rpm
Work done factor; τ = 0.92
Isentropic effjciency; η = 0.9
Stagnation temperature rise; ΔT_s = 22 K
i) Formula for Stagnation temperature is given as;
T_01 = T1 + C1/(2Cp)
Thus,making T1 the subject, we havw;
T1 = T_01 - C1/(2Cp)
Plugging in the relevant values, we have;
T1 = 290 - (145/(2 × 1005))
T1 = 289.93 K
Formula for the mach number relative to the tip is given by;
M = V1/√(γRT1)
Where V1 is relative velocity at the tip and R is a gas constant with a value of 287 J/Kg.K
Thus;
V1 = M√(γRT1)
V1 = 0.96√(1.4 × 287 × 289.93)
V1 = 0.96 × 341.312
V1 = 327.66 m/s
Now, tip speed is gotten from the velocity triangle in the image attached by the formula;
U_t = √(V1² - C1²)
U_t = √(327.66² - 145²)
U_t = √86336.0756
U_t = 293.83 m/s
Now relationship between tip speed and tip radius is given by;
U_t = (2πN/60)r_t
Where r_t is tip radius.
Thus;
r_t = (60 × U_t)/(2πN)
r_t = (60 × 293.83)/(2π × 5500)
r_t = 0.5101 m
ii) Now mean radius from derivations is; r_m = 1.5h
While relationship between mean radius and tip radius is;
r_m = r_t - h/2
Thus;
1.5h = 0.5101 - 0.5h
1.5h + 0.5h = 0.5101
2h = 0.5101
h = 0.5101/2
h = 0.2551
So, r_m = 1.5 × 0.2551
r_m = 0.3827 m
Formula for the area is;
A = 2πr_m × h
A = 2π × 0.3827 × 0.2551
A = 0.6134 m²
Isentropic relationship between pressure and temperature gives;
P1 = P_01(T1/T_01)^(γ/(γ - 1))
P1 = 1(289.93/290)^(1.4/(1.4 - 1))
P1 = 0.9992 bar = 0.9992 × 10^(5) N/m²
Formula for density is;
ρ1 = P1/(RT1)
ρ1 = 0.9992 × 10^(5)/(287 × 289.93)
ρ1 = 1.2 kg/m³
Mass flow rate at compressor inlet is;
m' = ρ1 × A × C1
m' = 1.2 × 0.6134 × 145
m' = 106.73 kg/s
iii) stagnation pressure ratio is given as;
R_s = (1 + ηΔT_s/T_01)^(γ/(γ - 1))
R_s = (1 + (0.9 × 22/290))^(1.4/(1.4 - 1))
R_s = 1.26
Work is;
W = C_p × ΔT_s
W = 1005 × 22
W = 22110 J/Kg
Power is;
P = W × m'
P = 22110 × 106.73
P = 2359800.3 W
P = 2359.8 kW
iv) We want to find the rotor angle.
now;
Tan β1 = U_t/C1
tan β1 = 293.83/145
tan β1 = 2.0264
β1 = tan^(-1) 2.0264
β1 = 63.73°
Formula for Stagnation pressure rise is given by;
ΔT_s = (τ•U_t•C1/C_p) × tan(β1 - β2)
Plugging in the relevant values;
22 = (0.92 × 293.83 × 145/1005) × (tan 63.73 - tan β2)
(tan 63.73 - tan β2) = 0.5641
2.0264 - 0.5641 = tan β2
tan β2 = 1.4623
β2 = tan^(-1) 1.4623
β2 = 55.63°
Related Questions
Water enters a centrifugal pump axially at atmospheric pressure at a rate of 0.12 m3
/s and at a
velocity of 7 m/s, and leaves in the normal direction along the pump casing, as shown in Figure.
Determine the force acting on the shaft (which is
also the force acting on the bearing of the shaft) in
the axial direction.
Answers
Answer:
Water enters a centrifugal pump axially at atmospheric pressure at a rate of 0.12 m3/s and at a velocity of 7 m/s, and leaves in the normal direction along the pump casing, as shown in Fig. PI3-39. Determine the force acting on the shaft (which is also the force acting on the bearing of the shaft) in the axial direction.
Step-by-step solution:
Step 1 of 5
Given data:-
The velocity of water is .
The water flow rate is.
please help i have no xlue
Answers
Here a quote from Britannica to support this answer:
“Earthquakes generate two main types of seismic, or shock, waves: body waves and surface waves. “
I need help with simply science
Answers
Answer:
mountain ranges may be
If you make a mistake in polarity when measuring the value of DC voltage in a circuit with a digital VOM, what will happen? A. The meter will be damaged. B. The meter will read positive voltage only. C. The meter will display a negative sign. D. The meter will display OL which states an overload condition.
Answers
Answer:
C. The meter will display a negative sign.
Explanation:
If you use an analog voltmeter and you measure voltage with reverse polarity you will damage it. But in this case we are using a digital multimeter. This kind of multimeter is designed to be able to deal with positive and negative voltages
A hair dryer is basically a duct in which a few layers of electric resistors are placed. A small fan pulls the air in and forces it to flow over the resistors, where it is heated. Air enters a 1400-W hair dryer at 100 kPa and 22°C and leaves at 47°C. The cross-sectional area of the hair dryer at the exit is 60 cm2. Neglect the power consumed by the fan and the heat losses through the walls of the hair dryer. The gas constant of air is R = 0.287 kPa·m3/kg·K. Also, cp = 1.007 kJ/kg·K for air at room temperature.
determine
(a) the volume flow rate of air at the inlet and
(b) the velocity of the air at the exit.
Answers
Answer:
a) volume flow rate of air at the inlet is 0.0471 m³/s
b) the velocity of the air at the exit is 8.517 m/s
Explanation:
Given that;
The electrical power Input W_elec = -1400 W = -1.4 kW
Inlet temperature of air T_in = 22°C
Inlet pressure of air p_in = 100 kPa
Exit temperature T_out = 47°C
Exit area of the dyer is A_out = 60 cm²= 0.006 m²
cp = 1.007 kJ/kg·K
R = 0.287 kPa·m3/kg·K
Using mass balance
m_in = m_out = m_air
W _elec = m_air ( h_in - h_out)
we know that h = CpT
so
W _elec = m_air.Cp ( T_in - T_out)
we substitute
-1.4 = m_air.1.007 ( 22 - 47 )
-1.4 = - m_air.25.175
m_air = -1.4 / - 25.175
m_ air = 0.0556 kg/s
a) volume flow rate of air at the inlet
we know that
m_air = P_in × V_in
now from the ideal gas equation
P_in = p_in / RT_in
we substitute our values
= (100×10³) / ((0.287×10³)(22+273))
= 100000 / 84665
P_in = 1.18 kg/m³
therefore inlet volume flowrate will be;
V_in = m_air / P_in
= 0.0556 / 1.18
= 0.0471 m³/s
the volume flow rate of air at the inlet is 0.0471 m³/s
b) velocity of the air at the exit
the mass flow rate remains unchanged across the duct
m_ air = P_in.A_in.V_in = P_out.A_out.V_out
still from the ideal gas equation
P_out = p_out/ RT_out ( assume p_in = p_out)
P_out = (100×10³) / ((0.287×10³)(47+273))
P_out = 1.088 kg/m³
so the exit velocity will be;
V_out = m_air / P_out.A_out
we substitute our values
V_out = 0.0556 / ( 1.088 × 0.006)
= 0.0556 / 0.006528
= 8.517 m/s
Therefore the velocity of the air at the exit is 8.517 m/s
7. The surface finish for the cylinder walls usually depends on the
O A. type of engine oil used.
O B. sharpness of the cylinder bore edges.
O C.type of piston rings used
O D. cylinder wall-to-piston clearance.
Answers
A spring with an m-kg mass and a damping constant 3 (kg/s) can be held stretched 0.5 meters beyond its natural length by a force of 1.5 newtons. If the spring is stretched 1 meters beyond its natural length and then released with zero velocity, find the mass that would produce critical damping.
Answers
Answer:
0.75 kg
Explanation:
c = Damping coefficient = 3 kg/s
x = Displacement of spring = 0.5 m
F = Force = 1.5 N
From Hooke's law we get
[tex]F=kx\\\Rightarrow k=\dfrac{F}{x}\\\Rightarrow k=\dfrac{1.5}{0.5}\\\Rightarrow k=3\ \text{N/m}[/tex]
In the case of critical damping we have the relation
[tex]c^2-4mk=0\\\Rightarrow m=\dfrac{c^2}{4k}\\\Rightarrow m=\dfrac{3^2}{4\times 3}\\\Rightarrow m=0.75\ \text{kg}[/tex]
Mass that would produce critical damping is 0.75 kg.
0.75 kg is the mass that would produce critical damping. As spring with an m-kg mass and a damping constant 3 (kg/s) can be held stretched 0.5 meters beyond its natural length by a force of 1.5 newtons.
What is zero velocity?A change in time and position is referred to as an object's velocity. When there is no movement of the object, the velocity of the object is said to be 0.
For any body in planar motion, the velocity is always instantaneously 0 at some point in the plane of motion (if it were rigidly connected to the body). This place is known as the instantaneous center of zero velocity, or IC.
Example: The gravitational pull of the earth pushes the ball away from the thrower when a ball is thrown upwards on Earth at a constant speed. The speed of the ball increases until it reaches its maximum, at which point it starts to plummet.
Thus, it is 0.75 kg.
For more information about zero velocity, click here:
https://brainly.com/question/18634403
#SPJ2
I need help please thank for the help on the last one <3
Answers
Which of the following is an example of a tax
Answers
Answer:
A tax is a monetary payment without the right to individual consideration, which a public law imposes on all taxable persons - including both natural and legal persons - in order to generate income. This means that taxes are public-law levies that everyone must pay to cover general financial needs who meet the criteria of tax liability, whereby the generation of income should at least be an auxiliary purpose. Taxes are usually the main source of income of a modern state. Due to the financial implications for all citizens and the complex tax legislation, taxes and other charges are an ongoing political and social issue.
I dont know I asked this to
Explanation:
Which method of freezing preserves the quality and taste of food?
Answers
Answer:
commercial freezing
Explanation:
smaller ice crystals are formed this causes less damage to cell membranes so the quality is less effected
Tubular centrifuge is used for recovering cells 60% of the cells or recover data flow rate of 12 l/min with a rotational speed of 4000 RPM what is the RPM to increase the recovery rate of the cells to 95% at the same flow rate
Answers
Answer:
The RPM to increase the recovery rate of the cells to 95% at the same flow rate is 6,333.3 RPM.
Explanation:
If the tubular centrifuge rotates at about 4,000 revolutions per minute to recover 60% of the cells, in case of wanting to recover 95% of the cells, the following calculation must be carried out to determine the required number of revolutions per minute:
60 = 4,000
95 = X
((95 x 4,000) / 60)) = X
(380,000 / 60) = X
6,333.3 = X
Therefore, as the calculation emerges, the tubular centrifuge will need to rotate at about 6,333.3 revolutions per minute to recover 95% of the cells in the same time.
A system samples a sinusoid of frequency 230 Hz at a rate of 175 Hz and writes the sampled signal to its output without further modification. Determine the frequency that the sampling system will generate in its output.
a. 120
b. 55
c. 175
d. 230
Answers
Which of the following is not one of the common classifications of product liability defects? A. Manufacture B. Materials C. Packaging D. Both "Materials" and "Packaging" E. Design
Answers
Answer:
D. Both "Materials" and "Packaging"
Explanation:
Product liability may refer to the manufacturer or the seller being held responsible or liable for providing any defective product into the hands of the consumer or the customer. Responsibility or liability for a defective product which causes injuries lies with all the sellers of the product from the manufacturer to the distributor to the seller.
There are majorly three product defects. They are :
1. Manufacturing defect
2. Design defect
3. Marketing defect
The structure of PF3(C6H5)2 is trigonal bipyramidal, with one equatorial and two axial F atoms which interchange positions when heated. Describe the low- and high- temperature 31P and 19F NMR spectra.
Answers
Answer:
For 31 P NMR spectra
low temperature
there is two types of 19f seen in low temperature and they are
one at equitorial one at axialtherefore in low temperature the 31p couples with the two types of 19F seen ( [tex]b_{f} and c_{f}[/tex]to form a triplet and this couples more with [tex]a_{f}[/tex] to form a doublet. i.e. one (1) peak
High temperature
At High temperature The exchange is fast here therefore the 31p spectra sees all 19p at once and in the same environment leading to the formation of one (1) peak
For 19 P NMR spectra
low temperature
In low temperature [tex]a_{f}, b_{f} , c_{f}[/tex] is fixed and the environment where [tex]b_{f} and c _{f}[/tex] is the same hence a peak is formed and another peak is formed by [tex]a_{f}[/tex] that makes the number of peaks = 2 peaks
High temperature
In high temperature [tex]a_{f}, b_{f} , c_{f}[/tex] exchange very fast therefore one peak is formed for all, since the fast exchanges makes NMR machine to take an average and produce just one peak for all
Explanation:
For 31 P NMR spectra
low temperature
there is two types of 19f seen in low temperature and they are
one at equitorial one at axialtherefore in low temperature the 31p couples with the two types of 19F seen ( [tex]b_{f} and c_{f}[/tex]to form a triplet and this couples more with [tex]a_{f}[/tex] to form a doublet. i.e. one (1) peak
High temperature
At High temperature The exchange is fast here therefore the 31p spectra sees all 19p at once and in the same environment leading to the formation of one (1) peak
For 19 P NMR spectra
low temperature
In low temperature [tex]a_{f}, b_{f} , c_{f}[/tex] is fixed and the environment where [tex]b_{f} and c _{f}[/tex] is the same hence a peak is formed and another peak is formed by [tex]a_{f}[/tex] that makes the number of peaks = 2 peaks
High temperature
In high temperature [tex]a_{f}, b_{f} , c_{f}[/tex] exchange very fast therefore one peak is formed for all, since the fast exchanges makes NMR machine to take an average and produce just one peak for all
The big ben clock tower in london has clocks on all four sides. If each clock has a minute hand that is 11.5 feed in length, how far does the tip of each hand travel in 52 minutes?
Answers
Answer:
Updated question
The big ben clock tower in London has clocks on all four sides. If each clock has a minute hand that is 11.5 feet in length, how far does the tip of each hand travel in 52 minutes?
The distance traveled by the tip of the minute hand of the clock would be 62.59 ft
Explanation:
Let us assume the shape of the clock is circular.
the minute hand is equal to the radius = 11.5 ft
Diameter = radius x 2
Diameter = 11.5 x 2 = 23 ft
The distance traveled by the tip of the minute hand can be calculated thus;
the fraction of the circumference traveled by the minute hand would be;
52/60 = 0.8667
Circumference of the clock would be;
C = pi x d
where C is the circumference
pi is a constant
d is the diameter
C = 3.14 x 23
C = 72.22 ft
Therefore the fraction of the circumference covered by the minute hand would be;
72.22 ft x 0.8667 = 62.59 ft
Therefore the distance traveled by the tip of the minute hand of the clock would be 62.59 ft
An equal-tangent sag vertical curve is designed with a PVC as station 109+00 and elevation 950ft, the PVI has a station of 110+77 and elevation of 947.34ft, and the low point at station 110+50. Determine the design speed of the cure.
Answers
Answer:
K = 96 and the design speed of the curve = 50mph
Explanation:
109+00 = 10900
Elevation = 950ft
110+77 = 11077
Elevation = 947.34ft
Station of low point = 110+50 = 11050
To get grade of curve
Gi = 947.34-950/11077-10900
= -2.66/177
= -0.015x100
= -1.5%
Locate of low point (XL)
= 11050-10900
= 150
To get the value of K
XL = |GL| x K
When we substitute values
150 = 1.5 x K
150 = 1.5K
K = 150/1.5
K = 100
The suitable and most nearest value is K = 96
Then we use the standard chart to get the design speed for K = 96
On this chart, the design speed for the curve = 50mph
Therefore K = 96 and speed = 50mph
Refrigerant-22 absorbs heat from a cooled space at 50°F as it flows through an evaporator of a refrigeration system. R-22 enters the evaporator at 10°F at a rate of 0.08 lbm/s with a quality of 0.3 and leaves as a saturated vapor at the same pressure. Determine:
a. The rate of cooling provided, in Btu/h.
b. The rate of exergy destruction in the evaporator.
c. The second-law efficiency of the evaporator.
Take T0 = 77°F. The properties of R-22 at the inlet and exit of the evaporator are: h1 = 107.5 Btu/lbm, s1 = 0.2851 Btu/lbm·R, h2 = 172.1 Btu/ lbm, s^2 = 0.4225 Btu/lbm·R.
Answers
Answer:
a) the rate of cooling provided is 18604.8 Btu/h
b) the rate of exergy destruction in the evaporator is 0.46 Btu/Ibm
c) the second-law efficiency of the evaporator is 37.39%
Explanation:
Given that;
Temperature of sink TL = 50°F = 510 R
Temperature at evaporator inlet TI = 10°F = 470 R
mass flow rate m" = 0.08 lbm/s
quality of refrigerant at evaporator inlet x1 = 0.3
quality of refrigerant at evaporator exit x2 = 1.0
T₀ = 77°F = 537 R
h1 = 107.5 Btu/lbm
s1 = 0.2851 Btu/lbm·R,
h2 = 172.1 Btu/ lbm,
s2 = 0.4225 Btu/lbm·R.
a) rate of cooling provided, in Btu/h.
QL = m"( h2 - h1)
we substitute
QL = 0.08( 172.1 - 107.5
= 0.08 × 64.6
= 5.168 Btu/s
we convert to Btu/h
5.168 × 60 × 60
QL = 18604.8 Btu/h
Therefore the rate of cooling provided is 18604.8 Btu/h
b) The rate of exergy destruction in the evaporator
Entropy generation can be expressed as;
S_gen = m"(s2 - s1) - QL/TL
so we substitute
S_gen = 0.08( 0.4225 - 0.2851 ) - 5.168 / 510
= 0.010992 - 0.01013
S_gen = 0.00086 Btu/ibm.R
now the energy destroyed expressed as;
X_dest = T₀ × S_gen
so
X_dest = 537 × 0.00086
X_dest = 0.46 Btu/Ibm
Therefore the rate of exergy destruction in the evaporator is 0.46 Btu/Ibm
c) The second-law efficiency of the evaporator.
Energy expended is expressed as;
X_exp = m"(h1 - h2) - m"T₀(s1 - s2)
we substitute
= 0.08( 107.5 - 172.1 ) - [0.08 × 537 ( 0.2851 - 0.4225 )
= -5.168 - [ - 5.9027)
= -5.168 + 5.9027
= 0.7347 Btu/s
Now second law efficiency is expressed as;
nH = 1 - (X_dest / X_esp)
= 1 - ( 0.46 / 0.7347 )
= 1 - 0.6261
= 0.3739
nH = 37.39%
Therefore the second-law efficiency of the evaporator is 37.39%
An unknown impedance Z is connected across a 380 V, 60 Hz source. This causes a current of 5A to flow and 1500 W is consumed. Determine the following: a. Real Power (kW) b. Reactive Power (kvar) c. Apparent Power (kVA) d. Power Factor e. The impedance Z in polar and rectangular form
Answers
Answer:
a) Real Power (kW) = 1.5 kW
b) Reactive Power (kvar) is 1.1663 KVAR
c) Apparent Power (kVA) is 1.9 KVA
d) the Power Factor cos∅ is 0.7894
e) the impedance Z in polar and rectangular form is 76 ∠ 37.87° Ω
Explanation:
Given that;
V = 380v
i = 5A
P = 1500 W
determine;
a) Real Power (kW)
P = 1500W = 1.5 kW
therefore Real Power (kW) = 1.5 kW
b) Reactive Power (kvar)
p = V×i×cos∅
cos∅ = p / Vi
cos∅ = 1500 / ( 380 × 5 ) = 0.7894
∅ = cos⁻¹ (0.7894)
∅ = 37.87°
Q = VIsin∅
Q = 380 × 5 × sin( 37.87° )
Q = 1.1663 KVAR
Therefore Reactive Power (kvar) is 1.1663 KVAR
c) Apparent Power (kVA)
S = P + jQ
= ( 1500 + J 1166.3 ) VA
S = 1900 ∠ 37.87° VA
S = 1.9 KVA
Therefore Apparent Power (kVA) is 1.9 KVA
d) Power Factor
p = V×i×cos∅
cos∅ = p / Vi
cos∅ = 1500 / ( 380 × 5 ) = 0.7894
Therefore the Power Factor cos∅ is 0.7894
e) The impedance Z in polar and rectangular form
Z = 380 / ( S∠-37.87) = V/I
Z = ( 60 + j 46.647) Ω
Z = 76 ∠ 37.87° Ω
Therefore the impedance Z in polar and rectangular form is 76 ∠ 37.87° Ω
Consider the string length equal to 7. This string is distorted by a function f (x) = 2 sin(2x) - 10sin(10x). What is the wave formed in this string? a. In=12cos (nit ) sin(max) b. 2cos(2t)sin (2x) - 10cos(10t ) sin(10x) c. n 2 sin 2x e' – 10sin 10x e
Answers
Answer:
hello your question has a missing part below is the missing part
Consider the string length equal to [tex]\pi[/tex]
answer : 2cos(2t) sin(2x) - 10cos(10t)sin(10x)
Explanation:
Given string length = [tex]\pi[/tex]
distorted function f(x) = 2sin(2x) - 10sin(10x)
Determine the wave formed in the string
attached below is a detailed solution of the problem
Which kind of fracture (ductile or brittle) is associated with each of the two crack propagation mechanisms?
Answers
dutile is the correct answer
A tube of diameter 3 cm and length 3 m has a water flow of 100 cm3/s. If the pollutant concentration in the water is constant at 2 mg/L, find the mass flux (mg/cm2-s) of pollutant through the tube due to advection.
Answers
Answer: the mass flux of pollutant through the tube due to advection is 0.0283 mg/cm².s
Explanation:
Given that;
Diameter of tube = 3 cm, radius r = 1.5 cm
water flow is 100 cm³/s
pollutant concentration = 2 mg/L
first we find the rate of flow of pollutant
we know that
1 L = 1000 cm³
xL = 100 cm³
100Lcm³ = xL1000cm³
xL = 100/1000
xL = 1/10 L
so 100cm³ = 1/10 L
now pollutant concentration in 100 cm³ = 1/10L × 2mg/L = 0.2 mg
Rate of flow of pollutant = 0.2 mg/s
Mass flux density is the pollutant mass per unit time per unit area
so Area of tube = πr² = 3.14 × 1.5² = 7.065 cm²
So
Mass flux = 0.2 / 7.065
Mass flux = 0.0283 mg/cm².s
Therefore, the mass flux of pollutant through the tube due to advection is 0.0283 mg/cm².s
A roadway is to be designed on a level terrain. The roadway id 500 ft. Five cross-sections have been selected at 0 ft, 125 ft, 250 ft, 375 ft, and 500 ft. the cross sections have areas of 130 ft^2, 140 ft^2, 60 ft^2, 110 ft^2, and 120 ft^2. What is the volume needed along this road assuming a 6% shrinkage?
Answers
Answer:
51112.5 ft^3
Explanation:
Determine the volume needed along the road when we assume a 6% shrinkage
shrinkage factor = 1 - shrinkage = 1 - 0.06 = 0.94
first we have to calculate the volume between the cross sectional areas (i.e. A1 ---- A5 ) using average end area method
Volume between A1 - A2
= (125 ft - 0 ft) * [(130 ft^2 + 140 ft^2) / 2]
= 125 ft * 135 ft^2
= 16875 ft^3
Volume between A2 - A3
= (250 ft - 125 ft) * [(140 ft^2 + 60 ft^2) / 2]
= 125 ft * (200 ft^2 / 2)
= 12500 ft^3
Volume between A3 - A4
= (375 ft - 250 ft) * [(60 ft^2 + 110 ft^2) / 2]
= 125 ft * (170 ft^2 / 2)
= 10625 ft^3
Volume between A4 - A5
(500 ft - 375 ft) * [(110 ft^2 + 120 ft^2) / 2]
= 125 ft * 115 ft^2
= 14375 ft3
Hence the total volume along the 500 ft road
= ∑ volumes between cross sectional areas
= 16875 ft^3 + 12500 ft^3 + 10625 ft^3 + 14375 ft^3 = 54375 ft^3
Finally the volume needed along this road is calculated as
Total volume * shrinkage factor
= 54375 * 0.94 = 51112.5 ft^3
Instead of running blood through a single straight vessel for a distance of 2 mm, one mammalian species uses an array of 100 tiny parallel pipes of the same total cross-sectional area, 4.0 mm2. Volume flow is 1000 mm3/s. The pressure drop for fluid passing through the single pipe is lower than that through the 100 vessel array by a factor of:_______.
A. 10
B. 100
C. 1000
Answers
Solution:
Given that :
Volume flow is, [tex]$Q_1 = 1000 \ mm^3/s$[/tex]
So, [tex]$Q_2= \frac{1000}{100}=10 \ mm^3/s$[/tex]
Therefore, the equation of a single straight vessel is given by
[tex]$F_{f_1}=\frac{8flQ_1^2}{\pi^2gd_1^5}$[/tex] ......................(i)
So there are 100 similar parallel pipes of the same cross section. Therefore, the equation for the area is
[tex]$\frac{\pi d_1^2}{4}=1000 \times\frac{\pi d_2^2}{4} $[/tex]
or [tex]$d_1=10 \ d_2$[/tex]
Now for parallel pipes
[tex]$H_{f_2}= (H_{f_2})_1= (H_{f_2})_2= .... = = (H_{f_2})_{10}=\frac{8flQ_2^2}{\pi^2 gd_2^5}$[/tex] ...........(ii)
Solving the equations (i) and (ii),
[tex]$\frac{H_{f_1}}{H_{f_2}}=\frac{\frac{8flQ_1^2}{\pi^2 gd_1^5}}{\frac{8flQ_2^2}{\pi^2 gd_2^5}}$[/tex]
[tex]$=\frac{Q_1^2}{Q_2^2}\times \frac{d_2^5}{d_1^5}$[/tex]
[tex]$=\frac{(1000)^2}{(10)^2}\times \frac{d_2^5}{(10d_2)^5}$[/tex]
[tex]$=\frac{10^6}{10^7}$[/tex]
Therefore,
[tex]$\frac{H_{f_1}}{H_{f_2}}=\frac{1}{10}$[/tex]
or [tex]$H_{f_2}=10 \ H_{f_1}$[/tex]
Thus the answer is option A). 10
What is the Bernoulli formula?
Answers
Answer:
P1+1/2pv2/1+pgh1=P2+1/2pv2/2+pgh2
What is computer programming
Answers
Answer:
Computer programming is where you learn and see how computers work. People do this for a living as a job, if you get really good at it you will soon be able to program/ create a computer.
Explanation:
Hope dis helps! :)
Using the following data, determine the percentage retained, cumulative percentage retained, and percent passing for each sieve.
Sieve size Weight retained (g) No. 4 59.5 No. 8 86.5 No. 16 138.0 No. 30 127.8 No. 50 97.0 No. 100 66.8 Pan 6.3
Answers
Solution :
Sieve Size (in) Weight retain(g)
3 1.62
2 2.17
[tex]$1\frac{1}{2}$[/tex] 3.62
[tex]$\frac{3}{4}$[/tex] 2.27
[tex]$\frac{3}{8}$[/tex] 1.38
PAN 0.21
Given :
Sieve weight % wt. retain % cumulative % finer
size retained wt. retain
No. 4 59.5 10.225% 10.225% 89.775%
No. 8 86.5 14.865% 25.090% 74.91%
No. 16 138 23.7154% 48.8054% 51.2%
No. 30 127.8 21.91% 70.7154% 29.2850%
No. 50 97 16.6695% 87.3849% 12.62%
No. 100 66.8 11.4796% 98.92% 1.08%
Pan 6.3 1.08% 100% 0%
581.9 gram
Effective size = percentage finer 10% ([tex]$$D_{20}[/tex])
0.149 mm, N 100, % finer 1.08
0.297, N 50 , % finer 12.62%
x , 10%
[tex]$y-1.08 = \frac{12.62 - 1.08}{0.297 - 0.149}(x-0.149)$[/tex]
[tex]$(10-1.08) \times \frac{0.297 - 0.149}{12.62 - 1.08}+ 0.149=x$[/tex]
x = 0.2634 mm
Effective size, [tex]$D_{10} = 0.2643 \ mm$[/tex]
Now, N 16 (1.19 mm) , 51.2%
N 8 (2.38 mm) , 74.91%
x, 60%
[tex]$60-51.2 = \frac{74.91-51.2}{2.38-1.19}(x-1.19)$[/tex]
x = 1.6317 mm
[tex]$\therefore D_{60} = 1.6317 \ mm$[/tex]
Uniformity co-efficient = [tex]$\frac{D_{60}}{D_{10}}$[/tex]
[tex]$Cu= \frac{1.6317}{0.2643}$[/tex]
Cu = 6.17
Now, fineness modulus = [tex]$\frac{\Sigma \text{\ cumulative retain on all sieve }}{100}$[/tex]
[tex]$=\frac{\Sigma (10.225+25.09+48.8054+70.7165+87.39+98.92+100)}{100}$[/tex]
= 4.41
which lies between No. 4 and No. 5 sieve [4.76 to 4.00]
So, fineness modulus = 4.38 mm
A person holds her hand out of an open car window while the car drives through still air at 65 mph. Under standard atmospheric conditions, what is the maximum pressure on her hand? What would be the maximum pressure if the "car" were an Indy 500 racer traveling 200 mph?
Answers
Answer:
[tex]10.8\ \text{lb/ft^2}[/tex]
[tex]101.96\ \text{lb/ft}^2[/tex]
Explanation:
[tex]v_1[/tex] = Velocity of car = 65 mph = [tex]65\times \dfrac{5280}{3600}=95.33\ \text{ft/s}[/tex]
[tex]\rho[/tex] = Density of air = [tex]0.00237\ \text{slug/ft}^3[/tex]
[tex]v_2=0[/tex]
[tex]P_1=0[/tex]
[tex]h_1=h_2[/tex]
From Bernoulli's law we have
[tex]P_1+\dfrac{1}{2}\rho v_1^2+h_1=P_2+\dfrac{1}{2}\rho v_2^2+h_2\\\Rightarrow P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 95.33^2\\\Rightarrow P_2=10.8\ \text{lb/ft^2}[/tex]
The maximum pressure on the girl's hand is [tex]10.8\ \text{lb/ft^2}[/tex]
Now [tex]v_1[/tex] = 200 mph = [tex]200\times \dfrac{5280}{3600}=293.33\ \text{ft/s}[/tex]
[tex]P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 293.33^2\\\Rightarrow P_2=101.96\ \text{lb/ft}^2[/tex]
The maximum pressure on the girl's hand is [tex]101.96\ \text{lb/ft}^2[/tex]
What is difference between a backdoor, a bot, a keylogger, and psyware,a nd a rootkit? Can they all present in the same malware?
Answers
Answer:
Yes, they can all be present in the same malware because each of them perform slightly differing functions.
Explanation:
Backdoor is a software which when placed into your computer will permit hackers to easily gain reentry into your computer. This can happen even after you have already patched the flaw that they have used to hack your system before.
A bot is a program that does the same task in a continuous manner akin to when you use a blender by pressing the button to blend what you have put into it.
A keylogger is a part of a hidden software that monitors and records everything you type on your computer keyboard after which it writes it onto a file, with the hopes of capturing relevant information such as your bank account number and even passwords and other sensitive means of identification.
A Spyware is somehow similar to a keylogger just that it steals information from your computer and sends it to someone else.
A root kit is a bad software that is capable of modifying the operating system or other privileged access devices in order to gain continuous access into your system for the purpose of gathering of information and/or reducing the system’s functionality.
Yes, they can all be present in the same malware because each of them perform slightly differing functions.
The structure of a house is such that it loses heat at a rate of 3800 kJ/h per C di erence between the indoors and outdoors. A heat pump that requires a power input of 4 kW is used to maintain this house at 24C. Determine the lowest outdoor temperature for which the heat pump can meet the heating requirements of this house.
Answers
Answer:
-9.5° C
Explanation:
See attachment for calculations.
On the concluding parts, from the attachment, we have that
√[(297000 * 4)/(1056)] = 297 - T(l), and solving further, we get
297 - T(l) = √(1188000/1056)
297 - T(l) = √1125
297 - T(l) = 33.5
T(l) = 297 - 33.5
T(l) = 263.5
When you convert back to °C, we have
263.5 - 273 = -9.5° C
A smooth ceramic sphere (SG 5 2.6) is immersed in a fl ow of water at 208C and 25 cm/s. What is the sphere diameter if it is encountering (a) creeping motion, Red 5 1 or (b) transition to turbulence, Red 5 250,000
Answers
Answer:
a. 4[tex]\mu m[/tex]
b. 1 m
Explanation:
According to the question, the data is as follows
The Density of water at 20 degrees celcius is 1000 kg/m^3
Viscosity is 0.001kg/m/.s
Velocity V = 25 cm/s
V = 0.25 m/s
Now
a. The creeping motion is
As we know that
Reynold Number = (Density of water × V × d) ÷ (Viscosity)
1 = (1,000 × 0.25 × d) ÷ 0.0001
d = (1 × 0.001) ÷ (1,000 × 0.25)
= 4E - 06^m
= 4[tex]\mu m[/tex]
b. Now the sphere diameter is
Reynold Number = (Density of water × V × d) ÷ (Viscosity)
250,000 = (1,000 × 0.25 × d) ÷ 0.0001
d = (250,000 × 0.001) ÷ (1,000 × 0.25)
= 1 m