Based on the available information, the the estimated towing power required is:
(a) For the parallel orientation: 655 kW
(b) For the normal orientation: 4,116.75 kW
Given:
- Diameter of the cylinder: 1.5 m
- Length of the cylinder: 22 m
- Towing speed (U): 5 m/s
- Water temperature: 20°C
(a) Cylinder parallel to the tow direction:
Step 1: Calculate the drag force for the parallel orientation.
The drag force for a submerged cylinder in parallel orientation is given by the formula:
F_D = 0.5 × ρ × C_D × A × U^2
Where:
- ρ is the density of fresh water at 20°C, which is approximately 998 kg/m³.
- C_D is the drag coefficient for a cylinder in parallel orientation, which is approximately 0.82.
- A is the cross-sectional area of the cylinder, which is π × D × L = π × 1.5 m × 22 m = 103.87 m².
Calculating the drag force:
F_D = 0.5 × 998 kg/m³ × 0.82 × 103.87 m² × (5 m/s)² = 131,000 N
Step 2: Calculate the towing power for the parallel orientation.
Towing power = Drag force × Towing speed
Towing power = 131,000 N × 5 m/s = 655,000 W = 655 kW
(b) Cylinder normal to the tow direction:
Step 3: Calculate the drag force for the normal orientation.
The drag force for a submerged cylinder in normal orientation is given by the formula:
F_D = 0.5 × ρ × C_D × A × U^2
Where:
- C_D is the drag coefficient for a cylinder in normal orientation, which is approximately 1.2.
- A is the cross-sectional area of the cylinder, which is π × D × D/4 = π × 1.5 m × 1.5 m/4 = 1.77 m².
Calculating the drag force:
F_D = 0.5 × 998 kg/m³ × 1.2 × 1.77 m² × (5 m/s)² = 823,350 N
Step 4: Calculate the towing power for the normal orientation.
Towing power = Drag force × Towing speed
Towing power = 823,350 N × 5 m/s = 4,116,750 W = 4,116.75 kW
Therefore, the estimated towing power required is:
(a) For the parallel orientation: 655 kW
(b) For the normal orientation: 4,116.75 kW
Consider a subsonic engine inlet (i.e., a diffuser) with an inlet area of 1.5 m² and exit-to-inlet area ratio of 1.44. Air enters the diffuser flowing at 5 kg/s with a static pressure of 0.350 bar and a velocity of 110 m/s. The gases exit the diffuser at a static pressure of 0.375 bar and a velocity of 80 m/s. Assume that the ambient (external) pressure on the walls of the diffuser is 0.340 bar.
(a) For these conditions, what is the force (in units of lbf and N) transmitted to the structure holding the diffuser? Provide both the direction and magnitude of the force and use a picture to help present your answer.
(b) Based solely on your results, comment on whether the force acting on the diffuser would help speed up or slow down a vehicle that used this diffuser as part of a jet propulsion system.
Answer:
Ai=2300 in² , Ao=Ai*1.44=3312 in²
m=10 lbm/s
Pi=5 psia , Po=5.4 psia , Pa=5.5 psia
Vi=120 m/s , Vo=78 m/s
a) Force =m(Vo-Vi) = -190.5 N = -42.82 lbf (towards the inlet)
b) since force is negative it will slow down the system.
Explanation:
The velocity field of a flow is given by where x and y are in feet. Determine the fluid speed at points along the x axis; along the y axis. What is the angle between the velocity vector and the x axis at points 15, 52, and 10, 52
There is a part of the question missing and it says;
The velocity field of a flow is given by V = [20y/(x² + y²)^(1/2)]î − [20x/(x² + y²)^(1/2)] ĵ ft/s, where x and y are in feet.
Answer:
A) At (1,5),angle is -11.31°
B) At (5,2),angle is -68.2°
C) At (1,0), angle is -90°
Explanation:
From the question ;
V = [20y/(x² + y²)^(1/2)]î − [20x/(x² + y²)^(1/2)] ĵ ft/s
Let us assume that u and v are the flow velocities in x and y directions respectively.
Thus we have the expression;
u = [20y/(x² + y²)^(1/2)]
and v = - [20x/(x² + y²)^(1/2)]
Thus, V = √(u² + v²)
V = √[20y/(x² + y²)^(1/2)]² + [-20x/(x² + y²)^(1/2)]²
V = √[400y²/(x²+y²)] +[400x²/(x²+y²)
V = √(400x² + 400y²)/(x²+y²)
Now for the angle;
tan θ = opposite/adjacent
And thus, in this question ;
tan θ = v/u
tan θ = [-20x/(x² + y²)^(1/2)]/ [20y/(x² + y²)^(1/2)]
Simplifying this, we have;
tan θ = - 20x/20y = - x/y
so the angle is ;
θ = tan^(-1)(-x/y)
So let's now find the angle at the various coordinates.
At, 1,5
θ = tan^(-1)(-1/5) = tan^(-1)(-0.2)
θ = -11.31°
At, 5,2;
θ = tan^(-1)(-5/2) = tan^(-1)(-2.5)
θ = -68.2°
At, 1,0;
θ = tan^(-1)(-1/0) = tan^(-1)(-∞)
θ = -90°
At,
Water flows in a tube that has a diameter of D= 0.1 m. Determine the Reynolds number if the average velocity is 10 diameters per second. (b) Repeat the calculations if the tube is a nanoscale tube with a dimeter of D= 100 nm.
Answer:
a) [tex]Re_{D} = 111896.745[/tex], b) [tex]Re_{D} = 1.119\times 10^{-7}[/tex]
Explanation:
a) The Reynolds number for the water flowing in a circular tube is:
[tex]Re_{D} = \frac{\rho\cdot v\cdot D}{\mu}[/tex]
Let assume that density and dynamic viscosity at 25 °C are [tex]997\,\frac{kg}{m^{3}}[/tex] [tex]0.891\times 10^{-3}\,\frac{kg}{m\cdot s}[/tex], respectively. Then:
[tex]Re_{D}=\frac{(997\,\frac{kg}{m^{3}} )\cdot (1\,\frac{m}{s} )\cdot (0.1\,m)}{0.891\times 10^{-3}\,\frac{kg}{m\cdot s} }[/tex]
[tex]Re_{D} = 111896.745[/tex]
b) The result is:
[tex]Re_{D}=\frac{(997\,\frac{kg}{m^{3}} )\cdot (10^{-6}\,\frac{m}{s} )\cdot (10^{-7}\,m)}{0.891\times 10^{-3}\,\frac{kg}{m\cdot s} }[/tex]
[tex]Re_{D} = 1.119\times 10^{-7}[/tex]
A fire hose nozzle has a diameter of 1.125 in. According to some fire codes, the nozzle must be capable of delivering at least 250 gal/min. If the nozzle is attached to a 3-in.-diameter hose, what pressure must be maintained just upstream of the nozzle to deliver this flowrate?
Answer:
[tex]P_{1} = 403,708\,kPa\,(58.553\,psi)[/tex]
Explanation:
Let assume that changes in gravitational potential energy can be neglected. The fire hose nozzle is modelled by the Bernoulli's Principle:
[tex]\frac{P_{1}}{\rho\cdot g} = \frac{P_{2}}{\rho \cdot g} + \frac{v^{2}}{2\cdot g}[/tex]
The initial pressure is:
[tex]P_{1} = P_{2}+ \frac{1}{2}\cdot \rho v^{2}[/tex]
The speed at outlet is:
[tex]v=\frac{\dot Q}{\frac{\pi}{4}\cdot D^{2}}[/tex]
[tex]v=\frac{(250\,\frac{gal}{min} )\cdot (\frac{3.785\times 10^{-3}\,m^{3}}{1\,gal} )\cdot(\frac{1\,min}{60\,s} )}{\frac{\pi}{4}\cdot [(1.125\,in)\cdot(\frac{0.0254\,m}{1\,in} )]^{2} }[/tex]
[tex]v\approx 24.592\,\frac{m}{s}\,(80.682\,\frac{ft}{s} )[/tex]
The initial pressure is:
[tex]P_{1} = 101.325\times 10^{3}\,Pa+\frac{1}{2}\cdot (1000\,\frac{kg}{m^{3}} )\cdot (24.592\,\frac{m}{s} )^{2}[/tex]
[tex]P_{1} = 403,708\,kPa\,(58.553\,psi)[/tex]
Answer:
P1 = 42.93 psi
Explanation:
For incompressible fluid, we know that;
A1V1 = A2V2
Making V1 the subject, we obtain;
V1 = A2V2/A1
Now A2V2 is the volumetric flow rate (V') .
Thus; V1 = V'/A1
A1 = πD²/4
Thus, V1 = 4V'/πD²
V' = 250 gal/min
But the diameter is in inches, let's convert to inches³/seconds.
Thus, V' = 250 x 3.85 = 962.5 in³/s
Substituting the relevant values to obtain,
V1 = (4 x 962.5)/(π x 3²) = 136.166 in/s.
Now let's convert to ft/s;
V1 = 136.166 x 0.0833 = 11.34 ft/s
Also for V2;
V2 = (4 x 962.5)/(π x 1.125²) = 968.29 in/s.
Now let's convert to ft/s;
V2 = 968.29 x 0.0833 = 80.66 ft/s
Setting bernoulli equation between the hose and the exit, we obtain;
(p1/γ) + (V1²/2g) = V2²/2g
Where V1 and V2 are intial and final velocities and γ is specific weight of water which is 62.43 lb/ft³ and g i acceleration due to gravity which is 32.2 ft/s²
Making p1 the subject, we obtain;
p1 = (γ/2g)(V2² - V1²)
p1 = (62.43/(2x32.2))(80.66² - 11.34²)
p1 = 6182.35 lb/ft²
So Converting to psi, we have;
p1 = 6182.35/144 = 42.93 psi
A cylindrical specimen of some metal alloy 11.2 mm (0.4409 in.) in diameter is stressed elastically in tension. A force of 15600 N (3507 lbf) produces a reduction in specimen diameter of 5 × 10-3 mm (1.969 × 10-4 in.). Compute Poisson's ratio for this material if its elastic modulus is 100 GPa (14.5 × 106 psi).
Answer:
attached below
Explanation:
Consider the general form of the Reynolds transport theorem (RTT) given by dBsys dt = d dt ∫CV rhob dV + ∫CS rhobV› r·n › dA where V › r is the velocity of the fluid relative to the control surface. Let Bsys be the mass m of a closed system of fluid particles. We know that for a system, dm/dt = 0 since no mass can enter or leave the system by definition. Use the given equation to derive the equation of conservation of mass for a control volume.
Answer:
Explanation:
note:
solution is attached due to error in mathematical equation. please find the attachment
In a CNC milling machine, the axis corresponding to the feed rate uses a dc servomotor as the drive unit and a rotary encoder as the feedback sensing device. The motor is geared to a leadscrew with a 10:1 reduction (10 turns of the motor for each turn of the leadscrew). If the leadscrew pitch is 6 mm, and the encoder emits 60 pulses per revolution, determine (a) the rotational speed of the motor and (b) pulse rate of the encoder to achieve a feed rate of 300 mm/min.
Answer:
a) 500 rev/min.
b) 50 Hz.
Explanation:
See the attached pictures.
. A two-dimensional fluid motion is represented by a point vortex of strength Γ set at distance c from an infinite straight solid boundary. Write expressions for the velocity potential and stream function in Cartesian coordinates. Derive an expression for velocity on the boundary. Draw the streamlines and plot the velocity distribution on the boundary when Γ = π and c = 1.
Answer:
The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.
Explanation:
A closed tank contains ethyl alcohol to a depth of 66 ft. Air at a pressure of 23 psi fills the gap at the top of the tank. Determine the pressure at a closed valve attached to the tank 10 ft above its bottom
Answer:
639.4psi
Explanation:
Pressure at the closed valve = Air pressure+ (density*gravity*height)
=23psi+(49.27lb/ft^3*32.17ft/s^2*56ft^3)
=23psi+88760.89psft
=23psi+(88760.89/144)psi
=23+616.4
=639.4psi
A road has a crest curve, where the PVI station is a 71 35. The road transitions from a 2.1% grade to a -3.4% grade. The highest point of the curve is at station 74 10. What are the PVC and PVT stations
Answer:
Stat PVC = Stat(82+98.5)
Stat PVT = Stat(59+71.5)
Explanation
PVI = 71 + 35
Let G1 = Grade 1; G2 = Grade 2
G1 = +2.1% ; G2 = -3.4%
Highest point of curve at station = 74 + 10
General equation of a curve:
[tex]y = ax^{2} +bx+c\\dy/dx=2ax+b\\[/tex]
At highest point of the curve [tex]dy/dx=o[/tex]
[tex]2ax+b=0\\x=-b/2a\\x=G1L/(G2-G1)\\x=L/2 +(stat 74+10)-(stat 71+35)\\x=L/2 + 275[/tex]
[tex]-G1L/(G2-G1) = (L/2 + 275)/100\\L = -2327 ft\\Station PVC = Stat(71+35)+(-2327/2)\\\\Stat PVC = 7135-1163.5\\Stat PVC = Stat(82+98.5)\\[/tex]
Station PVT
[tex]Station PVT = Stat PVI + (L/2)\\Station PVT = Stat(71+35)+(-2327/2)\\Station PVT = 7135-1163.5\\Stat PVT = Stat(59+71.5)[/tex]
Water is to be boiled at sea level in a 30-cm-diameter stainless steel pan placed on top of a 3-kW electric burner. If 60 percent of the heat generated by the burner is transferred to the water during boiling, determine the rate of evaporation of water.
Answer:
mevaporation=˙Qhfg=1. 8 kJ /s2269. 6 kJ /kg=0 . 793×10−3kg/ s=2. 855 kg /h
Explanation:
The properties of water at 1 atm and thus at the saturation temperature of 100C are hfg =2256.4 kJ/kg (Table A-4). The net rate of heat transfer to the water is ˙Q=0 . 60×3 kW=1 . 8 kWNoting that it takes 2256.4 kJ of energy to vaporize 1 kg of saturated liquid water, therate of evaporation of water is determined to be mevaporation=˙Qhfg=1. 8 kJ /s2269. 6 kJ /kg=0 . 793×10−3kg/ s=2. 855 kg /h
The rate of evaporation will be "2.871 Kg/hour".
Evaporation of water:According to the question,
Rate of heat supplied:
= 60% of 3 kW
= 1.8 kW
= 1.8 KJ/s
Vaporization of water, [tex]\Delta H = 2257 \ KJ/Kg[/tex]
Time taken will be:
= [tex]\frac{2257}{1.8}[/tex]
= [tex]1254 \ s[/tex]
= [tex]\frac{1254}{3600} \ hour[/tex]
= [tex]0.3482 \ hour[/tex]
hence,
The rate of evaporation,
= [tex]\frac{Mass \ of \ water}{Time}[/tex]
= [tex]\frac{1}{0.3482}[/tex]
= [tex]2.871 \ Kg/hour[/tex]
Thus the above answer is right.
Find out more information about evaporation here:
https://brainly.com/question/4406110
A one-dimensional plane wall of thickness 2L=100mm experiences uniform thermal energy generation of q dot=1000 W/m^3 and is convectively cooled at x=+-50mm by an ambient fluid characterized by T infinity=20degreesC. If the steady-state temperature distribution within the wall is T(x)=a(L^2-x^2)+b where a=10 degrees C/m^2 and b=30 degrees C, what is the thermal conductivity of the wall? What is the value of the convection heat transfer coefficient, h?
Answer:
note:
solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment
Answer:
A) Thermal conductivity of wall = 50 W/m.c
B) value of the convection heat transfer coefficient, h = 5 W/m².C
Explanation:
I've attached all explanations
A three-phase, 600 MVA, 13.8KV AC generator has a synchronous reactance of 2.0 per unit. The generator is connected to a system for which the specified bases are 100MVA and 345KV. a) Find the per-unit value of the generator synchronous reactance on the specified bases. b) Find the ohmic value of the synchronous reactance. Problem 4 A single-phase source is connected to an electrical load. The load
Answer:
(a) 0.00053
(b) 0.1 mΩ
Explanation:
New per-unit reactance is given as:New Per-unit reactance = [tex]2* \frac{100}{600} * (\frac{13.8}{345}) ^{2} = 0.00053[/tex]
Ohmic reactance : [tex]\frac{13.8^2}{600} = 0.31 ohm[/tex]Ohmic per unit : 0.31 * 0.00053 = 0.1 mΩGiven Information:
Zpu_old = 2 pu
Sbase_new = 100 MVA
Sbase_old = 600 MVA
kV_old = 13.8 kV
kV_new = 345 kV
Required Information:
Zpu_new = ?
ZΩ = ?
Answer:
Zpu_new = 0.000533 pu
ZΩ = 0.634 Ω
Explanation:
a) Find the per-unit value of the generator synchronous reactance on the specified bases.
When the base kVA and base kV are changed then we use following relation to update the per unit values.
Zpu_new = Zpu_old*(Sbase_new/Sbase_old)*(kV_old/kV_new)²
Zpu_new = 2*(100x10⁶/600x10⁶)*(13.8x10³/345x10³)²
Zpu_new = 0.000533 pu
b) Find the ohmic value of the synchronous reactance.
ZΩ = Zbase*Zpu_new
Where Zbase is calculated as
Zbase = (kVbase)²/Sbase
Zbase = (345x10³)²/100x10⁶
Zbase = 1190.25 Ω
ZΩ = Zbase*Zpu_new = 1190.25*0.000533 = 0.634 Ω
A vertical piston-cylinder device contains water and is being heated on top of a range. During the process, 10 kJ of heat is transferred to the water, and heat losses from the side walls amount to 80 J. The piston rises as a result of evaporation, and 2 J of work is done by the vapor. Determine the change in the energy of the water for this process.
Answer: 9.9KJ
Explanation: Q = U + W + losses
Q is heat transfered to the water
U is the change in energy of the system
W is work done by the system = 2J
Losses = 80J
Heat into system is 10kJ = 10000KJ
Therefore
U = Q - W - losses
U = 10000 - 2 - 80 = 9990J
= 9.9kJ
A load P is applied horizontally while the other end is fixed to a structure. where P = 185 N You have already examined the axial stresses inside the plates. In this problem, the bolted connections will be analyzed. Failure means that the pieces come apart because the bolts cannot hold the force. If two bolts are connecting the two plates, how much transverse force V must each bolt resist at each sheare joint?
Answer:
Explanation:
The bolt is under double shear because it connects 3 plates.
Now the shear force resisted by each shear joint(V):
V=P/(No. of shear area)=185/2=92.5N.
Write a program with two inputs, current price and last month's price (both integers). Then, output a summary listing the price, the change since last month, and the estimated monthly mortgage computed as (currentPrice * 0.045) / 12. in coral
Complete Question
Sites like Zillow get input about house prices from a database and provide nice summaries for readers. Write a program with two inputs, current price and last month's price (both integers). Then, output a summary listing the price, the change since last month, and the estimated monthly mortgage computed as (currentPrice * 0.045) / 12. Ex: If the input is 200000 210000, the output is: This house is $200000. The change is $-10000 since last month. The estimated monthly mortgage is $750.
Use Coral Programming Language
Answer:
// Program is written in Coral Programming Language
// Comments are used for explanatory purpose
// Program starts here
// Variable declaration
int currentprice
int prevprice
int change
float mortgage
Put "Enter current price to output" to output
currentprice = Get next input
Put "Enter last month price to output" to output
prevprice = Get next input
// Calculate Change since last month
change = currentprice - prevprice
// Calculate Monthly Mortgage
mortgage = currentprice * 0.045 / 12
// Print Results
Put "This house is $" to output
Put currentprice to output
Put "\n" to output
Put "This change is $" to output
Put change to output
Put "\n" to output
Put "This house is $" to output
Put currentprice to output
Put "since last month\n" to output
Put "This estimated monthly mortgage is $" to output
Put mortgage to output
// End of Program
A 3-kg block rests on top of a 2-kg block supported by, but not attached to, a spring of constant 40 N/m. The upper block is suddenly removed. Determine (a) the maximum speed reached by the 2-kg block, (b) the maximum height reached by the 2-kg block.
This Physics question is directed towards High School students and pertains to the maximum speed and height a 2-kg block achieves when released from a compressed spring, following the removal of a 3-kg block initially resting on it.
Explanation:The subject of this question is Physics, and it is appropriate for a High School grade level. The question is regarding the oscillatory motion and energy conversion of a block attached to a spring system. Specifically, it involves understanding the concepts of potential energy stored in a spring, conservation of mechanical energy, and kinematics of simple harmonic motion.
To solve part (a) of the problem for the 2-kg block, you'll need to use the conservation of energy principle. Initially, when the 3-kg block is also on the spring, the potential energy stored in the compressed spring is equal to the kinetic energy the 2-kg block will have when it reaches its maximum speed. After the upper block is removed, the spring force will only accelerate the 2-kg block.
To find the maximum speed reached by the 2-kg block, you would calculate the kinetic energy equivalent to the potential energy stored in the spring and set it equal to \\((1/2)mv^2\\). For part (b), you could use the conservation of energy principle again to find the maximum height the 2-kg block reaches by equating the initial kinetic energy to the potential energy at the maximum height (\\(mgh\\)).
Determine if each of the following statements is true or false:
1. Larger atoms are better nucleophiles due to polarizability.
2. The identity of the nucleophile affects the rate of an SN1 reaction.
3. SN2 reactions proceed via frontside attack.
4. Bimolecular reactions tend to be stereoselective.
5. SN2 reactions invert all stereocenters in a haloalkane.
6. Cl-, OH-, and H- are good all leaving groups.
7. Good bases tend to be good nucleophiles
8. Branching adjacent to a reacting carbon slows SN2 reactions due to steric hindrance.
9. SN1 reactions are slowed by hyperconjugation of the carbocation intermediate.
10. The rate determining step for SN1 reactions is the same as the rate determining step for E1 reactions.
Answer:
Explanation:
1) true ( reason is due to polarizability given in statement)
2)false ( nucleophile does not involve in rate determining step)
3)false ( it is an inverse attack , attacks from backside)
4) true
5) false ( it inverts stereo centres which are being attacked)
6) false (not H-)
7) true ( strong bases are good nucleophiles)
8) true ( because of steric hindrance)
9)false ( ther are stabilized by hyper conjugation)
10) false
Answer:
1) True
2) False
3) False
4)
5)
6) True
7)True
8) True
9) False
10) True
Explanation:
1) Generally, polarization increases down the column of the periodic table.
The tangent function is defined as tan(theta) = sin(theta)/cos(theta). This expression can be evaluated to solve for the tangent as long as the magnitude of cos(theta) is no too near to 0. Assume that theta is given in degrees, and write the MATLAB statements to evaluate tan(theta) as long as the magnitude of cos(theta) is greater than or equal to 10e-2. If the magnitude of cos(theta) is less than 10e-2, write out an error message instead.
Answer:
The code is as attached here.
Explanation:
The code is as given below
theta = input(' Enter the value of theta?');
y = sin(theta*pi()/180);
z = cos(theta*pi()/180);
if z < 0.01
fprintf('The value of theta is very low')
else
t=round(y/z,2);
disp(['The value of tangent theta is ',num2str(t)]);
end
In MATLAB, evaluate tangent of theta if the magnitude of cosine is
[tex]> = 10[/tex]⁻²; otherwise, display an error message.
Here are the MATLAB statements to evaluate [tex]\( \tan(\theta) \)[/tex] as long as the magnitude of [tex]\( \cos(\theta) \)[/tex] is greater than or equal to [tex]\( 10^{-2} \)[/tex], and display an error message if the magnitude of [tex]\( \cos(\theta) \)[/tex] is less than [tex]\( 10^{-2} \)[/tex]:
```matlab
% Define theta in degrees
theta_deg = input('Enter the value of theta in degrees: ');
% Convert theta to radians
theta_rad = deg2rad(theta_deg);
% Calculate cosine of theta
cos_theta = cos(theta_rad);
% Check if the magnitude of cos(theta) is greater than or equal to 10^-2
if abs(cos_theta) >= 1e-2
% Evaluate tangent of theta
tan_theta = sin(theta_rad) / cos_theta;
disp(['tan(theta) = ', num2str(tan_theta)]);
else
% Display error message
disp('Error: The magnitude of cos(theta) is too small. Cannot evaluate tan(theta).');
end
```
This script prompts the user to enter the value of [tex]\( \theta \)[/tex] in degrees. It then converts [tex]\( \theta \)[/tex] to radians and calculates [tex]\( \cos(\theta) \)[/tex]. If the magnitude of [tex]\( \cos(\theta) \)[/tex] is greater than or equal to [tex]\( 10^{-2} \)[/tex], it evaluates [tex]\( \tan(\theta) \)[/tex] using the given formula and displays the result. Otherwise, it displays an error message indicating that the magnitude of [tex]\( \cos(\theta) \)[/tex] is too small to evaluate [tex]\( \tan(\theta) \).[/tex]
In a typical transmission line, the current I is very small and the voltage V is very large. A unit length of the line has resistance R.
For a power line that supplies power to 10 000 households, we can conclude that:
a) ????V < ????^2 R
b) ????^2 R = 0
c) ????V = ????^2 R
d) ????V > I^2 R
e) ???? = V/R
Question:
In a typical transmission line, the current I is very small and the voltage V is very large. A unit length of the line has resistance R.
For a power line that supplies power to 10 000 households, we can conclude that
a) IV < I²R
b) I²R = 0
c) IV = I²R
d) IV > I²R
e) I = V/R
Answer:
d) IV > I²R
Explanation:
In a typical transmission line, the current I is very small and the voltage V is very high as to minimize the I²R losses in the transmission line.
The power delivered to households is given by
P = IV
The losses in the transmission line are given by
Ploss = I²R
Therefore, the relation IV > I²R holds true, the power delivered to the consumers is always greater than the power lost in the transmission line.
Moreover, losses cannot be more than the power delivered. Losses cannot be zero since the transmission line has some resistance. The power delivered to the consumers is always greater than the power lost in the transmission.
A freshwater jet boat takes in water through side vents and ejects it through a nozzle of diameter D = 75 mm; the jet speed is Vj. The drag on the boat is given by Fdrag =kV2, where V is the boat speed. Find an expression for the steady speed, V, in terms of water density rho, flow rate through the system of Q, constant k, and jet speed Vj. A jet speed Vj = 15 m=s produces a boat speed of V = 10 m=s. (a) Under these conditions, what is the new flow rate Q? (b) Find the value of the constant k. (c) What speed V will be produced if the jet speed is increased to Vj = 25 m=s? (d) What will be the new flow rate?
To find an expression for the jet boat's steady speed, equate the water jet force (ρQVj) with the drag force (kV^2). Solve for V to get the equation for speed in terms of ρ, Q, k, and Vj. Then use known values for Vj and V to find Q and k and repeat the process for a different Vj to get new V and Q values.
Explanation:To find an expression for the steady speed V of a freshwater jet boat in terms of the water density ρ, flow rate Q, constant k, and jet speed Vj, we can apply Newton's second law, assuming that the force provided by the water jet equals the drag force on the boat when at steady speed. The water jet force can be expressed as the rate of change of momentum of the water, which is ρQVj (since Q is the mass flow rate ρQ is the momentum flow rate), and the drag force is given as kV^2. Equating these two forces gives us:
ρQVj = kV^2
Solving for V will give us the steady speed expression we are looking for:
V = √(ρQVj / k)
To calculate the new flow rate Q given Vj = 15 m/s and V = 10 m/s, we plug these values into the expression obtained from above:
Q = kV^2 / (ρVj)
With given values, we would have:
Q = k * 10^2 / (ρ * 15)
The constant k can be determined using the known conditions and solving for k.
k = ρQVj / V^2
For (c) and (d), the same equations can be applied with the jet speed Vj changed to 25 m/s to find the new boat speed and flow rate.
Calculate the magnitude of the force FB in the back muscles that is needed to support the upper body plus the box and compare this with his weight. The mass of the upper body is 55.0 kg and the mass of the box is 30.0 kg.
Answer:
807.5N
Explanation:
The combined mass (m) on the back muscle is 55kg + 30kg = 85kg
Acceleration due to gravity (g) = 9.8m/s²
Therefore the force FB = ma = 85*9.8
FB= 807.5N
9.43 An ideal air-standard Brayton cycle operates at steady state with compressor inlet conditions of 300 K and 100 kPa and a fixed turbine inlet temperature of 1700 K. For the cycle, (a) determine the net work developed per unit mass flowing, in kJ/kg, and the thermal efficiency for a compressor pressure ratio of 8. (b) plot the net work developed per unit mass flowing, in kJ/kg, and the thermal efficiency, each versus compressor pressure ratio ranging from 2 to 50.
Answer:
The net work output from the cycle = 520.67 [tex]\frac{KJ}{kg}[/tex]
The efficiency of Brayton cycle = 0.448
Explanation:
Compressor inlet temperature [tex]T_{1}[/tex] = 300 K
Turbine inlet temperature [tex]T_{3}[/tex] = 1700 K
Pressure ratio [tex]r_{p}[/tex] = 8
For the compressor the temperature - pressure relation is given by the formula,
⇒ [tex]\frac{T_{2} }{T_{1} }[/tex] = [tex]r_{p}^{\frac{\gamma - 1}{\gamma} }[/tex]
⇒ [tex]\frac{T_{2} }{300} = 8^{\frac{1.4 - 1}{1.4} }[/tex]
⇒ [tex]T_{2}[/tex] = 543.42 K
This is the temperature at compressor outlet.
Same relation for turbine we can write this as,
⇒ [tex]\frac{T_{3} }{T_{4} }[/tex] = [tex]r_{p}^{\frac{\gamma - 1}{\gamma} }[/tex]
⇒[tex]\frac{1700 }{T_{4} }[/tex] = [tex]8^{0.2857}[/tex]
⇒ [tex]T_{4}[/tex] = 938.5 K
This is the temperature at turbine outlet.
Now the work output from the turbine [tex]W_{T}[/tex] = [tex]m C_{p} (T_{3} - T_{4} )[/tex]
Put all the values in above formula we get,
⇒ [tex]W_{T}[/tex] = 1 × 1.005 × ( 1700 - 938.5 )
[tex]W_{T} = 765.3 \frac{KJ}{kg}[/tex]
This is the work output from the turbine.
Now the work input to the compressor is [tex]W_{C}[/tex] = [tex]m C_{p} (T_{2} - T_{1} )[/tex]
Put all the values in above formula we get,
⇒ [tex]W_{C}[/tex] = 1 × 1.005 × ( 543.42 - 300 )
⇒ [tex]W_{C}[/tex] = 244.63 [tex]\frac{KJ}{kg}[/tex]
This is the work input to the compressor.
Net work output from the cycle [tex]W_{net} = W_{T} - W_{C}[/tex]
⇒ [tex]W_{net}[/tex] = 765.3 - 244.63
[tex]W_{net} = 520.67\frac{KJ}{kg}[/tex]
This is the net work output from the cycle.
The thermal efficiency is given by
[tex]E_{cycle} =1 - \frac{1}{r_{p}^{\frac{\gamma - 1}{\gamma} } }[/tex]
[tex]E_{cycle} =1 - \frac{1}{8^{\frac{1.4 - 1}{1.4} } }[/tex]
[tex]E_{cycle} = 0.448[/tex]
This is the efficiency of Brayton cycle.
(b). the graph between plot the net work developed per unit mass flowing and the thermal efficiency, each versus compressor pressure ratio ranging from 2 to 50 is shown in the image below.
George and Ellen Rottweiler encourage their adult daughter Guinevere to break her engagement and continue living in their home, saying, "You're so bright and attractive; you can find a better guy than this." A family systems theorist would term this as a(n) ________.
Answer:
Negative feedback
Explanation:
In Biology, negative feedback refers to the counteraction of an effect by its own influence on the process producing it. For instance, the presence of a high level of a particular hormone in the blood may inhibit further secretion of that hormone.
In other words, in negative feedback, the result of a certain action may inhibit further performance of that action
An air conditioner using refrigerant-134a as the working fluid and operating on the ideal vapor-compression refrigeration cycle is to maintain a space at 22°C while operating its condenser at 1000 kPa. Determine the COP of the system when a temperature difference of 2°C is allowed for the transfer of heat in the evaporator.
Answer:
note:
solution is attached due to error in mathematical equation. please find the attachment
Answer:
COP = 13.31
Explanation:
We have an allowed temperature difference of 2°C, thus, let's make use of temperature of 20°C in the evaporator.
Now, looking at table A-11 i have attached and looking at temperature of 20°C, we will see that the enthalpy(h1) = 261.59 Kj/Kg
While the enthropy(s1) = 0.92234 Kj/KgK
Now, the enthalpy at the second state will be gotten from the given condenser pressure under the condition s2 = s1.
Thus, looking at table A-13 which i have attached, direct 20°C is not there, so when we interpolate between the enthalpy values at 15.71°C and 21.55°C, we get an enthalpy of 273.18 Kj/Kg.
Now, the enthalpy at the third and fourth states is again obtained from interpolation between values at temperatures of 18.73 and 21.55 of the saturated liquid value in table A-12 i have attached.
Thus, h3=h4 = 107.34 Kj/kg
Formula for COP = QL/w = (h1- h4) / (h2 - h1)
COP = (261.59 - 107.34)/( 273.18 - 261.59) = 13.31
Write a program to calculate the property tax. Property tax is calculated on 92% of the assessed value of the property. For example, if the assessed value is $200,00.00, the property tax is on 184,000.00. Assume that the property tax rate is $1.05 for each $100 of the assessed value. Your program should prompt the user to enter the assessed value of the property. Store the output in a file in the following format. (Here is a sample output.
Answer:
#include<stdio.h>
void main()
{
// using file pointer to print output to txt file
FILE *fptr;
float assessedValue, taxableAmount, taxRate = 1.05, propertyTax;
/* open for writing */
fptr = fopen("output.txt", "w");
if (fptr == NULL)
{
printf("File does not exists \n");
return;
}
// prompting user to enter assessed value and storing it in assessedValue variable
printf("Enter the Assessed Value of property : ");
scanf("%f", &assessedValue);
//writing assessed value to output.txt file using fprintf file i/o function
fprintf(fptr, "AssessedValue : $ %.2f\n", assessedValue);
//calculating taxableAmount based on given condition in the question
taxableAmount = (assessedValue * 0.92);
//writing taxable Amount to output.txt file using fprintf file i/o function
fprintf(fptr, "TaxableAmount: $ %.2f\n", taxableAmount);
//writing tax Rate to output.txt file using fprintf file i/o function
fprintf(fptr, "Tax Rate for each $100.00: $ %.2f\n", taxRate);
//calculating propertyTax based on given condition in the question
propertyTax = ((taxableAmount/100)*taxRate);
//writing property Tax Amount to output.txt file using fprintf file i/o function
fprintf(fptr, "propertyTax: $ %.2f\n", propertyTax);
//closing file using fclose function
fclose(fptr);
}
Explanation :
I used Turbo C compiler to compile and run the C program. The below program compiles and at the run time, automatically, prints output to a file called output.txt.
When you compile the program, remember to check the BIN folder in Turbo c folder of C drive where your turbo c has been installed.
Output:
Assessed value: $200000
Taxable amount: $184000
Tax Rate for each $100.00: $1.05
While it is important to gain information on the prospect, it is relatively unimportant to gain any information on the prospect’s organization prior to initiating sales dialogue.ANS:FPTS:1DIF:Difficulty: EasyREF:p. 124-125OBJ:LO: 5-5
Answer:
FALSE.
Explanation:
This statement can be considered false, because it is important to know essential aspects of the prospect before starting the sales dialogue.
In addition to obtaining basic information about the client, it is relevant for the company to know the prospect's organization.
We can define the prospect as the one who, for the company, has the ideal customer profile, but who has not yet shown interest in consuming its products and services.
Therefore, knowing information about it, about its basic and complex characteristics will help the organization to develop sales and marketing strategies aimed at attracting the prospect.
Determine the average and rms values for the function, y(t)=25+10sino it over the time periods (a) 0 to 0.1 sec and (b) 0 to 1/3 sec. Discuss which case represents the long-term behavior of the signal (Hint, consider the period of the signal).
Answer:
Explanation:
AVERAGE: the average value is given as
[tex]\frac{1}{0.1} \int\limits^\frac{1}{10} _0 {25+10sint} \, dt = \frac{1}{0.1} [ 25t- 10cos\ t]_0^{0.1}[/tex]
=[tex]\frac{1}{0.1} ([2.5-10]-10)=-175[/tex]
RMS= [tex]\sqrt{\int\limits^\frac{1}{3} _0 {y(t)^2} \, dt }[/tex]
[tex]y(t)^2 = (25 + 10sin \ t)^2 = 625 +500sin \ t + 10000sin^2 \ t[/tex]
[tex]\frac{1}{\frac{1}{3} } \int\limits^\frac{1}{3} _0 {y(t)^2} \, dt =3[ 625t -500cos \ t + 10000(\frac{t}{2} - \frac{sin2t}{4} )]_0^{\frac{1}{3} }[/tex]
=[tex]3[[\frac{625}{3} - 500 + 10000(\frac{1}{6} - 0.002908)] + 500] = 2845.92\\[/tex]
therefore, RMS = [tex]\sqrt{2845.92} = 53.3[/tex]
Water at 60°F passes through 0.75-in-internal diameter copper tubes at a rate of 1.2 lbm/s. Determine the pumping power per ft of pipe length required to maintain this flow at the specified rate.
The density and dynamic viscosity of water at 70°F are rho = 62.30 lbm/ft^3 and μ = 6.556 x 10^-4 lbm/ft*s. The roughness of copper tubing is 5 x 10^-6 ft.
The pumping power per ft of pipe length required to maintain this flow at the specified rate = _________ W (per ft length)
Answer:
The pumping power per ft of pipe length required to maintain this flow at the specified rate 0.370 Watts
Explanation:
See calculation attached.
- First obtain the properties of water at 60⁰F. Density of water, dynamic viscosity, roughness value of copper tubing.
- Calculate the cross-sectional flow area.
- Calculate the average velocity of water in the copper tubes.
- Calculate the frictional factor for the copper tubing for turbulent flow using Colebrook equation.
- Calculate the pressure drop in the copper tubes.
- Then finally calculate the power required for pumping.
Two sections of a pressure vessel are to be held together by 5/8 in-11 UNC grade 5 bolts. You are told that the length of the bolts is 1.5 in, the length of the threaded portion of the bolts is 0.75 in, and that their elastic modulus is E=30 Mpsi. The total load on the joint is 36 kip and the stiffness of the members is given as km=8.95 Mlbf/in. What is the minimum number of bolts that should be used to guard against excess proof strength with a factor of safety of np=1.2? Be sure to make an estimate for the preload.
Answer:
The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.
Explanation: