Answer:
a. 46.15%
b. 261.73 kg/s
c. 54.79 kW/K
Explanation:
a. State 1
The parameters given are;
T₁ = 560°C
P₁ = 12 MPa = 120 bar
Therefore;
h₁ = 3507.41 kJ/kg, s₁ = 6.6864 kJ/(kg·K)
State 2
p₂ = 1 MPa = 10 bar
s₂ = s₁ = 6.6864 kJ/(kg·K)
h₂ = (6.6864 - 6.6426)÷(6.6955 - 6.6426)×(2828.27 - 2803.52) + 2803.52
= (0.0438 ÷ 0.0529) × 24.75 = 2824.01 kJ/kg
State 3
p₃ = 6 kPa = 0.06 bar
s₃ = s₁ = 6.6864 kJ/(kg·K)
sg = 8.3291 kJ/(kg·K)
sf = 0.52087 kJ/(kg·K)
x = s₃/sfg = (6.6864- 0.52087)/(8.3291 - 0.52087) = 0.7896
(h₃ - 151.494)/2415.17 = 0.7896
∴ h₃ = 2058.56 kJ/kg
State 4
Saturated liquid state
p₄ = 0.06 bar= 6000 Pa, h₄ = 151.494 kJ/kg, s₄ = 0.52087 kJ/(kg·K)
State 5
Open feed-water heater
p₅ = p₂ = 1 MPa = 10 bar = 1000000 Pa
s₄ = s₅ = 0.52087 kJ/(kg·K)
h₅ = h₄ + work done by the pump on the saturated liquid
∴ h₅ = h₄ + v₄ × (p₅ - p₄)
h₅ = 151.494 + 0.00100645 × (1000000 - 6000)/1000 = 152.4944113 kJ/kg
Step 6
Saturated liquid state
p₆ = 1 MPa = 10 bar
h₆ = 762.683 kJ/kg
s₆ = 2.1384 kJ/(kg·K)
v₆ = 0.00112723 m³/kg
Step 7
p₇ = p₁ = 12 MPa = 120 bar
s₇ = s₆ = 2.1384 kJ/(kg·K)
h₇ = h₆ + v₆ × (p₇ - p₆)
h₇ = 762.683 + 0.00112723 * (12 - 1) * 1000 = 775.08253 kJ/kg
The fraction of flow extracted at the second stage, y, is given as follows
[tex]y = \dfrac{762.683 - 152.4944113 }{2824.01 - 152.4944113 } = 0.2284[/tex]
The turbine control volume is given as follows;
[tex]\dfrac{\dot{W_t}}{\dot{m_{1}}} = \left (h_{1} - h_{2} \right ) + \left (1 - y \right )\left (h_{2} - h_{3} \right )[/tex]
= (3507.41 - 2824.01) + (1 - 0.22840)*(2824.01 - 2058.56) = 1274.02122 kJ/kg
For the pumps, we have;
[tex]\dfrac{\dot{W_p}}{\dot{m_{1}}} = \left (h_{7} - h_{6} \right ) + \left (1 - y \right )\left (h_{5} - h_{4} \right )[/tex]
= (775.08253 - 762.683) + (1 - 0.22840)*(152.4944113 - 151.494)
= 13.17 kJ/kg
For the working fluid that flows through the steam generator, we have;
[tex]\dfrac{\dot{Q_{in}}}{\dot{m_{1}}} = \left (h_{1} - h_{7} \right )[/tex]
= 3507.41 - 775.08253 = 2732.32747 kJ/kg
The thermal efficiency, η, is given as follows;
[tex]\eta = \dfrac{\dfrac{\dot{W_t}}{\dot{m_{1}}} -\dfrac{\dot{W_p}}{\dot{m_{1}}}}{\dfrac{\dot{Q_{in}}}{\dot{m_{1}}}}[/tex]
η = (1274.02122 - 13.17)/2732.32747 = 0.4615 which is 46.15%
(762.683 - 152.4944113)/(2824.01 - 152.4944113)
b. The mass flow rate, [tex]\dot{m_{1}}[/tex], into the first turbine stage is given as follows;
[tex]\dot{m_{1}} = \dfrac{\dot{W_{cycle}}}{\dfrac{\dot{W_t}}{\dot{m_{1}}} -\dfrac{\dot{W_p}}{\dot{m_{1}}}}[/tex]
[tex]\dot{m_{1}}[/tex] = 330 *1000/(1274.02122 - 13.17) = 261.73 kg/s
c. From the entropy rate balance of the steady state form, we have;
[tex]\dot{\sigma }_{cv} = \sum_{e}^{}\dot{m}_{e}s_{e} - \sum_{i}^{}\dot{m}_{i}s_{i} = \dot{m}_{6}s_{6} - \dot{m}_{2}s_{2} - \dot{m}_{5}s_{5}[/tex]
[tex]\dot{\sigma }_{cv} = \dot{m}_{6} \left [s_{6} - ys_{2} - (1 - y)s_{5} \right ][/tex]
= 261.73 * (2.1384 - 0.2284*6.6864 - (1 - 0.2284)*0.52087 = 54.79 kW/K
A piston–cylinder device contains 0.78 kg of nitrogen gas at 140 kPa and 37°C. The gas is now compressed slowly in a polytropic process during which PV1.3 = constant. The process ends when the volume is reduced by one-half. Determine the entropy change of nitrogen during this process. The gas constant of nitrogen is R = 0.2968 kJ/kg·K. The constant volume specific heat of nitrogen at room temperature is cv = 0.743 kJ/kg·K. (Round the final answer to five decimal places.)
Answer:
The entropy change of nitrogen during this process. is - 0.32628 kJ/K.
Explanation:
Solution
Given that:
A piston cylinder device contains =0.78 kg of nitrogen gas
Temperature = 37°C
The nitrogen gas constant of R = 0.2968 kJ/kg.K
At room temperature cv = 0.743 kJ/kg.K
Now,
We assume that at specific condition the nitrogen can be treated as an ideal gas
Nitrogen has a constant volume specific heat at room temperature.
Thus,
From the polytropic relation, we have the following below:
T₂/T₁ =(V₁/V₂)^ n-1 which is,
T₂ = T₁ ((V₁/V₂)^ n-1
= (310 K) (2)^1.3-1 = 381.7 K
So,
The entropy change of nitrogen is computed as follows:
ΔSN₂ = m (cv₁ avg ln T₂/T₁ + ln V₂/V₁)
= (0.78) ((0.743 kJ/kg .K) ln 381.7 K/310K + (0.2968 kJ/kg. K) ln (0.5))
= 0.57954 * 0.2080 + (-0.2057)
= 0.12058 + (-0.2057) = -0.32628
Therefore the entropy change of nitrogen during this process. is - 0.32628 kJ/K.
What is the transfer function of a filter? What is the transfer function of a filter? The transfer function shows how a filter affects the phase of input components as a function of frequency. It is defined as the output phasor divided by the input phasor as a function of frequency. The transfer function shows how a filter affects the amplitude and phase of input components as a function of frequency. It is defined as the input phasor divided by the output phasor as a function of frequency. The transfer function shows how a filter affects the amplitude and phase of input components as a function of frequency. It is defined as the output phasor divided by the input phasor as a function of frequency. The transfer function shows how a filter affects the phase of input components as a function of frequency. It is defined as the input phasor divided by the output phasor as a function of frequency.
Answer: Provided in the explanation section
Explanation:
A precise explanation is provided here to make it easy for understanding
(a). The transfer option shows how a filter affects the amplitude and phase of input components as a function of frequency. It is defined as the output phasor divided by the input phasor as a functoon of frequency.
H(jw) = V₀ (jw) / Vin(jw)
(b). To determine the transfunction of a filter, we connect a sinusoidal source to be the input part, measure the amplitude and phases of both the input signal and resulting output signal using voltmeters.
Oscilloscope or other instruments, and divide the Output phasor by the input phasor. This is is repeated for all frequencies.
cheers i hope this helped !!
The strength and stability of tires may be enhanced by heating both sides of the rubber ( 0.14 W/m·K, 6.35 × 10^-8m^2/s) in a steam chamber for which T[infinity]= 200°C. In the heating process, a 20-mm-thick rubber wall (assumed to be untreaded) is taken from an initial temperature of 35°C to a midplane temperature of 170°C. If steam flow over the tire surfaces maintains a convection coefficient of 200 W/m^2·K. How long will it take to achieve the desired midplane temperature?
Answer:
[tex]\mathbf{t_f = 1436.96 \ sec }[/tex]
Explanation:
Given that :
The strength and stability of tires may be enhanced by heating both sides of the rubber ( 0.14 W/m·K, 6.35 × 10^-8m^2/s)
i.e
k = 0.14 W/mK
∝ = 6.35 × 10⁻⁸ m²/s
L = 0.01 m
[tex]B_1 = \dfrac{hL}{k} \\ \\ B_1 = \dfrac{200*0.01}{0.14} \\ \\ B_1 = 14.2857[/tex]
We cannot use the model of Lumped Capacitance; SO Let assume that Fourier Number [tex]F_o > 0.2[/tex]
⇒ [tex]\dfrac{T_o - T_ \infty }{T_i - T_ \infty} = C_1 exp (- \zeta_i^2 *F_o)[/tex]
From Table 5.1 ; at [tex]B_1[/tex] = 14.2857
[tex]C_1 = 1.265 \\ \\ \zeta_1 = 1.458 \ rad[/tex]
[tex]\dfrac{170-200}{35-200} = 1.265 exp [ - (1.458)^2* \dfrac{ \alpha t_f}{L^2}][/tex]
[tex]In ( \dfrac{0.1818}{1.265}) = \dfrac{-1.458^2*6.35*10^{-8}*t_f}{0.01^2}[/tex]
[tex]-1.9399=-0.001350 *t_f[/tex]
[tex]t_f = \dfrac{-1.9399}{-0.001350}[/tex]
[tex]\mathbf{t_f = 1436.96 \ sec }[/tex]
A heat engine with a thermal efficiency of 25% is connected to an electric generator with an efficiency of 95%. A liquid fuel providing heat is consumed at 2.29 litres per hour. What is the power output in kW from the generator?
Heat of Combustion: 43.7 MJ/kg
Density of the fuel: 0.749 g/ml
Answer:
4.94 kW
Explanation:
The heat energy produced by the fuel in one hour is ...
(2.29 L/h)(0.749 kg/L)(43.7 MJ/kg) = 74.954677 MJ/h
Then the power output is ...
(74.954677 MJ/h)(1 h)/(3600 s) = 20.8207 kJ/s
Multiplying this heat energy by the efficiencies of the processes involved, the output power is ...
(20.8207 kW)(0.25)(0.95) = 4.94 kW
Describe with an example how corroded structures can lead to environment pollution?
Determine the largest load P that can be a applied to the frame without causing either the average normal stress or the average shear stress at section a-a to exceed σ = 160 MPa and τ = 60 MPa , respectively. Member CB has a square cross section of 26 mm on each side.
Answer:
(The diagram of the question is given in Attachment 1)
The largest load which can be applied is:
P=67.62 kN
Explanation:
Make a Free body Diagram:All the forces are shown in the diagram in Attachment 2.
Analyze the equilibrium of Joint C in Figure (a):∑ F(y)= 0 (Upwards is positive)
[tex]F_{BC}sin\theta-P=0\\\frac{4}{5}F_{BC} - P=0\\F_{BC}=\frac{5}{4}P\\\\F_{BC}=1.25P[/tex]
Substitute F(BC) in Figure (b):
∑ F(x)= 0 (Towards Right is positive)
[tex]N_{a-a} - F_{BC}cos\theta=0\\N_{a-a}-1.25P(\frac{3}{5})=0\\N_{a-a}=0.75P[/tex]
∑ F(y)= 0 (Upwards is positive)
[tex]F_{BC}sin\theta- V_{a-a}= 0\\(\frac{4}{5})1.25P-V_{a-a}=0\\V_{a-a}=P[/tex]
Find Cross Sectional Area:
The cross sectional area of a-a:
[tex]A_{a-a}= \frac{(0.026)(0.026)}{3/5}\\A_{a-a}= 1.127\cdot10^{-3}[/tex]
Find P from Normal Stress Equation:σ = N(a-a)/A(a-a)
Substitute values:
[tex]160\cdot10^6=\frac{0.75P}{1.127\cdot10^{-3}}\\P=240.42\cdot10^3 N\\P=240.42 kN[/tex]
Find P from Shear Stress Equation:Т= V(a-a)/A(a-a)
Substitute values:
[tex]60\cdot10^6=\frac{P}{1.127\cdot10^{-3}}\\P=67.62\cdot10^{3}N\\P=67.62kN[/tex]
Results:To satisfy both the condition, we have to choose the lower value of P.
P=67.62 kN
During one eight-hour shift, 750 non-defective parts are desired from a fabrication operation. The standard time for the operation is 15 minutes. Because the machine operators are unskilled, the actual time it takes to perform the operation is 20 minutes, and ,on average, one-fifth of the parts that begin fabrication are scrapped. Assuming that each of the machines used for this operation will not be available for one hour of each shift, determine the number of machines required.
Answer:
No. of Machines Required = 47
Explanation:
First, we calculate the number of parts that can be manufactured by one machine during the shift. For that purpose, we use following formula:
No. of Parts Manufactured by 1 Machine = Total Operating Time/Time taken to perform Operation
where,
Total Operating Time = 8 h - 1 h = (7 h)(60 min/h) = 420 min
Time taken to perform operation = 20 min
Therefore,
No. of Parts Manufactured by 1 Machine = 420 min/20 min
No. of Parts Manufactured by 1 Machine = 21
Now, we will calculate the no. of non-defective parts manufactured by 1 machine. Since, it is given that one-fifth of the parts manufactured by machine are defective. Therefore, the non-defective parts will be: 1 - 1/5 = 4/5 (four-fifth).
No. of Non- Defective Parts Manufactured by 1 Machine = N = (4/5)(21)
N = 16.8 = 16 (Since, the 17th part will not be able to complete in time)
So, the no. of machines required to produce 750 non-defective parts is given by:
No. of Machines Required = No. of non-defective parts required/N
No. of Machines Required = 750/16
No. of Machines Required = 46.9
No. of Machines Required = 47 (Since, 46 machines will not be able complete the job)
21.Why are throttling devices commonly used in refrigeration and air-conditioning
applications?
Answer is given below
Explanation:
we know that some common types of throttling devices are
Hard -throttling devices Capillary valve Constant pressure throttling devicesThermostatic expansion valve Float expansion valveso here throttling devices commonly used in refrigeration and air-conditioning because
To reduce the coolant pressure, the high pressure of the refrigerant from the condenser is necessary to reduce the evaporation to obtain evaporation at the right temperature To meet the refrigerated load, the throttling valve flows through the coolant to cool the load at high temperatures.A .02 kg of R-134a fills a 0.14-m3 weighted piston–cylinder device at a temperature of –26.4°C. The container is now heated until the temperature is 100°C. Determine the final volume of R-134a. Use data from the refrigerant tables.
Answer:
The answer is 0.06027 m³
Explanation:
Solution
Given that:
The first step to take is to determine the initial state of the volume for R-134a refrigerant
Now,
v₁ =V/m
V = the volume of weighted piston cylinder device at the normal state
m = the mass of the R-134a refrigerant
Thus,
We substitute the values 0.14 for V and 0.2 kg for m
Which results in
v₁ = 0.14/0.2
v₁ = 0.7 m₃/kg
The next step is to find the saturated pressure of the R-134a refrigerant from the temperature table of saturated refrigerant R-134a which is equivalent to the normal temperature of 26.4°C.
Thus, by applying the method of interpolation we have the following
P₁ =101.73 - ((101.73-92.76) * (-26-(-26.4)/-26- (-28))
P₁ = 99.936 kPa
So,
The refrigerant in the weighted piston–cylinder device is then heated until the temperature gets to a 100°C
Hence, the temperature and pressure at a state of two becomes
P₂ = 99.936 kPa
T₂ = 100°C
The next step is to determine the specific volume of the refrigerant R-134a at a final state from the super heated refrigerant R-134a which is equivalent to the pressure of 99.936 kPa
v₂ =0.30138 m³ /kg
Now,
we now calculate the final state of the weighted piston cylinder device
V₂ = mv₂
V₂ = 0.2 * 0.30138
V₂ = 0.06027 m³
Hence ,the final volume of the weighted cylinder piston device is 0.06027 m³
A Transmission Control Protocol (TCP) connection is in working order and both sides can send each other data. What is the TCP socket state?
Answer:
ESTABLISHED
Explanation:
What is TCP?
A Transmission Control Protocol (TCP) is a communication protocol which allows the exchange of data between computers in a network.
When a Transmission Control Protocol connection is up and running meaning that both sides can send and receive data then the corresponding TCP socket states is known as "ESTABLISHED".
The most common socket states are:
LISTEN:
Before a TCP connection is made, there needs to be a server with a listener that will listen on incoming connection request.
ESTABLISHED:
When a TCP connection is up and running meaning that both sides can send and receive data.
CLOSED:
The CLOSED state means that there is no TCP connection.
There are a total of 11 TCP socket states:
1. LISTEN
2. SYN-SENT
3. SYN-RECEIVED
4. ESTABLISHED
5. FIN-WAIT-1
6. FIN-WAIT-2
7. CLOSE-WAIT
8. CLOSING
9. LAST-ACK
10. TIME-WAIT
11. CLOSED
Pressure in U.S. customary units is measured in psi (pound per square inch). In SI metric units pressure is measured in Pa (N/m2). Write a user-defined MATLAB function that converts pressure given in units of psi to pressure in units of Pa. For the function name and arguments, use [Pa] = Psi-ToPa(psi). The input argument psi is the pressure in units of psi to be converted, and the output argument Pa is the converted pressure in units of Pa (rounded to the nearest integer). Use the function in the Command Window to: (a) Convert 120 psi to units of Pa. (b) Convert 3,000 psi to units of Pa.
Answer:
Check below for answers
Explanation:
Matlab code:
function[Pa] = Psi-ToPa(psi)
Pa = psi * 6894.75728;
end
a) To convert 120 psi to units of Pa, just call the function Pa using the command:
Psi-ToPa(120)
ans =
8.2737e+05
b) To convert 3000 psi to units of Pa, just call the function Pa using the command:
Psi-ToPa(3000)
ans =
2.0684e+07
For a statically indeterminate axially loaded member, Group of answer choices The total deflection between end A and end B of an axially loaded member must not be zero. The summation of the forces is not zero. The summation of the reaction forces is equal to the applied load. The applied load must exceed the total reaction forces. The compatibility condition cannot be satisfied, since it is indeterminate.
Answer: The summation of the reaction forces is equal to the applied load.
Explanation:
When solving an indeterminate structure, it is important for one to satisfy the force-displacement requirements, compatibility, and the equilibrium of the structure.
With regards to the above question, for a statically indeterminate axially loaded member, the summation of the reaction forces will be equal to applied load. Here, the summation of the reactive forces will then be zero. This is a condition that is necessary to solve the unknown reaction forces.
difference between a pillar drill and radial arm pillar
drill
Answer:
Pillar drill are small and used for woodworking while Radial arm pillar is mounted on a very large column.
1. The 1992 Ford Crown Victoria came with a 190 horsepower engine. It has a drag coefficient of 0.33 and a frontal projected area of 26.36 ft2. Assuming that all of the available power from the engine is used to overcome the drag force, then what would the maximum speed (in miles per hour) be for this car
Answer:
1.50 miles/hour is the correct answer to the given question
Explanation:
As mention in the question engine power = 190 horsepower engine.......eq(1)
we have to converted into the watt as we know that 1 horse power=746 watt
putting this value in eq(1) we get
[tex]Engine \ power\ =190 *746\\\ Engine \ power =146216\ NM/S[/tex]
Drag coefficient = 0.33
Projected area = 26.36 [tex]ft^{2}[/tex]............eq(2)
we have to converted into the [tex]m^{2}[/tex] as we know that [tex]1m^{2}\ =\ 10.7639\ ft^{2}[/tex]
putting this value in eq(2) we get
Projected area = [tex]2.448921\ m^{2}[/tex]
Now we to calculated the Drag Force
[tex]F\ =\frac{\ Drag \ coefficient\ *P*\ V^{2}\ *A }{2}[/tex]....................eq(3)
Putting the above value in eq(3) we get
[tex]F=\ \frac{0.33\ * 1.2 * \ V^{2 \ * 2.44821\ *N} }{2} \\F=\ 0.4848V^{2}[/tex]
As we know that
[tex]F=0.4848\ V^{2}\ *V \\F= \frac{\ 0.4848V^{3} NM}{S}[/tex]
As mention in the question
Power of Engine =Drag free force
[tex]146216\ =\ 0.4848\ V^{3}\\ V^{3} =301550.889\\V=67.05M/S[/tex]
We have to converted this value into the miles/hour
1 M/S =2.22369 M/H
Putting this value into the V we get
V=67.05 * 2.2234 M/H
V=150 M/H
Steam flows steadily through a turbine at a rate of 45,000 lbm/h, entering at 1000 psia and 9008F and leaving at 5 psia as saturated vapor. If the power generated by the turbine is 4 MW, determine the rate of heat loss from the steam.
QUESTIONS
DOK S STANDARD RST.1
MS-PS2-2
A car driving at a constant speed of 20 m/s(meters per second)
turns right, down a street.
1. Is the car accelerating during the turn? Why or why not?
2. Would you feel any of the movement during this event? Why
or why not?
ANSWER:
1) The car is not accelerating, because for the the car to turn while accelerating down the street, it has to deaccelerate (reduce speed) due to CENTRIPETAL FORCE that is acting on the car during the turn. This force acts to bring the car in the direction where it has turned to.
2) You will feel the movement as everyone in the car will tend to fall to the left, while the car makes a right turn. This is because of CENTRIFUGAL FORCE, which is acting to take the car to a left direction, while the car is turning to the right direction.
The wheel and the attached reel have a combined weight of 50lb and a radius of gyration about their center of 6 A k in = . If pulley B that is attached to the motor is subjected to a torque of M ft lb = − 50 , determine the velocity of the 200lb crate after the pulley has turned 5 revolutions. Neglect the mass of the pulley
The complete question is;
The wheel and the attached reel have a combined weight of 50 lb and a radius of gyration about their center of ka = 6 in. If pulley B that is attached to the motor is subjected to a torque of M = 50 lb.ft, determine the velocity of the 200lb crate after the pulley has turned 5 revolutions. Neglect the mass of the pulley.
The image of this system is attached.
Answer:
Velocity = 11.8 ft/s
Explanation:
Since the wheel at A rotates about a fixed axis, then;
v_c = ω•r_c
r_c is 4.5 in. Let's convert it to ft.
So, r_c = 4.5/12 ft = 0.375 ft
Thus;
v_c = 0.375ω
Now the mass moment of inertia about of wheel A about it's mass centre is given as;
I_a = m•(k_a)²
The mass in in lb, so let's convert to slug. So, m = 50/32.2 slug = 1.5528 slug
Also, let's convert ka from inches to ft.
So, ka = 6/12 = 0.5
So,I_a = 1.5528 × 0.5²
I_a = 0.388 slug.ft²
The kinetic energy of the system would be;
T = Ta + Tc
Where; Ta = ½•I_a•ω²
And Tc = ½•m_c•(v_c)²
So, T = ½•I_a•ω² + ½•m_c•(v_c)²
Now, m_c is given as 200 lb.
Converting to slug, we have;
m_c = (200/32.2) slugs
Plugging in the relevant values, we have;
T = (½•0.388•ω²) + (½•(200/32.2)•(0.375ω)²)
This now gives;
T = 0.6307 ω²
The system is initially at rest at T1 = 0.
Resolving forces at A, we have; Ax, Ay and Wa. These 3 forces do no work.
Whereas at B, M does positive work and at C, W_c does negative work.
When pulley B rotates, it has an angle of; θ_b = 5 revs × 2π rad/revs = 10π
While the wheel rotates through an angle of;θ_a = (rb/ra) • θ_b
Where, rb = 3 in = 3/12 ft = 0.25 ft
ra = 7.5 in = 7.5/12 ft = 0.625 ft
So, θ_a = (0.25/0.625) × 10π
θ_a = 4π
Thus, we can say that the crate will have am upward displacement through a distance;
s_c = r_c × θ_a = 0.375 × 4π
s_c = 1.5π ft
So, the work done by M is;
U_m = M × θ_b
U_m = 50lb × 10π
U_m = 500π
Also,the work done by W_c is;
U_Wc = -W_c × s_c = -200lb × 1.5π
U_Wc = -300π
From principle of work and energy;
T1 + (U_m + U_Wc) = T
Since T1 is zero as stated earlier,
Thus ;
0 + 500π - 300π = 0.6307 ω²
0.6307ω² = 200π
ω² = 200π/0.6307
ω² = 996.224
ω = √996.224
ω = 31.56 rad/s
We earlier derived that;v_c = 0.375ω
Thus; v_c = 0.375 × 31.56
v_c = 11.8 ft/s
Transactional Vs Transformational Leadership. Using the Internet, each member of your team should read at least 2 articles each on Transactional Vs Transformational Leadership. Summarize the articles in 300 words or more. Provide appropriate reference. Combine each summarize in one paper but do not change the wording of the original summary. As a term, write a comprehensive summary of the articles. Present a discussion of what your team learned from this exercise?
Answer: Provided in the explanation section
Explanation:
Transactional Leadership - This leadership style is mainly focused on the transactions between the leader and employees. If the employees work hard, achieve the objectives and deliver the results, they are rewarded through bonuses, hikes, promotions etc. If the employees fail to achieve the desired results, they are punished by awarding lower ratings in the performance appraisal, denying opportunities etc.
In this style, leader lays emphasis on the relation with the followers.
It is a reactive style where the growth of the employee in the organization completely depends on the performance with respect to the activities and deliverables.
It is best suited for regular operations and for a settled environment by developing the existing organizational culture which is not too challenging.
It is a bureaucratic style of leadership where the leader concentrates on planning and execution rather than innovation and creation.
A transactional leader is short-term focused and result oriented. He/she doesn't consider long-term strategic objectives regarding the organization's future.
cheers i hope this helped !!
A sedimentation basin in a water treatment plant has a length = 48 m, width = 12 m, and depth = 3 m. The flow rate = 4 m 3 /s; particle specific gravity = 1.1; water density = 10 3 kg/m 3 ; and dynamic viscosity = 1.30710 -3 N.sec/m 2 . What is the minimum particle diameter that is removed at 85%?
Answer:
The minimum particle diameter that is removed at 85% is 1.474 * 10 ^⁻4 meters.
Solution
Given:
Length = 48 m
Width = 12 m
Depth = 3m
Flow rate = 4 m 3 /s
Water density = 10 3 kg/m 3
Dynamic viscosity = 1.30710 -3 N.sec/m
Now,
At the minimum particular diameter it is stated as follows:
The Reynolds number= 0.1
Thus,
0.1 =ρVTD/μ
VT = Dp² ( ρp- ρ) g/ 10μ²
Where
gn = The case/issue of sedimentation
VT = Terminal velocity
So,
0.1 = Dp³ ( ρp- ρ) g/ 10μ²
This becomes,
0.1 = 1000 * dp³ (1100-1000) g 0.1/ 10 *(1.307 * 10 ^⁻3)²
= 3.074 * 10 ^⁻6 = dp³ (.g01 * 10^6)
dp³=3.1343 * 10 ^⁻12
Dp minimum= 1.474 * 10 ^⁻4 meters.
One piece of evidence that supports the Theory of Plate Tectonics is the discovery of what in both South America and Africa? The ancient atmosphere in both places was identical. The rates of weathering of rock are similar. Fossil remains of the same land-dwelling animal. Plants on both continents have similar flowers.
Answer: Fossil remains of the same land-dwelling animal.
Explanation: Fossil remains which were found to belong to same land dwelling animals, in South America and Africa was used as evidence to help support the theory of Tectonics plates, what this theory simply means is that the whole continents of earth were once fused together until a tectonic plate caused it’s division. Since same remains were found in Africa and South America this shows that both continents were once fused together.
Answer:
Fossil remains of the same land-dwelling animal
Explanation:
Fossil remains tell us where the animals once lived and how by the movement of plate spearated their remaind that was burried thousands of years ago.
In water and wastewater treatment processes a filtration device may be used to remove water from the sludge formed by a precipitation reaction. The initial concentration of sludge from a softening reaction (Chapter 4) is 3 2 percent (20,000 mg/L) and the volume of sludge is 100 m . After filtra- tion the sludge solids concentration is 35 percent. Assume that the sludge does not change density during filtration, and that liquid removed from the sludge contains no sludge. Using the mass balance method, determine the volume of sludge after filtration.
Answer:
The volume of sludge after filtration is 0.914 m
Explanation:
Solution
Given that:
We have to find the Volume of sludge after filtration
Now,
The Sludge concentration is = 32%,
The sludge volume = 100 m,
The sludge concentration after filtration = 35%
Then,
The mass balance equation is stated below
Cin∀in = Cout∀out
Now,
We Solve for ∀out
∀out =Cin∀in/Cout
By substituting the values
∀out = (0.32)(100 m)/(0.35) = 0.914 m
A cylindrical bar of metal having a diameter of 20.2 mm and a length of 209 mm is deformed elastically in tension with a force of 50500 N. Given that the elastic modulus and Poisson's ratio of the metal are 65.5 GPa and 0.33, respectively, determine the following: (a) The amount by which this specimen will elongate in the direction of the applied stress. (b) The change in diameter of the specimen. Indicate an increase in diameter with a positive number and a decrease with a negative number.
Answer:
A) ΔL = 0.503 mm
B) Δd = -0.016 mm
Explanation:
A) From Hooke's law; σ = Eε
Where,
σ is stress
ε is strain
E is elastic modulus
Now, σ is simply Force/Area
So, with the initial area; σ = F/A_o
A_o = (π(d_o)²)/4
σ = 4F/(π(d_o)²)
Strain is simply; change in length/original length
So for initial length, ε = ΔL/L_o
So, combining the formulas for stress and strain into Hooke's law, we now have;
4F/(π(d_o)²) = E(ΔL/L_o)
Making ΔL the subject, we now have;
ΔL = (4F•L_o)/(E•π(d_o)²)
We are given;
F = 50500 N
L_o = 209mm = 0.209m
E = 65.5 GPa = 65.5 × 10^(9) N/m²
d_o = 20.2 mm = 0.0202 m
Plugging in these values, we have;
ΔL = (4 × 50500 × 0.209)/(65.5 × 10^(9) × π × (0.0202)²)
ΔL = 0.503 × 10^(-3) m = 0.503 mm
B) The formula for Poisson's ratio is;
v = -(ε_x/ε_z)
Where; ε_x is transverse strain and ε_z is longitudinal strain.
So,
ε_x = Δd/d_o
ε_z = ΔL/L_o
Thus;
v = - [(Δd/d_o)/(ΔL/L_o)]
v = - [(Δd•L_o)/(ΔL•d_o)]
Making Δd the subject, we have;
Δd = -[(v•ΔL•d_o)/L_o]
We are given v = 0.33; d_o = 20.2mm
So,
Δd = -[(0.33 × 0.503 × 20.2)/209]
Δd = -0.016 mm
A 60-m-long steel wire is subjected to a 6-kN tensile load. Knowing that E = 200 GPa and that the length of the rod increases by 48 mm, determine a) the smallest diameter that can be selected for the wire.b) the corresponding normal stress.
Answer:
(a) 6.91 mm (b) 160 MPa
Explanation:
Solution
Given that:
E = 200 GPa
The rod length = 48 mm
P =P¹ = 6 kN
Recall that,
1 kN = 10^3 N
1 m =10^3 mm
I GPa = 10^9 N/m²
Thus
The rod deformation is stated as follows:
δ = PL/AE-------(1)
σ = P/A----------(2)
Now,
(a) We substitute the values in equation and obtain the following:
48 * 10 ^⁻3 m = (6 * 10³ N) (60 m)/A[ 200 * 10^9 N/m^2]
Thus, we simplify
A = (6 * 10³) (60)/ ( 200 * 10^9) (48 * 10 ^⁻3)m²
A =0.0375 * 10 ^⁻3 m²
A =37.5 mm²
A = π/4 d²
Thus,
d² = 4A /π
After inserting the values we have,
d = √37.5 * 4/3.14 mm
= 6.9116 mm
or d = 6.91 mm
Therefore, the smallest that should be used is 6.91 mm
(B) To determine the corresponding normal stress that is caused by the tensile force, we input the values in equation (2)
Thus,
σ = P/A
σ= 6 * 10 ^ 3 N/ 37. 5 * 10 ^⁻6 m²
σ= 160 MPa
Note: I MPa = 10^6 N/m²
Hence the the corresponding normal stress is σ= 160 MPa
What's resistance in an electrical circuit? 1) Opposition to the flow of electricity 2) The ability of electricity to do work 3) The ability to make current flow
A construction company distributes its products by trucks loaded at its loading station. A backacter in conjunction with trucks are used for this purpose. If it was found out that on an average of 12 trucks per hour arrived and the average loading time was 3 minutes for each truck. A truck must queue until it is loaded. The backacter’s daily all-in rate is GH¢ 1000 and that of the truck is GH¢ 400.
a) Compute the operating characteristics: L, Lq, W, Wq, and P.
b) The company is considering replacing the backacter with a bigger one which will have an average service rate of 1.5 minutes to serve trucks waiting to have their schedules improved. As a manager, would you recommend the new backacter if the daily all-in rate is GH¢ 1300.
c) The site management is considering whether to deploy an extra backwater to assist the existing one. The daily all-in-rate and efficiency of the new backwater is assumed to be the same as that of the existing backwater. Should the additional backwater be deployed?
Answer:
a) [tex]L = 1.5[/tex]
[tex]L_q = 0.9[/tex]
[tex]W = \dfrac{1 }{8 } \, hour[/tex]
[tex]W_q = \dfrac{3}{40 } \, hour[/tex]
[tex]P = \dfrac{3}{5 }[/tex]
b) The new backacter should be recommended
c) The additional backacter should not be deployed
Explanation:
a) The required parameters are;
L = The number of customers available
[tex]L = \dfrac{\lambda }{\mu -\lambda }[/tex]
μ = Service rate
[tex]L_q[/tex] = The number of customers waiting in line
[tex]L_q = p\times L[/tex]
W = The time spent waiting including being served
[tex]W = \dfrac{1 }{\mu -\lambda }[/tex]
[tex]W_q[/tex] = The time spent waiting in line
[tex]W_q = P \times W[/tex]
P = The system utilization
[tex]P = \dfrac{\lambda }{\mu }[/tex]
From the information given;
λ = 12 trucks/hour
μ = 3 min/truck = 60/3 truck/hour = 20 truck/hour
Plugging in the above values, we have;
[tex]L = \dfrac{12 }{20 -12 } = \dfrac{12 }{8 } = 1.5[/tex]
[tex]P = \dfrac{12 }{20 } = \dfrac{3}{5 }[/tex]
[tex]L_q = \dfrac{3}{5 } \times \dfrac{3}{2 } = \dfrac{9}{10 } = 0.9[/tex]
[tex]W = \dfrac{1 }{20 -12 } = \dfrac{1 }{8 } \ hour[/tex]
[tex]W_q = \dfrac{3}{5 } \times \dfrac{1}{8 } = \dfrac{3}{40 } \, hour[/tex]
(b) The service rate with the new backacter = 1.5 minutes/truck which is thus;
μ = 60/1.5 trucks/hour = 40 trucks/hour
[tex]P = \dfrac{12 }{40 } = \dfrac{3}{10}[/tex]
[tex]W = \dfrac{1 }{40 -12 } = \dfrac{1 }{38 } \, hour[/tex]
[tex]W_q = \dfrac{3}{10 } \times \dfrac{1}{38 } = \dfrac{3}{380 } \, hour[/tex]
λ = 12 trucks/hour
Total cost = [tex]mC_s + \lambda WC_w[/tex]
m = 1
[tex]C_s[/tex] = GH¢ = 1300
[tex]C_w[/tex] = 400
Total cost with the old backacter is given as follows;
[tex]1 \times 1000 + 12 \times \dfrac{1}{8} \times 400 = \$ 1,600.00[/tex]
Total cost with the new backacter is given as follows;
[tex]1 \times 1300 + 12 \times \dfrac{1}{38} \times 400 = \$ 1,426.32[/tex]
The new backacter will reduce the total costs, therefore, the new backacter is recommended.
c)
Here μ = 3 min/ 2 trucks = 2×60/3 truck/hour = 40 truck/hour
[tex]\therefore W = \dfrac{1 }{40 -12 } = \dfrac{1 }{38 } \, hour[/tex]
Total cost with the one backacter is given as follows;
[tex]1 \times 1000 + 12 \times \dfrac{1}{8} \times 400 = \$ 1,600.00[/tex]
Total cost with two backacters is given as follows;
[tex]2 \times 1000 + 12 \times \dfrac{1}{38} \times 400 = \$ 2,126.32[/tex]
The additional backacter will increase the total costs, therefore, it should not be deployed.
You built an android that has a subcircuit containing a power supply, a tactile sensor, and a fuse where safe operation should keep current below 250 mA. You measured that your sensor is dissipating 12 W, the power supply is pro- viding 12.2 W, and the voltage drop across the fuse is 500 mV. Is your circuit properly protected?
Answer:
a co mam zroic!!
Explanation:
Sea B = 5.00 m a 60.0°. Sea C que tiene la misma magnitud que A y un ángulo de dirección mayor que el de A en 25.0°. Sea A ⦁ B = 30.0 m2 y B ⦁ C = 35.0 m2 . Encuentre A.
Answer:
[tex]\| \vec A \| = 6.163\,m[/tex]
Explanation:
Sean A, B y C vectores coplanares tal que:
[tex]\vec A = (\| \vec A \|\cdot \cos \theta_{A},\| \vec A \|\cdot \sin \theta_{A})[/tex], [tex]\vec B = (\| \vec B \|\cdot \cos \theta_{B},\| \vec B \|\cdot \sin \theta_{B})[/tex] y [tex]\vec C = (\| \vec C \|\cdot \cos \theta_{C},\| \vec C \|\cdot \sin \theta_{C})[/tex]
Donde [tex]\| \vec A \|[/tex], [tex]\| \vec B \|[/tex] y [tex]\| \vec C \|[/tex] son las normas o magnitudes respectivas de los vectores A, B y C, mientras que [tex]\theta_{A}[/tex], [tex]\theta_{B}[/tex] y [tex]\theta_{C}[/tex] son las direcciones respectivas de aquellos vectores, medidas en grados sexagesimales.
Por definición de producto escalar, se encuentra que:
[tex]\vec A \,\bullet\, \vec B = \|\vec A \| \| \vec B \| \cos \theta_{B}\cdot \cos \theta_{A} + \|\vec A \| \| \vec B \| \sin \theta_{B}\cdot \sin \theta_{A}[/tex]
[tex]\vec B \,\bullet\, \vec C = \|\vec B \| \| \vec C \| \cos \theta_{B}\cdot \cos \theta_{C} + \|\vec B \| \| \vec C \| \sin \theta_{B}\cdot \sin \theta_{C}[/tex]
Asimismo, se sabe que [tex]\| \vec B \| = 5\,m[/tex], [tex]\theta_{B} = 60^{\circ}[/tex], [tex]\vec A \,\bullet \,\vec B = 30\,m^{2}[/tex], [tex]\vec B\, \bullet\, \vec C = 35\,m^{2}[/tex], [tex]\|\vec A \| = \| \vec C \|[/tex] y [tex]\theta_{C} = \theta_{A} + 25^{\circ}[/tex]. Entonces, las ecuaciones quedan simplificadas como siguen:
[tex]30\,m^{2} = 5\|\vec A \| \cdot (\cos 60^{\circ}\cdot \cos \theta_{A} + \sin 60^{\circ}\cdot \sin \theta_{A})[/tex]
[tex]35\,m^{2} = 5\|\vec A \| \cdot [\cos 60^{\circ}\cdot \cos (\theta_{A}+25^{\circ}) + \sin 60^{\circ}\cdot \sin (\theta_{A}+25^{\circ})][/tex]
Es decir,
[tex]30\,m^{2} = \| \vec A \| \cdot (2.5\cdot \cos \theta_{A} + 4.330\cdot \sin \theta_{A})[/tex]
[tex]35\,m^{2} = \| \vec A \| \cdot [2.5\cdot \cos (\theta_{A}+25^{\circ})+4.330\cdot \sin (\theta_{A}+25^{\circ}})][/tex]
Luego, se aplica las siguientes identidades trigonométricas para sumas de ángulos:
[tex]\cos (\theta_{A}+25^{\circ}) = \cos \theta_{A}\cdot \cos 25^{\circ} - \sin \theta_{A}\cdot \sin 25^{\circ}[/tex]
[tex]\sin (\theta_{A}+25^{\circ}) = \sin \theta_{A}\cdot \cos 25^{\circ} + \cos \theta_{A} \cdot \sin 25^{\circ}[/tex]
Es decir,
[tex]\cos (\theta_{A}+25^{\circ}) = 0.906\cdot \cos \theta_{A} - 0.423 \cdot \sin \theta_{A}[/tex]
[tex]\sin (\theta_{A}+25^{\circ}) = 0.906\cdot \sin \theta_{A} + 0.423 \cdot \cos \theta_{A}[/tex]
Las nuevas expresiones son las siguientes:
[tex]30\,m^{2} = \| \vec A \| \cdot (2.5\cdot \cos \theta_{A} + 4.330\cdot \sin \theta_{A})[/tex]
[tex]35\,m^{2} = \| \vec A \| \cdot [2.5\cdot (0.906\cdot \cos \theta_{A} - 0.423 \cdot \sin \theta_{A})+4.330\cdot (0.906\cdot \sin \theta_{A} + 0.423 \cdot \cos \theta_{A})][/tex]
Ahora se simplifican las expresiones, se elimina la norma de [tex]\vec A[/tex] y se desarrolla y simplifica la ecuación resultante:
[tex]30\,m^{2} = \| \vec A \| \cdot (2.5\cdot \cos \theta_{A} + 4.330\cdot \sin \theta_{A})[/tex]
[tex]35\,m^{2} = \| \vec A \| \cdot (4.097\cdot \cos \theta_{A} +2.865\cdot \sin \theta_{A})[/tex]
[tex]\frac{30\,m^{2}}{2.5\cdot \cos \theta_{A}+ 4.330\cdot \sin \theta_{A}} = \frac{35\,m^{2}}{4.097\cdot \cos \theta_{A} + 2.865\cdot \sin \theta_{A}}[/tex]
[tex]30\cdot (4.097\cdot \cos \theta_{A} + 2.865\cdot \sin \theta_{A}) = 35\cdot (2.5\cdot \cos \theta_{A}+4.330\cdot \sin \theta_{A})[/tex]
[tex]122.91\cdot \cos \theta_{A} + 85.95\cdot \sin \theta_{A} = 87.5\cdot \cos \theta_{A} + 151.55\cdot \sin \theta_{A}[/tex]
[tex]35.41\cdot \cos \theta_{A} = 65.6\cdot \sin \theta_{A}[/tex]
[tex]\tan \theta_{A} = \frac{35.41}{65.6}[/tex]
[tex]\tan \theta_{A} = 0.540[/tex]
Ahora se determina el ángulo de [tex]\vec A[/tex]:
[tex]\theta_{A} = \tan^{-1} \left(0.540\right)[/tex]
La función tangente es positiva en el primer y tercer cuadrantes y tiene un periodicidad de 180 grados, entonces existen al menos dos soluciones del ángulo citado:
[tex]\theta_{A, 1} \approx 28.369^{\circ}[/tex] y [tex]\theta_{A, 2} \approx 208.369^{\circ}[/tex]
Ahora, la magnitud de [tex]\vec A[/tex] es:
[tex]\| \vec A \| = \frac{35\,m^{2}}{4.097\cdot \cos 28.369^{\circ} + 2.865\cdot \sin 28.369^{\circ}}[/tex]
[tex]\| \vec A \| = 6.163\,m[/tex]
With a reservoir pressure of 1.0 MPa and temperature of 750 K, air enters a converging-diverging nozzle, in a steady fashion. Flow is isentropic and k=1.4. If exit Mach number is 2 and throat area is 20 cm2 , find (a) the throat conditions (static pressure, temperature, density, and mach number), (b) the exit plane conditions i
Answer:
a) P* = 0.5283 MPa , T* = 624.75 K , ρ* = 2.945 kg/m^3 , V* = 501.023 m/s
b) Pe = 0.1278 MPa , Te = 416.7 K , ρe = 1.069 kg/m^3 , Ve = 818.36 m/s, Ae = 33.75 cm^2
c) m' = 2.915 kg/s
Explanation:
Given:-
- The inlet pressure, Pi = 1.0 MPa
- The inlet temperature, Ti = 750 K
- Inlet velocity is negligible
- Steady, Isentropic Flow
- The specific heat ratio of air, k = 1.4
- Exit Mach number, Mae = 2
- The throat area, Ath = 20 cm^2
- Gas constant of air, R = 0.287 KJ / kg.K
Find:-
(a) the throat conditions (static pressure, temperature, density, and mach number)
b) the exit plane conditions
c) the mass flow rate
Solution:-
- For this problem we will assume air to behave like an ideal gas with constant specific heat at RTP. Also the flow of air through the nozzle is assumed to be steady, one dimensional, and Isentropic with constant specific heat ratio ( k ).
- First we will scrutinize on the exit conditions. We have a Mach number of 2 at the exit. The flow at the exit of converging-diverging nozzle is in super-sonic region this is only possible only if sonic ( Ma = 1 ) conditions are achieved by the flow at the throat area ( minimum cross-sectional area ).
- Moreover, the flow is almost still at the inlet. Hence, we can assume that the flow has negligible velocity ( vi = 0 m/s ) at the inlet and the reservoir temperature and pressure can be assumed to be stagnation temperature and pressures as follows:
[tex]P_o = 1.0 MPa\\\\T_o = 750 K[/tex]
- Using the ideal gas law we can determine the stagnation density ( ρo ) as follows:
[tex]p_o = \frac{P_o}{RT_o} = \frac{1000}{0.287*750} = 4.64576\frac{kg}{m^3}[/tex]
- We will use the already developed results for flow which has reached sonic velocity ( Ma = 1 ) at the throat region. Use Table A - 13, to determine the critical static values at the throat region:
[tex]\frac{P^*}{P_o} = 0.5283\\\\P^* = 0.5283*1 = 0.5283 MPa\\\\\frac{T^*}{T_o} = 0.8333\\\\T^* = 0.8333*750 = 624.75 K\\\\ \frac{p^*}{p_o} = 0.6339\\\\p^* = 0.6339*4.64576 = 2.945 \frac{kg}{m^3} \\\\[/tex]
[tex]V^* = \sqrt{kRT^*} =\sqrt{1.4*287*624.75} = 501.023 \frac{m}{s}[/tex]
- Similarly, we will again employ the table A - 13 to determine the exit plane conditions for ( Ma = 2 ) as follows:
[tex]\frac{P_e}{P_o} = 0.1278 \\\\P_e = 0.1278*1.0 = 0.1278 MPa\\\\\frac{T_e}{T_o} = 0.5556 \\\\T_e = 0.5556*750 = 416.7 K\\\\\frac{p_e}{p_o} = 0.23 \\\\p_e = 0.23*4.64576 = 1.069 \frac{kg}{m^3} \\\\\frac{A_e}{A_t_h} = 1.6875 \\\\A_e =1.6875*20 = 33.75 cm^2\\[/tex]
- The velocity at the exit plane ( Ve ) can be determined from the exit conditions as follows:
[tex]V_e = Ma_e*\sqrt{kRT_e} = 2*\sqrt{1.4*287*416.7} = 818.36 \frac{m}{s}[/tex]
- For steady flows the mass flow rate ( m' ) is constant at any section of the nozzle. We will use the properties at the throat section to determine the mass flow rate as follows:
[tex]m' = p^* A_t_h V^*\\\\m' = 2.945*20*10^-^4*501.023\\\\m' = 2.951 \frac{kg}{s}[/tex]
(25) Consider the mechanical system below. Obtain the steady-state outputs x_1 (t) and x_2 (t) when the input p(t) is the sinusodal force given by p(t) = P sin ωt. All positions are measured from equilibrium. Use m_1=1.5 kg, m_2=2 kg, k=7 N/m, b=3.2 (N∙s)/m, P=15 N, =12 rad/sec. Hint: first create the state space model for the system. Then use SS2TF to make the two transfer functions and then the two Bode plots (include with submission). Use the plots to find the steady-state equations.
A sandy soil has a natural water content of 12% and bulk unit weight of 18.8 kN/m3 . The void ratios corresponding to the densest state (emin) and loosest state (emax) of this soil are 0.48 and 0.88, respectively. Find the relative density and degree of saturation for this soil.
Answer:
The value of relative density is 75 % while that of degree of saturation is 54.82%.
Explanation:
Given Data:
Bulk density of Sandy Soil=[tex]\gamma_b=18.8\ kN/m^3[/tex]
Void Ratio in Densest state=[tex]e_{max}=0.88[/tex]
Void Ratio in Loosest state=[tex]e_{min}=0.48[/tex]
Water content=[tex]w=12\%[/tex]
To Find:
Relative Density=[tex]D_R=\dfrac{e_{max}-e}{e_{max}-e_{min}} \times 100 \%[/tex]
Degree of Saturation=[tex]S=\dfrac{w\times G_s}{e}[/tex]
Now all the other values are given except e. e is calculated as follows
e is termed as In situ void ratio and is given as
[tex]e=\dfrac{\gamma_w \times G_s-\gamma_d}{\gamma_d}[/tex]
Here
γ_w is the density of water whose value is 1
G_s is the constant whose value is 2.65
γ_d is the dry density of the sandy soil which is calculated as follows:
[tex]\gamma_d=\dfrac{\gamma_b}{1+\dfrac{w}{100}}[/tex]
Putting values
[tex]\gamma_d=\dfrac{18.8}{1+\dfrac{12}{100}}\\\gamma_d=16.78\ kN/m^3=1.678 g/cc \\[/tex]
Putting this value in the equation of e gives
[tex]e=\dfrac{1 \times 2.65-1.678}{1.678}\\e=0.579=0.58[/tex]
So the value of Relative density is given as
[tex]D_R=\dfrac{e_{max}-e}{e_{max}-e_{min}} \times 100 \%\\D_R=\dfrac{0.88-0.58}{0.88-0.48} \times 100 \%\\D_R=75 \%[/tex]
So the value of relative density is 75 %
Now the value of degree of saturation is given as
[tex]S=\dfrac{w\times G_s}{e}\\S=\dfrac{12\times 2.65}{0.58}\\S=54.82 \%[/tex]
The value of degree of saturation is 54.82%.
Answer:
The relative density = 0.83 which is equivalent to 83%
The degree of saturation, S = 0.58 which gives 58% saturation
Explanation:
The parameters given are;
Water content W% = 12%
Bulk unit weight, γ = 18.8 kN/m³
Void ratio of [tex]e_{min}[/tex] = 0.48
Void ratio of [tex]e_{max}[/tex] = 0.88
[tex]G_S[/tex] = Constant (As learnt from an answer to the question on the current page) = 2.65 for Sandy soil
[tex]\gamma =\dfrac{W}{V} = \dfrac{W_{w}+W_{s}}{V}[/tex]
Where, V = 1 m³
W = 18.8 KN
Bulk unit weight, γ = [tex]\gamma_d[/tex] × (1 + W)
∴ 18.8 = [tex]\gamma_d[/tex] × (1 + 0.12)
[tex]\gamma_d[/tex] = 18.8/ (1.12) = 16.79 kN/m³
[tex]\gamma_d =\dfrac{W_s}{V} = \dfrac{W_{s}}{1} = 16.79 \, kN/m^3[/tex]
[tex]W_s[/tex] = 16.79 kN
∴ [tex]W_w[/tex] = 18.8 kN - 16.79 kN = 2.01 kN
[tex]m_w = 2.01/9.81 = 0.205 \, kg[/tex]
Volume of water = 0.205 m³
[tex]\gamma = \dfrac{GS \times \gamma _{w}\times \left (1+w \right )}{1 + e} = \dfrac{GS \times 9.81\times \left (1+0.12 \right )}{1 + e} =18.8[/tex]
e + 1 = 0.58×GS = 0.58×2.65 =
e = 1.54 - 1 = 0.55
The relative density is given by the relation;
[tex]Relative \ density, Dr=\dfrac{e_{max} - e}{e_{max} - e_{min}}[/tex]
[tex]Relative \ density, Dr=\dfrac{0.88 - e}{0.88 - 0.48} = \dfrac{0.88 - 0.55}{0.4} = 0.83[/tex]
The relative density = 0.83
The relative density in percentage = 0.83×100 = 83%
S·e = GS×w = 0.12·2.65
S×0.55 = 0.318
The degree of saturation, S = 0.58
The degree of saturation, S in percentage = 58%.
