Answer:
[tex]\eta_{th} = 30\,\%[/tex], [tex]\eta_{th,max} = 40\,\%[/tex], [tex]\Delta S = \frac{1}{3}\,\frac{kJ}{K}[/tex], The cycle is irreversible.
Explanation:
The real cycle efficiency is:
[tex]\eta_{th} = \frac{1000\,kJ-700\,kJ}{1000\,kJ} \times 100\,\%[/tex]
[tex]\eta_{th} = 30\,\%[/tex]
The theoretical cycle efficiency is:
[tex]\eta_{th,max} = \frac{500\,K-300\,K}{500\,K} \times 100\,\%[/tex]
[tex]\eta_{th,max} = 40\,\%[/tex]
The reversible and real versions of the power cycle are described by the Clausius Inequalty:
Reversible Unit
[tex]\frac{1000\,kJ - 600kJ}{300\,K}= 0[/tex]
Real Unit
[tex]\Delta S = \frac{1000\,kJ-600\,kJ}{300\,K} -\frac{1000\,kJ-700\,kJ}{300\,K}[/tex]
[tex]\Delta S = \frac{1}{3}\,\frac{kJ}{K}[/tex]
The cycle is irreversible.
The cycle efficiency using clausius inequality is;
σ_cycle = 0.333 kJ/kg and is internally irreversible
For the cycle, we know that efficiency is;η = 1 - Q_c/Q_h
Thus;
Q_c = (1 - η)Q_h
Now, the cycle efficiency is derived from the integral;σ_cycle = -∫(dQ/dt)ₐ
Thus; σ_cycle = -[(Q_h/T_h) - (Q_c/T_c)]
We are given;
Q_h = 1000 kJ
T_h = 500 k
T_c = 300 k
Q_c = 700 kJ
Thus;σ_cycle = -[(1000/500) - (700/300)]
σ_cycle = -(2 - 2.333)
σ_cycle = 0.333 kJ/kg
Since σ_cycle > 0, then the cycle is internally irreversible
Read more about cycle efficiency at; https://brainly.com/question/16014998
The rainfall rate in a certain city is 20 inches per year over an infiltration area that covers 33000 acres. Twenty percent of the rainfall percolates into the groundwater, with the remaining 80% running off into the river. The city uses 83000 acre-ft per year, some of which comes from the river and the remainder from groundwater. The groundwater volume of fresh water is currently at 1.1 × 105 acre-ft and is expected to last for 30 years before being completely depleted of fresh water (assume uniform withdrawal each). Assuming the groundwater lasts exactly as expected, determine the rate at which water is being withdrawn from the river by the city.
Answer:
The rate at which water is being withdrawn from the river by the city is 57353 acre-ft/y
Explanation:
Please look at the solution in the attached Word file
A pendulum is made of two rods that are firmly welded together to make a cross. The rod AB is 2 m long with a mass of 20 kg. The rod CD is 1 m long with a mass of 10 kg. Recall that for a rod the moment of inertia about its center of mass is given by IG = 1/12 m L2 cross2 What is the moment of inertia about A for member AB?
Answer:
IA = 80/3 kgm^2
Explanation:
Given:-
- The mass of rod AB, m1 = 20 kg
- The length of rod AB, L1 = 2m
- The mass of rod CD, m2 = 10 kg
- The length of rod CD, L2 = 1m
Find:-
What is the moment of inertia about A for member AB?
Solution:-
- The moment of inertia About point "O" the center of rod AB is given as:
IG = 1/12*m1*L^2
- To shift the axis of moment of inertia for any object at a distance "d" from the center of mass of that particular object we apply the parallel axis theorem. The new moment of inertia about any arbitrary point, which in our case A end of rod AB is:
IA = IG + m1*d^2
- Where the distance "d" from center of rod AB to its ends is 1/2*L1 = 1 m.
So the moment of inertia for rod AB at point A would be:
IA = 1/12*m1*L^2 + m1*0.5*L1^2
IA = 1/3 * m1*L1^2
IA = 1/3*20*2^2
IA = 80/3 kgm^2
List the RTL (Register Transfer Language) sequence of micro-operations needed to execute the instruction
STORE X
from the MARIE instruction set architecture. Then write the corresponding signal sequence to perform these micro-operations and to reset the clock cycle counter.
You may refer to the provided "MARIE Architecture and Instruction Set" file in the Front Matter folder.
Answer:
So these are the RTL representation:
MAR<----X
MBR<-----AC
M[MAR]<------MBR
Control signal sequence are:
P3T0:MAR<----X
P2P3 P4T1:MBR<-----AC
P0P1 P3T2:M[MAR]<------MBR
Explanation:
STORE X instruction is used for storing the value of AC to the memory address pointed by X. This operation can be done by using the Register Transfers at System Level and this can be represented by using a notation called Register Transfer Language RTL. Let us see what are the register transfer operations happening at the system level.
1. First of all the address X has to be tranfered on to the Memory Address Register MAR.
MAR<----X
2. Next we have to tranfer the contents of AC into the Memory Buffer Register MBR
MBR<-----AC
3. Store the MBR into memory where MAR points to.
M[MAR]<------MBR
So these are the RTL representation:
MAR<----X
MBR<-----AC
M[MAR]<------MBR
Control signal sequence are:
P3T0:MAR<----X
P2P3 P4T1:MBR<-----AC
P0P1 P3T2:M[MAR]<------MBR
You are not changing the carrier signal, only the message. In the scope the center Page 5 of 6 frequency should remain at 14kHz, the frequency span at 20kHz and the dB/div at 15dB. Here are the AM signal in time domain and its frequency spectrum. Explain what you see. (Hint: Remember the Fourier analysis of the square wave performed in Lab #3.)
Answer:
We will see a waveform displayed on the screen and it will be PWM(pulse width modulation) of sinusoidal wave,this wave should have a frequency of 14KHz and it will form a vibration spectrum.
6msection of 150lossless line is driven by a source with vg(t) = 5 cos(8π × 107t − 30◦ ) (V) and Zg = 150 . If the line, which has a relative permittivity r = 2.25, is terminated in a load ZL = (150 − j50) , determine: (a) λ on the line. ∗ (b) The reflection coefficient at the load. (c) The input impedance. (d) The input voltage Vi. (e) The time-domain input voltage vi(t).
Answer:
a. 5m
b. r = 0.16 e^-80.5◦
c. Zpn = (115.7 + j27.4) ohms
d. Vi = 2.2e^-j22.56◦ volts
e. Vi(t) = 2.2 cos (8π × 107t − 22.56◦ ) Volts
Explanation:
In this question, we are tasked with calculating a series of terms.
Please check attachment for complete solution and step by step explanation
A square hole is to be cut using ECM through a plate of low alloy steel that is 12 mm thick. The hole is 35 mm on each side, but the electrode used to cut the hole is slightly less than 35 mm on its sides to allow for overcut, and its shape includes a hole in its center to permit the flow of electrolyte and to reduce the area of the cut. This will produce a center core that can be removed after the tool breaks through. This tool design results in a frontal area of 245 mm2 . The applied current = 1200 amps.
Using an efficiency of 95%, how long will it take to cut the hole?
It will take approximately 0.0238 s to cut the hole.
To calculate the time required to cut the hole using Electrochemical Machining (ECM), we can use the following formula:
[tex]\text { Time }=\frac{\text { Volume of Material to be Removed }}{\text { Material Removal Rate (MRR) }}[/tex]
First, let's calculate the volume of material to be removed:
Volume of Material=Frontal Area × Thickness of Plate
Given:
[tex]Frontal Area =245 \mathrm{~mm}^2\\Thickness of Plate $=12 \mathrm{~mm}$[/tex]
Volume of Material = [tex]245 \mathrm{~mm}^2 \times 12 \mathrm{~mm}=2940 \mathrm{~mm}^3[/tex]
Next, we need to calculate the Material Removal Rate (MRR). MRR is usually given in units of volume removed per unit time per unit current density. Let's assume it is given as [tex]X \mathrm{~mm}^3 / \mathrm{min} / \mathrm{A} / \mathrm{mm}^2 .[/tex] Given the current density [tex]\text { Current Density }=\frac{\text { Applied Current }}{\text { Frontal Area }}[/tex], we can calculate the MRR as:
MRR=X×Current Density
Given:
Applied Current = 1200 A
Efficiency = 95%
Frontal Area = 245 [tex]mm^2[/tex]
[tex]\begin{aligned}& \text { Current Density }=\frac{\text { Applied Current }}{\text { Frontal Area }}=\frac{1200 \mathrm{~A}}{245 \mathrm{~mm}^2} \\& \text { Current Density } \approx 4.897 \mathrm{~A} / \mathrm{mm}^2\end{aligned}[/tex]
Let's assume X=0.2[tex]mm^3[/tex] /s/A/[tex]mm^2[/tex](hypothetical value).
[tex]\mathrm{MRR}=0.2 \times 4.897 \approx 0.979 \mathrm{~mm}^3 / \mathrm{s}[/tex]
To convert mm³/s to g/s, we need to know the density of the material. Let's assume the density of low alloy steel is approximately [tex]7.85 \mathrm{~g} / \mathrm{cm}^3[/tex], which is [tex]7.85 \times 10^{-6} \mathrm{~g} / \mathrm{mm}^3 .[/tex]
Now, we can calculate the MRR in grams per second:
[tex]MRR in $\mathrm{g} / \mathrm{s}=\mathrm{MRR} \times$ Density of material\\MRR in $\mathrm{g} / \mathrm{s}=0.979 \mathrm{~mm}^3 / \mathrm{s} \times 7.85 \times 10^{-6} \mathrm{~g} / \mathrm{mm}^3$\\MRR in $\mathrm{g} / \mathrm{s} \approx 0.00000768515 \mathrm{~g} / \mathrm{s}$[/tex]
Now, we can use these values to calculate the time required:
[tex]\text { Time }=\frac{2940 }{0.00000768515} \\& \text { Time } \approx 0.022594341[/tex]
However, we need to consider the efficiency. Since the efficiency is 95%, the actual time required will be:
[tex]\begin{aligned}& \text { Actual Time }=\frac{\text { Time }}{\text { Efficiency }} \\& \text { Actual Time }=\frac{0.022594341}{0.95}\end{aligned}[/tex]
≈ 0.0238 s
A turbine operates at steady state, and experiences a heat loss. 1.1 kg/s of water flows through the system. The inlet is maintained at 100 bar, 520 Celsius. The outlet is maintained at 10 bar, 280 Celsius. A rate of heat loss of 60 kW is measured. Determine the rate of work output from the turbine, in kW.
Answer:
[tex]\dot W_{out} = 399.47\,kW[/tex]
Explanation:
The turbine is modelled after the First Law of Thermodynamics:
[tex]-\dot Q_{out} -\dot W_{out} + \dot m\cdot (h_{in}-h_{out}) = 0[/tex]
The work done by the turbine is:
[tex]\dot W_{out} = \dot m \cdot (h_{in}-h_{out})-\dot Q_{out}[/tex]
The properties of the water are obtained from property tables:
Inlet (Superheated Steam)
[tex]P = 10\,MPa[/tex]
[tex]T = 520\,^{\textdegree}C[/tex]
[tex]h = 3425.9\,\frac{kJ}{kg}[/tex]
Outlet (Superheated Steam)
[tex]P = 1\,MPa[/tex]
[tex]T = 280\,^{\textdegree}C[/tex]
[tex]h = 3008.2\,\frac{kJ}{kg}[/tex]
The work output is:
[tex]\dot W_{out} = \left(1.1\,\frac{kg}{s}\right)\cdot \left(3425.9\,\frac{kJ}{kg} -3008.2\,\frac{kJ}{kg}\right) - 60\,kW[/tex]
[tex]\dot W_{out} = 399.47\,kW[/tex]
The fins are 90% efficient. Determine the minimum free-stream velocity the fan needs to supply to avoid overheating. Assume the flow is laminar over the entire finned surface of the transformer and check your assumption at the end. Assume the same convection heat transfer coefficient for the finned and unfinned area. Use 40 [C] for the film temperature.
Answer:
The minimum required free stream velocity required to dissipate 12W via fins is V∞ = 0.0020378 m/s = 2.0378 mm/s.
Explanation:
Given:-
- The dimension of transformer surface ( L , w , H ) = ( 10 cm long , 6.2 cm wide, 5 cm high )
- The dimensions of the fin : ( l , h , t ) = (10 cm long , 5 mm high, 2 mm thick )
- The total number of fins, n = 7
- The convection heat transfer coefficient of the finned and unfinned area = h.
- The efficiency of fins, ε = 0.9 ( 90% )
- The transformer fin base temperature, Tb = 60°C
- The free temperature of air, T∞ = 25°C
- The free stream velocity of air = U∞
Find:-
Determine the minimum free-stream velocity the fan needs to supply to avoid overheating. ( U∞ )
Solution:-
- Since the convection heat transfer coefficient of the finned and unfinned area i.e the fins and the transformer base are at the same temperature (Tb).
- The theoretical heat transfer ( Q_th ) rate from the fins can be calculated from the following convection cooling relation.
Q_th = h*As*[ Tb - T∞ ]
Where,
As : The total available surface area available for heat transfer.
Surface area of the fins (As1)As1 = n * { 2* [ ( l * h ) + ( t * h ) ] + ( l * t ) }
As1 = 7* { 2* [ ( 0.5 * 10 ) + ( 0.2 * 0.5 ) + ( 10 * 0.5 ) }
As1 = 106.4 cm^2 .... 0.01064 m^2
Surface area of the unfinned part of base (As2)As2 = Total base area - Finned top plane area
As2 = ( L * w ) - n* ( l * t ) = ( 10 * 6.2 ) - 7* ( 10 * 0.5 )
As2 = 27 cm^2 .... 0.0027 m^2
- Therefore, the total available surface area (As) is:
As = As1 + As2
As = 0.01064 + 0.0027
As = 0.01334 m^2
- The heat transfer coefficient (h) using convection heat transfer relation:
Q* ε = h*As*[ Tb - T∞ ]
h = Q* ε / [As*[ Tb - T∞ ] ]
h = (12*0.9) / [ 0.01334*( 60 - 25 ) ]
h = 23.13129 W/m^2K
- The air properties at film temperature:
T = 40 C
Viscosity ν = 1.6982 m^2 / s
Thermal conductivity, k = 0.027076 W/mK
Prandlt Number Pr = 0.71207
- The Nusselt number for the convection heat transfer for the transformer along the fins (Assumed flat plate):
Nu = h*L / k
Nu = 23.13129*0.1 / 0.027076
Nu = 85.43
- The correlation for Nusselt number between flow conditions and viscosity effects of the flow (Re & Pr) for a isothermal flat plate - Laminar Flow is given:
[tex]Nu = 0.664*Re^\frac{1}{2} *Pr^\frac{1}{3} \\\\Re^\frac{1}{2} = \frac{Nu}{0.664*Pr^\frac{1}{3}} \\\\\\Re = \sqrt{\frac{Nu}{0.664*Pr^\frac{1}{3}}} \\\\\\Re = \sqrt{\frac{85.43063}{0.664*0.71207^\frac{1}{3}}}\\\\Re = 12.00[/tex]
- The reynold number denotes the characteristic of the flow by the following relation:
Re = V∞*L / ν
V∞ = Re*ν / L
V∞ = 12*1.6982*10^-5 / 0.1
V∞ = 0.0020378 m/s .... = 2.0378 mm/s
Consider a rectangular fin that is used to cool a motorcycle engine. The fin is 0.15m long and at a temperature of 250C, while the motorcycle is moving at 80 km/h in air at 27 C. The air is in parallel flow over both surfaces of the fin, and turbulent flow conditions may be assumed to exist throughout. What is the rate of heat removal per unit width of the fin?
Answer:
q' = 5826 W/m
Explanation:
Given:-
- The length of the rectangular fin, L = 0.15 m
- The surface temperature of fin, Ts = 250°C
- The free stream velocity of air, U = 80 km/h
- The temperature of air, Ta = 27°C
- Parallel flow over both surface of the fin, assuming turbulent conditions through out.
Find:-
What is the rate of heat removal per unit width of the fin?
Solution:-
- Assume steady state conditions, Negligible radiation and flow conditions to be turbulent.
- From Table A-4, evaluate air properties (T = 412 K, P = 1 atm ):
Dynamic viscosity , v = 27.85 * 10^-6 m^2/s
Thermal conductivity, k = 0.0346 W / m.K
Prandlt number Pr = 0.69
- Compute the Nusselt Number (Nu) for the - turbulent conditions - the appropriate relation is as follows:
[tex]Nu = 0.037*Re_L^\frac{4}{5} * Pr^\frac{1}{3}[/tex]
Where, Re_L: The average Reynolds number for the entire length of fin:
[tex]Re_L = \frac{U*L}{v} \\\\Re_L = \frac{80*\frac{1000}{3600} * 0.15}{27.85*10^-^6} \\\\Re_L = 119688.80909[/tex]
Therefore,
[tex]Nu = 0.037*(119688.80909)^\frac{4}{5} * 0.69^\frac{1}{3}\\\\Nu = 378[/tex]
- The convection coefficient (h) can now be determined from:
[tex]h = \frac{k*Nu}{L} \\\\h = \frac{0.0346*378}{0.15} \\\\h = 87 \frac{W}{m^2K}[/tex]
- The rate of heat loss q' per unit width can be determined from convection heat transfer relation, Remember to multiply by (x2) because the flow of air persists on both side of the fin:
[tex]q' = 2*[h*L*(T_s - T_a)]\\\\q' = 2*[87*0.15*(250 - 27)]\\\\q' = 5826\frac{W}{m}[/tex]
- The rate of heat loss per unit width from the rectangular fin is q' = 5826 W/m
- The heat loss per unit width (q') due to radiation:
[tex]q' = 2*a*T_s^4*L[/tex]
Where, a: Stefan boltzman constant = 5.67*10^-8
[tex]q' = 2*5.67*10^-^8*(523)^4*0.15\\\\q' = 1273 \frac{W}{m}[/tex]
- We see that radiation loss is not negligible, it account for 20% of the heat loss due to convection. Since the emissivity (e) of the fin has not been given. So, in the context of the given data this value is omitted from calculations.
Two monitoring wells were constructed in an unconfined aquifer. The wells are separated by a distance of 250 ft. The water surface elevations in the up-gradient and down-gradient wells were 101.00 ft and 100.85 ft, respectively. The aquifer hydraulic conductivity is 5 ft/day. The fluid velocity (ft/day) in the aquifer is most nearly:
Answer:
0.003
Explanation:
? → ? = −??ℎ??→ ?
= ??ℎ??? = ?ℎ2− ℎ1∆?→ ? = 5????.101 ?? − 100.85 ??250 ??= 0.003???
1. A spur gear made of bronze drives a mid steel pinion with angular velocity ratio of 13 /2 : 1. Thepressure angle is 14 1/2° . It transmits 5 kW at 1800 r.p.m. of pinion. Considering only strength, design the smallest diameter gears and find also necessary face width. The number of teeth should not be less than 15 teeth on either gear. The elastic strength of bronze may be taken as 84 MPa and of steel as 105 MPa. Lewis factor for 14 1/2° pressure angle may be taken as y =0.6840.124 –No. of teeth.
Answer:
Given data
w1/w2=6.5/1
Power=5 KW
wp=1800 rpm
angle=14 degrees
Based on above values,the minimum diameter=30 mm
Water flows steadily through a fire hose and nozzle. The hose is 75 mm inside diameter, and the nozzle tip is 25 mm inside diameter; water gage pressure in the hose is 510 kPa, and the stream leaving the nozzle is uniform. The exit speed 32 m/s and pressure is atmospheric. Determine the force transmitted by the coupling between the nozzle and hose. (25 points)
Answer:
R = 1804 N
Explanation:
Given:-
- The density of water, ρ = 997 kg/m^3
- The inside diameter of the hose, dh = 75 mm
- The gauge pressure of water in the hose, P1 = 510 KPa
- The exit speed of the water, V2 = 32 m/s
- The inside diameter of the nozzle tip, dn = 25 mm
- The atmospheric pressure (gauge), P2 = 0 KPa ... P = 1 atm (Absolute).
Find:-
Determine the force transmitted by the coupling between the nozzle and hose.
Solution:-
- We will first develop a control surface at the hose-nozzle interface.
- Assuming steady and one dimensional flow - (x-direction).
- Since there are no fictitious unbalanced forces acting on the fluid flow due to roughness of hose any any losses of energy from the fluid are negligible.
- The use of conservation of momentum of fluid flow is valid for an isolated system, where the flow of fluid into the control volume is denoted by (-) and the flow of fluid going out of the control volume is denoted by (+):
- The principle of conservation of momentum, the pair of equal force (Newton's third law) act on the control volume at (nozzle-hose) interface:
R = ρ*Q*(V2 - V1) + (P2*A2 - P1*A1)
Where, Q: Flow rate
V1: The velocity of fluid in hose
A1: Cross sectional area of the hose
A2: Cross sectional area of the nozzle exit
- We see that the reaction force (R) that acts on nozzle-hose interface is due to changes in dynamic and hydrostatic pressures.
- Compute the required quantities Q, A1 and A2 and V1 using the given data:
- The flow rate Q for any flow in the hose can be given, where the cross sectional area of hose (A1)is:
[tex]A1 =\pi\frac{d_h^2}{4} = \pi\frac{0.075^2}{4} \\\\A1 = 0.00441 m^2\\\\\\[/tex]
- The cross sectional area of the nozzle tip with diameter dn = 25 mm is:
[tex]A2 =\pi\frac{d_n^2}{4} = \pi\frac{0.025^2}{4} \\\\A2 = 0.00049 m^2\\\\\\[/tex]
- The flow rate (Q) can now be calculated:
[tex]Q = A2*V2\\\\Q = (0.00049)*(32)\\\\Q = 0.01570 \frac{m^3}{s}[/tex]
- Since, the density of the water does not vary along the direction of flow, the flow rate (Q) remains constant throughout. So from continuity equation we have:
[tex]Q = A2*V2 = A1*V1\\\\V1 = \frac{Q}{A1} = \frac{0.0157}{0.00441} \\\\V1 = 3.56189 \frac{m}{s}[/tex]
- Now use the calculated quantities and compute the pair of reaction force at the nozzle-hose interface:
R = ρ*Q*(V2 - V1) + (P2*A2 - P1*A1)
R = (997)*(0.01570)*(32-3.56189) + (0 - 510*0.00441)*1000
R = 445.13889 - 2,249.1
R = - 1803.961 ≈ -1,804 N
- Here the negative sign denotes the direction of in which the force (R) is exerted. Since, (-) denotes into the control volume it acts opposite to the flow of water.
The coupling between the nozzle and hose is -1.81N
This question relates to flow rate of a liquid
Data given:
The density of water = 997kg/m^3
The inside diameter of the hose = 75mm = 0.0075m
The gauge pressure of water in the hose = 510kPa
The exit speed of the water = 32m/s
The inside diameter of the nozzle tip = 25mm = 0.0025m
The atmospheric pressure = 0kPa or 1atm
Let's calculate the inlet velocity
[tex]v_1=v_2=A_2/A_1\\v_1=V_2(\frac{d_2}{d_1})^2\\v_1=32(\frac{25}{75})^2\\v_1=3.50m/s[/tex]
Calculating the force transmitted by coupling between the nozzle and hose
[tex]R_x+p_1gA_1=v_1[-|pv_1A_1|]+v_2[|pv_2A_2|]\\[/tex]
μ[tex]_1[/tex]=[tex]v_1[/tex] and μ[tex]_2[/tex] =[tex]v_2[/tex]
[tex]R_x=-p_1gA_1-v_1pv_1A_1+v_2pv_2A_2\\R_x=-p_1gA+pv_2A_2(v_2-v_1)\\R_x=-510*10^3N/m^3*\frac{\pi }{4}(0.075m)^2+997kg/m^3*32m/s*\frac{\pi }{4} (0.025m)^2(32-3.50)=-1805=-1.81kN[/tex]
The force between the nozzle and hose is -1.81
Learn more about flow rate;
https://brainly.com/question/17151453
A pipe 300 m long has a slope of 1 in 100 and tapers from 1.2 m diameter at the high end to 0.6 m diameter at the low end. Quantity of water flowing is 5400 liters per minute. If the pressure at the high end is 68.67 kPa, find the pressure at the low end. Neglect head loss.
Answer:
P = 98052.64 Pa or 98.05 kPa
Explanation:
Using Bernoulli's equation;
P+rho*v^2/2+ rho*g*z = constant
at high end;
dia= 1.2 m
flow, Q = 5400 L/min = 0.09 m^3/s
therefore, velocity at the high end, v1 = Q/A =0.09/(pi()*(1.2/2)^2) = 0.08 m/s
pressure, P = 68.67 kPa
Solving for elevation, z
assume lower end is reference line. then higher end will be 'x' m high wrt to lower end.
x= 300*sin(tan^-1(1/100)) = 3 m
that means higher end will be 3 m above with respect to lower end
Similarly for lower end;
dia= 0.6 m
flow, Q = 5400 L/min = 0.09 m^3/s
therefore, velocity at the high end, v2 = Q/A =0.09/(pi()*(0.6/2)^2) = 0.318 m/s
assume pressure, P
z=0
put all values in the formula, we get;
68.67*10^3+1000*0.08^2/2+ 1000*9.81*3 =P+1000*0.318^2/2+ 0
solving this, we get;
P = 98052.64 Pa or 98.05 kPa
According to Bernoulli's principle, a lower pressure than expected at the
lower end because the velocity of the fluid is higher.
The pressure at the low end is approximately 97.964 kPa.Reasons:The given parameter are;
Length of the pipe, L = 300 m
Slope of the pipe = 1 in 100
Diameter at the high end, d₁ = 1.2 m
Diameter at the low end, d₂ = 0.6 m
The volume flowrate, Q = 5,400 L/min
Pressure at the high end, P = 68.67 kPa
Required:Pressure at the low end
Solution:The elevation of the pipe, z₁ = [tex]300 \, m \times \frac{1}{100} = 3 \, m[/tex]
z₂ = 0
The continuity equation is given as follows;
Q = A₁·v₁ = A₂·v₂
[tex]\sqrt[n]{x} \displaystyle Q = 5,400 \, L/min = 5,400 \, \frac{L}{min} \times \frac{1 \, m^3}{1,000 \, L} \times \frac{1 \, min}{60 \, seconds} = \mathbf{ 0.09 \, m^3/s}[/tex]
Therefore;
[tex]\displaystyle 0.09 = \frac{\pi}{4} \times 1.2^2 \times v_1 = \mathbf{\frac{\pi}{4} \times 0.6^2 \times v_2}[/tex]
[tex]\displaystyle v_1 = \frac{0.09}{\frac{\pi}{4} \times 1.2^2 } \approx 0.0796[/tex]
[tex]\displaystyle v_2 =\frac{0.09}{\frac{\pi}{4} \times 0.6^2 } \approx \mathbf{0.318}[/tex]
The Bernoulli's equation is given as follows;[tex]\displaystyle \frac{p_1}{\rho \cdot g} + \frac{v_1^2}{2 \cdot g} + z_1 = \mathbf{ \frac{p_2}{\rho \cdot g} + \frac{v_2^2}{2 \cdpt g} + z_2}[/tex]
Therefore;
[tex]\displaystyle p_2 = \left(\frac{p_1}{\rho \cdot g} + \frac{v_1^2}{2 \cdot g} + z_1 - \left( \frac{v_2^2}{2 \cdot g} + z_2 \right)\right) \times \rho \cdot g = p_1 + \frac{\rho}{2} \cdot \left(v_1^2} -v_2^2 \right)+ \rho \cdot g \left(z_1 - z_2\right)[/tex]
The density of water, ρ = 997 kg/m³
Which gives;
[tex]\displaystyle p_2 = 68670 + \frac{997 }{2} \times \left(0.0796^2- 0.318^2 \right) + 997 \times 9.81 \times \left( 3 - 0 \right) \approx \mathbf{97,964.46}[/tex]The pressure at the low end, p₂ ≈ 97,964.46 Pa ≈ 97.964 kPaLearn more about Bernoulli's principle here:
https://brainly.com/question/6207420
Each of the two drums and connected hubs of 13-in. radius weighs 210 lb and has a radius of gyration about its center of 30 in. Calculate the magnitude of the angular acceleration of each drum. Friction in each bearing is negligible.
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The angular acceleration for first drum [tex]\alpha = 0.792 rad/s^2[/tex]
The angular acceleration for the second drum is [tex]\alpha =1.262[/tex]
Explanation:
From the question we are told that
Their radius of the drum is [tex]r = 13 in = \frac{13}{12} ft = 1.083ft[/tex] each
The weight is [tex]W = 210 lb[/tex]
The mass is [tex]M = \frac{210 lb}{32.2 ft /s^2} = 6.563\ lb s^2 ft^{-1}[/tex]
Their radius of gyration is [tex]z=30 in= \frac{30 }{12} = 2.5 ft[/tex]
The free body diagram of a drum and its hub and 30lb and in the case the weight is connect to the hub separately is shown on the second uploaded image
The T in the diagram is the tension of the string
Now taking moment about the center of the the drum P we have
[tex]\sum M_p = I_p \alpha[/tex]
=> [tex]T * r = Mz^2 * \alpha[/tex]
Where r is the radius ,z is the radius of gyration about the center O , M is the mass of the drum including the hub, and [tex]\alpha[/tex] is the angular acceleration
Inputting
[tex]T * 1.083 = 6.563 * 2.5^2 \alpha[/tex]
=> [tex]T = 37.87\alpha[/tex]
Considering the force equilibrium in the vertical direction (Looking at the second free body diagram now )
The first on is
[tex]\sum F_y = ma[/tex]
=> [tex]30lb - T = m(r \alpha )[/tex]
Where m is the mass of the hanging block which has a value of
[tex]m = \frac{30lb}{32.2 ft/s^2} = 0.9317 \ lb ft^{-1} s^2[/tex]
a is the acceleration of the hanging block
inputting values we have
[tex]30- 37.87 \alpha = 0.9317* 1.083 \alpha[/tex]
[tex]30 = 37.87\alpha + \alpha[/tex]
[tex]\alpha = \frac{30}{38.87 }[/tex]
[tex]\alpha = 0.792 rad/s^2[/tex]
So the angular acceleration for first drum [tex]\alpha = 0.792 rad/s^2[/tex]
The free body diagram of a drum and its hub when the only on the string is 30lb is shown on the third uploaded image
So here we would take the moment about O
[tex]\sum M_o = I_O \alpha[/tex]
So [tex]\sum M_o = 30* 1.083[/tex]
and [tex]I = M z^2[/tex]
Therefore we will have
[tex]30 * 1.083 = (Mz^2 )\alpha[/tex]
inputting values
[tex]30 * 1.083 = 6.563 * 2.5^2 \alpha[/tex]
[tex]32.49=41.0\alpha[/tex]
[tex]\alpha =\frac{41}{32.49}[/tex]
[tex]\alpha =1.262[/tex]
So the angular acceleration for the second drum is [tex]\alpha =1.262[/tex]
"A northbound freeway segment is on a 4% upgrade from station 430+20 to 450+00 and has two 11-ft wide lanes, a 5-ft right shoulder, and has a ramp density of 1 per mile in the 3 miles before and after station 440+10. The peak-hour factor is 0.9. Northbound traffic during the peak hour is 2550 cars, 300 STs, and 300 TTs. Determine the density and LOS of the freeway segment."
Answer: D = 23.09 pc/mi/h
LOS = C
Explanation:
we will begin by solving this with a step by step process for easy understanding;
given that the freeway has two lanes = 11 ft wide
and has a width of 5 ft right shoulder.
ramp density = 1 per mile in 3m
substituting the free flow speed value gives us;
Free flow speed = 75 - 1.9 - 0.6 - (3.22 × 1∧0.84) = 69.28 mph
we have that the length of the road = (1980 ft) × (1 mile / 5280 ft)
L = 0.375 mile
The next thing we will do is to calculate the proportion of bus and the truck
Pt = 300 + 300 / (2400 + 300 + 300)
Pt = 600/3000 = 20%
following up, we will consider the length and percentage of the buses and the trucks
Et = 2.0, Pr = 0, Er = 0
to calculate the percentage of truck
Ft = 1 / 1 + Pt (Et -1 ) + Pr (Er -1)
Ft = 1 / 1+0.2 (2-1) + 0 = 0.833
Truck percentage = 0.833
To the determine the traffic volume,
Vt = V / PHF × N × Ft × Fp
Vt = 2400/ (0.9×2×0.833×1) = 1600 pc/lane/hour
But the Density of the freeway is given thus;
D = Vp / FFS .............(1)
but to get the FFS, we will consider the graph of flow rate vs speed and show the level of service
FFS = 69.28 mi/h
From the above expression in (1) we have that
D = 1600/69.28 = 23.09 pc/mi/h
D = 23.09 pc/mi/h
now we have that the the density of the freeway segment is 23.09 pc/mi/h, we can thus safely say that the level of service (los) = C
cheers i hope this helps
Consider a refrigeration truck traveling at 55 mph at a location where the airtemperature is 80F. The refrigerated compartment of the truck can be considered to the a 9-ft-wide,8-ft-high, and 20-ft-long rectangular box. The refrigeration system of the truck removes heat at arate of 600 Btu/min. The outer surface of the truck is coated with a low-emissivity material, andthus radiation heat transfer is very small. Determine the averagetemperature of the outer surface ofthe refrigeration compartment of the truck. Assume the air flow over the entire outer surface to beturbulent, and the heat transfer coefficient at the front and rear surfaces to be equal to that on sidesurfaces.
The question is an engineering problem that involves calculating the average temperature of a truck's refrigeration compartment's outer surface. A precise answer would require additional thermal information and properties, which are not provided in the question.
Explanation:The question is related to the field of thermodynamics and heat transfer, specific to engineering. It requires determining the average temperature of the outer surface of the refrigeration compartment of a truck. However, to provide a precise answer, additional data would be required, including the thermal properties of the truck's material, the heat transfer coefficient, and the difference between the interior and exterior temperatures. Typically, such a problem would involve calculations using the concepts of convection and possibly conduction, considering the truck as a heat exchanger with heat being removed by the refrigeration system and added from the external environment. Without specific values for heat transfer coefficients and material properties, a calculation cannot be accurately made.
5. Steel balls 50 mm in diameter are annealed by heating to 1200 K and then slowly cooling to 450 K in an air environment for which the ambient temperature is 300 K and h = 20 W/m2 ·K. Assuming the properties of the steel to be k = 40 W/m·K, rho = 7800 kg/m3 , and c = 600 J/kg·K. Estimate the time required for this cooling process.
Answer:
time required for cooling process = 0.233 hours
Explanation:
In Transient heat conduction of a Sphere, the formula for Biot number is;
Bi = hL_c/k
Where L_c = radius/3
We are given;
Diameter = 12mm = 0.012m
Radius = 0.006m
h = 20 W/m²
k = 40 W/m·K
So L_c = 0.006m/3 = 0.002m
So,Bi = 20 x 0.002/40
Bi = 0.001
The formula for time required is given as;
t = (ρVc/hA)•In[(T_i - T_(∞))/(T - T_(∞))]
Where;
A is Area = πD²
V is volume = πD³/6
So,
t = (ρ(πD³/6)c/h(πD²))•In[(T_i - T_(∞))/(T - T_(∞))]
t = (ρDc/6h)•In[(T_i - T_(∞))/(T - T_(∞))]
We are given;
T_i = 1200K
T_(∞) = 300K
T = 450K
ρ = 7800 kg/m³
c = 600 J/kg·K
Thus, plugging in relevant values;
t = (7800 x 0.012 x 600/(6 x20) )•In[(1200 - 300)/(450 - 300)]
t = 468•In6
t = 838.54 seconds
Converting to hours,
t = 838.54/3600
t = 0.233 hours
The diameter of an extruder barrel = 85 mm and its length = 2.00 m. The screw rotates at 55 rev/min, its channel depth = 8.0 mm, and its flight angle = 18°. Head pressure at the die end of the barrel = 10.0(10^6) Pa. Viscosity of the polymer melt = 100 Pa- s
(a) Find the volume flow rate of plastic at the die end of the barrel.
________ x 10^-6 m^3/s
Answer:
Qx = 9.10[tex]9.10^5 \times 10^{-6}[/tex] m³/s
Explanation:
given data
diameter = 85 mm
length = 2 m
depth = 9mm
N = 60 rev/min
pressure p = 11 × [tex]10^6[/tex] Pa
viscosity n = 100 Pas
angle = 18°
so Qd will be
Qd = 0.5 × π² ×D²×dc × sinA × cosA ..............1
put here value and we get
Qd = 0.5 × π² × ( 85 [tex]\times 10^{-3}[/tex] )²× 9 [tex]\times 10^{-3}[/tex] × sin18 × cos18
Qd = 94.305 × [tex]10^{-6}[/tex] m³/s
and
Qb = p × π × D × dc³ × sin²A ÷ 12 × n × L ............2
Qb = 11 × [tex]10^{6}[/tex] × π × 85 [tex]\times 10^{-3}[/tex] × ( 9 [tex]\times 10^{-3}[/tex] )³ × sin²18 ÷ 12 × 100 × 2
Qb = 85.2 × [tex]10^{-6}[/tex] m³/s
so here
volume flow rate Qx = Qd - Qb ..............3
Qx = 94.305 × [tex]10^{-6}[/tex] - 85.2 × [tex]10^{-6}[/tex]
Qx = 9.10[tex]9.10^5 \times 10^{-6}[/tex] m³/s
• Build upon the results of problem 3-85 to determine the minimum factor of safety for fatigue based on infinite life, using the modified : Goodman criterion. The shaft CD rotates at a constant speed, has a constant diameter of 1.13 in, and is made from cold-drawn AISI 1018 steel. From problem 3-85, the critical stress element in shaft CD experiences a completely reversed bending stress due to the rotation, as well as steady torsional and axial stresses. Thus, a,bend= 12 kpsi, Om, bend= 0 kpsi, Oa,axial= 0 kpsi, om, axial= -0.9 kpsi, Ta = 0 kpsi, and Im = 10 kpsi. The minimum factor of safety for fatigue is
Answer:
minimum factor of safety for fatigue is = 1.5432
Explanation:
given data
AISI 1018 steel cold drawn as table
ultimate strength Sut = 63.800 kpsi
yield strength Syt = 53.700 kpsi
modulus of elasticity E = 29.700 kpsi
we get here
[tex]\sigma a[/tex] = [tex]\sqrt{(\sigma a \times kb)^2+3\times (za\times kt)^2}[/tex] ...........1
here kb and kt = 1 combined bending and torsion fatigue factor
put here value and we get
[tex]\sigma a[/tex] = [tex]\sqrt{(12 \times 1)^2+3\times (0\times 1)^2}[/tex]
[tex]\sigma a[/tex] = 12 kpsi
and
[tex]\sigma m[/tex] = [tex]\sqrt{(\sigma m \times kb)^2+3\times (zm\times kt)^2}[/tex] ...........2
put here value and we get
[tex]\sigma m[/tex] = [tex]\sqrt{(-0.9 \times 1)^2+3\times (10\times 1)^2}[/tex]
[tex]\sigma m[/tex] = 17.34 kpsi
now we apply here goodman line equation here that is
[tex]\frac{\sigma m}{Sut} + \frac{\sigma a}{Se} = \frac{1}{FOS}[/tex] ...................3
here Se = 0.5 × Sut
Se = 0.5 × 63.800 = 31.9 kspi
put value in equation 3 we get
[tex]\frac{17.34}{63.800} + \frac{12}{31.9} = \frac{1}{FOS}[/tex]
solve it we get
FOS = 1.5432
The attached program (studentsGpa.cpp) uses dynamic allocation to create an array of strings. It asks the user to enter a number and based on the entered number it allocates the array size. Then based on that number it asks the user that many times to enter student’s names. What you need to do:Add another array of doubles to store the gpa of each student as you enter themYou need to display both the student’s name and the gpa.NOTE: must use pointer notation not array subscript. Any submission that uses array subscript won’t be graded
Question: The program was not attached to your question. Find attached of the program and the answer.
Answer:
See the explanation for the answer.
Explanation:
#include <iostream>
using namespace std;
int main()
{
cout << "How many students will you enter? ";
int n;
cin>>n;
string *name = new string[n];
double *gpa = new double[n];
for(int i=0;i<n;i++){
cout<<"Enter student"<<(i+1)<<"'s name: ";
cin>>name[i];
cout<<"Enter student"<<(i+1)<<"'s gpa: ";
cin>>gpa[i];
}
cout<<"The list students"<<endl;
cout<<"Name GPA"<<endl;
cout<<"----------------------"<<endl;
for(int i=0;i<n;i++){
cout<<name[i]<<" "<<gpa[i]<<endl;
}
return 0;
}
OUTPUT : See the attached file.
5/25 At the instant under consideration, the hydraulic cylinder AB has a length L = 0.75 m, and this length is momentarily increasing at a constant rate of 0.2 m/s. If vA = 0.6 m/s and θ = 35°, determine the velocity of slider B.
The image of the question is missing, so i have attached it
Answer:
Velocity of slider B; = - 0.176 m/s
Explanation:
We are given;
Length of (AB) = 0.75 m
Rate of increase of length; (AB)' = 0.2 m/s
vA = 0.6 m/s
θ = 35°
We want to find vB;
Looking at the image attached, we can use the trigonometric ratio to find OA
Thus;
Sin θ = (OA)/(AB)
So, Sin 35° = (OA)/(AB)
(OA) = (AB)Sin 35°
(OA) = 0.75•Sin 35°
(OA) = 0.75•0.5736
(OA) = 0.43 m
Also, we can use the same system to find (OB)
Thus;
Cos θ = (OB)/(AB)
Cos 35° = (OB)/(AB)
(OB) = (AB)Cos 35°
(OB) = 0.75•Cos 35°
(OB) = 0.75•0.8192
(OB) = 0.6144 m
We apply Pythagoras' theorem as follows
(AB)² = (OA)² + (OB)²
We derive the equation;
2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB
Divide through by 2 to give;
(AB)*(AB)' = (OA)*vA + (OB)*vB
vB = ((AB)*(AB)' - (OA)*vA) / (OB)
We now have ;
vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s)/(0.614 m)
vB = - 0.176 m/s
An offset h must be introduced into a metal tube of 0.75-in. outer diameter and 0.08-in. wall thickness. Knowing that the maximum stress after the offset is introduced must not exceed 4 times the stress in the tube when it is straight, determine the largest offset that can be used
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The largest offset that can be used is [tex]h = 0.455 \ in[/tex]
Explanation:
From the question we are told that
The diameter of the metal tube is [tex]d_m = 0.75 \ in[/tex]
The thickness of the wall is [tex]D = 0.08 \ in[/tex]
Generally the inner diameter is mathematically evaluated as
[tex]d_i = d_m -2D[/tex]
[tex]= 0.75 - 2(0.08)[/tex]
[tex]= 0.59 \ in[/tex]
Generally the tube's cross-sectional area can be evaluated as
[tex]a = \frac{\pi}{4} (d_m^2 - d_i^2)[/tex]
[tex]= \frac{\pi}{4} (0.75^2 - 0.59^2)[/tex]
[tex]= 0.1684 \ in^2[/tex]
Generally the maximum stress of the metal is mathematically evaluated as
[tex]\sigma = \frac{P}{A}[/tex]
[tex]\sigma = \frac{P}{ 0.1684}[/tex]
The diagram showing when the stress is been applied is shown on the second uploaded image
Since the internal forces in the cross section are the same with the force P and the bending couple M then
[tex]M = P * h[/tex]
Where h is the offset
The maximum stress becomes
[tex]\sigma_n = \frac{P}{A} + \frac{M r_m }{I}[/tex]
Where [tex]r_m[/tex] is the radius of the outer diameter which is evaluated as
[tex]r_m = \frac{0.75}{2}[/tex]
[tex]r_m = 0.375 \ in[/tex]
and I is the moment of inertia which is evaluated as
[tex]I = \frac{\pi}{64} (d_m^4 - d_i^4 )[/tex]
[tex]= \frac{\pi}{64}(0.75^4 - 0.59^4)[/tex]
[tex]= 0.009583 \ in^4[/tex]
So the maximum stress becomes
[tex]\sigma' = \frac{P}{0.1684} + \frac{Phr}{0.009583}[/tex]
Now the question made us to understand that the maximum stress when the offset was introduced must not exceed the 4 times the original stress
So
[tex]\sigma ' = 4 \sigma[/tex]
=> [tex]\frac{P}{0.1684} + \frac{Phr_m }{0.009583} = 4 [\frac{P}{0.1684} ][/tex]
The P would cancel out
[tex]\frac{1}{0.1684} + \frac{h(0.375)}{0.009583} = \frac{4}{0.1684}[/tex]
[tex]5.94 + 39.13h = 23.753[/tex]
[tex]39.13h = 17. 813[/tex]
[tex]h = 0.455 \ in[/tex]
How does the resistance in the circuit impact the height and width of the resonance curve? (If the resistance were to increase would the height change? Would the width? If so, how?)
Answer:
The reactances vary with frequency, with large XL at high frequencies and large Xc at low frequencies, as we have seen in three previous examples. At some intermediate frequency fo, the reactances will be the same and will cancel, giving Z = R; this is a minimum value for impedance and a maximum value for Irms results. We can get an expression for fo by taking
XL=Xc
Substituting the definitions of XL and XC,
2[tex]\pi[/tex]foL=1/2[tex]\pi[/tex]foC
Solving this expression for fo yields
fo=1/2[tex]\pi[/tex][tex]\sqrt{LC}[/tex]
where fo is the resonant frequency of an RLC series circuit. This is also the natural frequency at which the circuit would oscillate if it were not driven by the voltage source. In fo, the effects of the inductor and capacitor are canceled, so that Z = R and Irms is a maximum.
Explanation:
Resonance in AC circuits is analogous to mechanical resonance, where resonance is defined as a forced oscillation, in this case, forced by the voltage source, at the natural frequency of the system. The receiver on a radio is an RLC circuit that oscillates best at its {f} 0. A variable capacitor is often used to adjust fo to receive a desired frequency and reject others is a graph of current versus frequency, illustrating a resonant peak at Irms at fo. The two arcs are for two dissimilar circuits, which vary only in the amount of resistance in them. The peak is lower and wider for the highest resistance circuit. Thus, the circuit of higher resistance does not resonate as strongly and would not be as selective in a radio receiver, for example.
A current versus frequency graph for two RLC series circuits that differ only in the amount of resistance. Both have resonance at fo, but for the highest resistance it is lower and wider. The conductive AC voltage source has a fixed amplitude Vo.
You are asked by your college crew to estimate the skin friction drag in their eight-seat racing shell. The hull of the shell may be approximated as half a circular cylinder with 450 mm diameter and 7.32 m length. The speed of the shell through the water is 6.71 m/s. Estimate the location of the transition from laminar to turbulent flow in the boundary layer on the hull of the shell. Calculate the thickness of the turbulent boundary layer at the rear of the hull. Determine the total skin friction drag on the hull under the given conditions
Answer:
The total skin friction drag on the hull under these conditions is 276N
Explanation:
In this question, we are asked to determine the total skin friction drag on the hull under the given conditions.
Please check attachment for complete solution and step by step explanation
Suppose we have a database for an investment firm, consisting of the following attributes: B (broker), O (office of a broker), I (investor), S (stock), Q (quantity of stock owned by an investor), and D (dividend paid by a stock), with the following functional dependencies: SD, I B.IS a Find a key for the relation scheme R(B,O,S,Q,I,D). b. Find a decomposition of R into third normal form, having a lossless join and preserving dependencies c.Find a lossless join decomposition of R into Boyce-Codd normal form.
Answer:
Given, FDs are:
S -> D
I -> B
IS -> Q
B -> O
a)
"I" and "S" must be there in any candidate key because they do not appear on the right side of any functional dependency.
The only candidate key is: IS
IS -> ISBDQO
b)
Decomposition of R into 3NF: (I, B), (S, D), (B, O), (I, S, Q)
c)
Decomposition of R into BCNF:
Decompose R by I → B into R1 = (I, B) and R2 = (I, O, S, Q, D).
R1 is in BCNF
Decompose R2 by S → D into R21 = (S, D) and R22 = (O, I, S, Q).
R21is in BCNF
Decompose R22 by I → O into R221 = (I, O) and R222 = (I, S, Q).
R221 is in BCNF.
R222 is in BCNF.
The decomposition is: (I, B), (S, D), (I, O), (I, S, Q)
We can also write it as: (I, B), (S, D), (B, O), (I, S, Q)
Explanation:
The answer above is rendered in a very explanatory way.
The density of a liquid is to be determined by an old 1-cm-diameter cylindrical hydrometer whose division marks are completely wiped out. The hydrometer is first dropped in water, and the water level is marked. The hydrometer is then dropped into the other liquid, and it is observed that the mark for water has risen 1.4 cm (hw) above the liquid–air interface. If the height of the original water mark is 23 cm (hl + hw), determine the density of the liquid.
Answer:
1064.8 kg/m³
Explanation:
Weight of the hydrometer = ρghA where ρ is the density, g is acceleration due to gravity, h is the submerged height and A is the cross sectional area.
W in water = ρwghwA
W in liquid = (ρliq)g hliq A where the cross sectional area is constant
W in water = W in liquid
(ρw)ghwA = (ρliq)g hliq A where ρw is density of water, ρliq is the density of liquid and hw and hliq are the heights of the liquid and that water. g acceleration due to gravity cancel on both sides as well as the constant A
pliq = [tex]\frac{hw}{hliq}[/tex] × 1000 kg /m³ ( density of water) =( [tex]\frac{23}{23-1.4}[/tex]) × 1000 = 1064.8 kg/m³
The high electrical conductivity of copper is an important design factor that helps improve the energy efficiency of electric motors. This is important because motors and motor-driven systems are significant consumers of electricity, accounting for 43% - 46% of all global electricity consumption and 69% of all electricity used by industry. Inefficient motors waste electrical energy and are indirect contributors to greenhouse gas emissions. ElectroSpark, Inc. has been developing a new copper die-cast rotor technology specifically for premium efficiency motors, replacing the standard aluminum rotor. There are multiple reasons for doing so, including the possibility that the motor will consume less energy. They designed an experiment to test their idea in a common ¾ Horse power (HP) motor that is normally manufactured with an aluminum rotor. They designed a copper rotor that fit in their ¾ HP motor housing and ran a production line for a day producing the motors. They randomly selected 20 copper-rotor motors from that output and 20 aluminum-rotor motors produced from the same line the day before. These 40 motors were all run for 8 hours a day for 30 days and the energy consumed was measured in total Kilowatt Hours (example data below, using alpha=.05):
Copper: 560.145 539.673 556.834 559.873
Aluminium: 564.674 573.912 553.385 574.078
What is the correct hypothesis to test the problem described in this scenario?
A. H0: μD (copper-aluminum) ≥ 0; H1: μD (copper-aluminum) < 0
B. H0: μ_copper – μ_aluminum ≥ 0; H1: μ_copper – μ_aluminum < 0
C. H0: μD (copper-aluminum) ≥ 0; H1: μD (copper-aluminum) > 0
D. H0: μ_copper – μ_Aluminum ≤ 0; H1: μ_copper – μ_aluminum > 0
Answer:
B
Explanation:
This is a two sample t-test and not a matched pair t-test
null hypothesis(H0) will be that mean energy consumed by copper rotor motors is greater than or equal to mean energy consumed by aluminium rotor motors
alternate hypothesis(H1) will be that mean energy consumed by copper rotor motors is less than or equal to mean energy consumed by aluminium rotor motors.
So, option D is rejected
The hypothesis will not compare mean of differences of values of energy consumed by copper rotor motor and aluminium rotor motor.
Option A and C are also rejected
For the following transfer function, derive expressions for the real and imaginary part for s = jω in terms of the frequency variable ω. Then write a MATLAB script to plot the imaginary part versus the real part (and its reflection about the real axis) for a frequency ω range of 10−2 to 102 radians per second. Verify that your plots match the output of the nyquist function in MATLAB. • G(s) = 1 /(s+0.5)(s+1)(s+2) Suppose G represents an open-loop plant transfer function. Use your plot to determine the Gain Margin for the closed-loop system, i.e., determine how much the loop gain could be increased before the closed-loop becomes unstable
Answer:
See all solutions attached as picture.
Explanation:
It is well explanatory
A short-circuit experiment is conducted on the high-voltage side of a 500 kVA, 2500 V/250 V, single-phase transformer in its nominal frequency. The short-circuit voltage is found as 100 V and the short-circuit current and power are 110 A and 3200 W, respectively. Find the series impedance of the transformer referred to its low voltage side.
Given Information:
Primary secondary voltage ratio = 2500/250 V
Short circuit voltage = Vsc = 100 V
Short circuit current = Isc = 110 A
Short circuit power = Psc = 3200 W
Required Information:
Series impedance = Zeq = ?
Answer:
Series impedance = 0.00264 + j0.00869 Ω
Step-by-step explanation:
Short Circuit Test:
A short circuit is performed on a transformer to find out the series parameters (Z = Req and jXeq) which in turn are used to find out the copper losses of the transformer.
The series impedance in polar form is given by
Zeq = Vsc/Isc < θ
Where θ is given by
θ = cos⁻¹(Psc/Vsc*Isc)
θ = cos⁻¹(3200/100*110)
θ = 73.08°
Therefore, series impedance in polar form is
Zeq = 100/110 < 73.08°
Zeq = 0.909 < 73.08° Ω
or in rectangular form
Zeq = 0.264 + j0.869 Ω
Where Req is the real part of Zeq and Xeq is the imaginary part of Zeq
Req = 0.264 Ω
Xeq = j0.869 Ω
To refer the impedance of transformer to its low voltage side first find the turn ratio of the transformer.
Turn ratio = a = Vp/Vs = 2500/250 = 10
Zeq2 = Zeq/a²
Zeq2 = (0.264 + j0.869)/10²
Zeq2 = (0.264 + j0.869)/100
Zeq2 = 0.00264 + j0.00869 Ω
Therefore, Zeq2 = 0.00264 + j0.00869 Ω is the series impedance of the transformer referred to its low voltage side.
A completely reversible heat pump produces heat at a rate of 100 kW to warm a house maintained at 25℃. The exterior air, which is at 0℃, serves as the source. Calculate the rate of entropy change of the two reservoirs and determine if this heat pump satisfies the second law according to the increase of entropy principle.
Answer:
1) 0. 03 kW/K
2) the value is grater than zero so it satisfies the second law of thermodynamic (states that rate of entropy change must be equal to or greater than zero) .
Explanation:
Rate of entropy change S = dQ/dT
= Q(1/T1 - 1/T2)
T2 = 25°C = 298 K
T1 = 0°C = 273 K
Q = 100 kW
S = 100( 1/273 - 1/298)
S = 100(0.0003) = 0. 03 kW/K