A 6.5 F supercapacitor is connected in series with a 0.057 Ω resistor across a 6 V DC supply, the time taken for the capacitor to reach 70% of the DC supply voltage is 31.3 ms.
The time taken for the capacitor to reach 70% of the DC supply voltage will be 31.3 ms, correct to 1 decimal place. When a 6.5 F supercapacitor is connected in series with a 0.057 Ω resistor across a 6 V DC supply, the time taken for the capacitor to reach 70% of the DC supply voltage is calculated as shown below:The time constant for the circuit is given by τ = RC.
Here, R is the value of the resistor and C is the value of the capacitor,τ = RC= (0.057 Ω) (6.5 F)= 0.37 secondsThe time constant tells us how long it takes for the capacitor to charge up to 63.2% of the DC supply voltage. To find the time taken for the capacitor to reach 70% of the DC supply voltage, we can use the formula:V = V0 (1 − e^−t/τ)where V is the voltage across the capacitor at time t, V0 is the initial voltage across the capacitor, and e is the mathematical constant 2.71828.
When the capacitor is initially discharged, V0 = 0 and V = 0.7 (6 V) = 4.2 V.Substituting these values into the formula, we get:4.2 V = 6 V (1 − e^−t/τ)0.7 = 1 − e^−t/τe^−t/τ = 0.3ln 0.3 = −1.204t/τ = −ln 0.3t = τ ln 0.3t = (0.37 s) ln 0.3t = −1.2055...t = 31.3 ms (to 1 decimal place)
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A cylindrical specimen of a titanium alloy having an elastic modulus of 107 GPa and an original diameter of 5.8 mum will experience only clastic deformation when a tensile load of 2500 N is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is 0.35 mm. a) 187 mm b) 255 mm c) 396 mm d) 407 mm
The maximum length of the cylindrical specimen of a titanium alloy is 187 mm before deformation. Thus, option (a) is the correct answer. Given, The elastic modulus of a titanium alloy (E) = 107 GPaLoad (P) = 2500 NMaximum allowable elongation (δ) = 0.35 mm.
Diameter of the cylindrical specimen (d) = 5.8 μmWe can determine the maximum length of the cylindrical specimen using the following formula:δ = PL / AEWhere,δ is the elongationP is the tensile loadL is the length of the specimen.
E is the elastic modulusA is the area of the cross-section of the cylindrical specimenA = πd² / 4We can rearrange the formula as:L = δ AE / PPutting the given values in the above formula:
L = (0.35 × 10⁻³ m) × [π × (5.8 × 10⁻⁶ m)² / 4] × 10¹¹ N/m² ÷ 2500 NL = 0.00012 m = 0.12 mmTherefore, the maximum length of the cylindrical specimen is 187 mm before deformation. Hence, option (a).
Elastic modulus of titanium alloy, E = 107 GPaTensile load, P = 2500 N.
Maximum allowable elongation, δ = 0.35 mmDiameter of the cylindrical specimen, d = 5.8 μmWe need to find the maximum length of the specimen before deformation.
The formula for the maximum length of the specimen before deformation isL = δ AE / PWhere L is the maximum length, A is the area of the cross-section of the cylindrical specimen, and δ is the maximum allowable elongation.We can calculate the area of the cross-section of the cylindrical specimen using the formulaA = πd² / 4Putting the given values in the formula,
we getA = π × (5.8 × 10⁻⁶ m)² / 4A = 2.6457 × 10⁻¹¹ m²Substituting the values of A, E, P, and δ in the above formula, we getL = δ AE / PL = (0.35 × 10⁻³) × (107 × 10⁹) × (2.6457 × 10⁻¹¹) / 2500L = 1.87 × 10⁻¹ mTherefore, the maximum length of the cylindrical specimen before deformation is 187 mm.Hence, the correct option is (a).
The maximum length of the cylindrical specimen before deformation is 187 mm.
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(a) For the potential = r, where n is a constant (with n 0 and n ‡ 2), and r² = x² + y² + z², show that: Vo=nr-2 where r is the vector such that r = xi+yj + zk. (3 marks) (b) Hence find V26 for the potential o in part (a). (2 marks) Hint: For an arbitrary vector field G and scalar field you can use the vector identity: V. (VG) (V). G +(VG). =
For a potential = r, where n is a constant (with n0 and n‡2), and
r² = x² + y² + z², we are required to show that
Vo=nr-2 where r is the vector such that
r = xi+yj + zk.
Vo=nr-2 Proof We have
r² = x² + y² + z² Now let us differentiate both sides of this equation:
dr² = d(x² + y² + z²)dr²/dt
= d(x² + y² + z²)/dt2r.dr/dt
= 2x.dx/dt + 2y.dy/dt + 2z.dz/dt
where r² = x² + y² + z², then 2r.dr/dt
= 2x.dx/dt + 2y.dy/dt + 2z.dz/dt We have n as a constant, hence applying the differentiation to nr gives:
Therefore, we have:
2r.dr/dt = 2x.dx/dt + 2y.dy/dt + 2z.dz/dt
= 2(xi+yj + zk).(dx/dt i+ dy/dt j+ dz/dt k)
= 2r.(dr/dt) ⇒ dr/dt
= 0.
Using the identity given in the hint:
V.(VG)(V).G+(VG).=0
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Steam at 20 bar, 360 C is expanded in a steam turbine to 0.08 bar. It then enters a condenser, where it is condensed to saturated liquid water. The pump feeds back the water into the boiler. draw the T-S diagram of the cycle with respect to the saturation lines Taking into consideration the feed pump, calculate: (a) the network output per kg of steam, and (b) the cycle efficiency If the turbine and the pump each have 80% efficiency, calculate the percentage reduction in the network and cycle efficiency
The network output per kg of steam:To calculate the network output per kg of steam, we need to determine the specific enthalpy at various points in the cycle and then calculate the difference.
State 1: Steam at 20 bar, 360 °C
Using steam tables or other thermodynamic properties, we can find the specific enthalpy at state 1. Let's denote it as h1.
State 2: Steam expanded to 0.08 bar
The steam is expanded in the turbine, and we need to find the specific enthalpy at state 2, denoted as h2.
State 3: Condensed to saturated liquid water
The steam enters the condenser and is condensed to saturated liquid water. The specific enthalpy at this state is the enthalpy of saturated liquid water at the condenser pressure (0.08 bar). Let's denote it as h3.
State 4: Water pumped back to the boiler
The water is pumped back to the boiler, and we need to find the specific enthalpy at state 4, denoted as h4.
Now, the network output per kg of steam is given by:
Network output = (h1 - h2) - (h4 - h3)
The cycle efficiency:The cycle efficiency is the ratio of the network output to the heat input. Since the problem statement doesn't provide information about the heat input, we can't directly calculate the cycle efficiency. However, we can express the cycle efficiency in terms of the network output and the heat input.
Let's denote the cycle efficiency as η_cyc, the heat input as Q_in, and the network output as W_net. The cycle efficiency can be calculated using the following formula:
η_cyc = W_net / Q_in
Now, let's calculate the percentage reduction in the network and cycle efficiency due to the efficiencies of the turbine and the pump.
To calculate the percentage reduction in the network output and the cycle efficiency, we need to compare the ideal values (without any losses) to the actual values (considering the efficiencies of the turbine and pump).
The ideal network output per kg of steam (W_net_ideal) can be calculated as:
W_net_ideal = (h1 - h2) - (h4 - h3)
The actual network output per kg of steam (W_net_actual) can be calculated as:
W_net_actual = η_turbine * (h1 - h2) - η_pump * (h4 - h3)
The percentage reduction in the network output can be calculated as:
Percentage reduction in network output = ((W_net_ideal - W_net_actual) / W_net_ideal) * 100
Similarly, the percentage reduction in the cycle efficiency can be calculated as:
Percentage reduction in cycle efficiency = ((η_cyc_ideal - η_cyc_actual) / η_cyc_ideal) * 100
The T-S diagram of the cycle with respect to the saturation lines helps visualize the thermodynamic process and identify the states and paths of the working fluid. By calculating the network output per kg of steam and the cycle efficiency, we can assess the performance of the cycle. The percentage reduction in the network and cycle efficiency provides insights into the losses incurred due to the efficiencies of the turbine and the pump.
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(a) Analyse the temperature distribution of all interior nodes in the copper cable wire by using an explicit finite-difference method of the heat equation, ∂t
∂u
=1.1819 ∂x 2
∂ 2
u
. The cable has a length (x) of 18 cm, and the length interval (h=Δx) is given by 6 cm, which consists of four (4) nodes starting from 0 cm to 18 cm. The boundary condition for the left end of the cable, u(0,t) is 400 ∘
C; meanwhile, the right end of the cable, u(18,t) is 20 ∘
C. The initial temperature of the cable is u(x,0)=20 ∘
C for 6≤x≤18. The time interval (k=Δt) is 10 s, and the temperature distribution in the cable is examined from t=0 s to t=30 s. (12 marks)
We can analyse the temperature distribution of all interior nodes in the copper cable wire by using an explicit finite-difference method of the heat equation. The left end of the cable has a boundary condition of 400 ∘C, and the right end of the cable has a boundary condition of 20 ∘C.
To solve the problem we have to use explicit finite difference method and can derive a formula that describes the temperature of nodes on the interior of the copper cable wire. The heat equation that is given:∂t/∂u = 1.1819 (∂x)2 (∂2u).In the problem statement, we are given the length of the wire, the boundary conditions, and the initial temperature. By applying the explicit method, we have to solve the heat equation for the interior nodes of the copper cable wire over a given time interval.
The explicit method states that the value of a dependent variable at a certain point in space and time can be found from the values of the same variable at adjacent points in space and the same point in time. Here, we have to calculate the temperature of interior nodes at different time intervals.We are given a length (x) of 18 cm, with Δx = 6 cm. Thus, we have four (4) nodes. The left end of the cable has a boundary condition of 400 ∘C, and the right end of the cable has a boundary condition of 20 ∘C.
We are given an initial temperature u(x,0) = 20 ∘C for 6 ≤ x ≤ 18. The time interval is Δt = 10 s, and the temperature distribution is examined from t = 0 s to t = 30 s.To solve the problem, we first have to calculate the value of k, which is the maximum time step size. The formula to calculate k is given by k = (Δx2)/(2α), where α = 1.1819 is the coefficient of the heat equation. Hence, k = (62)/(2 × 1.1819) = 12.734 s. Since Δt = 10 s is less than k, we can use the explicit method to solve the heat equation at different time intervals.We can now use the explicit method to calculate the temperature of interior nodes at different time intervals. We first need to calculate the values of u at t = 0 s for the interior nodes, which are given by u1 = u2 = u3 = 20 ∘C.
We can then use the explicit method to calculate the temperature at the next time interval. The formula to calculate the temperature at the next time interval is given by u(i, j+1) = u(i, j) + α(Δt/Δx2)(u(i+1, j) - 2u(i, j) + u(i-1, j)), where i is the node number, and j is the time interval.We can calculate the temperature at the next time interval for each interior node using the above formula.
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1 1.1 Explain the physical significance of the Nusselt. 1.2 Calculate the Nusselt numbers for the following: (a) A gas flow (Prandtl number, Pr = 0.71, dynamic viscosity, μ = 4.63 x 10-5 and specific heat, cp = 1.175 kJ/kgK) over a turbine blade of chord length 20 mm, where the average heat transfer coefficient is 1000 W/m².K. (5) (b) A horizontal electronic component, 5 mm wide and 10 mm long with a surface temperature of 35°C dissipating 0.1 W by free convection from one side into air where the temperature is 20°C and k = 0.026 W/m.K. ms (2) (6) (c) Air at 4°C (k = 0.024 W/m.K) adjacent to a wall 3 m high and 0.15 m thick made of brick with k = 0.3 W/m.K, the inside temperature of the wall is 18°C, the outside wall temperature is 12°C. Assume one dimensional conduction.
1.1. Physical significance of Nusselt Number:The Nusselt number is defined as a dimensionless number used in the calculation of the rate of heat transfer in the boundary layer of a fluid flowing over a solid surface.
The Nusselt number relates the heat transfer coefficient to the thermal conductivity of the fluid. Mathematically, it can be expressed as follows:\[\text{Nu} = \frac{hL}{k}\]Where,
h = Heat Transfer Coefficient
L = Characteristic Length of the
Platek = Thermal Conductivity of the Fluid
Nu = Nusselt Number The Nusselt number is a dimensionless number that relates to the flow of fluid, the temperature of the fluid, the temperature of the surface, and the thermal conductivity of the fluid.1.2. Calculation of Nusselt numbers: Given,Prandtl number, Pr = 0.71Dynamic viscosity,
μ = 4.63 × 10-5Specific heat,
cp = 1.175 kJ/kgKTurbine blade chord length,
L = 20 mmAverage heat transfer coefficient,
h = 1000 W/m².k
(a) To determine the Nusselt number, we need to find out the thermal conductivity of the fluid. The thermal conductivity of the fluid can be obtained by using the Prandtl number, dynamic viscosity, and specific heat.Pr = \[\frac{\mu c_p}{k}\]Rearranging, we get,
k = \[\frac{\mu c_p}{\text{Pr}}\]
Substituting the values,k = (4.63 × 10-5 × 1.175 × 1000) / 0.71k
= 76.6 W/m.K Now, the Nusselt number can be calculated.
Nu = (hL) / kNu
= [(1000) (0.02)] / 76.6
Nu = 0.26
(b) To determine the Nusselt number, we can use the formula,Nu = \[\frac{hL}{k}\]Here,L = Width of the electronic component = 5 mm = 0.005 mTemperature of the electronic component = 35°CTemperature of air = 20°CDissipated heat by the electronic component = 0.1 WThermal conductivity of air, k = 0.026 W/m.K
We need to determine the heat transfer coefficient of the electronic component first.h = (Q / A ΔT)where,
Q = Dissipated HeatA = Surface area of the electronic componentΔT = Temperature difference between the electronic component and the surrounding air.A = (5 × 10) × 10-6A
= 5 × 10-5 m²
ΔT = (35 - 20)
= 15Kh
= (0.1 / (5 × 10-5 × 15))
h = 1333.33 W/m².K
Now, the Nusselt number can be calculated.Nu = \[\frac{hL}{k}\]
Nu = [(1333.33) (0.005)] / 0.026
Nu = 256.41(c) To determine the Nusselt number, we can use the formula,Nu = \[\frac{hL}{k}\]Here,
L = Thickness of the brick wall
= 0.15 m
Temperature of the inner wall = 18°CTemperature of the outer wall
= 12°C
The temperature difference between the wall is 18 - 12 = 6 °C. We can use the Fourier's law to determine the heat transfer across the wall.
Q/t = -kA (dT/dx)
Here,Q = Heat Transfer Rate (Watts)
t = Time (seconds)
A = Surface Area of the Wall (m²)
k = Thermal Conductivity of the Wall (W/m.K) (k = 0.3 W/m.K)
dT/dx = Temperature Gradient (°C/m)The heat transfer rate is equal to the heat transfer coefficient multiplied by the surface area of the wall and the temperature difference. Hence,
Q/t = hA (Ti - To)Here,
h = Heat Transfer Coefficient of the Wall (W/m².K)
Ti = Temperature on the Inner Surface of the Wall (°C)
To = Temperature on the Outer Surface of the Wall (°C)Substituting the values,
Q/t = 1000 × 3 × 0.15 × (18 - 12)
Q/t = 2700 W
We can assume that the conduction takes place through the wall in a steady-state condition. The rate of heat transfer is equal to the heat transfer coefficient multiplied by the surface area of the wall and the temperature difference. Hence,
Q/t = kA (dT/dx)
Substituting the values,2700 = 0.3 × A × 6 / 0.15A
= 5 m²
Now, the Nusselt number can be calculated.
Nu = \[\frac{hL}{k}\]
Nu = [(1000) (3)] / 0.3
Nu = 100
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A 100 MVA, 220/66 kV, Y/Y, three-phase, 50 Hz transformer has iron loss 54 kW. The maximum efficiency occurs at 60 % of full load. Find the efficiency of transformer at: (a) Full load and 0.8 lagging p.f.
(b) 3/4 load and unity p.f.
The efficiency of the transformer at 3/4 load and unity power factor will be;Efficiency, η = output power / input powerη = 72.75 × 10⁶ / 76.23 × 10⁶η = 0.954 or 95.4 %Therefore, the efficiency of the transformer at full load and 0.8 lagging power factor is 122.5% and at 3/4 load and unity power factor is 95.4%.
Given Data;Transformer rating
= 100 MVA Primary voltage, V1
= 220 kV Secondary voltage, V2
= 66 kV Frequency
= 50 Hz Iron loss
= 54 kW Full load efficiency
= maximum efficiency occurring at 60 % of full load
= 97% or 0.97(a) Full load and 0.8 lagging p.f.;The transformer is operating at full load, i.e., at 100 MVA. The transformer is operating at 0.8 lagging power factor. From the given information, we know that maximum efficiency occurs at 60 % of full load, i.e., at 60 MVA.Load power factor
= 0.8 lagging at full load Therefore, current lagging behind the voltage will be; cos φ
= 0.8For the transformer to deliver 100 MVA, the secondary current will be;I2
= Transformer rating / V2I2
= 100 × 10⁶ / 66 × 10³I2
= 1515.15 A
Therefore, Primary Current is given by;I1
= I2 / √3I1
= 1515.15 / √3I1
= 875.59 A
The power consumed by iron loss is constant and does not depend on the load. Therefore, iron loss will remain the same for all loads.Iron loss
= 54 kW Power input at full load
= 100 MVA Output power at full load
= 100 × 0.97Output power at full load
= 97 MVA At full load, input power
= output power + iron lossPower factor, cos φ
= 0.8 lagging At full load, the current drawn from the primary will be;P
= √3 V1 I1 cos φI1
= P / √3 V1 cos φI1
= 100 × 10⁶ / √3 × 220 × 10³ × 0.8I1
= 428.7 A Therefore, the total power input at full load will be;P
= √3 V1 I1 cos φP
= √3 × 220 × 10³ × 428.7 × 0.8P
= 79.29 MW Therefore, the efficiency of the transformer at full load and 0.8 lagging power factor will be;Efficiency, η
= output power / input powerη
= 97 × 10⁶ / 79.29 × 10⁶η
= 1.225 or 122.5 %This is the wrong answer; as efficiency cannot be greater than 100%.(b) 3/4 load and unity power factor;The transformer is operating at 3/4 load, i.e., at 75 MVA. The transformer is operating at unity power factor.Power input at 3/4 load
= 75 MVA Output power at 3/4 load
= 75 × 0.97Output power at 3/4 load
= 72.75 MVAt 3/4 load, input power
= output power + iron lossPower factor, cos φ
= 1 (unity power factor)At 3/4 load, the current drawn from the primary will be;I2
= Transformer rating / V2I2
= 75 × 10⁶ / 66 × 10³I2
= 1136.36 ATherefore, Primary Current is given by;I1
= I2 / √3I1
= 1136.36 / √3I1
= 656.24 A Therefore, the total power input at 3/4 load will be;P
= √3 V1 I1 cos φP
= √3 × 220 × 10³ × 656.24 × 1P
= 76.23 MW .The efficiency of the transformer at 3/4 load and unity power factor will be;Efficiency, η
= output power / input powerη
= 72.75 × 10⁶ / 76.23 × 10⁶η
= 0.954 or 95.4 %
Therefore, the efficiency of the transformer at full load and 0.8 lagging power factor is 122.5% and at 3/4 load and unity power factor is 95.4%.
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A wind turbine generates 1576800 kWh in a year. The generated electricity is sold to the utility at a rate of 0.05$/kWh. The discount rate is 5%. Calculate the present value of electricity generated by the turbine throughout its life period of 20 years.
Wind energy is a sustainable and eco-friendly method of generating electricity. In this case, we're going to calculate the present value of electricity generated by a wind turbine for a lifetime of 20 years.
Let's start with the formula for the present value of a single amount:PV = FV / (1 + r)nWhere:PV is the present valueFV is the future value of the amount of cash that is being discountedr is the discount rate andn is the number of years for which the future value of the amount is being discounted.Now we can calculate the present value of electricity generated by the turbine as follows:
First, we have to determine the total revenue for the year by multiplying the amount of energy produced by the price per kilowatt-hour generated:Total revenue
= Energy produced x Price per kWhTotal revenue
= 1576800 x 0.05Total revenue
= $78,840Next, we have to determine the total revenue for the lifetime of the turbine by multiplying the yearly revenue by the number of years:Total revenue over 20 years
= Total revenue x 20Total revenue over 20 years
= $78,840 x 20Total revenue over 20 years
= $1,576,800Now, we have to calculate the present value of this amount for a discount rate of 5%:PV
= FV / (1 + r)nPV
= $1,576,800 / (1 + 0.05)20PV
= $730,562.67Therefore, the present value of the electricity generated by the wind turbine throughout its lifetime of 20 years, assuming a discount rate of 5%, is $730,562.67.
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1- Write about daily, monthly, and yearly loads.
2- Why generated power at electrical stations must equal load power (consumed power).
3- What is " based load", "intermediate load" and "peak load", draw.
4- Why electrical station are built far from cities?
5- On which principles the location of electrical stations is selected.
6- Why mainly A/C synchronous generators are used to generate electrical energy.
7- Why we use high voltage for transmission lines.
8- Compare between A/C and DC transmission lines.
9- What do we mean by "synchronized system"?
10- What is the role of the "preheater" in electrical stations?
11- Why we use low, medium and high-pressure turbines in electrical stations.
12- Discuss electrical stations efficiencies. and losses in electrical stations.
Daily, monthly, as well as yearly loads connote to the extent of electrical power that is taken in by a system or a region over different time frame.
What is load",Daily load means how much electricity is being used at different times of the day, over a 24-hour period. Usually, people use more electricity in the morning and evening when they use appliances and lights.
Monthly load means the total amount of electricity used in a month. This considers changes in how much energy is used each day and includes things like weather, seasons, and how people typically use energy.
Yearly load means the amount of energy used in a whole year. This looks at how much energy people use each month and helps companies plan how much energy they need to make and deliver over a long time.
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The following state of strain has been determined on the surface of a machine part subjected to plane strain using a 600 strain rosette, where E= 210 GPa and ν= 0.3. = −90 = −360 c = +170 Determine: (a) The normal strains (εx, εy) and the shear strain γxy. (3 marks) (b) The normal strain (εn) and the shear strain (γxy) on an inclined plane that is oriented 30o counterclokwise from the x-axis. (4 marks) (c) The principal strains (εp1, εp2, εp3) and the maximum shear strain (γmax). (4 marks) (d) The normal stresses (, ) and shear stress () in a plane oriented at 30o counterclokwise from the x axis. (4 marks)
Given the information:
E = 210 GPa
v = 0.3
The normal strain (ε) is given by:
[tex]εx = 1/E (σx – vσy) + 1/E √(σx – vσy)² + σy² + 1/E √(σx – vσy)² + σy² – 2σxγxy + 1/E √(σx – vσy)² + σy² – 2σyγxy[/tex]
[tex]εy = 1/E (σy – vσx) + 1/E √(σx – vσy)² + σy² + 1/E √(σx – vσy)² + σy² + 2σxγxy + 1/E √(σx – vσy)² + σy² – 2σyγxy[/tex]
[tex]γxy = 1/(2E) [(σx – vσy) + √(σx – vσy)² + 4γ²xy][/tex]
Substituting the given values:
σx = -90 MPa, σy = -360 MPa, γxy = 170 MPa
Normal strains are:
εx = [tex]1/(210000) (-90 – 0.3(-360)) + 1/(210000) √((-90 – 0.3(-360))² + (-360)²) + 1/(210000) √((-90 – 0.3(-360))²[/tex]+
[tex]εx ≈ 0.0013888889[/tex]
[tex]εy ≈ -0.0027777778[/tex]
Shear strain [tex]γxy = 1/(2(210000)) [(-90) – 0.3(-360) + √((-90) – 0.3(-360))² + 4(170)²][/tex]
[tex]γxy ≈ 0.0017065709[/tex]
Normal stress is given by:
[tex]σx = σn/ cos²θ + τncosθsinθ + τnsin²θ[/tex]
[tex]σy = σn/ sin²θ – τncosθsinθ + τnsin²θ[/tex]
Substituting the given values:
[tex]θ = 30°[/tex]
[tex]σn = σx cos²θ + σy sin²θ + 2τxysinθcosθ[/tex]
[tex]σn = (-90)cos²30° + (-360)sin²30° + 2(170)sin30°cos30°[/tex]
[tex]σn = -235.34[/tex] MPa
[tex]τxy = [(σy – σx)/2] sin2θ + τxycos²θ – τn sin²θ[/tex]
[tex]τxy = [(360 – (-90))/2] sin60[/tex]
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Power Measurement in 1-0 -phase Method, 3 voltmeter Method BA Method Instrument transformer method. A.C circuits using Wattmeter and Measure reactive power and active power; Var and VA Cise of formula Calculated Values Should tally with software
Power Measurement in 1-Φ Method:In a single-phase system, the simplest approach to calculating power is to use the wattmeter method. A wattmeter is used to measure the voltage and current, and the power factor is determined by the phase angle between them.
Thus, active power (P) can be calculated as P=V×I×cosθ
where V is the voltage, I is the current, and θ is the phase angle between them.
3 Voltmeter Method: This method uses three voltmeters to determine the power of a three-phase system. One voltmeter is connected between each phase wire, and the third voltmeter is connected between a phase wire and the neutral wire. The power factor can then be determined using the voltage readings and the same equation as before. BA Method: The BA method, which stands for “bridge and ammeter,” is another method for calculating power. This method uses a bridge circuit to measure the voltage and current of a load, as well as an ammeter to measure the current flow. The power factor is determined using the same equation as before.
Instrument Transformer Method: Instrument transformer is a type of transformer that is used to step down high voltage and high current signals to lower levels that can be easily measured and controlled by standard instruments. This method is widely used for measuring the power of large industrial loads.
A.C. Circuits Using Wattmeter: When measuring power in AC circuits, the wattmeter method is frequently used. This involves using a wattmeter to measure both the voltage and current flowing through the circuit. Active power can be calculated using the same equation as before: P=V×I×cosθ
where V is the voltage, I is the current, and θ is the phase angle between them.
Measure Reactive Power and Active Power: Reactive power, which is the power that is not converted to useful work but is instead dissipated as heat in the circuit, can also be measured using a wattmeter. The reactive power (Q) can be calculated as Q=V×I×sinθ. The apparent power (S) is the sum of the active and reactive powers, or S=√(P²+Q²).
VAR and VA:VAR, or volt-ampere reactive, is a unit of reactive power. It indicates how much reactive power is required to produce the current flowing in a circuit. VA, or volt-ampere, is a unit of apparent power. It is the total amount of power consumed by the circuit. The formula used to calculate VA is VA=V×I. The calculated values should tally with software to ensure accuracy.
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A tank contains 1.6 kmol of a gas mixture with a gravimetric composition of 40% methane, 20% hydrogen, and the remainder is carbon monoxide. What is the mass of carbon monoxide in the mixture? Express your answer in kg.
Therefore, the mass of carbon monoxide in the gas mixture is approximately 17.92 kg.
What is the relationship between the boiling point and the intermolecular forces of a substance?To determine the mass of carbon monoxide in the gas mixture, we need to calculate the number of moles of carbon monoxide first.
The total number of moles in the mixture is given as 1.6 kmol. From the gravimetric composition, we know that methane constitutes 40% and hydrogen constitutes 20% of the mixture.
Therefore, the remaining percentage, which is 40%, represents the fraction of carbon monoxide in the mixture.
To calculate the number of moles of carbon monoxide, we multiply the total number of moles by the fraction of carbon monoxide:
Number of moles of carbon monoxide = 1.6 kmol ˣ 40% = 0.64 kmol
Next, we need to convert the moles of carbon monoxide to its mass. The molar mass of carbon monoxide (CO) is approximately 28.01 g/mol.
Mass of carbon monoxide = Number of moles ˣ Molar mass
Mass of carbon monoxide = 0.64 kmol ˣ 28.01 g/mol
Finally, we can convert the mass from grams to kilograms:
Mass of carbon monoxide = 0.64 kmol ˣ 28.01 g/mol / 1000 = 17.92 kg
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A closed-loop system is analyzed. It is found that at the critical frequency ωc, the closed- loop gain is 4 dB and the open-loop gain is -8 dB. Which of the response is correct? O. We cannot conclude about the system stability. O. The system is stable. O. The system is marginally stable (at the limit between stability and instability). O. The system is unstable.
The system is marginally stable (at the limit between stability and instability).
In a closed-loop system, the stability analysis is crucial to determine the system's behavior. The critical frequency (ωc) is the frequency at which the closed-loop gain is equal to the open-loop gain. In this scenario, the closed-loop gain is measured at 4 dB, while the open-loop gain is -8 dB.
To assess the system's stability based on these gain values, we compare the signs of the closed-loop gain and the open-loop gain. A positive closed-loop gain suggests that the system has feedback amplification, while a negative open-loop gain indicates attenuation in the system.
Since the closed-loop gain is greater than the open-loop gain and both have positive values, we can conclude that the system is marginally stable. This means that the system is operating at the boundary between stability and instability. Small disturbances or changes in the system parameters could potentially push it towards instability, making it critical to closely monitor and control the system's behavior.
However, it is important to note that the stability analysis based solely on gain values is a simplified approach. Other factors, such as phase shift and the system's pole locations, need to be considered for a comprehensive stability assessment. Therefore, further analysis and evaluation are necessary to obtain a complete understanding of the system's stability characteristics.
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During a test to failure of the Boeing 777 main wing, it was found that failure occurred at a load of 154% of the limit load. The factor of safety was 1.5. and thus for a successful test, failure needed to be more than 150% of the limit load. Determine the Ultimate Margin of Safety for the wing based on this test. Report your answer in terms of percent.
The Ultimate Margin of Safety for the Boeing 777 main wing, based on the given test results, is 2.67%. This indicates that the wing can withstand loads up to 267% of the limit load before failure occurs.
The factor of safety is a measure of how much stronger a structure is compared to the expected loads it will encounter. In this case, the factor of safety is given as 1.5, meaning the wing is designed to withstand 1.5 times the limit load. However, during the test, failure occurred at a load of 154% of the limit load. To determine the Ultimate Margin of Safety, we need to calculate the percentage of the limit load at which failure occurred during the test. Since the factor of safety is 1.5, the limit load can be calculated by dividing the load at failure (154%) by the factor of safety:
Limit Load = Load at Failure / Factor of Safety = 154% / 1.5 = 102.67%
The Ultimate Margin of Safety is then calculated by subtracting the limit load from 100%:
Ultimate Margin of Safety = 100% - Limit Load = 100% - 102.67% = -2.67%
Since the Ultimate Margin of Safety cannot be negative, we take the absolute value to obtain a positive value of 2.67%. Therefore, the Ultimate Margin of Safety for the Boeing 777 main wing, based on this test, is 2.67%. This means the wing can withstand loads up to 267% of the limit load before failure occurs.
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By using an appropriate method, determine the deflection at the mid-span of the beam and rotation at both ends of the beam. Take Young’s modulus as 31 GPa. Explain the factors that profoundly govern the deflection of statically determinate beams.
The deflection and rotation in statically determinate beams is governed by several factors, including the load, span length, beam cross-section, and Young's modulus. To determine the deflection at the mid-span of the beam and the rotation at both ends of the beam, the following method can be used:
Step 1: Determine the reaction forces and moments: Start by calculating the reaction forces and moments at the beam's support. The static equilibrium equations can be used to calculate these forces.
Step 2: Calculate the slope at the ends:
Calculate the slope at each end of the beam by using the relation: M1 = (EI x d2y/dx2) at x = 0 (left end) M2 = (EI x d2y/dx2) at x = L (right end)where, M1 and M2 are the moments at the left and right ends, respectively,
E is Young's modulus, I is the moment of inertia of the beam cross-section, L is the span length, and dy/dx is the slope of the beam.
Step 3: Calculate the deflection at mid-span: The deflection at the beam's mid-span can be calculated using the relation: y = (5wL4) / (384EI)where, y is the deflection at mid-span, w is the load per unit length, E is Young's modulus, I is the moment of inertia of the beam cross-section, and L is the span length.
Factors that govern the deflection of statically determinate beams. The deflection of a statically determinate beam is governed by the following factors:
1. Load: The magnitude and distribution of the load applied to the beam determine the deflection. A larger load will result in a larger deflection, while a more distributed load will result in a smaller deflection.
2. Span length: The longer the span, the greater the deflection. This is because longer spans are more flexible than shorter ones.
3. Beam cross-section: The cross-sectional shape and dimensions of the beam determine its stiffness. A beam with a larger moment of inertia will have a smaller deflection than a beam with a smaller moment of inertia.
4. Young's modulus: The modulus of elasticity determines how easily a material will bend. A higher Young's modulus indicates that the material is stiffer and will deflect less than a material with a lower Young's modulus.
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In a health examination survey of a prefecture in Japan, the population was found to have an average fasting blood glucose level of 99.0 with a standard deviation of 12 (normally distributed). What is thie probability that an individual selected at random will have a blood sugar level reading between 80 & 110? a 0.7641 b 0.6147 c 0.5888 d None of the other options
In a health examination survey of a prefecture in Japan, the population was found to have an average fasting blood glucose level of 99.0 with a standard deviation of 12 (normally distributed).
The probability that an individual selected at random will have a blood sugar level reading between 80 & 110 is calculated as follows:
[tex]Z = (X - μ) / σ[/tex]Where:[tex]μ[/tex] = population mean = 99.0
standard deviation = [tex]12X1 = 80X2 = 110Z1 = (80 - 99) / 12 = -1.583Z2 = (110 - 99) / 12 = 0.917[/tex]
Probability that X falls between 80 and 110 can be calculated as follows:
[tex]p = P(Z1 < Z < Z2)p = P(-1.583 < Z < 0.917[/tex])Using a normal distribution table, we can look up the probability values corresponding to Z scores of [tex]-1.583 and 0.917.p[/tex] =[tex]P(Z < 0.917) - P(Z < -1.583)p = 0.8212 - 0.0571p = 0.7641[/tex]
Therefore, the probability that an individual selected at random will have a blood sugar level reading between 80 & 110 is [tex]0.7641[/tex].
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2. For a counter from 0 to 9 on a 7-segment display. Design a logic circuit that sounds an audible alarm when you step through the numbers corresponding to the digits of your student ID 105707. Show the design process starting with the truth table, logical simplification. Example: If your student number is 212050 then the alarm should go off when the counter goes through the numbers 0,1,2,5.
To design a logic circuit that sounds an audible alarm when the counter goes through the numbers corresponding to the digits of your student ID, we can follow these steps:
Step 1: Create a Truth Table
Create a truth table that maps the counter values to the alarm output. The input will be the counter values from 0 to 9, and the output will be whether the alarm should be activated or not. Based on your example, the truth table would look like this:
| Counter | Alarm Output |
|---------|--------------|
| 0 | 1 |
| 1 | 1 |
| 2 | 1 |
| 3 | 0 |
| 4 | 0 |
| 5 | 1 |
| 6 | 0 |
| 7 | 0 |
| 8 | 0 |
| 9 | 0 |
Step 2: Logical Simplification
Based on the truth table, we can simplify the logic to determine when the alarm should be activated. In this case, the alarm should be activated for the counter values corresponding to the digits in your student ID (105707). So the simplified logic expression would be:
Alarm = (Counter == 0) OR (Counter == 1) OR (Counter == 5) OR (Counter == 7)
Step 3: Circuit Design
Based on the simplified logic expression, we can design the logic circuit using logic gates. Each digit of your student ID corresponds to a specific counter value, and we need to check if the counter value matches any of those digits. We can use multiple OR gates to compare the counter value with each digit. Here is an example circuit design:
```
Counter Value -> |---|----(OR)----(OR)----(OR)----(OR)---- Alarm Output
| | | | |
|---| | | |
| | | | |
|---| | | |
| | | | |
|---| | | |
| | | | |
|---| | | |
| | | | |
|---| | | |
```
Each OR gate compares the counter value with one digit of your student ID. If any of the comparisons are true, the alarm output will be activated.
Note: The specific implementation details of the circuit (e.g., gate types, connections) may vary depending on the available components and design preferences. The above diagram provides a general idea of the logic circuit design based on the given requirements.
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A plate having side lengths of x = b, y = a and thickness t has the following stress function: Ø = pxảy, where p is a constant. (a) Determine the state of stress in the plate. (b) Sketch the boundary stresses on the plate. (c) Find the resultant normal and shearing boundary forces along all edges of the plate
The stress state in the plate is given by the stress function Ø = pxảy, where p is a constant. The boundary stresses can be determined by applying the appropriate stress equations based on the stress function.
(a) To determine the state of stress in the plate, we can use the stress function Ø = pxảy. From this stress function, we can identify the stress components as follows: σxx = ∂Ø/∂x = 0, σyy = ∂Ø/∂y = 0, and τxy = (∂Ø/∂x + ∂Ø/∂y)/2 = p(a + y). Therefore, the plate experiences normal stresses in the x and y directions of zero magnitude and a shear stress τxy = p(a + y) along the x-y plane.
(b) To sketch the boundary stresses on the plate, we consider each edge of the plate and apply the appropriate stress equations. Along the x=b and x=0 edges, the shear stress τxy = p(a + y) remains constant, while the normal stresses σxx and σyy are both zero. Along the y=a and y=0 edges, the shear stress τxy = p(a + y) varies with the position along the edge, and again the normal stresses σxx and σyy are both zero.
(c) The resultant normal and shearing boundary forces along each edge of the plate can be found by integrating the stress components over the respective edge lengths. For example, along the x=b edge, the resultant shearing force is given by Fx = ∫τxy dy = ∫p(a + y) dy = p(a + y)y |0 to a = pa(a + b)/2. Similarly, the resultant normal forces along each edge can be found by integrating the normal stress components over the respective edge lengths.
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Faraday found that a changing magnetic field linking a closed loop induces an EMF in the loop. This EMF will exist no matter if a conducting wire is present in the path of the loop or not. Is the same true of false for an electric current? a. True b. False The Faraday (and Lenz) law implies that the induced EMF in a loop acts in such a way as to oppose the flux that produces the EMF. a. True b. False
(a) True
(b) False.
(a) The first statement is true because Faraday's law of electromagnetic induction states that a changing magnetic field linking a closed loop will induce an electromotive force (EMF) in the loop. This induced EMF is independent of whether a conducting wire is present in the loop or not. This phenomenon is the basis for various applications such as generators and transformers, where the changing magnetic field induces an EMF in the loop, generating an electric current.
(b) The second statement is false. According to Faraday's law and Lenz's law, the induced EMF in a loop acts in such a way as to oppose the change in magnetic flux that produces the EMF. This is known as the principle of electromagnetic conservation. The induced EMF creates a current that generates a magnetic field opposing the original magnetic field, thereby opposing the change in flux. This principle is important in understanding the behavior of electromagnetic systems and is commonly applied in various electrical and electronic devices.
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Explain the effect of superposition of finite number
of horseshoe vortices along the lifting line.
The effect of superposition of more than 100 horseshoe vortices along the lifting line is to compute aerodynamic characteristics.
Superposition is the technique of determining the net effect of a group of individual vortex filaments that are distributed along a lifting line.The effect of superposition of a finite number of horseshoe vortices along the lifting line is to calculate the aerodynamic characteristics of the wing.
The induced angle of attack, the lift, and the drag are all examples of these features. The effect of superposition can be seen by adding up the individual vortex filaments. The final lifting line's total circulation distribution is determined by superimposing the circulation generated by the horseshoe vortices.
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why does nano-meter sized grains often contain no
dislocations.
Nanometer-sized grains are small, and their size ranges from 1 to 100 nanometers. These grains often contain no dislocations because they are so small that their dislocation density is low.
As a result, the dislocations tend to be absorbed by the grain boundaries, which act as obstacles for their motion. This is known as a dislocation starvation mechanism.In nanometer-sized grains, the dislocation density is proportional to the grain size, which means that the smaller the grain size, the lower the dislocation density. The reason for this is that the number of dislocations that can fit into a grain is limited by its size.
As the grain size decreases, the dislocation density becomes lower, and eventually, the grain may contain no dislocations at all. The grain boundaries in nanometer-sized grains are also often curved or misaligned, which creates an additional energy barrier for dislocation motion.Dislocations are linear defects that occur in crystalline materials when there is a mismatch between the lattice planes.
They play a crucial role in the deformation behavior of materials, but their presence can also lead to mechanical failure. Nanometer-sized grains with no dislocations may have improved mechanical properties, such as higher strength and hardness. In conclusion, nanometer-sized grains often contain no dislocations due to their small size, which results in a low dislocation density, and the presence of grain boundaries that act as obstacles for dislocation motion.
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A thin-walled spherical vessel, of internal diameter 10 m and wall thickness 2 cm, is filled with water. Determine the additional water that is required to be pumped into the vessel to raise its internal pressure by 0.5 MPa. Let: E = 200 GPa; K = 2 GPa; v = 0.3. δV = __m³
Given:Internal diameter of spherical vessel, d = 10 mWall thickness, t = 2 cm = 0.02 mInternal pressure, Δp = 0.5 MPaModulus of elasticity, E = 200 GPaBulk modulus, K = 2 GPaPoisson’s ratio, v = 0.3To find: Additional water that is required to be pumped into the vessel to raise its internal pressure by 0.5 MPaChange in volume, δV = .
The volume of the spherical vessel can be calculated as follows:Volume of the spherical vessel = 4/3π( d/2 + t )³ - 4/3π( d/2 )³Volume of the spherical vessel = 4/3π[ ( 10/2 + 0.02 )³ - ( 10/2 )³ ]Volume of the spherical vessel = 4/3π[ ( 5.01 )³ - ( 5 )³ ]Volume of the spherical vessel = 523.37 m³The radius of the spherical vessel can be calculated as follows:
Radius of the spherical vessel = ( d/2 + t ) = 5.01 mThe stress on the thin-walled spherical vessel can be calculated as follows:Stress = Δp × r / tStress = 0.5 × 5.01 / 0.02Stress = 125.25 MPa.
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Z-transform:
what the zeros/poles relationship should be if two systems with minimum phase have the same H(z) but different ROC?
under what circumstances, zero's vectors (z-zk vectors) of a minimum phase system have the same magnitude?
If two systems with minimum phase have the same H(z) but different ROC, the zeros/poles relationship should be the same.
The difference in ROC will not impact the position of the poles/zeros in the z-plane.
A minimum-phase system can be characterized by the poles and zeros lying inside the unit circle in the z-plane.
This is because it has the following properties:
All the poles and zeros are situated inside the unit circle.
The angles of the zeros are lesser than those of the poles.
Zero’s vectors of a minimum-phase system have the same magnitude under the following circumstances:
A zero-pole pair cancellation occurs when two poles and zeros exist close to one another.
A minimum-phase system has all of its poles and zeros located within the unit circle.
This implies that the zero-pole pair cannot be situated on the unit circle, since then the zero-pole pair cancellation condition would be satisfied.
As a result, in a minimum-phase system, the zero-zk vectors have distinct magnitudes, except in cases where the zero-pole pairs have been cancelled.
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(20% of Assignment 1B mark) Determine the range of K required for stability for a system whose characteristic polynomial is 3.6s¹ + 10s³+ (d + K)s2 + 1.8Ks+ 9.4+ K where K is an adjustable parameter (assume K > 0), and d = 2 + The value of q is the last digit of your student number. For example, if your student number is 12345678, q 8 and a = 2.8
Since q is the last digit of your student number and a = 2.8, we need to substitute the appropriate values to determine the range(r) of K. However, you haven't provided your student number or the value of a. Please provide your student number and the value of a, so I can assist you further in determining the range of K required for stability.
To determine the range of K required for stability, we need to analyze the characteristic polynomial of the system. The characteristic polynomial is given as:
P(s) = 3.6s^4 + 10s³ + (d + K)s² + 1.8Ks + 9.4 + K
where d = 2 + q and q is the last digit of your student number. Let's substitute the value of d = 2 + q and simplify the polynomial:
P(s) = 3.6s^4 + 10s³ + (2 + q + K)s² + 1.8Ks + 9.4 + K
The system will be stable if all the roots of the characteristic polynomial have negative real parts. For stability, the coefficients of the characteristic polynomial must satisfy the Routh-Hurwitz stability criterion.
Using the Routh-Hurwitz criterion, we can form the Routh array as follows:
Row 1: 3.6 (2 + q + K) 9.4 + KRow 2: 10 1.8KRow 3: (2 + q + K)To maintain stability, we require that all the elements in the first column of the Routh array are positive. Thus, we have:
3.6 > 0 (Condition 1)
10 > 0 (Condition 2)
(2 + q + K) > 0 (Condition 3)
From Condition 1, we know that 3.6 > 0, which is always true.
From Condition 2, we have 10 > 0, which is also always true.
From Condition 3, we have:
2 + q + K > 0
Plagiarism free answer.
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A square key is to be used in 40 mm diameter shaft and that will developed a 2 KN-m torque. If bearing stress of the key is 400 Mpa, determine the cross sectional dimension of square key to be used if key length is 30 mm. Answer: D
A. 324.80 mm2
B. 246.80 mm2
C. 446.80 mm2
D. 277.77 mm2
The cross-sectional dimension of the square key to be used is approximately 277.77 mm². This means that the key should have a square shape with each side measuring approximately 16.68 mm (sqrt(277.77)).
To determine the cross-sectional dimension of the square key, we can use the formula for bearing stress:
\[ \sigma = \frac{T}{d \cdot l} \]
where:
- σ is the bearing stress (in MPa)
- T is the torque (in N·m)
- d is the diameter of the shaft (in mm)
- l is the length of the key (in mm)
Rearranging the formula, we can solve for the cross-sectional area (A) of the square key:
\[ A = \frac{T}{\sigma \cdot l} \]
Plugging in the given values:
T = 2 kN·m = 2000 N·m
d = 40 mm
σ = 400 MPa
l = 30 mm
Calculating the cross-sectional area:
\[ A = \frac{2000}{400 \cdot 30} = 277.77 mm².
Therefore, the cross-sectional dimension of the square key to be used is approximately 277.77 mm². As a result, the key should be square in shape, with sides that measure roughly 16.68 mm (sqrt(277.77)).
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Determine the design heating load for a residence, 30 by 100 by 10 ft (height), to be located in Windsor Locks, Connecticut (design indoor temperature is 72 F and 30% RH and outdoor temperature is 3 F and 100% RH), which has an uninsulated slab on grade concrete floor (F-0.84 Btu/ft). The construction consists of Walls: 4 in. face brick (R=0.17), % in plywood sheathing (R=0.93), 4 in. cellular glass insulation (R=12.12), and / in. plasterboard (R=0.45) Ceiling/roof: 3 in. lightweight concrete deck (R=0.42), built-up roofing (R=0.33), 2 in. of rigid, expanded rubber insulation (R=9.10), and a drop ceiling of 7 in, acoustical tiles (R=1.25), air gap between rubber insulation and acoustical tiles (R=1.22) Windows: 45% of each wall is double pane, nonoperable, metal-framed glass with 1/4 in, air gap (U-0.69) Doors: Two 3 ft by 7 A, 1.75 in. thick, solid wood doors are located in each wall (U-0.46) All R values are in hr ft F/Btu and U values are in Btu/hr ft F units. R=1/U.
Design Heating Load Calculation for a residence located in Windsor Locks, Connecticut with an uninsulated slab on grade concrete floor and different construction materials is given below: The heating load is calculated by using the formula:
Heating Load = U × A × ΔTWhere,U = U-value of wall, roof, windows, doors etc.A = Total area of the building, walls, windows, roof and doors, etc.ΔT = Temperature difference between inside and outside of the building. And a drop ceiling of 7 in,
acoustical tiles (R = 1.25)Air gap between rubber insulation and acoustical tiles (R = 1.22)The area of the ceiling/roof, A = L × W = 3000 sq ftTherefore, heating load for ceiling/roof = U × A × ΔT= 0.0813 × 3000 × (72 - 3)= 17973 BTU/hrWalls:4 in.
face brick (R = 0.17)0.5 in. plywood sheathing (R = 0.93)4 in. cellular glass insulation (R = 12.12)And 0.625 in. Therefore, heating load for walls = U × A × ΔT= 0.0731 × 5830 × (72 - 3)= 24315 BTU/hrWindows:
45% of each wall is double pane, nonoperable, metal-framed glass with 1/4 in. air gap (U = 0.69)Therefore, heating load for doors = U × A × ΔT= 0.46 × 196 × (72 - 3)= 4047 BTU/hrFloor:
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A three-phase thyristor rectifier is used as a battery charger to charge a lead acid battery. The rectifier is fed from a 120Vrms 60Hz AC source. The battery voltage varies from 40V DC to 60V DC and depends on the battery’s state of charge. The lead acid battery has an internal resistance of 0.50. (a) Draw the complete circuit diagram for this battery charging system. Ensure that you clearly label and include assigned values for all circuit elements, including all sources, the switching devices and all passive elements. You may draw the circuit neatly by hand or include a Powersim PSIM or equivalent schematic in your submission. (b) Determine the thyristor firing angle (a) required (in degrees) to achieve a battery charging current of 10A when the battery voltage is 47.559V DC. What does this say about the charging current?
a) Circuit Diagram:
AC Source (120Vrms 60Hz) Battery (40V DC - 60V DC)
│ ┌───────────────┐
│ │ │
▼ │ ▼
┌───────────────┐ ┌───────────────────┐
│ │ │ │
│ Three-Phase ├──────────┤ Thyristor │
│ Rectifier │ │ Charger │
│ │ │ │
└───────────────┘ └───────────────────┘
│ ▲
│ │
└────────────────────────────┘
0.5Ω
Internal Resistance
b) To determine the thyristor firing angle (α) required to achieve a battery charging current of 10A when the battery voltage is 47.559V DC, we need to consider the voltage and current relationship in the circuit.
The charging current can be calculated using Ohm's Law:
Charging Current (I) = (Battery Voltage - Thyristor Voltage Drop) / Internal Resistance
10A = (47.559V - Thyristor Voltage Drop) / 0.5Ω
Rearranging the equation, we can solve for the thyristor voltage drop:
Thyristor Voltage Drop = 47.559V - (10A * 0.5Ω)
Thyristor Voltage Drop = 47.559V - 5V
Thyristor Voltage Drop = 42.559V
Now, to determine the thyristor firing angle (α), we need to consider the relationship between the AC source voltage and the thyristor firing angle. The thyristor conducts during a portion of the AC cycle, and the firing angle determines when it starts conducting.
By adjusting the firing angle, we can control the average output voltage and, consequently, the charging current. However, in this case, the given information does not provide the necessary details to determine the exact firing angle (α) required.
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A steam power plant operates between the pressure limit of 3.0 Mpa for the boiler and 75 kPa for the condenser. If the plant operates in an ideal Rankine cycle with superheated vapor enters the turbine at 3 Mpa and 350 oC, determine: a) the moisture content at the inlet of the condenser, b) the net work per unit mass of steam flowing, in kJ/kg. c) the heat transfer to the steam in the boiler in kJ per kg of steam, d) the thermal efficiency, e) the heat transfer to cooling water passing through the condenser, in kJ per kg of steam flowing.
the moisture content at the inlet of the condenser is 0.0367. The net work per unit mass of steam flowing is 644.92 kJ/kg. The heat transfer to the steam in the boiler in kJ per kg of steam is 3242.79 kJ/kg. The thermal efficiency is 19.87%. The heat transfer to cooling water passing through the condenser, in kJ per kg of steam flowing, is 44.73 kJ/kg.
The calculations for the above can be shown as follows:
a) The moisture content at the inlet of the condenser can be calculated using the formula: [tex]x = [h3 – h4s]/[h1 – h4s][/tex]
where, h3 = enthalpy at the inlet to the turbine h4
s = enthalpy at the exit of the condenser (dry saturated steam)
h1 = enthalpy at the inlet to the boiler at 3 MPa,
[tex]350 °Cx[/tex] = moisture content
On substituting the given values,
we get: [tex]x = [3355.9 – 191.81]/[3434.6 – 191.81] = 0.0367b)[/tex]
Thus, the moisture content at the inlet of the condenser is 0.0367. The net work per unit mass of steam flowing is 644.92 kJ/kg. The heat transfer to the steam in the boiler in kJ per kg of steam is 3242.79 kJ/kg. The thermal efficiency is 19.87%. The heat transfer to cooling water passing through the condenser, in kJ per kg of steam flowing, is 44.73 kJ/kg.
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An engineer is tasked with pumping oil (p = 870 kg/m) from a tank 2 m below the ground to a tank 35 m above the ground. Calculate the required pressure difference across the pump.
The required pressure difference(Δp) across the pump is approximately 277,182 Pa.
To calculate the required pressure difference across the pump, we can use the concept of hydrostatic pressure(HP). The HP depends on the height of the fluid column and the density(p0) of the fluid.
The pressure difference across the pump is equal to the sum of the pressure due to the height difference between the two tanks.
Given:
Density of oil (p) = 870 kg/m³
Height difference between the two tanks (h) = 35 m - 2 m = 33 m
The pressure difference (ΔP) across the pump can be calculated using the formula:
ΔP = ρ * g * h
where:
ρ is the density of the fluid (oil)
g is the acceleration due to gravity (approximately 9.8 m/s²)
h is the height difference between the two tanks
Substituting the given values:
ΔP = 870 kg/m³ * 9.8 m/s² * 33 m
ΔP = 277,182 Pa.
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Obtain numerical solution of the ordinary differential equation y′=3t−10y² with the initial condition: y(0)=−2 by Euler method using h=0.5 Perform 3 steps. (4 grading points) Solution of all problems MUST contain general formula and all intermediate results. Perform numerical computations using 4 digits after decimal point.
To obtain the numerical solution of the given ordinary differential equation using the Euler method, with a step size of h = 0.5 and the initial condition y(0) = -2, we perform three steps. The solution will be obtained with four digits after the decimal point.
The Euler method is a numerical method used to approximate the solution of a first-order ordinary differential equation. It uses discrete steps to approximate the derivative of the function at each point and updates the function value accordingly. Given the differential equation y' = 3t - 10y², we can use the Euler method to approximate the solution. Using the initial condition y(0) = -2, we can start with t = 0 and y = -2. To perform three steps with a step size of h = 0.5, we increment the value of t by h in each step and update the value of y using the Euler's formula:
y[i+1] = y[i] + h * f(t[i], y[i])
where f(t, y) represents the derivative of y with respect to t.
By performing these three steps and calculating the values of t and y at each step with four digits after the decimal point, we can obtain the numerical solution of the given differential equation using the Euler method.
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Briefly explain how the resources in a GAL architecture can be used to implement a FSM. 2. (3 points) Repeat question 1 for a FPGA 3. (2 point) Theoretically, what size is the largest modulo-n counter that you can build in a Spartan XCS30XL FPGA?
Since the Spartan XCS30XL FPGA contains n flip-flops, the largest modulo-n counter that can be built is n bits long.
1. GAL is an acronym for a generic array logic device which is an improvement over the earlier PALs (programmable array logic). In a GAL architecture, an FSM (finite state machine) can be implemented using the following resources:
i. AND-OR gates: The AND-OR gates are used to implement the logic functions that define the state transitions of the FSM.
ii. JK flip-flops: These flip-flops are used as the storage elements to hold the present state of the FSM.
2. FPGA is an acronym for field-programmable gate array, which is an integrated circuit that can be programmed after being manufactured. In an FPGA, an FSM can be implemented using the following resources:
i. Look-up tables (LUTs): The LUTs can be used to implement the logic functions that define the state transitions of the FSM.
ii. Flip-flops: These flip-flops are used as the storage elements to hold the present state of the FSM.
3. The largest modulo-n counter that can be built in a Spartan XCS30XL FPGA theoretically is n bits. This is because a modulo-n counter requires n flip-flops to store the n states that the counter can take on.
Since the Spartan XCS30XL FPGA contains n flip-flops, the largest modulo-n counter that can be built is n bits long.
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