The Net Positive Suction Head (NPSH) is a crucial parameter for centrifugal pump operation, representing the pressure available at the suction side to push the liquid into the pump and prevent cavitation.
The Net Positive Suction Head (NPSH) is a fundamental parameter used to determine the operating conditions and performance of a centrifugal pump. It represents the absolute pressure head available at the suction side of the pump, taking into account both the pressure exerted by the liquid and the vapor pressure of the fluid being pumped. In simple terms, it indicates how much pressure is available to push the liquid into the pump.
When a centrifugal pump operates, it creates a low-pressure zone at the suction inlet, which causes the liquid to flow towards the impeller. However, if the pressure at the suction side falls below a certain value, known as the Net Positive Suction Head Required (NPSHR), the liquid may start to vaporize or form bubbles. This phenomenon is called cavitation and can have detrimental effects on the pump's performance and lifespan.
The NPSH is calculated using the following equation:
NPSH = (P - Pv) / ρg
where:
P is the pressure at the pump suction,
Pv is the vapor pressure of the liquid being pumped,
ρ is the density of the liquid, and
g is the acceleration due to gravity.
Adequate NPSH is crucial to prevent cavitation and maintain optimal pump operation. Insufficient NPSH can lead to decreased pump efficiency, loss of flow rate, increased vibration and noise, erosion or damage to impellers, and even complete pump failure. Therefore, pump manufacturers specify the minimum NPSHR required for their pumps, and it is essential for operators to ensure that the NPSH available exceeds this value to avoid cavitation-related issues.
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For each of the transfer functions below, find the exact response of each system to a step input, using Laplace transform techniques.
a. T(s) = (s+3)(s+6) 10(s+7)
b. T(s) (s+10) (s+20) 20 c. T(s) s²+6s+144 s+2 d. T(s) s²+9 e. T(s) = s+5 (s+10)²
Step-by-step solutions for the given transfer functions are as follows a. T(s) = (s+3)(s+6) 10(s+7)For this transfer function, the response of the system to a step input can be obtained by using the following steps.
After obtaining the values of A, B, and C, the inverse Laplace of the transfer function will be as follows'(t) By putting the given values of A, B, C, and y(0), we get the exact response of the system to a step input as follows:
y(t) = (0.0833 e⁻⁷ᵗ) - (0.0268 e⁻³ᵗ) + (0.9435 e⁻⁶ᵗ) b.
T(s) (s+10) (s+20) 20For this transfer function, the response of the system to a step input can be obtained by using the following steps firstly, we need to convert the transfer function to a time domain function by taking the inverse Laplace transform.
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Which gate has its output equal 0 if and only if both inputs are 0 Select one: a. \( \mathrm{OR} \) b. AND c. NOT d. NAND
d. NAND gates have their output equal to 0 if and only if both inputs are 0; for all other input combinations, the output is 1.
The NAND gate, short for "NOT-AND," is a logic gate that performs the combination of an AND gate followed by a NOT gate. It has two inputs and one output. The output of a NAND gate is the logical negation of the AND operation performed on its inputs.
In the case of the NAND gate, if both inputs are 0 (logic low), the AND operation results in 0. Since the NAND gate also performs a logical negation, the output becomes 1 (logic high). However, for any other combination of inputs (either one or both inputs being 1), the AND operation results in 1, and the NAND gate's logical negation flips the output to 0.
The NAND gate has an output equal to 0 only when both of its inputs are 1. In all other cases, when at least one input is 0 or both inputs are 0, the NAND gate produces an output of 1. Therefore, the NAND gate has its output equal to 0 if and only if both inputs are 0.
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Question 6 (1 point) Listen If the rest of the sketch is correct, what will we see in the serial monitor when the following portion is executed (assuming there is no outer loop)? int x = 5; int y = 2; do { y = y + x; Serial.print(y); Serial.print(" "); } while(y > x && y < 22); // y is bigger than x and smaller than 22 O 7 12 17 O 27 12 17 O [Nothing. The program never enters this loop.] O 712 17 22
If the rest of the sketch is correct the thing that one see in the serial monitor when the following portion is executed is O 7 12 17
What is the loopA "do while" loop is a feature in computer programming that lets a section of code run over and over again until a certain condition is met. The do while method has a step and a rule.
Therefore, The do-while loop will keep going if y is greater than x and less than 22. At first, x equals 5 and y equals 2. The loop will run at least one time because the condition is true. In the loop, y gets bigger by adding x to it (y = y + x). This means that y becomes 7 the first time it's done.
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A high speed rotating machine weighs 1500 kg and is mounted on Insulator Springs with negligible mass. The static deflection of the springs as a result of the weight of the machine is 0.4 mm. The rotating part is unbalanced such that its equivalent unbalanced mass is 2.5 kg mass located at 500 mm from the axis of rotation. If the rotational speed of the machine is 1450 rpm I determine: a) The stiffness of the springs in N/m. b) The vertical vibration undamped natural frequency of the machine spring system, in rad/sec and Hz. c) The machine angular velocity in rad/s and centrifugal force in N resulting from the rotation of the unbalanced mass when the system is in operation
A rotating machine is mounted on insulator springs with negligible mass, and it weighs 1500 kg. As a result of the machine's weight, the static deflection of the springs is 0.4 mm.
The machine's rotating part is unbalanced such that the equivalent unbalanced mass is 2.5 kg mass located at 500 mm from the axis of rotation. If the rotational speed of the machine is 1450 rpm, the following items can be determined:
a) The stiffness of the springs in N/m.
b) The vertical vibration undamped natural frequency of the machine spring system, in rad/sec and Hz.
c) The machine angular velocity in rad/s and centrifugal force in N resulting from the rotation of the unbalanced mass when the system is in operation.
Given,Weight of machine, W = 1500 kg;Equivalent unbalanced mass, m = 2.5 kg;
Unbalanced mass eccentricity, e = 500 mm;
Rotational speed of machine, N = 1450 rpm = 1450/60 rad/s = 24.17 rad/s;
Static deflection of spring, δ = 0.4 mm = 0.4 × 10⁻³ m.
a) Stiffness of spring can be determined as;δ = W/k ⇒ k = W/δ = 1500/(0.4 × 10⁻³) = 3.75 × 10⁶ N/m.∴ The stiffness of the springs in N/m is 3.75 × 10⁶.
b) The natural frequency of a spring mass system is given as;f₀ = (1/2π) √(k/m) rad/s.f₀ = (1/2π) √(3.75 × 10⁶ /1500 + 2.5) = 11.38 rad/s.∴ The vertical vibration undamped natural frequency of the machine spring system is 11.38 rad/s and,Hz = f₀/2π = 1.81 Hz.
c) The angular velocity of the rotating mass is given as;ω = 2πN/60 rad/s.ω = 2π(1450)/60 = 241.02 rad/s.The centrifugal force due to the unbalanced mass can be calculated using the formula;
F = mω²e F = 2.5 × (241.02)² × 0.5 = 1.44 × 10⁵ N.
∴ The machine angular velocity in rad/s is 241.02 rad/s and the centrifugal force in N resulting from the rotation of the unbalanced mass when the system is in operation is 1.44 × 10⁵ N.
Therefore, the stiffness of the springs in N/m is 3.75 × 10⁶, the vertical vibration undamped natural frequency of the machine spring system is 11.38 rad/s and 1.81 Hz and, the machine angular velocity in rad/s is 241.02 rad/s and the centrifugal force in N resulting from the rotation of the unbalanced mass when the system is in operation is 1.44 × 10⁵ N.
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Consider a titanium alloy having shear modulus (modulus of rigidity, G=44,44 GPa). Calculate the shear stress, If a structure made of that material is subjected to an angular deformation a = 0.2º.
Select one: a. T = 17.21 MPa b. T = 80.43 MPa
c. T = 155.12 MPa d. T=40.11 MPa e. T-77.56 MPa
The shear stress in the titanium alloy is calculated to be 17.21 MPa when subjected to an angular deformation of 0.2º.
What is the significance of the Hubble Space Telescope in the field of astronomy and space exploration?To calculate the shear stress, we can use the formula:
Shear Stress (T) = Shear Modulus (G) * Angular Deformation (a)
Given that the shear modulus (G) is 44.44 GPa and the angular deformation (a) is 0.2º, we can substitute these values into the formula:
T = 44.44 GPa * 0.2º
To calculate the shear stress in MPa, we need to convert the shear modulus from GPa to MPa by multiplying it by 1000:
T = (44.44 GPa * 1000 MPa/GPa) * 0.2º
T = 44,440 MPa * 0.2º
T = 8,888 MPa * 0.2º
T = 1,777.6 MPa
Therefore, the shear stress is approximately 1,777.6 MPa. However, none of the given options match this value.
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Problem 2 Design a full return (fall) polynomial cam that satisfies the following boundary conditions (B.C): At 0=0°, y= h, y'= 0,4" = 0 = At 0= 5, y = 0, y = 0,4" = 0
A full return polynomial cam that satisfies the given boundary conditions can be designed by utilizing a suitable polynomial equation. The cam profile will have a height of 'h' at 0° with a slope of zero, and it will return to a height of zero at 5° with a slope of zero.
To design a full return polynomial cam, we can use a polynomial equation of the form y = a0 + a1θ + a2θ^2 + a3θ^3 + a4θ^4, where 'y' represents the cam height and 'θ' represents the angle of rotation. The coefficients 'a0', 'a1', 'a2', 'a3', and 'a4' need to be determined based on the given boundary conditions. At 0°, the cam height is 'h' and the slope is zero, which means y = h and y' = 0. Taking the derivative of the polynomial equation, we get y' = a1 + 2a2θ + 3a3θ^2 + 4a4θ^3. Setting θ = 0, we have a1 = 0. Since the slope should be zero, we can set a2 = 0 as well. At 5°, the cam height is zero and the slope is zero. Substituting θ = 5 and y = 0 into the polynomial equation, we get 0 = a0 + 25a3 + 625a4. To satisfy the condition y' = 0 at θ = 5, we take the derivative of the polynomial equation and set it to zero. This leads to a3 = -16a4. By solving these equations simultaneously, we can determine the values of the coefficients. With these coefficients, we can generate the cam profile that meets the given boundary conditions of returning to a height of zero at 5° with a slope of zero.
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A rigid tank contains 6 kg of saturated vapor steam at 100°C. The steam is cooled to the ambient temperature of 25°C. Determine the entropy change of the steam, in kJ/K. Use steam tables.
The entropy change of the steam is ___kJ/K
Given data are:Mass of steam m = 6kgTemperature of steam T1 = 100 °CTemperature of surrounding T2 = 25°CWe need to find entropy change of steam ∆S
.From steam table, we have:At 100°C, saturation pressure P1 = 1.013 bar Specific enthalpy of saturated vapour h1 = 2676.5 kJ/kgSpecific entropy of saturated vapour s1 = 6.828 kJ/kg KAt 25°C, saturation pressure P2 = 0.031 bar Specific enthalpy of saturated vapour h2 = 2510.1 kJ/kgSpecific entropy of saturated vapour s2 = 8.785 kJ/kg KThe entropy change of the steam is -0.116 kJ/K
In order to find the entropy change of steam, we will use the entropy formula. The entropy change of the steam can be calculated using the following formula:∆S = m * (s2 - s1)Where,m = Mass of steam = 6 kg.s1 = Specific entropy of saturated vapour at temperature T1.s2 = Specific entropy of saturated vapour at temperature T2.s1 and s2 values are obtained from steam tables.At 100°C,s1 = 6.828 kJ/kg KAt 25°C,s2 = 8.785 kJ/kg KNow, substituting the values in the formula, we get∆S = 6 * (8.785 - 6.828) = -0.116 kJ/KSo, the entropy change of the steam is -0.116 kJ/K.
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The entropy change of the steam is -40.902 kJ/K
How to determine the entropy changeUsing the steam tables, we have that the specific entropy values are;
At 100°C, the specific entropy of saturated vapor steam is s₁= 7.212 kJ/(kg·K).
At 25°C, the specific entropy of saturated liquid water is s₂= 0.395 kJ/(kg·K).
The formula for entropy change (Δs) is given as;
Δs = s₂ - s₁
Substitute the values from the steam table, we get;
Δs = 0.395 - 7.212
subtract the values
Δs = -6.817 kJ/(kg·K)
To calculate the total entropy change, we have;
Entropy change = Δs × mass
= -6.817 kJ/(kg·K) × 6 kg
Multiply the values
= -40.902 kJ/K
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A trapezoidal channel of bed width 10.0 m, side slope 3:2, longitudinal bed slope 10 cm/km, mean velocity 0.594 m/s, and Manning's coefficient 0.025. Determine: a) The average boundary shear stress acting on the channel wetted perimeter. b) The maximum boundary shear stress on the bed and sides. c) If the mean diameter of the material forming the channel bed and sides is 0.4 mm and the angle of repose is 35º, what is the maximum discharge that can pass in this channel without causing scour?
Bed width = 10.0 m Side slope = 3:2Longitudinal bed slope = 10 cm/km Mean velocity = 0.594 m/s Manning's coefficient = 0.025The formula for average boundary shear stress is:τb = (γ × R × S) / nwhere,γ = unit weight of waterR = hydraulic radius S = longitudinal bed slope n = Manning's coefficienta) The calculation of average boundary shear stress:
We can find the hydraulic radius using the given data. It is given by:R = (A / P)Where A is the cross-sectional area of the flow and P is the wetted perimeter of the channel. Here, the channel is trapezoidal. Therefore, A can be calculated using the formula:A = (b1 + b2) / 2 × ywhere b1 and b2 are the bottom widths of the trapezoidal channel and y is the depth of flow. P can be calculated using the formula:P = b1 + b2 + 2 × (y / sinθ)where θ is the angle between the horizontal and the side slope. Using the given data, we have:b1 = 10.0 mb2 = 3/2 × 10.0 = 15.0 my/s = 0.594 m/sn = 0.025S = 10 cm/kmγ = 9.81 kN/m³Now, we can use the values to calculate R as follows:Depth of flow:y = (4 / 3) × (b1 + b2) / (2 + 3) = 6.86 mCross-sectional area:A = (10.0 + 15.0) / 2 × 6.86 = 96.78 m²Wetted perimeter:P = 10.0 + 15.0 + 2 × (6.86 / sin(53.13º)) = 41.22 m Hydraulic radius:R = 96.78 / 41.22 = 2.345 mNow, we can calculate the average boundary shear stress.τb = (γ × R × S) / nτb = (9.81 × 2.345 × 0.1) / 0.025τb = 93.99 N/m²Therefore, the average boundary shear stress is 93.99 N/m².b) The calculation of the maximum boundary shear stress:We can use the following formula to calculate the maximum boundary shear stress:τmax = τb × Kcwhere Kc is the coefficient of contraction and its value is usually between 0.2 and 0.6.
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2) A piston-cylinder assembly is thermally insulated. There is saturated liquid water at 120°C with a mass of 1.8 kg. An electrical heater is placed inside of the cylinder and then switched on for a duration of 10 minutes. During this time, the volume of the cylinder increases by four times. Assume the piston is allowed to move while keeping pressure constant, also ignore kinetic and potential energies. Find the volume of the cylinder, the temperature of the final state, and the electrical power rating of the heater in kW.
Given:Mass of saturated liquid water = 1.8 kgInitial temperature of the water = 120°C The cylinder is thermally insulated.The piston is allowed to move while keeping the pressure constant.
The volume of the cylinder increases four times in 10 minutes.Ignore kinetic and potential energies.Now,The initial condition can be determined using the saturation table, we find the specific volume of saturated liquid water v1= 0.001074 m3/kg.
The initial volume of water in the cylinder will be V1 = m/v1 = 1.8/0.001074 = 1674.77 cm3 = 1.67477 LThe volume of the cylinder during the process is 4 V1 = 6.699 LFrom the steam tables, we find the saturation temperature at the final volume (V2 = 6.699 L) and find it to be 193.65°C.So, 193.65°C is the final temperature.
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Five miners must be lifted from a mineshaft (vertical hole) 100m deep using an elevator. The work required to do this is found to be 341.2kJ. If the gravitational acceleration is 9.75m/s^2, determine the average mass per person in kg.
The average mass per person in kg is given by;First, we will calculate the gravitational potential energy as;Gravitational potential energy = mass × g × h341.2 × 1000 = mass × 9.75 × 100
mass = (341.2 × 1000) / (9.75 × 100)mass = 350.26 kg
Therefore, the average mass per person in kg is 70.05 kg.
The problem requires the determination of the average mass per person in kg when five miners must be lifted from a mineshaft (vertical hole) 100m deep using an elevator given that the work required to do this is found to be 341.2kJ, and the gravitational acceleration is 9.75m/s^2. The gravitational potential energy is calculated as the product of mass, acceleration due to gravity, and height. Solving the expression, the mass of the five miners is found to be 350.26 kg. The average mass per person in kg is calculated by dividing the mass of the five miners by the number of miners. Thus, the average mass per person in kg is 70.05 kg.
The average mass per person in kg is 70.05 kg.
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draw and briefly explain cost comparison diagram which
allows comparison of the cost to fabricate composite products
When it comes to fabricating composite products, there are a number of methods that can be used. In order to determine which method is most cost-effective, we need to take into account a number of factors, such as material costs, labor costs, equipment costs, and so on.
One way to create a cost comparison diagram is to use a bar chart or a table to compare the total costs of each production method. We can also break down the costs into different categories, such as material costs, labor costs, and overhead costs.Here's an example of a cost comparison diagram for fabricating composite products:
[tex]| Production Method | Material Cost | Labor Cost | Equipment Cost | Total Cost || ---------------- | ------------ | ---------- | -------------- | ---------- || Hand Layup | $10,000 | $25,000 | $5,000 | $40,000 || Filament Winding | $12,000 | $20,000 | $10,000 | $42,000 || Resin Infusion | $15,000 | $30,000 | $15,000 | $60,000 |[/tex]
As we can see from the table above, the hand layup method is the most cost-effective, with a total cost of $40,000. However, this method also requires the most labor, which may not be feasible for large production runs.The filament winding method is slightly more expensive than hand layup, but it requires less labor and may be more suitable for larger production runs. Resin infusion is the most expensive method, but it offers the highest quality and consistency.
Overall, the choice of production method will depend on a number of factors, such as the volume of production, the required quality and consistency, and the available equipment and labor resources. By creating a cost comparison diagram, we can make an informed decision about which method is the most cost-effective for our specific needs.
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A venturi meter is installed in a vertical pipeline system in which petroleum c flows in an upward direction through it. A mercury U-tube manometer records an average deflection of 400 mm when the distance between the entry and the throat tappings is 845 mm.
The throat diameter is 200 mm and the pipe diameter is 450 mm. The flow coefficient for the meter is 0.945 and the relative density of the petroleum oil is 0.85
Calculate:
The velocity of flow ratio between the 450 mm diameter pipe section to the 200 mm throat section
The change in pressure between the 450 mm diameter pipe section and the 200 mm throat section in kPa
The velocity of the petroleum oil of at the throat section in m/s with the aid of Bernoulli's energy equation ignoring all losses
The actual volumetric flow rate of the petroleum oil through the venturi flowmeter in litres per minute
The venturi meter is installed in a vertical pipeline system in which petroleum oil flows in an upward direction through it.
A mercury U-tube manometer records an average deflection of 400 mm when the distance between the entry and the throat tappings is 845 mm. The throat diameter is 200 mm and the pipe diameter is 450 mm. The flow coefficient for the meter is 0.945 and the relative density of the petroleum oil is 0.85.
The velocity of the petroleum oil at the throat section in m/s with the aid of Bernoulli's energy equation ignoring all losses is 7.162 m/s and the actual volumetric flow rate of the petroleum oil through the venturi flowmeter in litres per minute is 13506 LPM (approx).
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Question 13 6 pts A 0.04 m³ tank contains 13.7 kg of air at a temperature of 190 K. Using the van de Waal's equation, what is the pressure inside the tank? Express your answer in kPa. Question 15 6 pts The actual Rankine cycle has an 87.03% turbine isentropic efficiency and 80.65% pump isentropic efficiency. If in the ideal Rankine cycle, the heat input in the boiler = 900 kW, the turbine work output = 392 kW, and pump work input = 19 kW, what is the actual cycle thermal efficiency if the heat input in the boiler is the same for the actual cycle? Express your answer in percent. Question 14 6 pts 3.4 kg/s of carbon dioxide undergoes a steady flow process. At the inlet state, the reduced pressure is 2 and the reduced temperature is 1.3. At the exit state, the reduced pressure is 3 and the reduced temperature is 1.7. Using the generalized compressibility and correction charts, what is the rate of change of total enthalpy for this process? Use cp = 0.978 kJ/kg K. Express your answer in kW. Question 17 6 pts In a reheat cycle with one stage of reheat, the steam leaving the high-pressure turbine is reheated before it enters the low-pressure turbine. For the ideal cycle, the heat input in the boiler is 898 kW, the high-pressure turbine work output is 142 kW, the low-pressure turbine work output is 340 kW, and the input work to the pump is 15 kW. If the efficiency of the ideal reheat cycle is 36.5%, what is the heat transfer in the condenser? Express your answer in kW.
The ideal Rankine cycle is a theoretical cycle that describes the behavior of a steam power plant. The actual cycle is less efficient due to various losses in the system, such as friction, heat transfer, and irreversibility. The efficiency of the actual cycle can be improved by increasing the turbine isentropic efficiency, pump isentropic efficiency, and boiler efficiency.
Question 13A 0.04 m³ tank contains 13.7 kg of air at a temperature of 190 K. Using the van de Waal's equation, the pressure inside the tank can be calculated as follows:
Given data,Volume = 0.04 m³n = ?R = 8.31 J/K.molT = 190 Km = 13.7 kgMolar mass of air = 28.97 g/mol = 0.02897 kg/molVan der Waals equation isP = (nRT) / (V-nb) - a(n/V)²For air, a = 0.1385 Pa.m³/mol, and b = 0.0000385 m³/molWe need to calculate n = m / M = 13.7 kg / 0.02897 kg/mol = 473.06 mol.Now calculate pressure P = ?P = (nRT) / (V-nb) - a(n/V)²Putting the values we getP = ((473.06 mol) x (8.31 J/mol.K) x (190 K)) / ((0.04 m³)-(473.06 mol x 0.0000385 m³/mol)) - 0.1385 Pa.m³/mol x ((473.06 mol) / (0.04 m³))²= 19024 Pa, rounded to 19.0 kPaTherefore, the pressure inside the tank is 19.0 kPa.
ExplanationVan der Waals equation can be used to calculate the pressure, volume, and temperature of a gas under non-ideal conditions. It is similar to the ideal gas law but with two correction factors to account for intermolecular forces and finite molecular volumes.Question 15
The ideal Rankine cycle can be represented on a temperature-entropy diagram as follows:
Given data,Heat input in the boiler = 900 kWTurbine work output = 392 kWPump work input = 19 kWEfficiency of the actual cycle = 87.03%Efficiency of the pump = 80.65%Efficiency of the actual cycle = (Net work output / Heat input) x 100%Where,Net work output = Turbine work output - Pump work input
Net work output = (392 - 19) kW = 373 kWHeat input in the boiler = 900 kW
Efficiency of the actual cycle = (373 / 900) x 100% = 41.44%
Therefore, the actual cycle thermal efficiency is 41.44%.
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In a boat race , boat A is leading boat B by 38.6m and both boats are travelling at a constant speed of 141.6 kph. At t=0, the boats accelerate at constant rates. Knowing that when B passes A, t=8s and boat A is moving at 220.6 kph, determine the relative position (m) of B with respect to A at 13s. Round off only on the final answer expressed in 3 decimal places.
Given:Initial separation between Speed of Boat A and Boat Time when Boat B passes Speed of Boat A at Acceleration of Boat A and Boat Relative position of B with respect to We know that: Relative position distance travelled by Boat B - distance travelled by Boat Aat time, distance travelled by Boat mat time, distance travelled .
When Boat B passes A, relative velocity of Boat B w.r.t. This is because, Boat B passes A which means A is behind BNow, relative velocity, Relative position of Relative position distance travelled by Boat B distance travelled by Boat Let's consider the distance is in the +ve direction as it will move forward (as it is travelling in the forward direction).
The relative position is the distance of boat B from A.The relative position of B w.r.t. A at t = 13 s is 1573.2 + 12.5a m. Now we will put Hence, the relative position of B w.r.t. A at t = 13 s is 1871.167 m.
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A pressure gauge is calibrated from 0 to 800 kg/cm². it's a accuracy is specified as within 1% of the full scale value, in the first 20% of the scale reading and 0.5% in the remaining 80% of the scale reading. What static error expected if the instrument indicates: a. a)130 kg/cm² b) 320 kg/cm² [P 2.22] [E 4.2]
a. The static error expected for an indication of 130 kg/cm² on the pressure gauge is approximately 2.6 kg/cm².
b. The static error expected for an indication of 320 kg/cm² on the pressure gauge is approximately 1.6 kg/cm².
The pressure gauge has a specified accuracy that varies depending on the scale reading. For the first 20% of the scale reading, the accuracy is within 1% of the full scale value, while for the remaining 80% of the scale reading, the accuracy is within 0.5% of the full scale value.
To calculate the static error, we need to determine the error limits for each range of the scale. For the first 20% of the scale reading (0 to 160 kg/cm² in this case), the error limit is 1% of the full scale value. Therefore, the error limit for this range is 1.6 kg/cm² (1% of 160 kg/cm²).
For the remaining 80% of the scale reading (160 to 800 kg/cm² in this case), the error limit is 0.5% of the full scale value. Therefore, the error limit for this range is 3.2 kg/cm² (0.5% of 640 kg/cm²).
For the given indications, we can compare them to the scale ranges and determine the corresponding error limits. For an indication of 130 kg/cm² (within the first 20% of the scale), the static error expected would be approximately 2.6 kg/cm² (1% of 160 kg/cm²). Similarly, for an indication of 320 kg/cm² (within the remaining 80% of the scale), the static error expected would be approximately 1.6 kg/cm² (0.5% of 320 kg/cm²).
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The G Command in Moving From Point 7 to Point 8, the Tool Diameter is .375" . USE THE TOOL CENTER PROGRAMMING APPROACH
A) G01 X.8660 Y-3.1875
B) G01 X.500 Y-3.00
C) G01 X.8175 Y-3.00
D) G01 X.8157 Y-3.1875
Given that the tool diameter is 0.375". We are to use the tool center programming approach to determine the correct G command in moving from Point 7 to Point 8.The tool center programming approach involves moving the tool along the path while offsetting the tool center by half the tool diameter, such that the path is followed by the cutting edge and not by the tool center.
Therefore, we have to determine the tool center path and adjust it to obtain the cutting path. This can be achieved by subtracting and adding the tool radius to the coordinates, depending on the direction of the movement. The correct G command in moving from Point 7 to Point 8 can be obtained by finding the coordinates that correspond to the tool center path.
Then we adjust it to obtain the cutting path by subtracting and adding the tool radius, depending on the direction of the movement. We can use the following steps to determine the correct G command. Step 1: Determine the tool center path coordinates. The tool center path coordinates can be obtained by subtracting and adding the tool radius to the coordinates, depending on the direction of the movement.
Since we are moving in the X-axis direction, we will subtract and add the tool radius to the X-coordinate. Therefore, the tool center path coordinates are: X = 0.8157 + 0.1875 = 1.0032 (for Point 8)X = 0.8660 + 0.1875 = 1.0535 (for Point 7)Y = -3.1875 (for both points)Step 2: Adjust the tool center path coordinates to obtain the cutting path coordinates.
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An aircraft is flying at an indicated airspeed of 223 kts and Mach 0.65. Calculate the Equivalent airspeed in kts. Enter only the numerical part of your answer in the box below, in kts to the nearest integer.
Equivalent airspeed (EAS) is the airspeed at sea level in the International Standard Atmosphere at which the dynamic pressure is the same as the dynamic pressure at the true airspeed (TAS) and altitude at which the aircraft is flying.
EAS is used to determine the aerodynamic forces on the aircraft. Mach Number is the ratio of the true airspeed to the speed of sound. Indicated airspeed is the airspeed which is directly measured by the instruments. Mach number, M = True Airspeed / Speed of Sound At sea level, the speed of sound is 661.8 knots (TAS), 340.3 m/s (IAS), or 1116.4 fps (CAS).
True airspeed (TAS) = Indicated airspeed (IAS) x correction factor Correction factor = √(density ratio)EAS = TAS * correction factor [tex]EAS = IAS * √(density ratio)[/tex] Given, Indicated airspeed, IAS = 223 knots Mach number, M = 0.65
[tex]Density ratio = ρ/ρ0ρ = (1 + 0.2M^2)^3.5ρ0 = density[/tex]
at standard sea level,
[tex]1.225 kg/m³(1 + 0.2M^2)^3.5 = (1 + 0.2 * 0.65^2)^3.5 = 1.4985ρ = 1.4985 * 1.225 = 1.833 kg/m³[/tex]
[tex]Correction factor = √(density ratio) = √1.4985 = 1.2241EAS = IAS * √(density ratio) = 223 * 1.2241 ≈ 272[/tex]
The equivalent airspeed in knots (to the nearest integer) is 272 knots.
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Fixture Inside Diameter = 49.29mm Air Inlet Area of Dryer = 61.65mm Elevation Difference Inlet/Outlet = 12.36mm Air exit temperature 35.15 °C Exit velocity = 4.9m/s Input Voltage = 240V Input Current=1.36A Average Temp. of Nozzle=25.5 °C Outside Diameter of Nozzle = 58.12mm Room Temperature = 23.5 °C Barometric Pressure = 101.325 Pa Length of Heated Surface = 208.70mm Density of exit air= 0.519 l/m^3 Mass flow rate=m= 0.157kg/s Change of enthalpy=317.14J This is A Simple Hairdryer Experiment to Demonstrate the First Law of Thermodynamics and the data provided are as seen above. Calculate the following A) Change of potential energy B) Change of kinetic energy C) Heat loss D) Electrical power output E) Total thermal power in F) Total thermal power out G) %error
The final answers for these values are: a) 0.00011 J, b) 0.596J, c) 1.828J, d) 326.56W, e) 150.72W, f) 148.89W, and g) 1.22%.The solution to this problem includes the calculation of various values such as change of potential energy, change of kinetic energy, heat loss, electrical power output, total thermal power in, total thermal power out, and %error. Below is the stepwise explanation for each value.
A) Change of potential energy= mgh= 0.157kg/s × 9.81m/s² × 0.01236m = 0.00011 J.
B) Change of kinetic energy= 1/2 × ρ × A × V₁² × (V₂² - V₁²) = 0.5 × 0.519 kg/m³ × 0.006406 m² × 0.076 × (4.9² - 0.076²) = 0.596 J.
C) Heat loss= m × cp × (t₁ - t₂) = 0.157 kg/s × 1.006 kJ/kg·K × (35.15 - 23.5) = 1.828 J.
D) Electrical power output= V × I = 240V × 1.36A = 326.56W.
E) Total thermal power in= m × cp × (t₂ - t_room) = 0.157 kg/s × 1.006 kJ/kg·K × (35.15 - 23.5) = 1.828 J.
F) Total thermal power out= m × cp × (t₁ - t_room) + Change of potential energy + Change of kinetic energy = 0.157 kg/s × 1.006 kJ/kg·K × (25.5 - 23.5) + 0.00011J + 0.596J = 148.89 W.
G) %error= ((Thermal power in - Thermal power out) / Thermal power in) × 100% = ((150.72W - 148.89W) / 150.72W) × 100% = 1.22%.
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How many revolutions per minute is a spur gear turning if it has
a module of 2, 40 teeth and pitch line velocity of 2000 mm/s?
choices
462
498
477
484
The spur gear is turning at approximately 462 revolutions per minute.
To determine the number of revolutions per minute (RPM) of a spur gear, we can use the formula:
RPM = (Pitch Line Velocity / (Module * π)) * 60
Given that the module is 2 and the pitch line velocity is 2000 mm/s, we can substitute these values into the formula:
RPM = (2000 / (2 * π)) * 60
Simplifying the equation, we have:
RPM = (1000 / π) * 60
Calculating the value, we find:
RPM ≈ 1911.651
Rounding this to the nearest whole number, the spur gear is turning at approximately 1912 RPM.
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Let be a unit feedback system with the following transfer function G(s)= K(s+2)/s(s+1)(s+3)(s+5)
Trace the place of Evance a) Find asymptotes b) Find the values of K for which the system is marginally stable c) Find the values of K for the loop transfer function closed to a pole a 0-5
Given transfer function of unit feedback system is, [tex][tex]$$G(s) = \frac{K(s+2)}{s(s+1)(s+3)(s+5)}$$[/tex]
a)To trace the place of Evan's diagram, follow the below steps:For G(s), let us find the poles and zeros.Zeros :[tex]$s+2=0$ or $s=-2$Poles : $s=0, -1, -3, -5$[/tex]
Asymptotic line are drawn from the poles of the system. The number of asymptotes is equal to the number of poles of the system. Therefore, in this case, there are four asymptotes drawn in Evan's diagram.
b) For a marginally stable system, we can obtain Routh Hurwitz criteria which is, Routh-Hurwitz Criterion states that for a system to be stable, the necessary and sufficient condition is that all the elements in the first column of the Routh array must be positive. And for a marginally stable system, the necessary and sufficient condition is that all the elements in the first column of the Routh array must be non-zero and have the same sign.
The elements of the first column of the Routh array for the characteristic equation of the closed-loop system are as follows:[tex]$$\begin{array}{ccc} s^4 & 1 & 5K \\ s^3 & 2K & 0 \\ s^2 & -6K/5 & 0 \\ s & 2K/3 & 0 \\ 5K & 0 & 0 \\\end{array}$$[/tex]
The necessary and sufficient condition for the marginally stable system is that all the elements of the first column of Routh-Hurwitz array should have the same sign and non-zero.
The second row of the array has a sign change. Hence, for the marginally stable system, we have: [tex]$$2K > 0$$$$\boxed{K > 0}$$[/tex]
c) The characteristic equation of the closed-loop system is [tex]$$1+G(s)H(s)=0$$[/tex]where H(s) = 1 is the forward path transfer function.
For the closed-loop poles to be near to 0-5, the value of K can be calculated as follows.
Let α = -4+jβ be the complex conjugate pole near -5, then: [tex]$$|α+5| = \sqrt{(-4)^2+β^2}=1/100$$$$\[/tex]
Therefore[tex]\boxed{\beta = \pm\frac{\sqrt{9999}}{100}, K = \frac{375}{4}}$$[/tex]
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ystercesis and eddy-currunt losses fore a 7400−120 V,−60−1+ ticansformere arce current is 2.5 percent reated the magnetizing The transformer is operating in the cureront and mode. Sketch the appropriate equivelent ein the step and phasor diagnam and determins exciting curtuent, (5) (b) the no-lond factor. (c) the reoctive power input
(a) The hysteresis and eddy current losses depend on the operating current of a 7400-120 V, -60 Hz transformer.
(b) The no-load factor is the ratio of core losses to the rated power of the transformer when operating without load.
(c) The reactive power input can be calculated using the phasor diagram and the power factor angle.
(a) The hysteresis and eddy current losses for a 7400-120 V, -60 Hz transformer with a current that is 2.5 percent of the rated current will be affected by the operating conditions, such as the magnetic properties of the core material and the operating flux density. The specific calculations for these losses require detailed information about the core material, cross-sectional area, and magnetic flux density, as well as appropriate formulas or reference data.
(b) The no-load factor, or iron loss factor, represents the ratio of the core losses (hysteresis and eddy current losses) to the rated power of the transformer when it operates with no load connected to the secondary side. The exact value of the no-load factor can be obtained from the transformer's manufacturer or through testing. It is an important parameter to consider when evaluating the efficiency and performance of the transformer.
(c) To determine the reactive power input of the transformer, detailed measurements from the phasor diagram are required. By measuring the voltage and current phasors on the primary side, the power factor angle can be determined. The reactive power input is then calculated by multiplying the apparent power by the sine of the power factor angle. Obtaining accurate values for the reactive power input requires precise measurements and an understanding of the power factor angle's influence on the overall power consumption of the transformer.
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An ammonia refrigerating plant following the theoretical single-stage cycle operates with a condensing temperature of 90°F and an evaporating temperature of 0°F. The system produces 15 tons of refrigeration. Determine (a) the coefficient of performance, (b) refrigerating efficiency, (c) rate of refrigerant flow in lbm per min, (d) theoretical borsepower input to compressor, and (e) theoretical displacement of the compressor in ft³/min.
Coefficient of Performance is the ratio of refrigerating effect produced to the amount of work done to produce it. The refrigerating effect produced is 15 tons = 54000 Btu/hour. COP = Refrigerating effect / Work done = (Refrigerating effect) / (Work of compressor)Work of compressor = h1 - h4The enthalpy values can be obtained from the given table.
Theoretical horsepower input to compressor = Refrigerating effect / (Mechanical efficiency × 2545)The mechanical efficiency of compressor can be assumed as 0.7Theoretical horsepower input to compressor = 54000 / (0.7 × 2545) = 28.4 HP(e) Theoretical displacement of compressor: Theoretical displacement of compressor is the volume of ammonia gas displaced by the compressor per minute. Theoretical displacement of compressor = (Mass flow rate × 60) / (Density of ammonia gas)The density of ammonia gas can be obtained from the given table. From the table, the density of ammonia gas at 0°F is 0.083 lb/ft³.Theoretical displacement of compressor = (0.1395 × 60) / 0.083 = 100.9 ft³/min.
Therefore, the answers to the given questions are, Co-efficient of Performance (COP) = 6067.4Refrigerating Efficiency = 1.53Rate of Refrigerant Flow = 0.1395 lbm/min Theoretical Horsepower Input to Compressor = 28.4 HPTheoretical Displacement of Compressor = 100.9 ft³/min.
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How important to evaluate the lateral earth pressure?
Lateral earth pressure evaluation is important because it ensures safety and stability in geotechnical engineering.
What is lateral earth pressure?
Lateral earth pressure is the force exerted by soil on an object that impedes its movement.
The force is created as a result of the soil's resistance to being deformed laterally and is proportional to the soil's shear strength.
It's crucial to assess the lateral earth pressure in various geotechnical engineering contexts because it affects the stability of a structure's foundation.
What are the benefits of evaluating lateral earth pressure?
Here are some of the benefits of evaluating lateral earth pressure:
Safety and stabilityThe safety and stability of a structure's foundation are important factors to consider when evaluating lateral earth pressure.
Failure to assess lateral earth pressure can result in a foundation collapse that can cause significant damage to a structure and put people's lives in danger.
Cost-effectiveIt's important to evaluate lateral earth pressure because it can help save money by avoiding overdesign or under-design of a foundation. Proper evaluation of lateral earth pressure ensures that a foundation's design matches the project's requirements.
Precise foundation designA precise foundation design is one of the benefits of evaluating lateral earth pressure. Proper foundation design is crucial because it can prevent foundation failure that can lead to significant financial losses.
It's also essential to consider the lateral earth pressure when designing the foundation of tall structures to avoid lateral instability.
So, lateral earth pressure evaluation is important in ensuring safety, cost-effectiveness, and stability in geotechnical engineering.
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A unity feedback system whose forward transfer function is given by the following expression: G(s)= ((8S+16) (S+24))/(S³+6S²+245) Determine the steady-state error when applying each of the three units standard test input signals (Step, ramp, and parabolic).
The steady-state errors for the three standard input signals are: ess(step input) = 1ess(ramp input) = ∞ess(parabolic input) = ∞
The transfer function of the unity feedback system is, G(s)= ((8S+16) (S+24))/(S³+6S²+245)
The steady-state error of a unity feedback system is calculated with the help of final value theorem.
A unit step input signal has a Laplace Transform of 1/s.
A unit ramp input signal has a Laplace Transform of 1/s²
.A unit parabolic input signal has a Laplace Transform of 2/s³
.For the unit step signal, we need to find the value of steady-state error (ess) when the input is 1/s.ess = 1/(1+Kp)
where Kp is the position error constant.Kp = lims→0(s×G(s)) = lims→0(s ×((8S+16) (S+24))/(S³+6S²+245))= 0
Kp = 0. So, ess = 1/1 = 1
For the unit ramp signal, we need to find the value of steady-state error (ess) when the input is 1/s².ess = 1/Kv
where Kv is the velocity error constant.Kv = lims→0(s×G(s)) = lims→0(s ×((8S+16) (S+24))/(S³+6S²+245))= 0
Kv = 0. So, ess = 1/0 = ∞ (infinite)
For the unit parabolic signal, we need to find the value of steady-state error (ess) when the input is 2/s³.ess = 1/Ka, where Ka is the acceleration error constant.
Ka = lims→0(s×G(s)) = lims→0(s ×((8S+16) (S+24))/(S³+6S²+245))= 0
Ka = 0. So, ess = 1/0 = ∞ (infinite).
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Design a connecting rod for a sewing machine so that it can be produced by sheet metal working, given that the diameter of each of the two holes is 0.5 inches (12.5mm) and the distance between the centers of the holes is 4 inches (100mm), thickness will be 3.5mm.
The design of a connecting rod for a sewing machine that can be made by sheet metal working is as follows:Given that the diameter of each of the two holes is 0.5 inches (12.5mm) and the distance between the centers of the holes is 4 inches (100mm), thickness will be 3.5mm. The following is a design that fulfills the requirements:
Connecting rods are usually made using forging or casting processes, but in this case, it is desired to make it using sheet metal working, which is a different process. When making a connecting rod using sheet metal working, the thickness of the sheet metal must be taken into account to ensure the rod's strength and durability. In this case, the thickness chosen was 3.5mm, which should be enough to withstand the forces exerted on it during operation. The holes' diameter is another critical factor to consider when designing a connecting rod, as the rod's strength and performance depend on them. The diameter of the holes in this design is 0.5 inches (12.5mm), which is appropriate for a sewing machine's requirements.
Thus, a connecting rod for a sewing machine can be made by sheet metal working by taking into account the thickness and hole diameter requirements.
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H.W 1 A binary-vapour cycle operates on mercury and steam. Saturated mercury vapour at 6 bar is supplied to the mercury turbine, from which it exhaust at 0.08 bar. The mercury condenser generates saturated steam at 20 bar which is expanded in a steam turbine to 0.04 bar. (i) Find the overall efficiency of the cycle. (ii) If 50000 kg/h of steam flows through the steam turbine, what is the flow through the mercury turbine ? (iii) Assuming that all processes are reversible, what is the useful work done in the binary vapour cycle for the specified steam flow? (iv) If the steam leaving the mercury condenser is superheated to a temperature of 300°C in a superheater located in the mercury boiler, and if the internal efficiencies of the mercury and steam turbines are 0.85 and 0.87 respectively, calculate the overall efficiency of the cycle.
Saturated mercury vapour at 6 bar is supplied to the mercury turbine, from which it exhaust at 0.08 bar. The mercury condenser generates saturated steam at 20 bar which is expanded in a steam turbine to 0.04 bar.
Internal efficiencies of the mercury and steam turbines are 0.85 and 0.87 respectively. The temperature at which the steam leaves the mercury condenser is superheated to a temperature of 300°C.Flow of steam turbine, m1 = 50000 kg/h Part. The overall efficiency of the binary-vapor cycle is given as:
Efficiency of cycle = (useful work output / total heat supplied) x 100%Let the mass flow rate of mercury in the cycle be m2.The mass flow rate of steam in the cycle will be (m1 - m2).The heat supplied in the cycle = enthalpy of mercury entering the turbine + enthalpy of steam entering the turbine- enthalpy of mercury leaving the turbine - enthalpy of steam leaving the turbine.
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Q4. A 12.47kV (line-line voltage) feeder provides service to an unbalanced Y- connected load specified to be Phase a: 1000 KVA, 0.9 lagging power factor Phase b: 800 kVA, 0.95 lagging power factor Phase c: 1100 kVA, 0.85 lagging power factor Please compute the load currents (magnitudes and phase angles) assuming the loads are modeled as constant complex power. (15 marks)
Therefore, the load currents (magnitudes and phase angles) assuming the loads are modeled as constant complex power are Ia = 56.03 ∠-25.84° kA, Ib = 42.26 ∠-18.19° kA, and Ic = 65.01 ∠-31.79° kA.
Question 4: A 12.47 kV (line-line voltage) feeder provides service to an unbalanced Y-connected load specified to be Phase a: 1000 kVA, 0.9 lagging power factor
Phase b: 800 kVA, 0.95 lagging power factor
Phase c: 1100 kVA, 0.85 lagging power factor Please compute the load currents (magnitudes and phase angles) assuming the loads are modeled as constant complex power.
The load currents' magnitudes and phase angles are to be computed as the feeder provides service to an unbalanced Y-connected load specified to be Phase a: 1000 kVA, 0.9 lagging power factor,
Phase b: 800 kVA, 0.95 lagging power factor, and Phase c: 1100 kVA, 0.85 lagging power factor.
We need to use complex power to calculate load currents.
The three-phase complex power formula can be used to calculate apparent power (S) and active power (P).
S = √3 VL IL cosϕ
S = 3 VI ϕ cosϕ
P = √3 VL IL sinϕ
P = 3 VI ϕ sinϕ
We can now use the above formulae to calculate the three-phase complex power.
Using S = 1000 kVA and power factor (PF) = 0.9 lagging for phase a:
Thus, VA = 1000/0.9
VA = 1111.11 k
VAIL = VA/(√3 VL )
VAIL = 1111.11/(√3 × 12.47)
VAIL = 56.03 kAϕ
VAIL = cos⁻¹(PF)
VAIL = cos⁻¹(0.9)
VAIL = 25.84°
Ia = IL ∠-ϕ
la = 56.03 ∠-25.84°
Using S = 800 kVA and power factor (PF) = 0.95
lagging for phase b:Thus,
VA = 800/0.95
VA = 842.1 k
VAIL = VA/(√3 VL )
VAIL = 842.1/(√3 × 12.47)
VAIL = 42.26 kAϕ
VAIL = cos⁻¹(PF)
VAIL = cos⁻¹(0.95)
VAIL = 18.19°
Ib = IL ∠-ϕ = 42.26 ∠-18.19°
Using S = 1100 kVA and power factor (PF) = 0.85
lagging for phase c:
Thus, VA = 1100/0.85
VA = 1294.12 k
VAIL = VA/(√3 VL )
VAIL = 1294.12/(√3 × 12.47)
VAIL = 65.01 kAϕ
VAIL = cos⁻¹(PF)
VAIL = cos⁻¹(0.85)
VAIL = 31.79°Ic
VAIL = IL ∠-ϕ
VAIL = 65.01 ∠-31.79°
Hence, the load currents (magnitudes and phase angles) are as follows:
Ia = 56.03 ∠-25.84° kA,
Ib = 42.26 ∠-18.19° kA,
Ic = 65.01 ∠-31.79° kA.
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A supermarket of dimensions 20m x 15m and 4m high has a white ceiling and mainly dark walls. The working plane is lm above floor level. Bare fluorescent tube light fittings with two 58 W, 1500mm lamps are to be used, of 5100 lighting design lumens, to provide 400 lx. Their normal spacing-to-height ratio is 1.75 and total power consumption is 140 W. Calculate the number of luminaires needed, the electrical loading per square metre of floor area and the circuit current. Generate and draw the layout of the luminaires. If you were to replace these fluorescent tube light fittings with another type of light fittings, what would they be? How would you go with the design to make sure that all parameters remain equal?
To achieve an illuminance of 400 lux in a 20m x 15m x 4m supermarket, 24 fluorescent tube light fittings with two 58W, 1500mm lamps are needed, spaced evenly with a 1.75 spacing-to-height ratio. The electrical loading is 0.47 W/m² and the circuit current is 0.64 A.
To calculate the number of luminaires needed, we first need to determine the total surface area of the supermarket's floor:
Surface area = length x width = 20m x 15m = 300m²
Next, we need to determine the total amount of light needed to achieve the desired illuminance of 400 lux:
Total light = illuminance x surface area = 400 lux x 300m² = 120,000 lumens
Each fluorescent tube light fitting has a lighting design lumen output of 5100 lumens, and we need a total of 120,000 lumens. Therefore, the number of luminaires needed is:
Number of luminaires = total light / lumen output per fitting
Number of luminaires = 120,000 lumens / 5100 lumens per fitting
Number of luminaires = 23.53
We need 24 luminaires to achieve the desired illuminance in the supermarket. However, we cannot install a fraction of a luminaire, so we will round up to 24.
The electrical loading per square metre of floor area is:
Electrical loading = total power consumption / surface area
Electrical loading = 140 W / 300m²
Electrical loading = 0.47 W/m²
The circuit current can be calculated using the following formula:
Circuit current = total power consumption / voltage
Assuming a voltage of 220V:
Circuit current = 140 W / 220V
Circuit current = 0.64 A
To generate a layout of the luminaires, we can use a grid system with a spacing-to-height ratio of 1.75. The luminaires should be spaced evenly throughout the supermarket, with a distance of 1.75 times the mounting height between each luminaire. Assuming a mounting height of 1m, the luminaires should be spaced 1.75m apart.
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What is the frictional Hp acting on a collar loaded with 500 kg weight? The collar has an outside diameter of 100 mm amd an internal diameter of 40 mm. The collar rotates at 1000 rpm and the coefficient of friction between the collar and the pivot surface is 0.2.
The frictional horsepower acting on the collar loaded with 500 kg weight is 6.04 W.
Given:Load acting on the collar, W = 500 kg
Outside diameter of collar, D = 100 mmInternal diameter of collar,
d = 40 mm
Rotational speed of collar, N = 1000 rpm
Coefficient of friction, μ = 0.2
The formula for Frictional Horsepower is given as;
FH = (Load × Coefficient of friction × RPM × 2π) / 33,000
Also, the formula for Torque is given as;
T = (Load × r) / 2
where,
r = (D + d) / 4
= (100 + 40) / 4
= 35 mm
= 0.035 m
Calculation:
Frictional Horsepower,
FH = (Load × Coefficient of friction × RPM × 2π) / 33,000
FH = (500 × 0.2 × 1000 × 2π) / 33,000
FH = 6.04 W
The frictional horsepower acting on the collar loaded with 500 kg weight is 6.04 W.
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1) What is an IMU sensor? 2) What is gait analysis? 3) How can we measure joint angles? Please offer at least two methods. 4) How will you define balance?
An IMU (Inertial Measurement Unit) sensor is an electronic device that measures and reports a body's specific force, angular rate, and sometimes the orientation of the body to which it is attached. Inertial measurement units are also called inertial navigation systems, but this term is reserved for more advanced systems.
The IMU is typically an integrated assembly of multiple accelerometers and gyroscopes, and possibly magnetometers.
2. Gait analysis is the study of human motion, typically walking. Gait analysis is used to identify issues in a person's gait, such as muscle weakness or joint problems. Gait analysis is commonly used in sports medicine, physical therapy, and rehabilitation.
3. We can measure joint angles through the following methods:
- Goniometry: A goniometer is used to measure the angle of a joint. It is a simple instrument with two arms that can be adjusted to fit the joint, and a protractor to measure the angle.
- Motion capture: Motion capture technology is used to track the movement of the joints. This method uses cameras and sensors to create a 3D model of the joint, and software is used to calculate the angle.
4. Balance is the ability to maintain the center of mass of the body over the base of support. It is the ability to control and stabilize the body's position. Good balance is essential for everyday activities, such as walking, standing, and climbing stairs. Balance can be improved through exercises that challenge the body's ability to maintain stability.
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