The rotation matrix BₐR, the displacement vector BₐPᴼᴿG, the Homogeneous Transformation Matrix BₐA, and the object coordinates in frame A( Aₙᴾ) are given below.
The rotation matrix BₐR:The rotation matrix BₐR is given byBₐR = [0.5 0.8660 0; -0.8660 0.5 0; 0 0 1]The displacement vector BₐPᴼᴿG:The displacement vector BₐPᴼᴿG is given byBₐPᴼᴿG = [5; -5; 7]The Homogeneous Transformation Matrix BₐA:The Homogeneous Transformation Matrix BₐA is given by:BₐA = [0.5 0.8660 0 5; -0.8660 0.5 0 -5; 0 0 1 7; 0 0 0 1]The object coordinates in frame A( Aₙᴾ):The object coordinates in frame A( Aₙᴾ) are given by Aₙᴾ = BₐT BₙᴾAₙᴾ = [6.20; -1.96; 7]. Therefore, the rotation matrix BₐR, the displacement vector BₐPᴼᴿG, the Homogeneous Transformation Matrix BₐA, and the object coordinates in frame A( Aₙᴾ) are calculated.
The rotation matrix BₐR is given by BₐR = [0.5 0.8660 0; -0.8660 0.5 0; 0 0 1], the displacement vector BₐPᴼᴿG is given by BₐPᴼᴿG = [5; -5; 7], the Homogeneous Transformation Matrix BₐA is given by BₐA = [0.5 0.8660 0 5; -0.8660 0.5 0 -5; 0 0 1 7; 0 0 0 1], and the object coordinates in frame A( Aₙᴾ) are given by Aₙᴾ = BₐT BₙᴾAₙᴾ = [6.20; -1.96; 7].
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E3: Air flows from a nozzle with a mass flow rate of 1.0 slugs /sec. If T 0=607 ∘F,p 0 = 120psia, and p exit =15psia, the Ma number at the exit is most nearly a. 2.55 b. 2.05 c. 3.15 d. 1.00 e. 1.60
The Mach number at the exit of the nozzle, given a mass flow rate of 1.0 slugs/sec, is option (a) 2.55, according to the provided parameters.
To determine the Mach number at the exit of the nozzle, we can use the isentropic flow equations and the given parameters.
Given:
Mass flow rate (ṁ) = 1.0 slugs/sec
Total temperature at the inlet (T₀) = 607 °F
Total pressure at the inlet (p₀) = 120 psia
Pressure at the exit (p_exit) = 15 psia
First, we need to convert the total temperature from Fahrenheit to Rankine:
T₀ = 607 °F + 459.67 °R = 1066.67 °R
Next, we can use the mass flow rate and the total pressure to find the exit velocity (V_exit):
V_exit = ṁ / (A_exit * ρ_exit)
To find the exit area (A_exit), we need to calculate the exit density (ρ_exit) using the ideal gas equation:
Ρ_exit = p_exit / (R * T_exit)
The gas constant R for air is approximately 1716.5 ft·lbf/(slug·°R).
Using the isentropic flow equations, we can find the exit temperature (T_exit) as follows:
(p_exit / p₀) = (T_exit / T₀) ^ (γ / (γ – 1))
Here, γ is the specific heat ratio for air, which is approximately 1.4.
Now, let’s calculate the exit temperature:
(T_exit / T₀) = (p_exit / p₀) ^ ((γ – 1) / γ)
(T_exit / 1066.67 °R) = (15 psia / 120 psia) ^ ((1.4 – 1) / 1.4)
(T_exit / 1066.67 °R) = 0.3272
T_exit = 0.3272 * 1066.67 °R = 349.96 °R
Now, we can calculate the exit density:
Ρ_exit = 15 psia / (1716.5 ft·lbf/(slug·°R) * 349.96 °R) ≈ 0.00624 slug/ft³
Next, let’s calculate the exit velocity:
V_exit = 1.0 slugs/sec / (A_exit * 0.00624 slug/ft³)
Now, we can use the mass flow rate equation to find the exit area (A_exit):
A_exit = 1.0 slugs/sec / (V_exit * 0.00624 slug/ft³)
Finally, we can calculate the Mach number at the exit:
M_exit = V_exit / (γ * R * T_exit)^0.5
Let’s plug in the values and calculate the Mach number:
A_exit = 1.0 slugs/sec / (V_exit * 0.00624 slug/ft³)
M_exit = V_exit / (1.4 * 1716.5 ft·lbf/(slug·°R) * 349.96 °R)^0.5
After performing the calculations, the most approximate Mach number at the exit is option (a) 2.55.
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To estimate the loss factor (η) of an elastomer in the circular disc shape shown, a 10 kg mass is mounted on its top. A vertical harmonic force excites the mass: F(t) = 900sin(10t) in N. The elastomer is placed on a flat rigid foundation, having a thickness d = 0.01 m and diameter D = 0.04 m. It is known that the response amplitude of the elastomer-mass system at resonance is Xrm = 0.003 m, and Xrm = F0 / ( a E η ) where F0 is the driving force amplitude, E = 1.5 x106 Pa is the Young’s modulus, stiffness of the system k = a E, a = π(D/2)2 / d is a constant governed by the shape of the elastomer. a) Determine the loss factor (η) of the elastomer-mass system. [10 marks] b) Calculate the stiffness of the system k. [3 marks] c) Find natural frequency of the system ωn. [3 marks] d) Describe 3 advantages and 1 disadvantage of using viscoelastic materials such as an elastomer for vibration isolation.
Viscoelastic materials have many advantages, including: It has very high damping capacity. High strength and elasticity. It is used to filter or remove unwanted frequencies. Although viscoelastic materials have many benefits, they also have some drawbacks, including :It has a limited operating range. Highly dependent on temperature .It may have a low natural frequency.
Loss factor (η) of the elastomer-mass system;
The response amplitude of the elastomer-mass system at resonance is Xrm = 0.003 m, and Xrm = F0 / ( a E η ) where F0 is the driving force amplitude, E = 1.5 x106 Pa is the Young’s modulus, stiffness of the system k = a E, a = π(D/2)2 / d is a constant governed by the shape of the elastomer.
At resonance, Xrm = F0/(aEη) η = F0/(aEXrm) = 900 / (π (0.02)2 x 1.5 x 106 x 0.003) = 0.24.
Stiffness of the system k
Stiffness of the system k = aE= π (0.02)2 / 0.01 x 1.5 x 106 = 1.26 N/mc)
Natural frequency of the system ωn.
Natural frequency of the system is given by, ωn = sqrt(k/m)
Here, m = mass = 10 kg; k = stiffness = 1.26 N/mωn = sqrt (1.26 / 10) = 0.4 rad/s.
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Q5
Question 5 What is the Australian standard number for tensile testing (i.e.) "metallic materials - tensile testing at ambient temperatures"?
An Australian standard number refers to a unique identification number assigned to a specific standard published by Standards Australia. The Australian standard number for tensile testing of metallic materials at ambient temperatures is AS 1391.
AS 1391 is the Australian standard that specifically addresses the tensile testing of metallic materials at ambient temperatures. This standard provides guidelines and requirements for conducting tensile tests on metallic materials to determine their mechanical properties.
Tensile testing is a widely used method for evaluating the mechanical behavior and performance of metallic materials under tensile forces. It involves subjecting a specimen of the material to a gradually increasing axial load until it reaches failure.
AS 1391 outlines the test procedures, specimen preparation methods, and reporting requirements for tensile testing at ambient temperatures. It ensures consistency and standardization in conducting these tests, allowing for accurate and reliable comparison of material properties across different laboratories and industries in Australia.
The Australian standard number for tensile testing of metallic materials at ambient temperatures is AS 1391. This standard provides guidelines and requirements for conducting tensile tests to evaluate the mechanical properties of metallic materials. Adhering to this standard ensures consistency and reliability in conducting tensile tests in Australia
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True or False: The convective heat transfer coefficient in
laminar flow over a flat plate INCREASES with distance
Explain your reasoning
The convective heat transfer coefficient in a fluid is directly proportional to the heat transfer surface area. This statement is False.
Convective heat transfer is the transfer of heat from one point to another in a fluid through the mixing of fluid particles. The convective heat transfer coefficient in a fluid is directly proportional to the fluid velocity, the fluid density, and the thermal conductivity of the fluid. The convective heat transfer coefficient is also indirectly proportional to the viscosity of the fluid. The heat transfer surface area only affects the total heat transfer rate. Therefore, the statement is false.
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1. (a) Let A and B be two events. Suppose that the probability that neither event occurs is 3/8. What is the probability that at least one of the events occurs? (b) Let C and D be two events. Suppose P(C)=0.5,P(C∩D)=0.2 and P((C⋃D) c)=0.4 What is P(D) ?
(a) The probability that at least one of the events A or B occurs is 5/8.
(b) The probability of event D is 0.1.
(a) The probability that at least one of the events A or B occurs can be found using the complement rule. Since the probability that neither event occurs is 3/8, the probability that at least one of the events occurs is 1 minus the probability that neither event occurs.
Therefore, the probability is 1 - 3/8 = 5/8.
(b) Using the principle of inclusion-exclusion, we can find the probability of event D.
P(C∪D) = P(C) + P(D) - P(C∩D)
0.4 = 0.5 + P(D) - 0.2
P(D) = 0.4 - 0.5 + 0.2
P(D) = 0.1
Therefore, the probability of event D is 0.1.
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Explain four characteristics that the SCR and the Triac have in common. Question#8: What is the purpose of the resistor in the snubbing circuit? Question#9: Design a snubbing network for the S4006LS SCR driving a 200 W light bulb from a 120 Vrms line voltage. Determine the: a. Capacitor's capacitance and peak voltage. b. Minimum resistor. c. Diode's peak current. d. Inductor's inductance
The SCR (Silicon Controlled Rectifier) and the Triac are two of the most frequently used types of thyristors (a type of semiconductor device).
The four characteristics that the SCR and the Triac have in common are as follows:
Characteristics that the SCR and the Triac have in common: Both devices are unidirectional switching devices. These devices are triggered by a gate current, and once they are triggered, they remain on until the anode-cathode current is reduced to a level below the holding current. These devices do not require continuous gate current to remain turned on. They can be turned on with a gate current, and once turned on, they remain on until the anode-cathode current falls below a particular level. Both devices are ideal for controlling the output power to a load from a variable voltage source in the phase-control application. Their primary applications are in power switching, motor control, and lighting applications.
8: The purpose of the resistor in the snubbing circuit is to limit the rate of rise of the voltage at the anode of the SCR (or TRIAC) when it switches from its ON to OFF state. The voltage across the capacitor rises slowly in a snubbing circuit due to the resistor, preventing voltage spikes from damaging the device. It also aids in the dissipation of stored energy by the capacitor and the inductor.
9: Snubbers are used to protect the SCR and load against overvoltage transients and/or to reduce electromagnetic interference. The Snubbing network for the S4006LS SCR driving a 200 W light bulb from a 120 Vrms line voltage is given below:
Snubbing Network Capacitor's Capacitance: The value of the capacitor is calculated using the following formula:
C=I×t/V
Where I is the current that flows through the load, t is the pulse width of the voltage across the load, and V is the maximum voltage that the capacitor should be able to withstand.
Here, I = P/Vrms
= 200/120 = 1.66 A,
t = 5 μs (Assuming), and V = 1.414 × 120 = 170.8 V.
Substituting the values, C = (1.66) × (5 × 10-6)/170.8 = 48.6 nF Peak Voltage Across Capacitor: The peak voltage across the capacitor can be determined using the following formula:
Vp = Vrms × √2 = 120 × 1.414 = 170.8 V Minimum Resistor: The minimum resistance of the resistor is calculated using the following formula:
R = Vp/I
Where I is the current that flows through the load, and Vp is the peak voltage across the capacitor.
Here, I = 1.66 A and Vp = 170.8 V.
Substituting the values, R = 102.8 Ω (Round off to the nearest standard value of 100 Ω).
Peak Current of Diode: The peak current of the diode can be determined using the following formula:
Ip = Vp/[(L × t) + R]
Where L is the value of the inductance. Here, L = 0.01 H,
t = 5 μs, and R = 102.8 Ω.
Substituting the values, Ip = 835 mA (Round off to the nearest standard value of 1 A).
Inductance of Inductor: The value of the inductance is calculated using the following formula:
L = [(Vp/I) - R] × t/2
Where I is the current that flows through the load, and Vp is the peak voltage across the capacitor.
Here, I = 1.66 A, Vp = 170.8 V,
t = 5 μs, and R = 102.8 Ω.
Substituting the values, L = 0.01 H.
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A plastic dry rubbing bearing runs over a 9 mm non-rotating steel axle. The radial wear rate of the bearing is approximately 0.086 mm/year. If the bearings are to be replaced when the diametral clearance reaches 555 μ m , calculate the minimum expected bearing life (in years). Give your answer to 2 decimal places. Hint: start by choosing a suitable hole basis fit . Then, calculate the maximum diametral clearance (using the fit you selected) and use this value to determine the bearing life.
The life expectancy of a plastic bearing running over a non-rotating steel axle can be calculated using the radial wear rate and the diametral clearance limit.
Firstly, we'll select a suitable hole basis fit. Given the conditions, a clearance fit seems suitable. However, without specific fit class information, it's hard to precisely determine the initial diametral clearance. Assuming zero initial clearance (or at least significantly less than the replacement limit), the maximum diametral clearance before replacement becomes 555 μm or 0.555 mm. This clearance will be twice the radial wear due to symmetry (diametral wear = 2*radial wear). The wear rate is given as 0.086 mm/year radially, which translates to 0.172 mm/year diametrically. Hence, the minimum expected life of the bearing will be the maximum diametral clearance divided by the diametral wear rate. This will provide the estimated bearing life in years, rounded to two decimal places.
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A 2mx 2m solar absorber plate is at 400 K while it is exposed to solar irradiation. The surface is diffuse and its spectral absorptivity is a = 0, for λ >1 >0.5 μm a=0.8, for 0.5 u m> > λ μm a = 0, for lym> > λ2um a =0.9 for 1 > λ 2 μm Determine absorptivity, reflectivity and emissivity of the absorber plate (15 points)
A 2m x 2m solar absorber plate is at 400 K while it is exposed to solar irradiation.
The surface is diffuse and its spectral absorptivity is as follows:a = 0, for λ >1 >0.5 μma = 0.8, for 0.5 µm > λ > 2 µma = 0, for λ > 2 µma =0.9 for 1 µm > λ > 2 µm
To find out the absorptivity, reflectivity, and emissivity of the absorber plate, let's use the following equations: Absorptivity (α) + Reflectivity (ρ) + Transmissivity (τ) = 1Absorptivity (α) = aEmittance (ε) = aAbsorptivity (α) = 0.9 (for 1 > λ > 2 µm) and 0.8 (for 0.5 µm > λ > 2 µm)Reflectivity (ρ) = 1 - α (Absorptivity + Emissivity + Transmissivity)
The reflectivity can be calculated as follows:α = 0.9 (for 1 > λ > 2 µm)ρ = 1 - αρ = 1 - 0.9ρ = 0.1α = 0.8 (for 0.5 µm > λ > 2 µm)ρ = 1 - αρ = 1 - 0.8ρ = 0.2α = 0 (for λ > 2 µm)ρ = 1 - αρ = 1 - 0ρ = 1
The reflectivity is calculated to be 0.1, 0.2, and 1, respectively, for the above wavelength ranges. The emissivity can be found using the following equation:ε = α = 0.9 (for 1 > λ > 2 µm)ε = α = 0.8 (for 0.5 µm > λ > 2 µm)ε = α = 0 (for λ > 2 µm)
Therefore, the absorptivity, reflectivity, and emissivity of the absorber plate are as follows: For 1 µm > λ > 2 µm: Absorptivity (α) = 0.9 Reflectivity (ρ) = 0.1 Emissivity (ε) = 0.9For 0.5 µm > λ > 2 µm: Absorptivity (α) = 0.8Reflectivity (ρ) = 0.2 Emissivity (ε) = 0.8For λ > 2 µm: Absorptivity (α) = 0 Reflectivity (ρ) = 1 Emissivity (ε) = 0.
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To design a simply supported RCC slab for a roof of a hall 4000x9000 mm inside dimension, with 250 mm wall all around, consider the following data: d= 150 mm, design load intensity=15 kN/m², M25, Fe415. a. Find the effective span and load per unit width of the slab. b. Find the ultimate moment per unit width of the slab. c. Find the maximum shear force per unit width of the slab. d. Find the effective depth required from ultimate moment capacity consideration and comment on the safety. e. Is it necessary to provide stirrups for such a section?
Stir rups are not necessary in this slab design.
How to solve the problemsa. The effective span of the slab is the longer dimension of the hall: 9000 mm or 9 m.
The load per unit width (w) is equal to the design load intensity: 15 kN/m.
b. The ultimate moment (Mu) per unit width of the slab can be found using the formula for a simply supported slab under uniformly distributed load: Mu = w*L²/8.
Mu = 15 kN/m * (9 m)² / 8
= 151.88 kNm/m.
c. The maximum shear force (Vu) per unit width of the slab can also be found using a formula for a simply supported slab under uniformly distributed load: Vu = w*L/2.
Vu = 15 kN/m * 9 m / 2
= 67.5 kN/m.
d. Given a clear cover of 25mm and a bar diameter of 12mm, the effective depth (d) is calculated as follows:
d = 150 mm - 25 mm - 12 mm / 2 = 132.5 mm.
The ultimate moment of resistance (Mr) provided by the slab can be given by Mr = 0.138 * f * (d)²,
where fc is 25 N/mm² for M25 concrete.
Mr = 0.138 * 25 N/mm² * (132.5 mm)² = 482.25 kNm/m.
e. Since Mr > Mu (482.25 kNm/m > 151.88 kNm/m), the slab is safe for the bending moment. Therefore, stir rups are not necessary in this slab design.
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Mention the name and function of ten (10) components/instruments are involved in a power system.
A power system involves several components and instruments that work together to provide a reliable and efficient supply of power. These components can be broadly categorized into four main categories: generation, transmission, distribution, and consumption.
1. Generators - they convert mechanical energy into electrical energy.
2. Transformers - they step up or step down the voltage levels to facilitate transmission and distribution of power.
3. Circuit Breakers - they are used to protect the system from overloads and short circuits by interrupting the flow of electricity when necessary.
4. Capacitors - they store electrical energy and are used to improve the power factor of the system.
5. Switches - they are used to control the flow of electricity in the system.
6. Meters - they measure the electrical parameters such as voltage, current, power, and energy consumption.
7. Protective Relays - they are used to detect abnormal conditions in the system and trigger the appropriate response to protect the system from damage.
8. Transmission Lines - they are used to transport electricity from the power generation station to the distribution substations.
9. Distribution Transformers - they are used to step down the voltage levels for distribution to consumers.
10. Loads - they are the electrical devices that consume power, such as light bulbs, motors, and electronic appliances.
These components and instruments are crucial in a power system, and their proper functioning ensures a reliable and efficient supply of power.
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Sketch a 1D, 2D, and 3D element type of your choice. (sketch 3 elements) Describe the degrees of freedom per node and important input data for each structural element. (Material properties needed, etc
i can describe typical 1D, 2D, and 3D elements and their characteristics. 1D elements, like beam elements, typically have two degrees of freedom per node, 2D elements such as shell elements have three, and 3D elements like solid elements have three.
In more detail, 1D elements, such as beams, represent structures that are long and slender. Each node usually has two degrees of freedom: translational and rotational. Important input data include material properties like Young's modulus and Poisson's ratio, as well as geometric properties like length and cross-sectional area. 2D elements, such as shells, model thin plate-like structures. Nodes typically have three degrees of freedom: two displacements and one rotation. Input data include material properties and thickness. 3D elements, like solid elements, model volume. Each node typically has three degrees of freedom, all translational. Input data include material properties.
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Prove that the transfer function is as follows using the block diagram reduction. 160K₁/s²+(32+160K₁K₂)s + 160K₁
The given transfer function is:
[tex]$$\frac{160K_1}{s^2+(32+160K_1K_2)s + 160K_1}$$[/tex]
We can represent the above transfer function in a block diagram form as follows: Block diagram The first block represents the gain of 160K₁, the second block represents the summing point, the third block represents the gain of K₂ and the fourth block represents the plant with a transfer function of
[tex]$$\frac{1}{s^2+32s+160K_1}$$[/tex]
Now, we will perform the block diagram reduction process. First, we will merge the second and third blocks using the summing point reduction technique. Summing point reduction technique Applying summing point reduction technique, we get a new transfer function:
[tex]$$\frac{160K_1}{s^2+(32+160K_1K_2)s + 160K_1} = \frac{K_2}{1+sT_1}\left(\frac{160K_1}{s^2+2\xi\omegas+\omega_n^2}\right)$$[/tex]
where[tex]$$\omega_n = 4\sqrt{10K_1}$$$$\xi = \frac{1}{8\sqrt{10K_1}}$$$$T_1 = \frac{1}{8\sqrt{10K_1}K_2}$$[/tex]
Hence, we have proved that the given transfer function can be represented using the block diagram reduction and we got a new transfer function.
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Work speed of circular electrode of 220 mm diameter for carrying out seam welding at 4 welds per cm on 1.6 mm thick mild steel tube.
Welding cycle consists of 3 cycles ‘on’ and 2 cycles ‘off’ Power supply is at 50 Hz. Calculate R.P.M. and energy requirement of the above circular electrodes assuming effective resistance between electrodes as 100 ohm
calculate the RPM (Revolutions Per Minute) and energy requirement of the circular electrodes for seam welding, we need to consider the welding speed, the number of welds per unit length, the thickness of the material, and the effective resistance.
First, let's calculate the welding speed (S) in centimeters per minute: S = WPC * f . S = 4 welds/cm * 50 Hz . S = 200 cm/min .Next, let's calculate the RPM (N) of the circular electrode: N = (S * 60) / (π * D) . N = (200 cm/min * 60) / (π * 22 cm) . N ≈ 172.52 RPM . Now, let's calculate the energy requirement (E) of the circular electrodes: E = (P * t) / (WPC * f * (3 + 2)) E = (P * t) / (4 welds/cm * 50 Hz * 5 cycles) E = (P * t) / 1000 where:
- P is the power in watts .
Since we are given the effective resistance (R), we can calculate the power (P) using the formula: P = (V^2) / R . Assuming a standard voltage of 220 volts: P = (220^2) / 100 , P = 48400 / 100 , P = 484 watts . Finally, let's calculate the energy requirement: E = (P * t) / 1000 . E = (484 watts * 0.016 meters) / 1000 , E = 7.744 joules . Therefore, the RPM of the circular electrode is approximately 172.52 RPM, and the energy requirement is approximately 7.744 joules.
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The actual Rankine cycle has an 89.07% turbine isentropic efficiency and 75.22% pump isentropic efficiency. If in the ideal Rankine cycle, the heat input in the boiler = 924 kW, the turbine work output = 398 kW, and pump work input = 20 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.
The actual cycle thermal efficiency is 31.83%. The thermal efficiency of the ideal Rankine cycle is given by:
η = (W_net / Q_in) x 100%
where:
W_net = turbine work output - pump work input
Q_in = heat input in the boiler
Given:
Heat input in the boiler (Q_in) = 924 kW
Turbine work output = 398 kW
Pump work input = 20 kW
Using the above values and the ideal Rankine efficiency formula, we can calculate the ideal Rankine cycle thermal efficiency:
W_net = Turbine work output - Pump work input
W_net = 398 kW - 20 kW
W_net = 378 kW
η_ideal = (W_net / Q_in) x 100%
η_ideal = (378 kW / 924 kW) x 100%
η_ideal = 40.98%
The thermal efficiency of the actual Rankine cycle is given by:
η_actual = (W_net_actual / Q_in) x 100%
where:
W_net_actual = turbine work output - pump work input losses
The pump and turbine isentropic efficiency values can be used to calculate pump and turbine work input losses as follows:
Pump work input losses = pump work input / pump isentropic efficiency
Pump work input losses = 20 kW / 0.7522
Pump work input losses = 26.57 kW
Turbine work output losses = turbine work output - actual turbine work output
Turbine work output losses = turbine work output - (turbine isentropic efficiency x turbine work output)
Turbine work output losses = 398 kW - (0.8907 x 398 kW)
Turbine work output losses = 398 kW - 354.6 kW
Turbine work output losses = 43.4 kW
Therefore,
W_net_actual = W_net - pump work input losses - turbine work output losses
W_net_actual = 378 kW - 26.57 kW - 43.4 kW
W_net_actual = 308.03 kW
η_actual = (W_net_actual / Q_in) x 100%
η_actual = (308.03 kW / 924 kW) x 100%
η_actual = 31.83%
The actual cycle thermal efficiency of the Rankine cycle is 31.83% given the ideal Rankine cycle parameters, 89.07% turbine isentropic efficiency, and 75.22% pump isentropic efficiency.
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The Reynolds number, pvD/u, is a very important parameter in fluid mechanics. Verify that the Reynolds number is dimensionless, using the MLT system for basic dimensions, and determine its value for ethyl alcohol flowing at a velocity of 3 m/s through a 2-in- diameter pipe.
Reynolds number is a dimensionless quantity which represents the ratio of inertial forces (ρvD) to the viscous forces (u).Here,ρ is the density of the fluidv is the velocity of the fluidD is the diameter of the pipemu is the dynamic viscosity of the fluid.
If the Reynolds number is very less than 2300, then the flow is laminar and if it is greater than 4000, then the flow is turbulent.If the Reynolds number lies between 2300 and 4000, then the flow is transitional. Ethyl alcohol is flowing through a 2-inch diameter pipe at a velocity of 3 m/s.
We have to find the Reynolds number value.Let's put the values in the formula,Re = ρvd/µRe = (7850 kg/m³ x 3 m/s x 0.0508 m) / (1.2 x 10⁻³ N s/m²)Re = 9,34,890.67Reynolds number value is more than 100 words.
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If the same turbine runs at two different rotation speeds but under dynamically similar conditions, which of the following is NOT true? The flow rate will be different. The work output will be different. The pressure ratio will be different The efficiency will be the same. 6 The flow coefficient will be the same.
Under dynamically similar conditions for a turbine running at two different rotation speeds, the statement that "the efficiency will be the same" is not true. Turbine efficiency is not solely dependent on dynamical similarity.
When a turbine operates under dynamically similar conditions at two different rotational speeds, most parameters like flow rate, work output, pressure ratio, and flow coefficient will differ. However, the statement "the efficiency will be the same" is not necessarily true. Turbine efficiency is influenced by several factors, including design, fluid properties, and operating conditions. While dynamical similarity tries to ensure a degree of correspondence between scenarios, the efficiency can still change with rotational speed. This variation results from influences like alterations in the Reynolds number, which could shift flow characteristics. Consequently, despite maintaining dynamical similarity.
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Q5 a. Describe the meaning of Tolerance Reference Hole System. b. Explain the meaning of the symbol "m 32 x 1.5-6g"
The Tolerance Reference Hole System refers to a standardized method for specifying the dimensions and tolerances of holes in engineering drawings.The symbol "m 32 x 1.5-6g" indicates the specifications of a metric thread.
The Tolerance Reference Hole System refers to a standardized method for specifying the dimensions and tolerances of holes in engineering drawings. It establishes a set of guidelines that ensure consistency and compatibility in hole size and fit. It includes reference values for the diameter, depth, and tolerance of the hole.
The symbol "m 32 x 1.5-6g" indicates the specifications of a metric thread. "M" refers to the metric system, "32" represents the major diameter of the thread in millimeters, "1.5" signifies the pitch (the distance between corresponding points on adjacent threads) in millimeters, and "-6g" denotes the tolerance class. In this case, the tolerance class is 6g, which means that the thread has a medium tolerance range.
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A biomass digester receives 23 kJ of energy in the form of heat. There is a mechanical mixer that keeps rotating and putting 1.4 kJ work into the system. Calculate the energy balance in the system in kJ, if there is a heat loss of 7 kJ. A small fan in a closed insulated room releases heat at the rate of 42 watts and pushes the air at the rate of 9 m/s with a mass flow rates of 1.2 kg/s. The room has a heater that produces heat at the rate of 0.30 kJ/s as well as a computer that produces 65 watts. Light bulbs in the room produce up to 125 watts. The room looses 0.32 kJ/s. Calculate the amount of heat maintained in the room.
The energy balance in the system can be calculated by summing up all the energy inputs and subtracting the energy losses. The energy balance is 23 kJ (heat input) + 1.4 kJ (work input) - 7 kJ (heat loss) = 17.4 kJ.
To calculate the amount of heat maintained in the room, we need to consider the various energy inputs and losses within the system.
Energy Inputs:
Heater: The heater produces heat at the rate of 0.30 kJ/s.
Small fan: The small fan releases heat at the rate of 42 watts (0.042 kJ/s) due to its operation.
Computer: The computer produces heat at the rate of 65 watts (0.065 kJ/s).
Light bulbs: The light bulbs produce heat up to 125 watts (0.125 kJ/s).
Energy Losses:
Heat loss: The room loses heat at the rate of 0.32 kJ/s.
To calculate the amount of heat maintained in the room, we sum up all the energy inputs and subtract the energy losses:
Total Energy Input = Heater + Small fan + Computer + Light bulbs
= 0.30 kJ/s + 0.042 kJ/s + 0.065 kJ/s + 0.125 kJ/s
= 0.532 kJ/s
Heat Maintained = Total Energy Input - Heat Loss
= 0.532 kJ/s - 0.32 kJ/s
= 0.212 kJ/s
Therefore, the amount of heat maintained in the room is 0.212 kJ/s.
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Question 5 (a) Draw the sketch that explain the changes occurs in the flow through oblique and normal shock waves? (5 marks) (b) The radial velocity component in an incompressible, two-dimensional flow (v, = 0) is: V, = 2r + 3r2 sin e Determine the corresponding tangential velocity component (ve) required to satisfy conservation of mass. (10 marks) (c) Air enters a square duct through a 1.0 ft opening as is shown in figure 5-c. Because the boundary layer displacement thickness increases in the direction of flow, it is necessary to increase the cross-sectional size of the duct if a constant U = 2.0 ft/s velocity is to be maintained outside the boundary layer. Plot a graph of the duct size, d, as a function of x for 0.0 SX S10 ft, if U is to remain constant. Assume laminar flow. The kinematic viscosity of air is v = 1.57 x 10-4 ft2/s. (10 marks) U= 2 ft/s 1 ft dux) 2 ft/s
Part a)The oblique shock wave occurs when a supersonic flow over a wedge or any angled surface. The normal shock wave occurs when a supersonic flow is blocked by a straight surface or an object.
The normal shock wave has a sharp pressure rise and velocity decrease downstream of the wave front, while the oblique shock wave has a gradual pressure rise and velocity decrease downstream of the wave front. The oblique shock wave can be calculated by the wedge angle and the Mach number of the upstream flow. The normal shock wave can be calculated by the Mach number of the upstream flow only. Part b)Given radial velocity component, V, = 2r + 3r2 sin e
Required tangential velocity component (v?) to satisfy conservation of mass. Here, u, = 0 and
v, = 2r + 3r2 sin e.
Conservation of mass is given by Continuity equation, in polar coordinates, as : r(∂u/∂r) + (1/r)(∂v/∂θ) = 0 Differentiating the given expression of u with respect to r we get, (∂u/∂r) = 0
Similarly, Differentiating the given expression of v with respect to θ, we get, (∂v/∂θ) = 6r sin θ
From continuity equation, we have r(∂u/∂r) + (1/r)(∂v/∂θ) = 0
Substituting the values of (∂u/∂r) and (∂v/∂θ), we get:r(0) + (1/r)(6r sin θ) = 0Or, 6 sin θ
= 0Or,
sin θ = 0
Thus, the required tangential velocity component (v?) to satisfy conservation of mass is ve = r(∂θ/∂t) = r(2) = 2r.
Part c)GivenU = 2.0 ft/s kinematic viscosity of air, v = 1.57 × 10-4 ft2/sAt x = 0
duct size, d1 = 1.0 ft
At x = 10 ft,
duct size, d2 = ?
Reynolds number for the laminar flow can be calculated as: Re = (ρUd/μ) Where, ρ = density of air = 0.0023769 slug/ft3μ = dynamic viscosity of air = 1.57 × 10-4 ft2/s
U = velocity of air
= 2.0 ft/s
d = diameter of duct
Re = (ρUd/μ)
= (0.0023769 × 2 × d/1.57 × 10-4)
For laminar flow, Reynolds number is less than 2300.
Thus, Re < 2300 => (0.0023769 × 2 × d/1.57 × 10-4) < 2300
=> d < 0.0726 ft or 0.871 inches or 22.15 mm
Assuming the thickness of the boundary layer to be negligible at x = 0, the velocity profile for the laminar flow in the duct at x = 0 is given by the Poiseuille’s equation:u = Umax(1 - (r/d1)2)
Here, Umax = U = 2 ft/s
Radius of the duct at x = 0 is r = d1/2 = 1/2 ft = 6 inches.
Thus, maximum velocity at x = 0 is given by:u = Umax(1 - (r/d1)2)
= 2 × (1 - (6/12)2)
= 0.5 ft/s
Let the velocity profile at x = 10 ft be given by u = Umax(1 - (r/d2)2)
The average velocity of the fluid at x = 10 ft should be U = 2 ft/s
As the boundary layer thickness increases in the direction of flow, it is necessary to increase the cross-sectional area of the duct for the same flow rate.Using the continuity equation,Q = A1 U1 = A2 U2
Where,Q = Flow rate of fluid
A1 = Area of duct at x
= 0A2
= Area of duct at x
= 10ftU1 = Velocity of fluid at x
= 0U2 = Velocity of fluid at x
= 10ft
Let d be the diameter of the duct at x = 10ft.
Then, A2 = πd2/4
Flow rate at x = 0 is given by,
Q = A1 U1 = π(1.0)2/4 × 0.5
= 0.3927 ft3/s
Flow rate at x = 10 ft should be the same as flow rate at x = 0.So,0.3927
= A2 U2
= πd2/4 × 2Or, d2
= 0.6283 ft = 7.54 inches
Thus, the diameter of the duct at x = 10 ft should be 7.54 inches or more to maintain a constant velocity of 2.0 ft/s.
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The mass of water steam in the cylinder, in kg, is A.1.082 B 0.526 C(1.003 D 0.972 0.972
The mass of water steam in the cylinder, in kg, is 0.972 (option D).The mass of water steam in the cylinder is 0.972 kg. To understand how this value was obtained.
we need to look at the problem in more detail. However, since no additional information is provided, let's make some assumptions that will help us answer the question. Let's assume that the water steam is in a cylinder under certain conditions.
We know that steam is produced when water is heated, so we can assume that the water in the cylinder is heated, causing it to evaporate and become steam.
We also know that the mass of water is usually measured in kilograms (kg), so we can assume that we are looking for the mass of steam in the cylinder in kg.Now, let's look at the options again. Option A is 1.082 kg, option B is 0.526 kg, option C is 1.003 kg, and option D is 0.972 kg.
We can immediately eliminate options A and C because they are greater than 1 kg, which is unlikely for the mass of steam in a cylinder. Option B is also too small, as it is less than 1 kg, and steam usually has a greater mass than water due to its gaseous state.
This leaves us with option D, which is 0.972 kg. This value is within the range of what we might expect for the mass of steam in a cylinder.
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An industrial plant absorbs 500 kW at a line voltage of 480 V with a lagging power factor of 0.8 from a three-phase utility line. The apparent power absorbed is most nearly O a. 625 KVA O b. 500 KVA O c. 400 KVA O d. 480 KVA
So, the most nearly apparent power absorbed is 625 KVA.Answer: The correct option is O a. 625 KVA.
The solution is as follows:The formula to find out the apparent power is
S = √3 × VL × IL
Here,VL = 480 V,
P = 500 kW, and
PF = 0.8.
For a lagging power factor, the apparent power is always greater than the real power; thus, the value of the apparent power will be greater than 500 kW.
Applying the above formula,
S = √3 × 480 × 625 A= 625 KVA.
So, the most nearly apparent power absorbed is 625 KVA.Answer: The correct option is O a. 625 KVA.
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For water at 3 MPa and 507 K, van der Waals equation of state predicts the following compressibility factors: 0.9231 0.0336 0.0650 Z= (a) (10 points) Using the Z values obtained from van der Waals equation, is water at saturation conditions? plain briefly
The compressibility factor, Z is utilized to account for the deviations from the ideal gas behavior that exist in real gases. The compressibility factor varies with pressure and temperature and is a function of the molar volume of a gas. The Van der Waals equation of state is one of the most commonly utilized equations of state that model real gases.
It is more accurate than the ideal gas law since it accounts for the volume occupied by the gas molecules and the attraction forces that exist between them. When the compressibility factor is equal to 1, the gas behaves ideally. If Z > 1, the gas behaves as if the attractive forces between the gas molecules are more powerful than those predicted by the ideal gas law.
If Z < 1, the gas behaves as if the attractive forces between the gas molecules are less powerful than those predicted by the ideal gas law. The compressibility factor, Z can be determined using the following equation:
Z = PV/RT where P is the pressure of the gas, V is its volume, R is the gas constant, and T is the temperature of the gas.
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In a food processing plant, hot liquid water is being transported in a pipe (k = 15 W/m-K, D₁= 2.5 cm, Do=3 cm, and L=10 m). The hot water flowing with a mass flow rate of 8.65Kg/s enters the pipe at 100°C and exits at 60°C. The plant supervisor thinks that since the hot water exits the pipe at 60°C, the pipe's outer surface temperature should be safe from thermal burn hazards. In order to prevent thermal burn upon accidental contact with skin tissue for individuals working in the vicinity of the pipe, the pipe's outer surface temperature should be kept below 45°C. Determine whether or not there is a risk of thermal burn on the pipe's outer surface. Assume the pipe outer surface temperature remains constant. The properties of water at the bulk mean temperature of Tb=(T₁+T₂)/2 = (100 + 60)/2 = 80°C are: Cp=4197 J/kg-K, k = 0.670 W/m-K, µ = 0.355 × 10-3 kg/m-s, and Pr = 2.22. The thermal conductivity of the pipe is given as Kpipe = 15 W/m-K. The outer pipe surface temperature is °C.
Where ρ is the density, v is the velocity of the water in the pipe, D is the diameter of the pipe, and µ is the viscosity of the water.
Re = (8.65/π(0.025)² × 0.355 × 10⁻³)
Re = 18,686.40
And the Nusselt number can be computed using the Reynolds number and Prandtl number.
Nu = 0.023R[tex]e^{(4/5)[/tex] P[tex]r^n[/tex]
Where n is the exponent of Prandtl number;
when the fluid is in turbulent flow, the exponent value is 0.4.
Nu = 0.023 × [tex](18,686.4)^{(4/5)[/tex] [tex](2.22)^{0.4[/tex]
Nu = 146.05
The Nusselt number is 146.05.
Q = πDL(U)(T₁-T₂)
Where L is the length of the pipe.
Q/πDL = U(T₁-T₂)
U = (Q/πDL)/(T₁-T₂)
U = (mCp(T₁-T₂)/πDL)/(T₁-T₂)
U = (mCp)/(πDL)
U = (8.65 × 4197)/(π × 0.03 × 10)
U = 11814.11 W/m²K
Substituting the calculated values into the expression for Q;
Q = (11814.11)(π × 0.03 × 10)(100-60)
Q = 21,165.41W
The expression for the outer pipe surface temperature is;
Tₒₑ = T₁ - Q/π
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3y+2x=z+1 3x+2z=8−5y
3z−1=x−2y
Determine the value of y by using Cramer's rule. 2.1.1 Sketch the graph of y= −2/ x
2.1.2 Is the graph of y= -2/ x continuous or discontinuous? 2.1.3 State any one line to which the graph is symmetrical to.
The value of y is 1 using Cramer's rule. The graph of y = -2/x is discontinuous because it has asymptotes at x = 0 and y = 0.
To determine the value of y using Cramer's rule, we need to solve the system of equations:
3y + 2x = z + 1
3x + 2z = 8 - 5y
3z - 1 = x - 2y
First, let's find the determinant of the coefficient matrix, D:
D = | 3 2 -1 |
| 2 0 2 |
| -2 -2 3 |
D = (3 * 0 * 3) + (2 * 2 * -2) + (-1 * 2 * -2) - (-1 * 0 * -2) - (3 * 2 * 2) - (2 * -2 * 3)
D = 0 + (-8) + 4 - 0 - 12 - (-12)
D = -4
Next, let's find the determinant of the matrix obtained by replacing the y-column with the constant terms:
Dy = | 3 2 -1 |
| 2 0 2 |
| -2 -2 3 |
Dy = (3 * 0 * 3) + (2 * 2 * -2) + (-1 * 2 * -2) - (-1 * 0 * -2) - (3 * 2 * 2) - (2 * -2 * 3)
Dy = -4
Now, let's find the determinant of the matrix obtained by replacing the x-column with the constant terms:
Dx = | 3 2 -1 |
| 8 0 2 |
| -2 -2 3 |
Dx = (3 * 0 * 3) + (2 * 2 * -2) + (-1 * 2 * -2) - (8 * 0 * 3) - (-2 * 2 * 3) - (2 * -2 * 3)
Dx = -4
Finally, let's find the determinant of the matrix obtained by replacing the z-column with the constant terms:
Dz = | 3 2 1 |
| 2 0 8 |
| -2 -2 -1 |
Dz = (3 * 0 * -1) + (2 * 8 * -2) + (1 * 2 * -2) - (-2 * 0 * -1) - (3 * 8 * -2) - (2 * -2 * -2)
Dz = -2
Now, we can calculate the value of y using Cramer's rule:
y = Dy / D
y = -4 / -4
y = 1
2.1.1 To sketch the graph of y = -2/x, we can plot some points and connect them. As x approaches infinity or negative infinity, y approaches 0. As x approaches 0, y approaches negative infinity. As x decreases, y increases, and as x increases, y decreases. The graph is a hyperbola that passes through the points (1, -2), (-1, 2), (2, -1), and (-2, 1), with asymptotes at y = 0 and x = 0.
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Paper Clip
For each component of the materials used in the paper clip, explore the following:
Emphasis on material selection, process selection, and cost of manufacturing.
Examine the advantages, disadvantages, and costs of the materials used?
Study the bending method for manufacturing the paper clip.
Determine at what stage of manufacture and by which method should the surface treatment be performed.
Determine the final coating of the paper clip.
The main answer:Materials used in the paper clip There are different types of materials used in the manufacturing of the paper clip. Some of the most commonly used materials include stainless steel, zinc-coated steel, plastic, and aluminum.The material selection is crucial in the manufacturing of the paper clip.
The material must be strong enough to hold papers together. Additionally, it must be flexible and malleable to allow the bending of the paper clip.Process selection is also an essential aspect of paper clip manufacturing. The production process involves wire drawing, heat treatment, wire forming, surface treatment, and finishing.Cost of manufacturing is another essential aspect of the paper clip. The manufacturing cost should be kept low to allow for a low-cost product. Advantages, disadvantages, and costs of materialsStainless steel is the most commonly used material for paper clip manufacturing. Its advantages include high durability, corrosion resistance, and high strength.
However, its main disadvantage is that it's expensive to manufacture.Zinc-coated steel is also another material used for paper clip manufacturing. Its advantages include low cost and rust resistance. However, its main disadvantage is that it's not as strong as stainless steel.Plastic is another material used for paper clip manufacturing. Its advantages include low cost and versatility. However, its main disadvantage is that it's not strong enough for heavy-duty use.Aluminum is another material used for paper clip manufacturing. Its advantages include high strength and lightweight. However, its main disadvantage is that it's expensive to manufacture.Bending method for manufacturing the paper clipThe bending method involves the use of a wire bender to shape the wire into a paper clip. The wire is first cut into a specific length and then fed into the bender, which shapes it into a paper clip.The bending method is fast and efficient and can produce paper clips in large quantities.
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Stickman has built a rocket sled. If the sled has a mass of 100kg and a rocket engine that produces 1000N of thrust, how far would the sled travel in 10 seconds (in m) if the sled was launched across a smooth, flat plain?
The rocket sled, with a mass of 100kg and a thrust of 1000N, would travel 500 meters in 10 seconds across a smooth, flat plain.
To calculate the distance the sled would travel, we can use Newton’s second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration. In this case, the net force is the thrust produced by the rocket engine, and the acceleration is the sled’s acceleration.
First, we need to determine the acceleration of the sled. We can use the formula:
Acceleration = Net Force / Mass
In this case, the net force is 1000N (thrust) and the mass is 100kg:
Acceleration = 1000N / 100kg = 10 m/s²
Now that we have the acceleration, we can use the kinematic equation to calculate the distance traveled:
Distance = Initial Velocity × Time + 0.5 × Acceleration × Time²
Since the sled starts from rest, the initial velocity is 0 m/s. Plugging in the values:
Distance = 0 × 10 + 0.5 × 10 × 10²
Distance = 0 + 0.5 × 10 × 100
Distance = 0 + 0.5 × 1000
Distance = 0 + 500
Distance = 500 meters
Therefore, the sled would travel a distance of 500 meters in 10 seconds across a smooth, flat plain.
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A single stage reciprocating compressor takes 1m of air per minute and 1.013 bar and 15°C and delivers at 7 bar. Assuming Adiabatic law (n=1.35) and no clearance. Calculate: 1.1. Mass flow rate (1.226 kg/min) 1.2. Delivery Temperature (475.4 K) 1.3. Indicated power (4.238 kW) This same compressor is now driven at 300 rpm, has a stroke to bore ratio of (1,5:1), it has a mechanical efficiency for the compressor of 85% and motor transmission efficiency of 90%. Calculate: 1.4. Volume per cycle (0.00333 m²/cycle) 1.5. Cylinder bore diameter (141.4 mm) 1.6. Power to the compressor (4.99 kW) 1.7. Motor power needed (5.54 kW) 1.8. The isothermal power (3.265 kW) 1.9. The isothermal efficiency (77%)
Therefore, the delivery temperature is 475.4 K.1.3. Calculation of Indicated Power The indicated power of the compressor can be calculated using the formula, Power = P * Q * n Where P is the pressure, Q is the flow rate, and n is the polytropic index.
Motor power = Power to compressor / η_tHere,
Power to compressor = 4.99 kW and
η_t = 0.90
So, the motor power needed is 5.54 kW.1.8. Calculation of Isothermal Power Isothermal Power can be calculated using the formula, P1V1/T1 = P2V2/T2 So, the isothermal power is 3.265 kW.1.9.
Calculation of Isothermal Efficiency The isothermal efficiency can be calculated using the formula, Isothermal efficiency = (Isothermal power / Indicated power) * 100 Substituting the values, we get,
Isothermal efficiency = (3.265 / 4.238) * 100 = 77%
Therefore, the isothermal efficiency is 77%.
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A vapor-compression refrigeration system utilizes a water-cooled intercooler with ammonia as the refrigerant. The evaporator and condenser temperatures are -10 and 40°C, respectively. The mass flow rate of the intercooler water is 0.35 kg/s with a change in enthalpy of 42 kJ/kg. The low-pressure compressor discharges the refrigerant at 700 kPa. Assume compression to be isentropic. Sketch the schematic and Ph diagrams of the system and determine: (a) the mass flow rate of the ammonia refrigerant, (b) the capacity in TOR, (c) the total compressor work, and (d) the COP.
In a vapor-compression refrigeration system with an ammonia refrigerant and a water-cooled intercooler, the goal is to determine the mass flow rate of the refrigerant, the capacity in TOR (ton of refrigeration), the total compressor work, and the coefficient of performance (COP).
To determine the mass flow rate of the ammonia refrigerant, we need to apply mass and energy balance equations to the system. The mass flow rate of the intercooler water and its change in enthalpy can be used to calculate the heat transfer in the intercooler and the heat absorbed in the evaporator. The capacity in TOR can be calculated by converting the heat absorbed in the evaporator to refrigeration capacity. TOR is a unit of refrigeration capacity where 1 TOR is equivalent to 12,000 BTU/hr or 3.517 kW.
The total compressor work can be calculated by considering the isentropic compression process and the pressure ratio across the compressor. The work done by the compressor is equal to the change in enthalpy of the refrigerant during compression. The COP of the refrigeration system can be determined by dividing the refrigeration capacity by the total compressor work. COP represents the efficiency of the system in providing cooling for a given amount of work input. Schematic and Ph diagrams can be sketched to visualize the system and understand the thermodynamic processes involved. These diagrams aid in determining the properties and states of the refrigerant at different stages of the cycle.
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A fluid in a fire hose with a 46.5 mm radius, has a velocity of 0.56 m/s. Solve for the power, hp, available in the jet at the nozzle attached at the end of the hose if its diameter is 15.73 mm. Express your answer in 4 decimal places.
Given data: Radius of hose
r = 46.5m
m = 0.0465m
Velocity of fluid `v = 0.56 m/s`
Diameter of the nozzle attached `d = 15.73 mm = 0.01573m`We are supposed to calculate the power, hp available in the jet at the nozzle attached to the hose.
Power is defined as the rate at which work is done or energy is transferred, that is, P = E/t, where E is the energy (J) and t is the time (s).Now, Energy E transferred by the fluid is given by the formula E = 1/2mv² where m is the mass of the fluid and v is its velocity.We can write m = (ρV) where ρ is the density of the fluid and V is the volume of the fluid. Volume of the fluid is given by `V = (πr²l)`, where l is the length of the hose through which fluid is coming out, which can be assumed to be equal to the diameter of the nozzle or `l=d/2`.
Thus, `V = (πr²d)/2`.Energy transferred E by the fluid can be expressed as Putting the value of V in the above equation, we get .Now, the power of the fluid P, can be written as `P = E/t`, where t is the time taken by the fluid to come out from the nozzle.`Putting the given values of r, d, and v, we get Thus, the power available in the jet at the nozzle attached to the hose is 0.3011 hp.
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A fan with 40% static efficiency has a capacity of 60,000 ft3/ hr at 60°F and barometer of 30 in Hg and gives a static pressure of 2 in WG on full delivery. What size electric motor should be used to drive this fan?
O 0.8 HP O 1.2 HP O 1.6 HP O 2 HP
To determine the size of the electric motor required to drive the fan, we need to consider the fan's capacity, static efficiency, and static pressure. The correct answer is: O 0.8 HP.
The capacity of the fan is given as 60,000 ft3/hr, which indicates the volume of air the fan can deliver in one hour. The static pressure is stated as 2 in WG (inches of water gauge), which represents the pressure the fan can generate against resistance.
The static efficiency of the fan is mentioned as 40%, which indicates the ratio of the actual work performed by the fan to the input power.
To calculate the power required to drive the fan, we can use the following formula:
Power (HP) = (Flow Rate * Static Pressure) / (Fan Efficiency * 6356)
Converting the given capacity of 60,000 ft3/hr to cubic feet per minute (CFM), we get:
Flow Rate (CFM) = 60,000 ft3/hr / 60 min/hr = 1000 ft3/min
Substituting the given values into the formula, we have:
Power (HP) = (1000 * 2) / (0.4 * 6356) ≈ 0.793 HP
Rounding off to the nearest option, the size of the electric motor required to drive this fan is 0.8 HP.
Therefore, the correct answer is: O 0.8 HP.
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