The short-circuit current is approximately 1443 A (b).
What is the short-circuit current of a generator with specific parameters?However, I can provide you with the explanations for the two questions you presented:
A generator rated 600 kVA, 2,400 V, 60 Hz, 3-phase, 6-poles, and wye-connected with 10% synchronous reactance. The short-circuit current can be calculated using the formula:
[tex]\[I_{\text{short-circuit}} = \frac{V_{\text{rated}}}{\sqrt{3}X_{\text{d}}}\][/tex]
where[tex]\(V_{\text{rated}}\)[/tex]is the rated voltage and [tex]\(X_{\text{d}}\)[/tex] is the synchronous reactance. Plugging in the given values, we have:
[tex]\[I_{\text{short-circuit}} = \frac{2400}{\sqrt{3}\times0.1} \approx 1443 \text{ A}\][/tex]
Therefore, the correct answer is (b) 1443 A.
A 200-Hp, 2,200 V, 3-phase star-connected synchronous motor with a synchronous impedance of 0.3+j302 per phase. To find the induced emf per phase, we can use the formula:
[tex]\[E_{\text{induced}} = V_{\text{rated}} + I_{\text{load}}Z_{\text{sync}}\][/tex]
where[tex]\(V_{\text{rated}}\)[/tex]is the rated voltage, [tex]\(I_{\text{load}}\)[/tex]is the load current, and \[tex](Z_{\text{sync}}\)[/tex] is the synchronous impedance. Since the motor operates at full load with a power factor of 0.8 leading and an efficiency of 94%, we can calculate the load current as follows:
[tex]\[P_{\text{load}} = \sqrt{3}V_{\text{rated}}I_{\text{load}}\cos\phi\][/tex]
where
[tex]e \(P_{\text{load}}\)[/tex]is the load power. Rearranging the equation, we find:
[tex]\[I_{\text{load}} = \frac{P_{\text{load}}}{\sqrt{3}V_{\text{rated}}\cos\phi}\][/tex]
Plugging in the given values, we get:
[tex]\[I_{\text{load}} = \frac{200 \times 746}{\sqrt{3} \times 2200 \times 0.8} \approx 114.15 \text{ A}\][/tex]
Now, substituting the values into the induced emf equation, we have:
[tex]\[E_{\text{induced}} = 2200 + 114.15 \times (0.3 + j302) \approx 1354 \text{ V}\][/tex]
Therefore, the correct answer is (a) 1,354 V.
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A packet between two hosts passes through 5 switches and 7 routers until it reaches its destination. Between the sending application and the receiving application, how often is it handled by the transport layer?
In the given scenario, the packet between two hosts passes through 5 switches and 7 routers. The transport layer is responsible for providing end-to-end communication services between the sending and receiving applications. Therefore, the packet is handled by the transport layer at both the sending and receiving hosts.
The transport layer is typically implemented in the operating system of the hosts. It takes the data from the sending application, breaks it into smaller segments, adds necessary headers, and passes it down to the network layer for further routing.
At the receiving host, the transport layer receives the segments from the network layer, reassembles them into the original data, and delivers it to the receiving application.
Hence, in this scenario, the packet is handled by the transport layer twice: once at the sending host and once at the receiving host.
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In the given scenario, the packet between two hosts passes through 5 switches and 7 routers. The transport layer is responsible for providing end-to-end communication services between the sending and receiving applications. Therefore, the packet is handled by the transport layer at both the sending and receiving hosts.
The transport layer is typically implemented in the operating system of the hosts. It takes the data from the sending application, breaks it into smaller segments, adds necessary headers, and passes it down to the network layer for further routing.
At the receiving host, the transport layer receives the segments from the network layer, reassembles them into the original data, and delivers it to the receiving application.
Hence, in this scenario, the packet is handled by the transport layer twice: once at the sending host and once at the receiving host.
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A ship displaces 9450 tons has a block coefficient of 0.7. The area of immersed midship is 106 square meters. If beam is 0.13 × Length = 2.1 × draft. Calculate the length of the ship and prismatic coefficient.
A ship 135m long, 18m beam and 7.6m draft has a displacement of 14,000 tons. Area of load water plane is 1925 square meters and area immersed midship is 130 square meters. Calculate Coefficient of midship, Waterplane coefficient, Block coefficient, Prismatic coefficient.
The Coefficient of midship is 0.956, the Waterplane coefficient is 0.796, the Block coefficient is 0.975, and the Prismatic coefficient is 0.424.
First problem: Given:
- Ship displacement: 9450 tons
- Block coefficient: 0.7
- Immersed midship area: 106 square meters
- Beam: 0.13 x Length
- Let's assume a rectangular midship section.
- The displacement volume of the ship can be calculated by dividing the displacement by the density of water. Since 1 tonne = 1 m^3 for seawater, the displacement volume is 9450 m^3.
- The block coefficient is the ratio of the volume of displacement to the volume of a rectangular block having the same length, beam, and draft. Hence, the volume of the block can be found by dividing the displacement volume by the block coefficient, which gives us 13500 m^3.
- The beam of the ship is 0.13 x in Length, which means the width of the rectangular midship section is 0.13L.
- We can then find the length of the midship section by dividing the volume of the block by the area of the midship section, which is V/A = L x B x T. Substituting the given values,
V/A = (0.13L) x L x T = 13500 m^3
T = 9450/(0.7 x 106) = 123.94 m
L = (13500/106)/123.94 = 0.989 km
- The prismatic coefficient is the ratio of the volume of the underwater portion of the hull to the volume of a similarly shaped prism having the same length as the waterline length. We can assume the midship section has a rectangular shape. Hence, the prismatic coefficient is the ratio of the immersed midship area to the product of the waterline length (LWL) and the beam, i.e., Cp = Am / (L x B).
- The length of the ship is 0.989 km and the prismatic coefficient is Am / (L x B) = 106 / (0.989 x 0.13) = 657.
Second problem:
Given:
- Length: 135 m
- Beam: 18 m
- Draft: 7.6 m
- Displacement: 14000 tons
- Load water plane area: 1925 square meters
- Immersed midship area: 130 square meters
- The block coefficient is the ratio of the volume of displacement to the volume of a rectangular block having the same length, beam, and draft. Hence, the volume of the block can be found by dividing the displacement volume by the block coefficient, which is V_block = D/ρ = 14000 x 1000 / 1025 = 13658.54 cubic meters.
- The midship coefficient is the ratio of the immersed midship area to the product of the beam and draft, which is given by C_mid = Am / (B x T) = 130 / (18 x 7.6) = 0.956.
- The waterplane coefficient is the ratio of the load waterplane area to the product of the waterline length (LWL) and the beam, which is given by C_wp = Awp / (L x B) = 1925 / (135 x 18) = 0.796.
- The prismatic coefficient is the ratio of the volume of the underwater portion of the hull to the volume of a similarly shaped prism having the same length as the waterline length. We can assume the midship section has a rectangular shape. Hence, the prismatic coefficient is the ratio of the immersed midship area to the product of the waterline length (LWL) and the beam, i.e.,
Cp = Am / (L x B) = 130 / (135 x 18) = 0.424.
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In flow measurements experiment using Venturi meter (D₁=20 mm,D₂=10 mm ) a reading of 2 liters were flow in 6 seconds. The head loss (cm) is a 86 b 93 c 54 d 75
The head loss in the flow measurements experiment using a Venturi meter with D₁=20 mm and D₂=10 mm, where 2 liters were flowing in 6 seconds, is 86 cm.
The head loss in a Venturi meter can be calculated using Bernoulli's equation. The formula for head loss (h) in a Venturi meter is given by h = (V₁² - V₂²) / (2g), where V₁ and V₂ are the velocities at sections 1 and 2 respectively, and g is the acceleration due to gravity.
To calculate the head loss, we need to determine the velocities at sections 1 and 2. Since the flow rate is given as 2 liters in 6 seconds, we can convert it to m³/s by dividing by 1000. Thus, the flow rate (Q) is 0.002 m³/s.
Using the equation of continuity, A₁V₁ = A₂V₂, where A₁ and A₂ are the cross-sectional areas at sections 1 and 2 respectively, we can find V₂ in terms of V₁.
Given that D₁=20 mm and D₂=10 mm, we can calculate the areas A₁ and A₂.
A₁ = π(D₁/2)² = π(0.02/2)² = 0.000314 m²
A₂ = π(D₂/2)² = π(0.01/2)² = 0.0000785 m²
By rearranging the equation of continuity, we find V₂ = (A₁/A₂)V₁.
Now, we can substitute the values into the head loss formula:
h = (V₁² - V₂²) / (2g).
Plugging in the values, we can solve for V₁.
By measuring the time of 6 seconds, we can calculate the average velocity (V₁) as V₁ = Q / A₁ = 0.002 / 0.000314 = 6.369 m/s.
Substituting the values of V₁ and V₂ into the head loss formula:
h = (6.369² - ((0.000314/0.0000785)*6.369)²) / (2 * 9.81)
≈ 86 cm.
The head loss in the given flow measurements experiment using a Venturi meter with D₁=20 mm and D₂=10 mm, where 2 liters have flowed in 6 seconds, is approximately 86 cm. This head loss is an important parameter to consider when analyzing fluid flow and pressure variations in the Venturi meter.
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