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A single phase overhead transmission line delivers a power of 5500 kW to a load at 11 kV. The receiving end voltage leads the current by 45°. The resistance and the inductive reactance of the transmission line are 10 Ω and 10 Ω respectively. The sending end voltages is
- a)12 kV
- b)15 kV
- c)21 kV
- d)22 kV
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A single phase overhead transmission line delivers a power of 5500 kW ...
Concept:
Short Transmission Line:
When the length of the overhead transmission line is up to about 50 km and the line voltage is comparatively up to 20 kV.
Due to smaller lengths and lower voltage, the capacitance effect is small and may be neglected.
Considered a short transmission line of resistance R and Reactance XL over a length.
Vs is the sending end voltage
VR is the receiving end voltage
I is the load current.
cos ϕR is receiving end power factor
and, cos ϕs is sending end power factor
The phasor diagram of the system can be drawn by taking load current as a reference,
From the phasor,
(OC)2 = (OD)2 + (DC)2
(OC)2 = (OE + ED)2 + (DB + BC)2
V2S = (VR cos ϕR + IR)2 + (VR sin ϕR + IXL)2
Calculation:
Given, P = 5500 kW
VR = 11 kV
R = 10 Ω
XL = 10 Ω
P = VR I cos ϕR
cos ϕR = cos (45°) = 0.707
From above concept,
sin ϕR = 0.707
(VR cos ϕR + IR) = (11000 × 0.707) + (707.21 × 10) = 14849 volt
(VR sin ϕR + IXL) = (11000 × 0.707) + (707.21 × 10) = 14849 volt
⇒ VS = 21 kV
Short Transmission Line:
When the length of the overhead transmission line is up to about 50 km and the line voltage is comparatively up to 20 kV.
Due to smaller lengths and lower voltage, the capacitance effect is small and may be neglected.
Considered a short transmission line of resistance R and Reactance XL over a length.
Vs is the sending end voltage
VR is the receiving end voltage
I is the load current.
cos ϕR is receiving end power factor
and, cos ϕs is sending end power factor
The phasor diagram of the system can be drawn by taking load current as a reference,
From the phasor,
(OC)2 = (OD)2 + (DC)2
(OC)2 = (OE + ED)2 + (DB + BC)2
V2S = (VR cos ϕR + IR)2 + (VR sin ϕR + IXL)2
Calculation:
Given, P = 5500 kW
VR = 11 kV
R = 10 Ω
XL = 10 Ω
P = VR I cos ϕR
cos ϕR = cos (45°) = 0.707
From above concept,
sin ϕR = 0.707
(VR cos ϕR + IR) = (11000 × 0.707) + (707.21 × 10) = 14849 volt
(VR sin ϕR + IXL) = (11000 × 0.707) + (707.21 × 10) = 14849 volt
⇒ VS = 21 kV
Most Upvoted Answer
A single phase overhead transmission line delivers a power of 5500 kW ...
Concept:
Short Transmission Line:
When the length of the overhead transmission line is up to about 50 km and the line voltage is comparatively up to 20 kV.
Due to smaller lengths and lower voltage, the capacitance effect is small and may be neglected.
Considered a short transmission line of resistance R and Reactance XL over a length.
Vs is the sending end voltage
VR is the receiving end voltage
I is the load current.
cos ϕR is receiving end power factor
and, cos ϕs is sending end power factor
The phasor diagram of the system can be drawn by taking load current as a reference,
From the phasor,
(OC)2 = (OD)2 + (DC)2
(OC)2 = (OE + ED)2 + (DB + BC)2
V2S = (VR cos ϕR + IR)2 + (VR sin ϕR + IXL)2
Calculation:
Given, P = 5500 kW
VR = 11 kV
R = 10 Ω
XL = 10 Ω
P = VR I cos ϕR
cos ϕR = cos (45°) = 0.707
From above concept,
sin ϕR = 0.707
(VR cos ϕR + IR) = (11000 × 0.707) + (707.21 × 10) = 14849 volt
(VR sin ϕR + IXL) = (11000 × 0.707) + (707.21 × 10) = 14849 volt
⇒ VS = 21 kV
Short Transmission Line:
When the length of the overhead transmission line is up to about 50 km and the line voltage is comparatively up to 20 kV.
Due to smaller lengths and lower voltage, the capacitance effect is small and may be neglected.
Considered a short transmission line of resistance R and Reactance XL over a length.
Vs is the sending end voltage
VR is the receiving end voltage
I is the load current.
cos ϕR is receiving end power factor
and, cos ϕs is sending end power factor
The phasor diagram of the system can be drawn by taking load current as a reference,
From the phasor,
(OC)2 = (OD)2 + (DC)2
(OC)2 = (OE + ED)2 + (DB + BC)2
V2S = (VR cos ϕR + IR)2 + (VR sin ϕR + IXL)2
Calculation:
Given, P = 5500 kW
VR = 11 kV
R = 10 Ω
XL = 10 Ω
P = VR I cos ϕR
cos ϕR = cos (45°) = 0.707
From above concept,
sin ϕR = 0.707
(VR cos ϕR + IR) = (11000 × 0.707) + (707.21 × 10) = 14849 volt
(VR sin ϕR + IXL) = (11000 × 0.707) + (707.21 × 10) = 14849 volt
⇒ VS = 21 kV
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Community Answer
A single phase overhead transmission line delivers a power of 5500 kW ...
Degrees. The resistance and inductive reactance of the line are 0.5 ohm and 1 ohm respectively.
To calculate the sending end voltage and the power loss in the transmission line, we can use the following formulas:
Sending end voltage:
V_s = V_r + I * (R + jX)
Power loss in the transmission line:
P_loss = 3 * I^2 * R
Given:
Power, P = 5500 kW
Load voltage, V_r = 11 kV
Phase angle, δ = 45 degrees
Resistance, R = 0.5 ohm
Inductive reactance, X = 1 ohm
First, we need to convert the power from kW to W and the load voltage from kV to V:
P = 5500 * 1000 = 5,500,000 W
V_r = 11 * 1000 = 11,000 V
We can calculate the current using the power and the load voltage:
P = 3 * V_r * I * cos(δ)
5,500,000 = 3 * 11,000 * I * cos(45)
I * cos(45) = 5,500,000 / (3 * 11,000)
I * cos(45) = 55/3
I = (55/3) / cos(45)
I ≈ 12,903.3 A
Next, we can calculate the imaginary part of the current using the power factor (sin(δ)):
I * sin(δ) = P / (3 * V_r * sin(δ))
I * sin(45) = 5,500,000 / (3 * 11,000 * sin(45))
I * sin(45) = 55/3
I = (55/3) / sin(45)
I ≈ 12,903.3 A
Now, we can calculate the sending end voltage using the load voltage, current, resistance, and reactance:
V_s = V_r + I * (R + jX)
V_s = 11,000 + 12,903.3 * (0.5 + j * 1)
V_s = 11,000 + 12,903.3 * (0.5 + j)
V_s = 11,000 + 6,451.65 + j * 12,903.3
V_s ≈ 17,451.65 + j * 12,903.3 V
Finally, we can calculate the power loss in the transmission line using the current and resistance:
P_loss = 3 * I^2 * R
P_loss = 3 * (12,903.3)^2 * 0.5
P_loss = 3 * 166,811,810.89 * 0.5
P_loss ≈ 249,217,716.335 W
Therefore, the sending end voltage is approximately 17,451.65 + j * 12,903.3 V and the power loss in the transmission line is approximately 249,217,716.335 W.
To calculate the sending end voltage and the power loss in the transmission line, we can use the following formulas:
Sending end voltage:
V_s = V_r + I * (R + jX)
Power loss in the transmission line:
P_loss = 3 * I^2 * R
Given:
Power, P = 5500 kW
Load voltage, V_r = 11 kV
Phase angle, δ = 45 degrees
Resistance, R = 0.5 ohm
Inductive reactance, X = 1 ohm
First, we need to convert the power from kW to W and the load voltage from kV to V:
P = 5500 * 1000 = 5,500,000 W
V_r = 11 * 1000 = 11,000 V
We can calculate the current using the power and the load voltage:
P = 3 * V_r * I * cos(δ)
5,500,000 = 3 * 11,000 * I * cos(45)
I * cos(45) = 5,500,000 / (3 * 11,000)
I * cos(45) = 55/3
I = (55/3) / cos(45)
I ≈ 12,903.3 A
Next, we can calculate the imaginary part of the current using the power factor (sin(δ)):
I * sin(δ) = P / (3 * V_r * sin(δ))
I * sin(45) = 5,500,000 / (3 * 11,000 * sin(45))
I * sin(45) = 55/3
I = (55/3) / sin(45)
I ≈ 12,903.3 A
Now, we can calculate the sending end voltage using the load voltage, current, resistance, and reactance:
V_s = V_r + I * (R + jX)
V_s = 11,000 + 12,903.3 * (0.5 + j * 1)
V_s = 11,000 + 12,903.3 * (0.5 + j)
V_s = 11,000 + 6,451.65 + j * 12,903.3
V_s ≈ 17,451.65 + j * 12,903.3 V
Finally, we can calculate the power loss in the transmission line using the current and resistance:
P_loss = 3 * I^2 * R
P_loss = 3 * (12,903.3)^2 * 0.5
P_loss = 3 * 166,811,810.89 * 0.5
P_loss ≈ 249,217,716.335 W
Therefore, the sending end voltage is approximately 17,451.65 + j * 12,903.3 V and the power loss in the transmission line is approximately 249,217,716.335 W.
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A single phase overhead transmission line delivers a power of 5500 kW to a load at 11 kV. The receiving end voltage leads the current by 45°. The resistance and the inductive reactance of the transmission line are 10 Ω and 10 Ω respectively. The sending end voltages isa)12 kVb)15 kVc)21 kVd)22 kVCorrect answer is option 'C'. Can you explain this answer?
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