Answer:
Explanation:
f = 50.0 Hz, L = 0.650 H, π = 3.14
C = 4.80 μF, R = 301 Ω resistor. V = 120volts
XL = wL = 2πfL
= 2×3.14×50* 0.650
= 204.1 Ohm
Xc= 1/wC
Xc = 1/2πfC
Xc = 1/2×3.14×50×4.80μF
= 1/0.0015072
= 663.48Ohms
1. Total impedance, Z = sqrt (R^2 + (Xc-XL)^2)= √ 301^2+ (663.48Ohms - 204.1 Ohm)^2
√ 90601 + (459.38)^2
√ 90601+211029.98
√ 301630.9844
= 549.209
Z = 549.21Ohms
2. I=V/Z = 120/ 549.21Ohms =0.218Ampere
3. P=V×I = 120* 0.218 = 26.16Watt
Note that
I rms = Vrms/Xc
= 120/663.48Ohms
= 0.18086A
4. I(max) = I(rms) × √2
= 0.18086A × 1.4142
= 0.2557
= 0.256A
5. V=I(max) * XL
= 0.256A ×204.1
=52.2496
= 52.250volts
6. V=I(max) × Xc
= 0.256A × 663.48Ohms
= 169.85volts
7. Xc=XL
1/2πfC = 2πfL
1/2πfC = 2πf× 0.650
1/2×3.14×f×4.80μF = 2×3.14×f×0.650
1/6.28×f×4.8×10^-6 = 4.082f
1/0.000030144× f = 4.082×f
1 = 0.000030144×f×4.082×f
1 = 0.000123f^2
f^2 = 1/0.000123048
f^2 = 8126.922
f =√8126.922
f = 90.14 Hz
Answer:
The work furnished by the compressor is 
The minimum work required for the state to change is 
Explanation:
The explanation to these solution is on the first, second , third and fourth uploaded image respectively
Answer:
The power developed by engine is 167.55 KW
Explanation:
Given that
Mean effective pressure = 6.4 bar
Speed = 2000 rpm
We know that power is the work done per second.
So
We have to notice one point that we divide by 120 instead of 60, because it is a 4 cylinder engine.
P=167.55 KW
So the power developed by engine is 167.55 KW
Answer:
15.8
0.0944
Explanation:
L = 1.5
B = 1.0
Speed of water = 15cm
Temperature = 20⁰C
At 20⁰C
Specific weight = 9790
Kinematic viscosity v = 1.00x10^-4m²/s
Dynamic viscosity u = 1.00x10^-3
Density p = 998kg/m²
Reynolds number
= 0.15x1.5/1.00x10^-4
= 225000
S = 5
5x1.5/225000^1/2
= 0.0158
= 15.8mm
Resistance on one side of plate
F = 0.664x1x1.0x10^-3x0.15x225000^1/2
= 0.04724N
Total resistance
= 2N
= 2x0.04724
= 0.0944N
