Which of the following units of measurement is denoted by a single apostrophe mark (')?

4.71 A full-wave rectifier circuit with a 1-kΩ operates from a 120-V (rms) 60-Hz household supply through a 6-to-1 transformer h

brilliants [131]

Answer:

$V_{p (load)} = 28,3 V - 0,7 V = 27,6 V$

$V_{p (load)} = 27,6 V\\V_{avg} = 17,57 V$

The upper diode conduces in the odd half cycles. The lower diode conduces in the even half cycles.

$I_{avg} = \frac{V_{avg}}{R_{load}} = \frac{17,57 V}{1000 Ω} = 17,57 mA$

Explanation:

The peak voltage after the 6 to 1 step down is $V_{p} = \frac{120}{6} \sqrt{2} = 28,3V$ . Then, the peak voltage of the rectified output is $V_{p} - [tex]V_{avg} = \frac{2V_{p (load)} }{\pi} = \frac{55,2 V}{\pi } = 17,6 V$ V_{d}[/tex] and according to the statement, the diodes can be modeled to be $V_{d} = 0,7 V$ . Then, the peak voltage in the load is $V_{p (load)} = 28,3 V - 0,7 V = 27,6 V$ .

The upper diode conduces in the odd half cycles. The lower diode conduces in the even half cycles.

The average output voltage is calculated as:

$V_{p (load)} = 27,6 V\\V_{avg} = 17,57 V$

The average current in the load is calculated as:

$I_{avg} = \frac{V_{avg}}{R_{load}} = \frac{17,57 V}{1000 Ω} = 17,57 mA$

4 0

3 years ago

Bulk wind shear is calculated by finding the vector difference between the winds at two different heights. Using the supercell w

ivanzaharov [21]

Answer:

See explaination

Explanation:

2. 0-1 km shear value: taking winds at 1000mb and 850 mb

15 kts south easterly and 50 kts southerly

Vector difference 135/15 and 180/50 will be 170/61 or southerly 61 kts

3. 0-6 km shear value: taking winds at 1000 mb and 500 mb

15 kts south easterly and 40 kts westerly

Vector difference 135/15 and 270/40 will be 281/51 kts

please see attachment

5 0

3 years ago

The bulk density of a compacted soil specimen (Gs = 2.70) and its water content are 2060 kg/m^3 and 15.3%, respectively. If the

yaroslaw [1]

Answer:

the saturated density should be

Explanation:

4 0

3 years ago

Oil with a density of 800 kg/m3 is pumped from a pressure of 0.6 bar to a pressure of 1.4 bar, and the outlet is 3 m above the i

Naddik [55]

Answer:

23.3808 kW

20.7088 kW

Explanation:

ρ = Density of oil = 800 kg/m³

P₁ = Initial Pressure = 0.6 bar

P₂ = Final Pressure = 1.4 bar

Q = Volumetric flow rate = 0.2 m³/s

A₁ = Area of inlet = 0.06 m²

A₂ = Area of outlet = 0.03 m²

Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s

Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s

Height between inlet and outlet = z₂ - z₁ = 3m

Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation

$\frac {P_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h=\frac {P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\\Rightarrow h=\frac{P_2-P_1}{\rho g}+\frac{V_2^2-V_1^2}{2g}+z_2-z_1\\\Rightarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+\frac{6.67_2^2-3.33^2}{2\times 9.81}+3\\\Rightarrow h=14.896\ m$

Work done by pump

$W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 14.896\\\Rightarrow W_{p}=23380.8\ W$

∴ Power input to the pump 23.3808 kW

Now neglecting kinetic energy

$h=\frac{P_2-P_1}{\rho g}+z_2-z_1\\\Righarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+3\\\Righarrow h=13.19\ m\\$

Work done by pump

$W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 13.193\\\Rightarrow W_{p}=20708.8\ W$

∴ Power input to the pump 20.7088 kW

6 0

3 years ago

What is the function maintenance? List some important steps for vibration monitoring based maintenance.

Mrrafil [7]

The maintenance is in charge of controlling that all the machines of a company are constantly running in order to avoid damages that cause loss of money when the machines fail.

The maintenance based on vibration monitoring allows to predict failures in some rotating machines such as:

1. worn bearings

2.alignment

3.balance

4. affected gears

5. bent shafts

6. rocks

7.gags

8. eccentricity

9. failures of electrical origin

4 0

2 years ago