Help me my life is depending on this question... 2+2

Two loads connected in parallel draw a total of 2.4 kW at 0.8 pf lagging from a 120-V rms, 60-Hz line. One load absorbs 1.5 kW a

stealth61 [152]

Answer: a) 0.948 b) 117.5µf

Explanation:

Given the load, a total of 2.4kw and 0.8pf

V= 120V, 60 Hz

P= 2.4 kw, cos θ= 80

P= S sin θ - (p/cos θ) sin θ

= P tan θ(cos^-1 (0.8)

=2.4 tan(36.87)= 1.8KVAR

S= 2.4 + j1. 8KVA

1 load absorbs 1.5 kW at 0.707 pf lagging

P= 1.5 kW, cos θ= 0.707 and θ=45 degree

Q= Ptan θ= tan 45°

Q=P=1.5kw

S1= 1.5 +1.5j KVA

S1 + S2= S

2.4+j1.8= 1.5+1.5j + S2

S2= 0.9 + 0.3j KVA

S2= 0.949= 18.43 °

Pf= cos(18.43°) = 0.948

b.) pf to 0.9, a capacitor is needed.

Pf = 0.9

Cos θ= 0.9

θ= 25.84 °

(WC) V^2= P (tan θ1 - tan θ2)

C= 2400 ( tan (36. 87°) - tan (25.84°)) /2 πf × 120^2

f=60, π=22/7

C= 117.5µf

7 0

3 years ago

Make a proposal to add a small pizza shop to a historical part of town. How could it be designed to “fit” into the area?

spin [16.1K]

It could be designed because if we had a small little pizza shop it would save our problems of going through all the trouble just for pizza

7 0

3 years ago

Read 2 more answers

8. Block A shown in the figure below weighs 2000 N. The chord attached to A passes over a

Kobotan [32]

Answer:

Read the passage. Then, answer the questions about the metaphor in boldface in the text.

Lately, I've been so overwhelmed with school and sports. There was a time when I enjoyed going to classes and going to practice every afternoon. Now, everything is piling up and wearing me down. Thankfully, I get to see you every day. You are truly the sunshine of my life. Thank you for making me laugh when I'm feeling down.

What is the context of the passage?

What is being compared in the metaphor?

What is the meaning of the metaphor?

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

An electric field is expressed in rectangular coordinates by E = 6x2ax + 6y ay +4az V/m.Find:a) VMN if point M and N are specifi

Fittoniya [83]

Answer:

a.) -147V

b.) -120V

c.) 51V

Explanation:

a.) Equation for potential difference is the integral of the electrical field from a to b for the voltage V_ba = V(b)-V(a).

b.) The problem becomes easier to solve if you draw out the circuit. Since potential at Q is 0, then Q is at ground. So voltage across V_MQ is the same as potential at V_M.

c.) Same process as part b. Draw out the circuit and you'll see that the potential a point V_N is the same as the voltage across V_NP added with the 2V from the other box.

Honestly, these things take practice to get used to. It's really hard to explain this.

3 0

4 years ago