What is the relationship between compressor work and COPR?

A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic

rusak2 [61]

Answer:

The answer is below

Explanation:

A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic field of the stator is rotating. 2- the speed of the rotor when the slip is 0.05. 3- the frequency of the rotor currents when the slip is 0.04. 4- the frequency of the rotor currents at standstill.

Given that:

number of poles (p) = 4, frequency (f) = 60 Hz

1) The synchronous speed of the motor is the speed at which the magnetic field of the stator is rotating. It is given as:

$n_s=\frac{120f}{p}=\frac{120*60}{4}=1800\ rpm$

2) The slip (s) = 0.05

The speed of the motor (n) is the speed of the rotor, it is given as:

$n=n_s-sn_s\\\\n=1800-0.05(1800)\\\\n=1800-90\\\\n=1710\ rpm$

3) s = 0.04

The rotor frequency is the product of the supply frequency and slip it is given as:

$f_r=sf\\\\f_r=0.04*60\\\\f_r=2.4\ Hz$

4) At standstill, the motor speed is zero hence the slip = 1:

$s=\frac{n_s-n}{n_s}\\ \\n=0\\\\s=\frac{n_s-0}{n_s}\\\\s=1$

The rotor frequency is the product of the supply frequency and slip it is given as:

$f_r=sf\\\\f_r=1*60\\\\f_r=60\ Hz$

4 0

3 years ago

The small washer is sliding down the cord OA. When it is at the midpoint, its speed is 28 m/s and its acceleration is 7 m/s 2 .

Neporo4naja [7]

Answer:

Velocity components

$V_r = -16.28 m/s$

$V_z = -22.8 m/s$

$V_q = 0 m/s$

For Acceleration components;

$a_r = -4.07m/s^2$

$a_z = -5.70m/s^2$

$a_q = 0m/s^2$

Explanation:

We are given:

$Speed v_o = 28 m/s$

$Acceleration a_o= 7 m/s^2$

We first need to find the radial position r of washer in x-y plane.

Therefore

$r = \sqrt{300^2 + 400^2}$

r = 500 mm

To find length along direction OA we have:

$L = \sqrt{500^2 + 700^2}L = 860 mm$

Therefore, the radial and vertical components of velocity will be given as:

$V_r = V_o*cos(Q)$

$V_z = V_o*sin(Q)$

Where Q is the angle between OA and vector r.

Therefore,

$V_r = 28 * \frac{r}{L} = > 28 * \frac{500}{860}$

$V_r = -16.28 m/s$

• $V_z = 28 * \frac{700}{860} = -22.8$

• $V_q = 0 m/s$

The radial and vertical components of acceleration will be:

$a_r = a_o*cos(Q)$

$a_z = a_o*sin(Q)$

Therefore we have:

• $a_r = 7* \frac{500}{860} = -4.07m/s^2$

• $a_z = 7 * \frac{700}{860} = -5.70 m/s^2$

• $a_q = 0 m/s^2$

Note : image is missing, so I attached it

3 0

3 years ago

A lake with constant volume 1.1 x 10^6 m^3 is fed by a stream with a non-conservative pollutant of 2.3 mg/L and flow rate 35 m^3

Jet001 [13]

Answer:

12.84 mg/L

Explanation:

We are given;

Volume of lake; V = 1.1 x 10^(6) m³

decay coefficient; K = 0.10/day = 0.1/(24 × 60 × 60) /s = 0.00000115741 /s

Factory rate: Q_f = 4.3 m³/s

Factory concentration: C_f = 100 mg/L

Stream rate: Q_s = 34 m³/s

Stream Concentration: C_s = 2.3 mg/L

Now, to find the steady state concentration of pollutant in the lake, we will use the formula;

(Q_s•C_s) + (Q_f•C_f) = (Q_f + Q_s)C_L + (KV•C_L)

Where C_L is the steady state concentration of pollutant in the lake.

Thus, making C_L the subject, we have;

C_L = [(Q_s•C_s) + (Q_f•C_f)]/(Q_f + Q_s + K•V)

Plugging in the relevant values gives;

C_L = ((34 × 2.3) + (4.3 × 100))/(4.3 + 34 + (0.00000115741 × 1.1 × 10^(6)))

C_L = 12.84 mg/L

4 0

3 years ago

Calculate the molar heat capacity of a monatomic non-metallic solid at 500K which is characterized by an Einstein temperature of

aleksandr82 [10.1K]

Answer:

Explanation:

Given

Temperature of solid $T=500\ K$

Einstein Temperature $T_E=300\ K$

Heat Capacity in the Einstein model is given by

$C_v=3R\left [ \frac{T_E}{T}\right ]^2\frac{e^{\frac{T_E}{T}}}{\left ( e^{\frac{T_E}{T}}-1\right )^2}$

$e^{\frac{3}{5}}=1.822$

Substitute the values

$C_v=3R\times (\frac{300}{500})^2\times (\frac{1.822}{(1.822-1)^2})$

$C_v=3R\times \frac{9}{25}\times \frac{1.822}{(0.822)^2}$

$C_v=0.97\times (3R)$

6 0

3 years ago

A ball A is thrown vertically upward from the top of a 30-m-high building with an initial velocity of 5 m>s. At the same inst

expeople1 [14]

Answer:

s= 20.4 m

Explanation:

First lets write down equations for each ball:

s=so+vo*t+1/2a_c*t^2

for ball A:

s_a=30+5*t+1/2*9.81*t^2

for ball B:

s_b=20*t-1/2*9.81*t^2

to find time deeded to pass we just put that

s_a = s_b

30+5*t-4.91*t^2=20*t-4.9*t^2

t=2 s

now we just have to put that time in any of those equations an get distance from the ground:

s = 30 + 5*2 -1/2*9.81 *2^2

s= 20.4 m

6 0

3 years ago