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2 years ago

3. Which instrument measures the height above the ground?

1 answer:

7 0

Altimeters are important navigation instruments for aircraft and spacecraft pilots who monitor their height above the Earth's surface. Skydivers and mountaineers also use altimeters to pinpoint their location in the sky or on the ground.

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If you replace the text value in an associative dimension, the text value will not change when the

Nostrana [21]

Answer:

Explanation:

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4 years ago

A insulated vessel s has two compartments separated by a membreane. On one side is 1kg of steam at 400 degC and 200 bar. The oth

Lilit [14]

Answer:

See explaination

Explanation:

See attachment for the detailed step by step solution of the given problem.

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3 years ago

A tool chest has 650 N weight that acts through the midpoint of the chest. The chest is supported by feet at A and rollers at B.

erica [24]

Answer:

the value of horizontal force P is 170.625 N

the value of horizontal force at P = 227.5 N is that the block moves to right and this motion is due to sliding.

Explanation:

The first diagram attached below shows the free body diagram of the tool chest when it is sliding.

Let start out by calculating the friction force

$F_f= \mu N_2$

where :

$F_f =$ friction force

$\mu$ = coefficient of friction

$N_2$ = normal friction

Given that:

$\mu$ = 0.3

$F_f =$ 0.3 $N_2$

Using the equation of equilibrium along horizontal direction.

$\sum f_x = 0$

P - $F_f =$ 0

P = 0.3 $N_2$ ----- Equation (1)

To determine the moment about point B ; we have the expression

$\sum M_B = 0$

0 = $N_2*70-W*35-P*100$

where;

P = horizontal force

$N_2$ = normal force at support A

W = self- weight of tool chest

Replacing W = 650 N

0 = $N_2*70-650*35-100*P$

$P = \frac{70 N_2-22750}{100} ----- equation (2)$

Replacing $\frac{70 N_2-22750}{100}$ for P in equation (1)

$\frac{70N_2 -22750}{100} =0.3 N_2$

$N_2 = \frac{22750}{40}$ $N_2 = 568.75 \ N$

Plugging the value of $N_2 = 568.75 \ N$ in equation (2)

$P = \frac{70(568.75)-22750}{100} \\ \\ P = \frac{39812.5-22750}{100} \\ \\ P = \frac{17062.5}{100}$

P =170.625 N

Thus; the value of horizontal force P is 170.625 N

b) From the second diagram attached the free body diagram; the free body diagram of the tool chest when it is tipping about point A is also shown below:

Taking the moments about point A:

$\sum M_A = 0$

-(P × 100)+ (W×35) = 0

P = $\frac{W*35}{100}$

Replacing 650 N for W

$P = \frac{650*35}{100}$

P = 227.5 N

Thus; the value of horizontal force P, when the tool chest tipping about point A is 227.5 N

We conclude that the motion will be impending for the lowest value when P = 170.625 N and when P= 227.5 N

However; the value of horizontal force at P = 227.5 N is that the block moves to right and this motion is due to sliding.

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4 years ago

A 50 Hz, four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8.0 MJ/MVA. (a) Find the stored energy in the

raketka [301]

Given Information:

Frequency = f = 60 Hz

Complex rated power = G = 100 MVA

Intertia constant = H = 8 MJ/MVA

Mechanical power = Pmech = 80 MW

Electrical power = Pelec = 50 MW

Number of poles = P = 4

No. of cycles = 10

Required Information:

(a) stored energy = ?

(b) rotor acceleration = ?

Answer:

(a) stored energy = 800 Mj

(b) rotor acceleration = 337.46 elec deg/s²

Explanation:

(a) Find the stored energy in the rotor at synchronous speed.

The stored energy is given by

$E = G \times H$

Where G represents complex rated power and H is the inertia constant of turbo-generator.

$E = 100 \times 8 \\\\E = 800 \: MJ$

(b) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find rotor acceleration, neglecting mechanical and electrical losses.

The rotor acceleration is given by

$P_a = P_{mech} - P_{elec} = M \frac{d^2 \delta}{dt^2}$

Where M is given by

$M = \frac{E}{180 \times f}$

$M = \frac{800}{180 \times 50}$

$M = 0.0889 \: MJ \cdot s/ elec \: \: deg$

So, the rotor acceleration is

$P_a = 80 - 50 = 0.0889 \frac{d^2 \delta}{dt^2}$

$30 = 0.0889 \frac{d^2 \delta}{dt^2}$

$\frac{d^2 \delta}{dt^2} = \frac{30}{0.0889}$

$\frac{d^2 \delta}{dt^2} = 337.46 \:\: elec \: deg/s^2$

(c) If the acceleration calculated in part(b) is maintained for 10 cycles, find the change in torque angle and rotor speed in revolutions per minute at the end of this period.

The change in torque angle is given by

$\Delta \delta = \frac{1}{2} \cdot \frac{d^2 \delta}{dt^2}\cdot (t)^2$