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Gre4nikov [31]
3 years ago
11

A heat engine is coupled with a dynamometer. The length of the load arm is 900 mm. The spring balance reading is 16. Applied wei

ght is 500 N. Rotational speed is 1774. How many kW of power will be developed?
Engineering
1 answer:
miss Akunina [59]3 years ago
4 0

Answer:

P = 80.922 KW

Explanation:

Given data;

Length of load arm is 900 mm = 0.9 m

Spring balanced  read 16 N

Applied weight is 500 N

Rotational speed is 1774 rpm

we know that power is given as

P = T\times \omega

T Torque = (w -s) L = (500 - 16)0.9 = 435.6 Nm

\omega angular speed =\frac{2 \pi N}{60}

Therefore Power is

P =\frac{435.6 \time 2 \pi \times 1774}{60} = 80922.65  watt

P = 80.922 KW

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Option B

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A 7-hp (shaft) pump is used to raise water to an elevation of 15 m. If the mechanical efficiency of the pump is 82 percent, dete
Natali [406]

The maximum volume flow rate of water is determined as 0.029 m³/s.

<h3>Power of the pump</h3>

The power of the pump is watt is calculated as follows;

1 hp = 745.69 W

7 hp = ?

= 7 x 745.69 W

= 5,219.83 W

<h3>Mass flow rate of water</h3>

η = mgh/P

mgh = ηP

m = ηP/gh

m = (0.82 x 5,219.83)/(9.8 x 15)

m = 29.12 kg/s

<h3>Maximum volume rate</h3>

V = m/ρ

where;

  • ρ is density of water = 1000 kg/m³

V = (29.12)/(1000)

V = 0.029 m³/s

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1 year ago
Describe how you would control employee exposure to excessive noise in a mining environment
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Answer:

1. Buy Quiet – select and purchase low-noise tools and machinery

2. Maintain tools and equipment routinely (such 3. as lubricate gears)

3. Reduce vibration where possible

4. Isolate the noise source in an insulated room or enclosure

5. Place a barrier between the noise source and the employee

6. Isolate the employee from the source in a room or booth (such as sound wall or window

Explanation:

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2 years ago
If a signal is transmitted at a power of 250 mWatts (mW) and the noise in the channel is 10 uWatts (uW), if the signal BW is 20M
Bess [88]

Answer:

C = 292 Mbps

Explanation:

Given:

- Signal Transmitted Power P = 250mW

- The noise in channel N = 10 uW

- The signal bandwidth W = 20 MHz

Find:

what is the maximum capacity of the channel?

Solution:

-The capacity of the channel is given by Shannon's Formula:

                            C = W*log_2 ( 1 + P/N)

- Plug the values in:

                            C = (20*10^6)*log_2 ( 1 + 250*10^-3/10)

                            C = (20*10^6)*log_2 (25001)

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                           C = 292 Mbps

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