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2 years ago

Discuss the importance of dust and fluff removal from spinning mills and how it is realised in air conditioning plants

Engineering

1 answer:

Alina [70]2 years ago

8 0

Answer:

Removal of dust and fluff from spinning mill is important as it has adverse and detrimental effects on the health of the workers in these industries. Tiny and microscopic particles of various substances present in the surrounding air is transferred from one place to another and these causes various respiratory diseases and pose health hazards for the workers and make work environment unhealthy and hazardous thus affecting the over all efficiency and productivity.

Cotton dust , the major pollutant, when breathed in affetcs the lungs badly and workers experience symptoms such as respiratory problems, coughing, tightness in chest, etc. Thus to ensure proper health of the workers spinning mills have been provided with powerful air conditioning to ensure purity of air, to maintain proper moisture levels and to ensure dust and fluff removal.

The dust and fluff laiden air is humidified, purified and then recirculated. Optimization of number of air changes/hour to clean air stream and prevent any health risk of the workers.

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Choose the correct word or phrase to complete the sentence to explain human intervention in a machine system.

maksim [4K]

Answer:

Fully Automated

Periodic Maintenance Activities

6 0

2 years ago

The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine-generator at a locati

Lynna [10]

Answer:

a) 0.76

b) 0.80

c) 1964 kW

Explanation:

GIVEN DATA:

$\dot m = 5000 kg/s$

Assume Mechanical energy at exist is negligible

A) Take lake bottom as reference, and then kinetic and potential energy are taken as zero.

change in mechanical energy is givrn as

$e_{in} - e_{out} = \frac{P}{\rho} - 0 = gh = 9.81 \times 50( \frac{1 kJ/kg}{1000 m^2/s^2}$

= 0.491 kJ/kg

$\Delta \dot E_{mec} = \dot m (e_{in} - e_{out}) = 5000 \times 0.491 = 2455 kW$

$\eta_{OVERALL} = \frac{\dot W}{\Delta \dot E_{mec}} = \frac{1862}{2455} = 0.76$

B) $\eta -{gen} = \frac{\eta_{overall}}{\eta_{gen}} = \frac{0.76}{0.95} = 0.80$

c) $\dot W_{shaft} = \eta_{overall} \left | \Delta \dot E_{mec} \right | = 0.80(2455)$

$\dot W_{shaft} = 1964 kW$

7 0

3 years ago

What is the mass of a brass axle that has a volume of 0.318 cm?

NeX [460]

Answer:

2.7g

Explanation:

the mass of a brass axle that has a volume of 0.318 cm is 2.7g.

8 0

2 years ago

The underground storage of a gas station has leaked gasoline into the ground. Among the constituents of gasoline are benzene, wi

vovangra [49]

Answer:

a) benzene = 910 days

b) toluene = 1612.67 days

Explanation:

Given:

Kd = 1.8 L/kg (benzene)

Kd = 3.3 L/kg (toluene)

psolid = solids density = 2.6 kg/L

K = 2.9x10⁻⁵m/s

pores = n = 0.37

water table = 0.4 m

ground water = 15 m

u = K/n = (2.9x10⁻⁵ * (0.4/15)) / 0.37 = 2.09x10⁻⁶m/s

a) For benzene:

$R=1+\frac{\rho * K_{d} }{n}, \rho = 2.6\\ R=1+\frac{2.6*1.8}{0.37} =13.65$

The time will take will be:

$t=\frac{xR}{a} , x=12,a=0.18\\t=\frac{12*13.65}{0.18} =910days$

b) For toluene:

$R=1+\frac{2.6*3.3 }{0.37} = 24.19$

$t=\frac{12*24.19}{0.18} =1612.67days$

6 0

2 years ago

Read 2 more answers

A water-filled manometer is used to measure the pressure in an air-filled tank. One leg of the manometer is open to atmosphere.

ddd [48]

Answer:

$P = 150.335\,kPa$ (Option B)

Explanation:

The absolute pressure of the air-filled tank is:

$P = 101.3\,kPa + \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{kg}{m^{3}} \right)\cdot (5\,m)\cdot \left(\frac{1\,kPa}{1000\,Pa} \right)$

$P = 150.335\,kPa$

4 0

3 years ago