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LenKa [72]
4 years ago
12

What is the best source of information about the world?

Engineering
1 answer:
iogann1982 [59]4 years ago
8 0
To be honest theres a lot...
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3 years ago
On some engines, after torquing cylinder head fasteners, you
djverab [1.8K]

Answer:

Re-torque the bolts as required while your engine is warm. But if you're using aluminum cylinder heads, you should wait until your engine is complete cooled until re-torquing

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3 years ago
Ammonia gas is diffusing at a constant rate through a layer of stagnant air 1 mm thick. Conditions are such that the gas contain
fiasKO [112]

Answer:

The solution to this question is 5.153×10⁻⁴(kmol)/(m²·s)

That is the rate of diffusion of ammonia through the layer is

5.153×10⁻⁴(kmol)/(m²·s)

Explanation:

The diffusion through a stagnant layer is given by

N_{A}  = \frac{D_{AB} }{RT} \frac{P_{T} }{z_{2} - z_{1}  } ln(\frac{P_{T} -P_{A2}  }{P_{T} -P_{A1} })

Where

D_{AB} = Diffusion coefficient or diffusivity

z = Thickness in layer of transfer

R = universal gas constant

P_{A1} = Pressure at first boundary

P_{A2} = Pressure at the destination boundary

T = System temperature

P_{T} = System pressure

Where P_{T} = 101.3 kPa P_{A2} =0, P_{A1} =y_{A}, P_{T} = 0.5×101.3 = 50.65 kPa

Δz = z₂ - z₁ = 1 mm = 1 × 10⁻³ m

R =  \frac{kJ}{(kmol)(K)} ,    T = 298 K   and  D_{AB} = 1.18 \frac{cm^{2} }{s} = 1.8×10⁻⁵\frac{m^{2} }{s}

N_{A} = \frac{1.8*10^{-5} }{8.314*295} *\frac{101.3}{1*10^{-3} }* ln(\frac{101.3-0}{101.3-50.65}) = 5.153×10⁻⁴\frac{kmol}{m^{2}s }

Hence the rate of diffusion of ammonia through the layer is

5.153×10⁻⁴(kmol)/(m²·s)

5 0
3 years ago
A BOD test is run using 30 mL of wastewater and 270 mL of dilution water. The initial DO of the mixture is 9.0 mg/L. After 5 day
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3 years ago
Your family has asked you to estimate the operating costs of your clothes dryer for the year. The clothes dryer in your home has
trasher [3.6K]

Answer:

The costs to run the dryer for one year are $ 9.03.

Explanation:

Given that the clothes dryer in my home has a power rating of 2250 Watts, and to dry one typical load of clothes the dryer will run for approximately 45 minutes, and in Ontario, the cost of electricity is $ 0.11 / kWh, to calculate the costs to run the dryer for one year the following calculation must be performed:

1 watt = 0.001 kilowatt

2250/45 = 50 watts per minute

45 x 365 = 16,425 / 60 = 273.75 hours of consumption

50 x 60 = 300 watt = 0.3 kw / h

0.3 x 273.75 = 82.125

82.125 x 0.11 = 9.03

Therefore, the costs to run the dryer for one year are $ 9.03.

8 0
3 years ago
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