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Helen [10]
3 years ago
10

Technician A says that the use of methanol in internal combustion engines has declined over the years. Technician B says that th

e use of MTBE as a gasoline additive has declined over the years. Who is correct?
Engineering
1 answer:
enyata [817]3 years ago
4 0

Answer:

Both Technician A and B are correct

Explanation:

Methanol is used in internal combustion engines. However, the use of methanol in internal combustion engines has decreased lately even though it was thought to lead to cleaner emissions than gasoline. Methanol internal combustion engines produce formaldehyde which is also an environmental pollutant. Also, the cost of methanol is slightly higher than that of good quality gasoline.

MTBE replaced tetraethyllead as a gasoline additive because the former lead to the emission of particulate lead from automobile exhausts which is a serious environmental pollutant. The use of MTBE has declined over the years due to environmental concerns. It has been banned because it has been found to be a significant groundwater pollutant if gasoline containing MTBE is spilled or leaked at gas stations.

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A mass of 5 kg of saturated water vapor at 100 kPa is heated at constant pressure until the temperature reaches 200°C.
Alex73 [517]

Answer: you can watch a video on how to solve this question on you tube

6 0
4 years ago
Tech A says never use a water hose to clean up dust after a repair. Tech B says never use a floor scrubber to clean up dust afte
DENIUS [597]

Answer:

Tech B

Explanation:

You ruin your mop that way. it will clump up and than you'll have a mud mess.

5 0
3 years ago
Air is compressed adiabatically from p1 1 bar, T1 300 K to p2 15 bar, v2 0.1227 m3 /kg. The air is then cooled at constant volum
sashaice [31]

Answer:

Work done for the adiabatic process = -247873.6 J/kg = - 247.9 KJ/kg

Heat transfer for the constant volume process = - 244.91 KJ/kg

Explanation:

For the first State,

P₁ = 1 bar = 10⁵ Pa

T₁ = 300 K

V₁ = ?

Second state

P₂ = 15 bar = 15 × 10⁵ Pa

T₂ = ?

V₂ = 0.1227 m³/kg

Third state

P₃ = ?

T₃ = 300 K

V₃ = ?

We require the workdone for step 1-2 (which is adiabatic)

And heat transferred for steps 2-3 (which is isochoric/constant volume)

Work done for an adiabatic process is given by

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

where γ = ratio of specific heats = 1.4 for air since air is mostly diatomic

K = PVʸ

Using state 2 to calculate for k

K = P₂V₂ʸ = (15 × 10⁵)(0.1227)¹•⁴ = 79519.5

We also need V₁

For an adiabatic process

P₁V₁ʸ = P₂V₂ʸ = K

P₁V₁ʸ = K

(10⁵) (V₁¹•⁴) = 79519.5

V₁ = 0.849 m³/kg

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

W = 79519.5 [(0.1227)⁻⁰•⁴ - (0.849)⁻⁰•⁴]/(1 - 1.4)

W = (79519.5 × 1.247)/(-0.4) = - 247873.6 J/kg = - 247.9 KJ/kg

To calculate the heat transferred for the constant volume process

Heat transferred = Cᵥ (ΔT)

where Cᵥ = specific heat capacity at constant volume for air = 0.718 KJ/kgK

ΔT = T₃ - T₂

We need to calculate for T₂

Assuming air is an ideal gas,

PV = mRT

T = PV/mR

At state 2,

V/m = 0.1227 m³/kg

P₂ = 15 bar = 15 × 10⁵ Pa

R = gas constant for air = 287.1 J/kgK

T₂ = 15 × 10⁵ × 0.1227/287.1 = 641.1 K

Q = 0.718 (300 - 641.1) = - 244.91 KJ/kg

7 0
3 years ago
A cylindrical drill with radius 4 is used to bore a hole through the center of a sphere of radius 5. Find the volume of the ring
ANTONII [103]

Answer:

The volume of the ring shaped solid that remains is 21 unit^3.

Explanation:

The total volume of the sphere is given as:

Volume of Sphere = (4/3)πr^3

where, r = radius of sphere

Volume of Sphere = (4/3)(π)(5)^3

Volume of Sphere = 523.6 unit^3

Now, we find the volume of sphere removed by the drill:

Volume removed = (Cross-sectional Area of drill)(Diameter of Sphere)

Volume removed = (πr²)(D)

where, r = radius of drill = 4

D = diameter of sphere = 2*5 = 10

Therefore,

Volume removed = (π)(4)²(10)

Volume removed = 502.6 unit^3

Therefore, the volume of ring shaped solid that remains will be the difference between the total volume of sphere, and the volume removed.

Volume of Ring = Volume of Sphere - Volume removed

Volume of Ring = 523.6 - 502.6

<u>Volume of Ring = 21 unit^3</u>

5 0
3 years ago
What would be the most likely scale factor to use for an n-gauge model train setup? (An n-gauge layout uses locomotives that are
slega [8]
A or b for sure brother
8 0
3 years ago
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