All categories

2 years ago

CNG is a readily available alternative to _________. A) Antifreeze B) Brake fluid C) Gasoline D) Oil

Engineering

2 answers:

Natali [406]2 years ago

7 0

CNG is a readily available alternative to (C) Gasoline.

Answer: C) Gasoline

Explanation

CNG is abbreviation for Compressed Natural Gas.

It is a type of fuel prepared on compressing methane at high pressure.

They are used as an alternative to diesel fuel, gasoline and LPG.

The CNG is environmental friendly as it prohibits the emission of harmful gases during combustion.

It can be synthesized and stored easily. So it acts as an alternative to diesel fuels and gasoline.

As it is obtained from natural gases, the reproducibility will be more and also the probability of blasting due to any reason will be less as CNG are lighter compared to conventional fuels.

balu736 [363]2 years ago

6 0

Answer:

C is the correct answer

Explanation:

CNG can be called that the 2nd option to make gasoline by compressing natural gas to less than 1% of its volume at normal atmospheric pressure

Hope this help you :3

You might be interested in

A six- lane freeway ( three lanes in each direction) in a scenic area has a measured free- flow speed of 55 mi/ h. The peak- hou

Novosadov [1.4K]

Answer:

0.867

Explanation:

The driver population factor ( $f_{p}$ )can be estimated using the equation below:

$f_{p} = \frac{V}{PHF*N*f_{HV}*v_{p}}$

The value of the heavy vehicle factor ( $f_{HV}$ ) is determined below:

The values of the $E_{T}$ = 2 and $E_{R}$ = 3 are gotten from the tables for the RVs, trucks and buses upgrades for passenger-car equivalents. Therefore:

$f_{HV}$ = 1/[1+0.08(2-1)+0.06(3-1)] = 1/[1+0.08+0.12] = 1/1.2 = 0.833

Furthermore, the vp is taken as 2250 pc/(h*In) from the table of LOS criteria for lane freeway using the 15 minutes flow rate. Therefore:

$f_{p}$ = 3900/[0.8*3*0.833*2250] = 3900/4498.2 = 0.867

6 0

2 years ago

The 8-mm-thick bottom of a 220-mm-diameter pan may be made from aluminum (k = 240 W/m ⋅ K) or copper (k = 390 W/m ⋅ K). When use

Artemon [7]

Answer:

For aluminum 110.53 C

For copper 110.32 C

Explanation:

Heat transmission through a plate (considering it as an infinite plate, as in omitting the effects at the borders) follows this equation:

$q = \frac{k * A * (th - tc)}{d}$

Where

q: heat transferred

k: conduction coeficient

A: surface area

th: hot temperature

tc: cold temperature

d: thickness of the plate

Rearranging the terms:

d * q = k * A * (th - tc)

$\frac{d * q}{k * A} = th - tc$

$th = \frac{d * q}{k * A} + tc$

The surface area is:

$A = \frac{\pi * d^2}{4}$

$A = \frac{\pi * 0.22^2}{4} = 0.038 m^2$

If the pan is aluminum:

$th = \frac{0.008 * 600}{240 * 0.038} + 110 = 110.53 C$

If the pan is copper:

$th = \frac{0.008 * 600}{390 * 0.038} + 110 = 110.32 C$

7 0

2 years ago

A pressure gage connected to a tank reads 50 psi at a location where the barometric reading is 29.1 inches Hg. Determine the abs

Effectus [21]

Answer:

Absolute pressure , P(abs)= 433.31 KPa

Explanation:

Given that

Gauge pressure P(gauge)= 50 psi

We know that barometer reads atmospheric pressure

Atmospheric pressure P(atm) = 29.1 inches of Hg

We know that

1 psi = 6.89 KPa

So 50 psi = 6.89 x 50 KPa

P(gauge)= 50 psi =344.72 KPa

We know that

1 inch = 0.0254 m

29.1 inches = 0.739 m

Atmospheric pressure P(atm) = 0.739 m of Hg

We know that density of Hg = $13.6\times 10^3\ kg/m^3$

P = ρ g h

P(atm) = 13.6 x 1000 x 9.81 x 0.739 Pa

P(atm) = 13.6 x 9.81 x 0.739 KPa

P(atm) =98.54 KPa

Now

Absolute pressure = Gauge pressure + Atmospheric pressure

P(abs)=P(gauge) + P(atm)

P(abs)= 344.72 KPa + 98.54 KPa

P(abs)= 433.31 KPa

3 0

2 years ago

A completely reversible heat pump produces heat at a rate of 300 kW to warm a house maintained at 24°C. The exterior air, which

Westkost [7]

Answer:

Entropy generation rate of the two reservoirs is approximately zero ( $\dot S_{gen} = 9.318 \times 10^{-4}\,\frac{kW}{K}$ ) and system satisfies the Second Law of Thermodynamics.

Explanation:

Reversible heat pumps can be modelled by Inverse Carnot's Cycle, whose key indicator is the cooling Coefficient of Performance, which is the ratio of heat supplied to hot reservoir to input work to keep the system working. That is:

$COP_{H} = \frac{\dot Q_{H}}{\dot W}$

The following simplification can be used in the case of reversible heat pumps:

$COP_{H,rev} = \frac{T_{H}}{T_{H} - T_{L}}$

Where temperature must written at absolute scale, that is, Kelvin scale for SI Units:

$COP_{H, rev} = \frac{297.15\,K}{297.15\,K-280.15\,K}$

$COP_{H, rev} = 17.479$

Then, input power needed for the heat pump is:

$\dot W = \frac{\dot Q}{COP_{H,rev}}$

$\dot W = \frac{300\,kW}{17.749}$

$\dot W = 16.902\,kW$

By the First Law of Thermodynamics, heat pump works at steady state and likewise, the heat released from cold reservoir is now computed:

$-\dot Q_{H} + \dot W + \dot Q_{L} = 0$

$\dot Q_{L} = \dot Q_{H} - \dot W$

$\dot Q_{L} = 300\,kW - 16.902\,kW$

$\dot Q_{L} = 283.098\,kW$

According to the Second Law of Thermodynamics, a reversible heat pump should have an entropy generation rate equal to zero. The Second-Law model for the system is:

$\dot S_{in} - \dot S_{out} - \dot S_{gen} = 0$

$\dot S_{gen} = \dot S_{in} - \dot S_{out}$

$\dot S_{gen} = \frac{\dot Q_{L}}{T_{L}} - \frac{\dot Q_{H}}{T_{H}}$

$\dot S_{gen} = \frac{283.098\,kW}{280.15\,K} - \frac{300\,kW}{297.15\,K}$

$\dot S_{gen} = 9.318 \times 10^{-4}\,\frac{kW}{K}$

Albeit entropy generation rate is positive, it is also really insignificant and therefore means that such heat pump satisfies the Second Law of Thermodynamics. Furthermore, $\dot S_{in} = \dot S_{out}$ .

5 0

2 years ago

print('Predictions are hard.") print(Especially about the future.) user_num = 5 print('user_num is:' user_num)

Bas_tet [7]

Answer:

The correct code is given below:-

print("Predictions are hard.")

print("Especially about the future.")

user_num = 5

print("user_num is:", user_num)

7 0

2 years ago