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2 years ago

How have the conditions under which auto fuel economy is measured changed over the years?

Engineering

1 answer:

Elina [12.6K]2 years ago

3 0

We have more technology so fuel has other factors not like before that it was more natural

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Proper handling of blueprints includes which of the following

marta [7]

Answer:

folding plans neatly after use

3 0

2 years ago

Air modeled as an ideal gas enters a well-insulated diffuser operating at steady state at 270 K with a velocity of 180 m/s and e

ZanzabumX [31]

Answer:

exit temperature 285 K

Explanation:

given data

temperature T1 = 270 K

velocity = 180 m/s

exit velocity = 48.4 m/s

solution

we know here diffuser is insulated so here heat energy is negleted

so we write here energy balance equation that is

0 = m (h1-h2) + m × $(\frac{v1^2}{2}-\frac{v2^2}{2})$ .....................1

so it will be

$h1 + \frac{v1^2}{2} = h2 + \frac{v2^2}{2}$ .....................2

put here value by using ideal gas table

and here for temperature 270K

h1 = 270.11 kJ/kg

$270.11 + \frac{180^2\times \frac{1}{1000}}{2} = h2 + \frac{48.4^2\times \frac{1}{1000}}{2}$

solve it we get

h2 = 285.14 kJ/kg

so by the ideal gas table we get

T2 = 285 K

4 0

3 years ago

A kernel-level thread wishes to acquire a mutex lock declared as global in the process. True or False: the function call used be

Alik [6]

Answer:

A kernel-level thread wishes to acquire a mutex lock declared as global in the process. "True"
The function call used below causes the thread to block at this point, if the mutex lock is already acquired by a different thread. int resultcode -pthread mutex_trylock (myMutex). "False"

7 0

3 years ago

Consider a simple ideal Rankine cycle and an ideal regenerative Rankine cycle with one open feedwater heater. The two cycles are

borishaifa [10]

Answer:

They both have the same efficiency.

Explanation:

The simple ideal Rankine cycle and an ideal regenerative Rankine cycle with one open feedwater heater would both have the same efficiency because the extraction steam would just create a mini cycle that recirculates. The energy given to the feedwater heater is proportional to the added heat in the boiler to the feedwater in the simple cycle to raise its temperature to the same boiler inlet condition.

Therefore in comparison, the efficiency is the same for both.

4 0

2 years ago

The diffusion coefficients for species A in metal B are given at two temperatures:

Kruka [31]

Answer:

a) 149 kJ/mol, b) 6.11*10^-11 m^2/s ,c) 2.76*10^-16 m^2/s

Explanation:

Diffusion is governed by Arrhenius equation

$D = D_0e^{\frac{-Q_d}{RT} }$

I will be using R in the equation instead of k_b as the problem asks for molar activation energy

I will be using

$R = 8.314\ J/mol*K$

and

°C + 273 = K

here, adjust your precision as neccessary

Since we got 2 difusion coefficients at 2 temperatures alredy, we can simply turn these into 2 linear equations to solve for a) and b) simply by taking logarithm

So:

$ln(6.69*10^{-17})=ln(D_0) -\frac{Q_d}{R*(1030+273)}$