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3 years ago

You can't mix conventional and syntheic oil

Engineering

1 answer:

elena-14-01-66 [18.8K]3 years ago

4 0

Answer:

There is no danger mixing synthetic and conventional motor oil. However, conventional oil will detract from the superior performance of synthetic oil and reduce its benefits.

Explanation:

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A DNS record that's used to redirect traffic from one domain name to another is known as a _______ record.

Marta_Voda [28]

Answer:

cname record

Explanation:

A Canonical Name record (cname record) is a form of resource record that links one domain name to another.

It is commonly used when you are to use different services in the same IP address. For example, one can map www.brainlyquestion.com and www.brainlyquery.com to the DNS entry for www.brainlyenquiry.com (these are just examples, not actual websites). This makes it very effective in redirecting traffic from one domain name to another.

3 0

3 years ago

) A brick of clay is being dried in a batch dryer using constant drying conditions. You have been asked to determine the amount

dimaraw [331]

Answer:

The drying time is calculated as shown

Explanation:

Data:

Let the moisture content be = 0.6

the free moisture content be = 0.08

total moisture of the clay = 0.64

total drying time for the period = 8 hrs

then if the final dry and wet masses are calculated, it follows that

t = (X0+ Xc)/Rc) + (Xc/Rc)* ln (Xc/X)

= 31.3 min.

4 0

2 years ago

A treated wastewater (with a flow of 0.415 m^3 /s, an ultimate BOD of 40 mg/L and the DO of 1.3 mg/L) is discharged into a strea

shutvik [7]

JUST STOP NOBODY LIKES YOU

LIKE GO AND GET A LIFE CAUSE YOU DONT HAVE ONE

8 0

3 years ago

Are there any companies that you can get a job at as an air craft engeer after university

stiks02 [169]

Explanation:

most big airports. my father has the same degree and works for southwest airlines

6 0

2 years ago

Consider a turbofan engine installed on an aircraft flying at an altitude of 5500m. The CPR is 12 and the inlet diameter of this

g100num [7]

Answer:

The thrust of the engine calculated using the cold air is 34227.35 N

Explanation:

For the turbofan engine, firstly the overall mass flow rate is considered. The mass flow rate is given as

$\dot{m}=\rho AV_a$

Here

ρ is the density which is given as $\dfrac{P}{RT}$

P is the pressure of air at 5500 m from the ISA whose value is 50506.80 Pa
R is the gas constant whose value is 286.9 J/kg.K
T is the temperature of the inlet which is given as 253 K

A is the cross-sectional area of the inlet which is given by using the diameter of 2.0 m
V_a is the velocity of the aircraft which is given as 250 m/s

So the equation becomes

$\dot{m}=\rho AV_a\\\dot{m}=\dfrac{P}{RT} AV_a\\\dot{m}=\dfrac{50506.80}{286.9\times 253} \times (\dfrac{\pi}{4}\times 2^2)\times 250\\\dot{m}=546.4981\ kgs^{-1}$

Now in order to find the flow from the fan, the Bypass ratio is used.

$\dot{m}_f=\dfrac{BPR}{BPR+1}\times \dot{m}$

Here BPR is given as 8 so the equation becomes

$\dot{m}_f=\dfrac{BPR}{BPR+1}\times \dot{m}\\\dot{m}_f=\dfrac{8}{8+1}\times 546.50\\\dot{m}_f=485.77\ kgs^{-1}$

Now the exit velocity is calculated using the total energy balance which is given as below:

$h_4+\dfrac{1}{2}V_a^2=h_5+\dfrac{1}{2}V_e^2$

Here

h_4 and h_5 are the enthalpies at point 4 and 5 which could be rewritten as $c_pT_4$ and $c_pT_5$ respectively.
The value of T_4 is the inlet temperature which is 253 K
The value of T_5 is the outlet temperature which is 233K
The value of c_p is constant which is 1005 J/kgK
V_a is the inlet velocity which is 250 m/s
V_e is the outlet velocity that is to be calculated.

So the equation becomes

$h_4+\dfrac{1}{2}V_a^2=h_5+\dfrac{1}{2}V_e^2\\c_pT_4+\dfrac{1}{2}V_a^2=c_pT_5+\dfrac{1}{2}V_e^2$

Rearranging the equation gives

$\dfrac{1}{2}V_e^2=c_pT_4-c_pT_5+\dfrac{1}{2}V_a^2\\\dfrac{1}{2}V_e^2=c_p(T_4-T_5)+\dfrac{1}{2}V_a^2\\V_e^2=2c_p(T_4-T_5)+V_a^2\\V_e=\sqrt{2c_p(T_4-T_5)+V_a^2}\\V_e=\sqrt{2\times 1005\times (253-233)+(250)^2}\\V_e=320.46 m/s$

Now using the cold air approach, the thrust is given as follows

$T=\dot{m}_f(V_e-V_a)\\T=485.77\times (320.46-250)\\T=34227.35\ N$

So the thrust of the engine calculated using the cold air is 34227.35 N

5 0

2 years ago