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3 years ago

What information does the api symbol or donut provide?.

Engineering

1 answer:

kondor19780726 [428]3 years ago

7 0

Answer:

The API "Donut" identifies oils that meet current API engine oil standards.

Explanation:

hope this helps :)

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Helium is used as the working fluid in a Brayton cycle with regeneration. The pressure ratio of the cycle is 8, the compressor i

Temka [501]

Answer:

Explanation:

Find the temperature at exit of compressor

$T_2=300 \times 8^{\frac{1.667-1}{1.667} }\\=689.3k$

Find the work done by the compressor

$\frac{W}{m} =c_p(T_2-T_1)\\\\=5.19(689.3-300)\\=2020.4kJ/kg$

Find the actual workdone by the compressor

$\frac{W}{m} =n_c(\frac{W}{m} )\\\\=1 \times 2020.4kJ/kg$

Find the temperature at exit of the turbine

$T_4=\frac{1800}{8^{\frac{1.667-1}{1.667} }} \\\\=787.3k$

Find the actual workdone by the turbine

$1 \times 5.19 (1800-783.3)\\=5276.6kJ/kg$

Find the temperature of the regeneration

$\epsilon = \frac{T_5-T_2}{T_4-T_2} \\\\0.75=\frac{T_5-689.3}{783.3-689.3} \\\\T_5=759.8k$

Find the heat supplied

$Q_i_n=c_p(T_3-T_5)\\\\=5.19(1800-759.8)\\\\=5388.2kJ/kg$

Find the thermal efficiency

$n_t_h=\frac{W_t-W_c}{Q_i_n} \\\\=\frac{5276.6-2020.4}{5388.2} \\\\n_t_h=60.4$

60.4%

Find the mass flow rate

$m=\frac{W_net}{P} \\\\\frac{60 \times 10^3}{5276.6-2020.4} \\\\=18.42$

Find the actual workdone by the compressor

$\frac{W_c}{m} =\frac{(\frac{W}{m} )}{n_c} \\\\=\frac{2020.4}{0.8} \\\\=2525.5kg$

Find the actual workdone by the turbine

$\frac{W_t}{m} =n_t(\frac{W}{m} )\\\\=0.8 \times5.19(1800-783.3)\\\\=4221.2kJ/kg$

Find the temperature of the compressor exit

$\frac{W_t}{m} =c_p(T_2_a-T_1)\\2525.5=5.18(T_2_a-300)\\T_2_a=787.5k$

Find the temperature at the turbine exit

$4221.2=5.18(1800-T_4_a)\\\\T_4_a=985k$

Find the temperature of regeneration

$\epsilon =\frac{T_5-T_2}{T_4-T_2}\\\\0.75=\frac{T_5-787.5}{985-787.5}\\\\T_5=935.5k$

6 0

3 years ago

Read 2 more answers

The disk of radius 0.4 m is originally rotating at ωo=4 rad/sec. If it is subjected to a constant angular acceleration of α=5 ra

Lina20 [59]

Answer:32.4m/ $s^2$

Explanation:

Given data

$radius\left ( r\right )$ =0.4m

Intial angular velocity $\left ( \omega_0\right )$ =4rad/s

angular acceleration $\left ( \alpha\right )$ =5rad/ $s^2$

angular velocity after 1 sec

$\omega$ = $\omega_0$ + $\alpha\times\t$

$\omega$ =4+5 $\left ( 1\right )$

$\omega$ =9rad/s

Velocity of point on the outer surface of disc $\left ( v\right )$ = $\omega_0\timesr$

v= $9\times0.4$ m/s=3.6m/s

Normal component of acceleration $\left ( a_c\right )$ = $\frac{v^2}{r}$

$a_c$ = $\frac{3.6\times3.6}{0.4}$ =32.4m/ $s^2$

3 0

4 years ago

On what kind of sectional drawing would you find the maximum length of a building?

Genrish500 [490]

Answer:

A longitudinal section

Explanation:

A longitudinal section is a section drawn along the length of an object, as opposed to a cross section, which is drawn across the width or diameter of an object.

5 0

2 years ago

If coolant is mixed 50-50 with water, what is the freezing point?

sashaice [31]

Answer:

-35 degrees F

When mixed in equal parts with water (50/50), antifreeze lowers the freezing point to -35 degrees F and raises the boiling temperature to 223 degrees F. Antifreeze also includes corrosion inhibitors to protect the engine and cooling system against rust and corrosion.

8 0

3 years ago

A manometer consists of an inclined glass tube which communicates with a metal cylinder standing upright; liquid fills the appar

Vanyuwa [196]

Answer:

48.61

Explanation:

See attached diagram.

The level rise in the tube is l sin α.

The level drop in the cylinder (let's call it y) is:

π/4 D² y = π/4 d² l

D² y = d² l

y = l (d/D)²

The elevation difference is the sum:

l sin α + l (d/D)²

l (sin α + (d/D)²)

From Bernoulli's principle:

P = ρgl (sin α + (d/D)²)

Divide both sides by density of water (ρw) and gravity:

P/(ρw g) = (ρ/ρw) l (sin α + (d/D)²)

h = S l (sin α + (d/D)²)

If we disregard the level change in the cylinder:

h = S l (sin α)

We want the percent error between these two expressions for h to be 0.1% when α = 25°.

[ S l (sin α + (d/D)²) − S l (sin α) ] / [ S l (sin α + (d/D)²) ] = 0.001

[ S l sin α + S l (d/D)² − S l sin α ] / [ S l (sin α + (d/D)²) ] = 0.001

[ S l (d/D)² ] / [ S l (sin α + (d/D)²)] = 0.001

(d/D)² / (sin α + (d/D)²) = 0.001

(d/D)² = 0.001 (sin α + (d/D)²)

(d/D)² = 0.001 sin α + 0.001 (d/D)²

0.999 (d/D)² = 0.001 sin α

d/D = √(0.001 sin α / 0.999)

When α = 25°:

d/D ≈ 0.02057

D/d ≈ 48.61

7 0

3 years ago