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2 years ago

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Engineering

2 answers:

kiruha [24]2 years ago

5 0

Answer:

Explanation:

hey be careful!

Just so everyone knows HEY PLS DON'T JOIN THE ZOOM CALL OF A PERSON WHO'S ID IS 825 338 1513 (I'M NOT SAYING THE PASSWORD) HE IS A CHILD PREDATOR AND A PERV. HE HAS LOTS OF ACCOUNTS ON BRAINLY BUT HIS ZOOM NAME IS MYSTERIOUS MEN.. HE ASKS FOR GIRLS TO SHOW THEIR BODIES AND -------- PLEASE REPORT HIM IF YOU SEE A QUESTION LIKE THAT. WE NEED TO TAKE HIM DOWN!!! PLS COPY AND PASTE THIS TO OTHER COMMENT SECTIONS, someone told me to do this

svetlana [45]2 years ago

4 0

Answer:

Explanation:

thxs

You might be interested in

A) Consider an air standard otto cycle that has a heat addition of 2800 kJ/kg of air, a compression ratio of 8 and a pressure an

Angelina_Jolie [31]

Answer:

a) i) The maximum pressure is approximately 122.37 bar

ii) The thermal efficiency is approximately 56.47%

iii) The mean effective pressure is approximately 20.974 bar

b) (b) Four types of internal combustion engine includes;

1) The diesel engine

2) The Otto engine

3) The Brayton engine

4) The Wankel engine

Explanation:

The parameters of the Otto cycle are;

The heat added, $Q_{in}$ = 2,800 kJ/kg

The compression ratio, r = 8

The beginning compression pressure, P₁ = 1 bar

The beginning compression temperature, T₁ = 300 K

Cp = 1.005 kJ/kg·K

Cv = 0.718 kJ/kg·K

R = 287 kJ/kg·K

K = Cp/Cv = 1.005 kJ/kg·K/(0.718 kJ/kg·K) ≈ 1.4

T₂ = T₁×r^(k - 1)

∴ T₂ = 300 K×8^(1.4 - 1) ≈ 689.219 K

$\dfrac{P_1\cdot V_1}{T_1} = \dfrac{P_2\cdot V_2}{T_2}$

$P_2 = \dfrac{P_1\cdot V_1 \cdot T_2}{T_1 \cdot V_2} = \dfrac{V_1}{V_2} \cdot \dfrac{P_1 \cdot T_2}{T_1 } = r \cdot \dfrac{P_1 \cdot T_2}{T_1 }$

∴ P₂ = 8 × 1 bar × (689.219K)/300 K ≈ 18.379 bar

$Q_{in}$ = m·Cv·(T₃ - T₂)

∴ $Q_{in}$ = 2,800 ≈ 0.718 × (T₃ - 689.219)

T₃ = 2,800/0.718 + 689.219 = 4588.94 K

P₃ = P₂ × (T₃/T₂)

P₃ = 18.379 bar × 4588.94K/(689.219 K) = 122.37 bar

The maximum pressure = P₃ ≈ 122.37 bar

(ii) The thermal efficiency, $\eta_{Otto}$ , is given as follows;

$\eta_{Otto} = 1 - \dfrac{1}{r^{k - 1}}$

Therefore, we have;

$\eta_{Otto} = 1 - \dfrac{1}{8^{1.4 - 1}} \approx 0.5647$

The thermal efficiency, $\eta_{Otto}$ ≈ 0.5647

Therefore, the thermal efficiency ≈ 56.47%

(iii) The mean effective pressure, MEP is given as follows;

$MEP = \dfrac{\left(P_3 - P_1 \cdot r^k \right) \cdot \left(1 - \dfrac{1}{r^{k-1}} \right)}{(k -1)\cdot (r - 1)}$

Therefore, we get;

$MEP = \dfrac{\left(122.37 - 1 \times 8^{1.4} \right) \cdot \left(1 - \dfrac{1}{8^{1.4-1}} \right)}{(1.4 -1)\cdot (8 - 1)} \approx 20.974$

The mean effective pressure, MEP ≈ 20.974 bar

(b) Four types of internal combustion engine includes;

1) The diesel engine; Compression heating is the source of the ignition, with constant pressure combustion

2) The Otto engine which is the internal combustion engine found in cars that make use of gasoline as the source of fuel

The Otto engine cycle comprises of five steps; intake, compression, ignition, expansion and exhaust

3) The Brayton engine works on the principle of the steam turbine

4) The Wankel it follows the pattern of the Otto cycle but it does not have piston strokes

8 0

2 years ago

What measurement is the usable area of conduit based on?

Mariana [72]

The cable outside diameter and the conduit inside diameter.

6 0

1 year ago

A sample of soil has a volume of 0.45 ft^3 and a weight of 53.3 lb. After being dried inan oven, it has a weight of 45.1 lb. It

topjm [15]

Answer:

a) the moisture content before it was placed in the oven is 18.18%

b) degree of saturation for soil is 72.19%

Explanation:

Given the data in the question;

Moisture Content = [(Weight of soil before dry - dry weight) / dry weight] × 100

so we substitute

Moisture content = [(53.3 - 45.1) / 45.1 ] × 100

= (8.2/45.1) × 100

= 18.18%

Therefore the moisture content before it was placed in the oven is 18.18%

Dry Unit Weight = dry weight / volume

Dry Unit Weight = 45.1 lb / 0.45 ft³

Dry Unit Weight = 100.22 lb/ft³

we know that;

dry unit weight = (Specific gravity × unit weight of water) / (1 + e)

we also know that; unit weight of water is 62.43 lbf/ft³

so we substitute

e = (2.70×62.43 / 100.22) - 1

e = 1.68 - 1

e = 0.68

so void ratio e = 0.68

Now we determine the degree of saturation using the equation;

degree of saturation = (Moisture content × specific gravity) / void ratio

we substitute

degree of saturation = ( 18.18% × 2.7) / 0.68

= 0.49086 / 0.68

= 0.7219 ≈ 72.19%

Therefore degree of saturation for soil is 72.19%

7 0

2 years ago

Consider two different versions of algorithm for finding gcd of two numbers (as given below), Estimate how many times faster it

juin [17]

Answer:

Explanation:

Step 1:

a) The formula for compute greatest advisor is

gcd(m,n) = gcd (n,m mod n)

the gcd(31415,14142) by applying Euclid's algorithm is

gcd(31,415,14,142) =gcd(14,142,3,131)

=gcd=(3,131, 1,618)

=gcd(1,618, 1,513)

=gcd(1,513, 105)

=gcd(105, 43)

=gcd(43, 19)

=gcd(19, 5)

=gcd(5, 4)

=gcd(4, 1)

=gcd(1, 0)

STEP 2

b) The number of comparison of given input with the algorithm based on checking consecutive integers and Euclid's algorithm is

The number of division using Euclid's algorithm =10 from part (a)

The consecutive integer checking algorithm:

The number of iterations =14,142 and 1 or 2 division of iteration.

14,142 ∠= number of division∠ = 2*14,142

Euclid's algorithm is faster by at least 14,142/10 =1400 times

At most 2*14,142/10 =2800 times.

5 0

2 years ago

Stream Piracy – Kaaterskill, NY. Check and double-click the Problem 15 folder. The dark blue and orange streams highlight the pr

baherus [9]

Answer:

b. The pirating streams are eroding headwardly to intersect more of the other streams’ drainage basins, causing water to be diverted down their steeper gradients.

Explanation:

From the Kaaterskill NY 15 minute map (1906), this shows two classic examples of stream capture.

The Kaaterskill Creek flow down the east relatively steep slopes into the Hudson River Valley. While, the Gooseberry Creek is a low gradient stream flowing down the west direction which in turn drains the higher parts of the Catskills in this area.

However, there is Headward erosion of Kaaterskill Creek which resulted to the capture of part of the headwaters of Gooseberry Creek.

The evidence for this is the presence of "barbed" (enters at obtuse rather than acute angle) tributary which enters Kaaterskill Creek from South Lake which was once a part of the Gooseberry Creek drainage system.

It should be noted again, that there is drainage divide between the Gooseberry and Kaaterskill drainage systems (just to the left of the word Twilight) which is located in the center of the valley.

As it progresses, this divide will then move westward as Kaaterskill captures more and more of the Gooseberry system.

5 0

2 years ago