A controlled variable is:

An ideal Diesel cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the (1) maximum air temperature and (2)

weqwewe [10]

Answer:

(1) The maximum air temperature is 1383.002 K

(2) The rate of heat addition is 215.5 kW

Explanation:

T₁ = 17 + 273.15 = 290.15

$\frac{T_2}{T_1} =r_v^{k - 1} =18^{0.4} =3.17767$

T₂ = 290.15 × 3.17767 = 922.00139

$\frac{T_3}{T_2} =\frac{v_3}{v_2} = r_c = 1.5$

Therefore,

T₃ = T₂×1.5 = 922.00139 × 1.5 = 1383.002 K

The maximum air temperature = T₃ = 1383.002 K

(2)

$\frac{v_4}{v_3} =\frac{v_4}{v_2} \times \frac{v_2}{v_3} = \frac{v_1}{v_2} \times \frac{v_2}{v_3} = 18 \times \frac{1}{1.5} = 12$

$\frac{T_3}{T_4} =(\frac{v_4}{v_3} )^{k-1} = 12^{0.4} = 2.702$

Therefore;

$T_4 = \frac{1383.002}{2.702} =511.859 \ k$

$Q_1 = c_p(T_3-T_2)$

Q₁ = 1.005(1383.002 - 922.00139) = 463.306 kJ/jg

Heat rejected per kilogram is given by the following relation;

$c_v(T_4-T_1)$ = 0.718×(511.859 - 290.15) = 159.187 kJ/kg

The efficiency is given by the following relation;

$\eta = 1-\frac{\beta ^{k}-1}{\left (\beta -1 \right )r_{v}^{k-1}}$

Where:

β = Cut off ratio

Plugging in the values, we get;

$\eta = 1-\frac{1.5 ^{1.4}-1}{\left (1.5 -1 \right )18^{1.4-1}}= 0.5191$

Therefore;

$\eta = \frac{\sum Q}{Q_1}$

$\therefore 0.5191 = \frac{150}{Q_1}$

Heat supplied = $\frac{150}{0.5191} = 288.978 \ hp$

Therefore, heat supplied = 215491.064 W

Heat supplied ≈ 215.5 kW

The rate of heat addition = 215.5 kW.

7 0

4 years ago

What is the objects acceleration

alekssr [168]

Answer:

Acceleration is a vector quantity which is defined as the rate at which an object changes its velocity. An object is accelerating if it is changing its velocity.

5 0

2 years ago

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The diagram shows a simplified periodic table. Each of the numbered blocks represents

sattari [20]

Answer:

F

Explanation:

Its a transition metal

5 0

3 years ago

Please help me with this question someone

Gekata [30.6K]

From the answers provided, I believe the possible answer would be the last option, silicon, oxygen, and one or more metals. Here's my reasoning: the most abundant mineral group found in the Earth's crust is the silicate group. The silicate materials contain both oxygen and silicon. Silicates are the most common minerals in the rock-formation process, and it has, in fact, been estimated that they make up 75 to 90 percent of the Earth's crust. From this piece of evidence, I can guess that the answer will possibly be D, silicon, oxygen, and one or more metals.
It should also be noted that the additional elements that combine with the silicon-oxygen tetrahedron are involved with the other elements commonly found in the Earth's crust and mantle. They are aluminum, calcium, iron, magnesium, potassium and sodium.

8 0

4 years ago

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Two planets are orbiting a star in a distant galaxy the first has a semimajor axis of 150 x 10^6 km an eccentricity of 0. 20 and

e-lub [12.9K]

For two planets orbiting a star in a distant galaxy, having the first semimajor axis of 150 x 10^6 km, the period of the second plane is mathematically given as

T=2.2earth year

<h3>What is the period of the second planet?</h3>

Generally, the equation for the time period is mathematically given as

$T^{2} \appox a^{3}$

Therefore, for planet one

$1^2\approx (150 x 10^6)^3$

In conclusion, For planet 2

T^2=(250/150)^3

T=√4.632

T=2.2earth year