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blondinia [14]
4 years ago
7

Air at 7°C enters a turbojet engine at a rate of 16 kg/s and at a velocity of 220 m/s (relative to the engine). Air is heated in

the combustion chamber at a rate 15,000 kJ/s, and it leaves the engine at 427°C. Determine the thrust produced by this turbojet engine.
Engineering
1 answer:
Readme [11.4K]4 years ago
6 0

Answer:

F=14.05 KN

Explanation:

Now from first law for open system

h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w

In Turbojet net work out put is zero because ,power produce by turbine is consumed by compressor.

So w=0

Given mass flow rate =16 kg/s

V_1=220 m/s

Heat input Q=15000 KW  (Q=937.5 KJ/kg)

h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}

1.005\times (273+70)+\dfrac{220^2}{2000}+937.5=1.005\times (273+427)+\dfrac{V_2^2}{2000}

So V_2=1098.1 m/s

So thrust

F=\dot{m_a}(V_2-V_1)

F=16(1098.32-220)

F=14.05 KN

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In a tempering process, glass plate, which is initially at a uniform temperature Ti, is cooled by suddenly reducing the temperat
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Answer:

See attached document for complete answers and also follow the step by step explanation to get given parameters

Explanation:

Given that:

In a tempering process, glass plate, which is initially at a uniform temperature Ti, is cooled by suddenly reducing the temperature of both surfaces to Ts. The plate is 10 mm thick, and the glass has a thermal diffusivity of 6 × 10−7 m2/s. (a) How long will it take for the midplane temperature to achieve 75% of its maximum possible temperature reduction? (b) If (Ti − Ts) = 300°C, what is the maximum temperature gradient in the glass at the time calculated in part (a)? (c) Plot the temperature distribution within the glass plate at t = 10s

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4 years ago
For some alloy, the yield stress is 345-MPa (50,000-psi) and the elastic modulus (E) is 103-GPa (15x106 psi). What is the maximu
OverLord2011 [107]

The maximum load that may be applied to a specimen with a cross-sectional area of 130 mm² is; 35535 N

<h3>How to find Elastic Modulus?</h3>

We are told that for an alloy, the yield stress is 345-MPa and the elastic modulus (E) is 103-GPa.

Now, we want to find the maximum load that may be applied to a specimen with a cross-sectional area of 130-mm² without plastic deformation. Thus;

We are given the parameters;

Yield Stress; σ = 345 Mpa = 345 * 10⁶ Pa

Elastic Modulus; E = 103 GPa = 103 * 10⁹ Pa

Cross sectional Area; A = 130 mm² = 103 * 10⁻⁶ m²

Formula for stress without Plastic deformation is;

σ = F_max/Area

where;

σ is stress

F_max is maximum force

Area is Area

Thus making maximum force the subject of the formula gives;

F_max = σ * A

Plugging in the relevant values for stress and area gives us;

F_max = 345 * 10⁶ * 103 * 10⁻⁶

F_max = 35535 N

The maximum load that may be applied to a specimen with a cross-sectional area of 130 mm² is gotten to be 35535 N

Read more about Elastic Modulus at; brainly.com/question/6864866

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