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blondinia [14]
3 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]3 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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Water is the working fluid in an ideal Rankine cycle. Saturated vapor enters the turbine at 12 MPa, and the condenser pressure i
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Answer:

\dot Q_{in} = 372.239\,MW

Explanation:

The water enters to the pump as saturated liquid and equation is modelled after the First Law of Thermodynamics:

w_{in} + h_{in}- h_{out} = 0

h_{out} = w_{in}+h_{in}

h_{out} = 12\,\frac{kJ}{kg} + 191.81\,\frac{kJ}{kg}

h_{out} = 203.81\,\frac{kJ}{kg}

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h_{out} = 2685.4\,\frac{kJ}{kg}

The rate of heat transfer in the boiler is:

\dot Q_{in} = \left(150\,\frac{kg}{s}\right)\cdot \left(2685.4\,\frac{kJ}{kg}-203.81\,\frac{kJ}{kg} \right)\cdot \left(\frac{1\,MW}{1000\,kW} \right)

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A cylindrical bar of metal having a diameter of 20.2 mm and a length of 209 mm is deformed elastically in tension with a force o
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Answer:

A) ΔL = 0.503 mm

B) Δd = -0.016 mm

Explanation:

A) From Hooke's law; σ = Eε

Where,

σ is stress

ε is strain

E is elastic modulus

Now, σ is simply Force/Area

So, with the initial area; σ = F/A_o

A_o = (π(d_o)²)/4

σ = 4F/(π(d_o)²)

Strain is simply; change in length/original length

So for initial length, ε = ΔL/L_o

So, combining the formulas for stress and strain into Hooke's law, we now have;

4F/(π(d_o)²) = E(ΔL/L_o)

Making ΔL the subject, we now have;

ΔL = (4F•L_o)/(E•π(d_o)²)

We are given;

F = 50500 N

L_o = 209mm = 0.209m

E = 65.5 GPa = 65.5 × 10^(9) N/m²

d_o = 20.2 mm = 0.0202 m

Plugging in these values, we have;

ΔL = (4 × 50500 × 0.209)/(65.5 × 10^(9) × π × (0.0202)²)

ΔL = 0.503 × 10^(-3) m = 0.503 mm

B) The formula for Poisson's ratio is;

v = -(ε_x/ε_z)

Where; ε_x is transverse strain and ε_z is longitudinal strain.

So,

ε_x = Δd/d_o

ε_z = ΔL/L_o

Thus;

v = - [(Δd/d_o)/(ΔL/L_o)]

v = - [(Δd•L_o)/(ΔL•d_o)]

Making Δd the subject, we have;

Δd = -[(v•ΔL•d_o)/L_o]

We are given v = 0.33; d_o = 20.2mm

So,

Δd = -[(0.33 × 0.503 × 20.2)/209]

Δd = -0.016 mm

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