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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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5 0
2 years ago
How might an operations manager alter operations to meet customer demand? Name at least two ways.
Citrus2011 [14]
One way is manager changes itself and the other one is the same thing i think.
4 0
3 years ago
A gas flows through a one-inlet, one-exit control volume operating at steady state. Considering an adiabatic control volume with
Hunter-Best [27]

Answer:

b. equal to the specific entropy of the gas at the inlet.

Explanation:

Isentropic process is the process in which the entropy of the system remains unchanged. The word isentropic is formed from the combination of the prefix "iso" which means "equal" and the word entropy.

If a process is completely reversible, without the need to provide energy in the form of heat, then the process is isentropic.

3 0
3 years ago
When block C is in position xC = 0.8 m, its speed is 1.5 m/s to the right. Find the velocity of block A at this instant. Note th
Amanda [17]

Answer:

The answer is "2 m/s".

Explanation:

The triangle from of the right angle:

\to (x_c-0.8)+(1.5+y_4) +\sqrt{x_c^2 + 1.5^2}= constant

Differentiating the above equation:

\to V_c +V_A+ \frac{X_cV_c}{\sqrt{x_c^2 +1}}=0\\\\\to 1-V_A+ \frac{0.8 \times 1.5}{\sqrt{ 0.8^2+1.5}}=0\\\\

\to V_A=  \frac{1.2}{\sqrt{ 0.64+1.5}}+1\\\\

        = \frac{1.2}{ 1.46}+1\\\\= \frac{1.2+ 1.46}{ 1.46}\\\\ = \frac{2.66}{1.46}\\\\= 1.82 \ \frac{m}{s}\\\\= 2 \ \frac{m}{s}

3 0
2 years ago
Air (cp = 1.005 kJ/kg·°C) is to be preheated by hot exhaust gases in a cross-flow heat exchanger before it enters the furnace. A
uysha [10]

Answer:

Q=67.95 W

T=119.83°C

Explanation:

Given that

For air

Cp = 1.005 kJ/kg·°C

T= 20°C

V=0.6 m³/s

P= 95 KPa

We know that for air

P V = m' R T

95 x 0.6 = m x 0.287 x 293

m=0.677 kg/s

For gas

Cp = 1.10 kJ/kg·°C

m'=0.95 kg/s

Ti=160°C   ,To= 95°C

Heat loose by gas = Heat gain by air

[m Cp ΔT] for air =[m Cp ΔT] for gas

by putting the values

0.677 x 1.005 ( T - 20)= 0.95 x 1.1 x ( 160 -95 )

T=119.83°C

T is the exit temperature of the air.

Heat transfer

Q=[m Cp ΔT] for gas

Q=0.95 x 1.1 x ( 160 -95 )

Q=67.95 W

7 0
3 years ago
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