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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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B tiny nano means small!!
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3 years ago
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Find the derivative of y = sin(ln(5x2 − 2x))
pickupchik [31]

Answer:

y = \cos[\ln x + \ln (5\cdot x - 2)]\cdot \left(\frac{1}{x} + \frac{5}{5\cdot x-2} \right)

Explanation:

Let y = \sin[\ln(5\cdot x^{2}-2\cdot x)] and we proceed to find the derivative by the following steps:

1) y = \sin[\ln(5\cdot x^{2}-2\cdot x)] Given

2) y = \sin [\ln[x\cdot (5\cdot x - 2)]] Distributive property

3) y = \sin[\ln x + \ln (5\cdot x - 2 )] \ln (a\cdot b) = \ln a + \ln b

4) y = \cos[\ln x + \ln (5\cdot x - 2)]\cdot \left(\frac{1}{x} + \frac{5}{5\cdot x-2} \right)  \frac{d}{dx} (\sin x) = \cos x/\frac{d}{dx}(\ln x) = \frac{1}{x}/\frac{d}{dx}(c\cdot x^{n}) = n\cdot c\cdot x^{n-1}/Rule of chain/Result

3 0
2 years ago
The pump of a water distribution system is powered by a 6-kW electric motor whose efficiency is 95 percent. The water flow rate
Sonja [21]

Answer:

a) Mechanical efficiency (\varepsilon)=63.15%  b) Temperature rise= 0.028ºC

Explanation:

For the item a) you have to define the mechanical power introduced (Wmec) to the system and the power transferred to the water (Pw).

The power input (electric motor) is equal to the motor power multiplied by the efficiency. Thus, Wmec=0.95*6kW=5.7 kW.

Then, the power transferred (Pw) to the fluid is equal to the flow rate (Q) multiplied by the pressure jump \Delta P. So P_W = Q*\Delta P=0.018m^3/s * 200x10^3 Pa=3600W.

The efficiency is defined as the ratio between the output energy and the input energy. Then, the mechanical efficiency is \varepsilon=3.6kW/5.7kW=0.6315=63.15\%

For the b) item you have to consider that the inefficiency goes to the fluid as heat. So it is necessary to use the equation of the heat capacity but in a "flux" way. Calling <em>H</em> to the heat transfered to the fluid, the specif heat of the water and \rho the density of the water:

[tex]H=(5.7-3.6) kW=\rho*Q*c*\Delta T=1000kg/m^3*0.018m^3/s*4186J/(kg \ºC)*\Delta T[/tex]

Finally, the temperature rise is:

\Delta T=2100/75348 \ºC=0.028 \ºC

7 0
2 years ago
1. The equilibrium number of vacancies in Ni at 1123 K is 4.7x1022m-3. The atomic weight and density of Ni at 1123 K are 58.69 g
Ahat [919]

Answer:

vacancy formation energy of Ni is 1.400 eV

Explanation:

given data

number of vacancies in Ni = 4.7 x 10^{22}  m^{-3}

atomic weight = 58.69 g/mol

density = 8.8 g/cm³  

solution

we get here N that is

N  = \frac{N_A \times \rho}{A}   ...........1

N = \frac{6.023\times 10^{23} \times 8.8 \times 10^6}{58.69}

N = 9.030 \times 10^{28}  

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Nv = N \times e^{\frac{-Qv}{kT}}  .................2

put here value

4.7 \times 10^{22} = 9.030 \times 10^28 \times e^{\frac{-Qv}{8.62\times 10^{-5}\times 1123}}  

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take ln both side

ln (10^{-7} \times 5.20487 ) = ln (e^{\frac{-Qv}{0.0968}})

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3 0
3 years ago
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Whitepunk [10]

Answer:

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Explanation:

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Thus,

VN²/19.62 = 700/9.81

So,

VN² =1400

VN =37.416 m/s

Note: (800 - 100) = 700

7 0
2 years ago
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