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IRISSAK [1]
3 years ago
7

Air at 7 deg Celcius enters a turbojet engine at a rate of 16 kg/s and at a velocity of 300 m/s (relative to engine). Air is hea

ted in the combustion chamber and it leaves the engine at 427 deg Celcius. Determine the thrust produced by this turbojet engine.
Engineering
1 answer:
Savatey [412]3 years ago
4 0

To solve this problem we will start using the given temperature values and then transform them to the Kelvin scale. From there through the properties of the tables, described for the Air, we will find the entropy. With these three data we can perform energy balance and find the speed of the fluid at the exit, which will finally help us find the total force:

V_i = 300m/s

T_1 = 7\°C = 280K

T_2 = 427\°C = 700K

\dot{m} 16kg/s

\dot{Q}_{in} = 15000kJ/s

Using the tables for gas properties of air we can find the enthalpy in this two states, then

T_1 = 280K \rightarrow h_1 = 280.13kJ/Kg\cdot K

T_2 = 700K \rightarrow h_2 = 713.27kJ/Kg\cdot K

Applying energy equation to the entire engine we have that

\frac{\dot{Q}_{in}}{\dot{m}}+h_1 +\frac{1}{2} V^2_{in} = h_2 + \frac{1}{2} V^2_{out}

\frac{15000}{16}*10^3+280.13*10^3+\frac{1}{2} 300^2 = 713.27*10^3+\frac{1}{2} V^2_{out}

V_{out} = 1048.198m/s

Finally the force in terms of mass flow and velocity is

F = \dot{m} (V_{out}-V_{in})

F = 16(1048.198-300)

F= 11971.168N

Therefore the thrust produced by this turbojet engine is  11971.168N

You might be interested in
An air-conditioning system operating on the reversed Carnot cycle is required to transfer heat from a house at a rate of 755 kJ/
Lyrx [107]

Answer:

There is 0.466 KW required to operate this air-conditioning system

Explanation:

<u>Step 1:</u> Data given

Heat transfer rate of the house = Ql = 755 kJ/min

House temperature = Th = 24°C = 24 +273 = 297 Kelvin

Outdoor temperature = To = 35 °C = 35 + 273 = 308 Kelvin

<u>Step 2: </u> Calculate the coefficient of performance o reversed carnot air-conditioner working between the specified temperature limits.

COPr,c = 1 / ((To/Th) - 1)

COPr,c = 1 /(( 308/297) - 1)

COPr,c = 1/ 0.037

COPr,c = 27

<u>Step 3:</u> The power input cna be given as followed:

Wnet,in = Ql / COPr,max

Wnet, in = 755  / 27

Wnet,in = 27.963 kJ/min

Win = 27.963 * 1 KW/60kJ/min  = 0.466 KW

There is 0.466 KW required to operate this air-conditioning system

3 0
3 years ago
I. Draw the velocity diagram for the instant shown and determine the velocity of
trapecia [35]

Answer:

?????????????????????????????????????

3 0
3 years ago
Problem 34.3 The elevation of the end of the steel beam supported by a concrete floor is adjusted by means of the steel wedges E
Natasha2012 [34]

The image is missing, so i have attached it.

Answer:

A) P = 65.11 KN

B) Q = 30 KN

Explanation:

We are given;

The end reaction of the beam; F = 100KN

Coefficient of static friction between two steel surfaces;μ_ss = 0.3

Coefficient of static friction between steel and concrete;μ_sc = 0.6

So, F1 = μ_ss•F =0.3 x 100 = 30 KN

F2 = μ_ss•N_EF = 0.3N_EF

From the screen shot, we see that the angle is 12°

Sum of forces in the Y-direction gives;

F2•sin12 - N_EF•cos12 + 100 = 0

Rearranging gives;

N_EF•cos12 - F2•sin12 = 100

Let's put 0.3N_EF for F2 to give;

N_EF•cos12 - 0.3N_EF•sin12 = 100

Thus;

N_EF(0.9158) - 0.1247 = 100

N_EF(0.9781) = 100 + 0.1247

N_EF = 100.1247/0.9158

N_EF = 109.33 KN

Thus, F2 = 0.3N_EF = 0.3 x 109.33 = 32.8 KN

Wedge will move if;

P = (F1 + F2cos12 + N_EFsin12)

Thus;

P = 10 + (32.8 x 0.9781) + (109.33 x 0.2079)

P ≥ 65.11 KN

B) For static equilibrium, Q = F1

Thus, Q = 30 KN

3 0
4 years ago
Based on experimental observations, the acceleration of a particle is defined by the relationa = -( 0.1 + sin(x/b) ),where a and
yKpoI14uk [10]

Answer:

a) v = +/- 0.323 m/s

b) x = -0.080134 m

c) v = +/- 1.004 m/s

Explanation:

Given:

                             a = - (0.1 + sin(x/b))

b = 0.8

v = 1 m/s @ x = 0

Find:

(a) the velocity of the particle when x = -1 m

(b) the position where the velocity is maximum

(c) the maximum velocity.

Solution:

- We will compute the velocity by integrating a by dt.

                           a = v*dv / dx =  - (0.1 + sin(x/0.8))

- Separate variables:

                           v*dv = - (0.1 + sin(x/0.8)) . dx

-Integrate from v = 1 m/s @ x = 0:

                          0.5(v^2) = - (0.1x - 0.8cos(x/0.8)) - 0.8 + 0.5

                          0.5v^2 =  0.8cos(x/0.8) - 0.1x - 0.3

- Evaluate @ x = -1

                          0.5v^2 = 0.8 cos(-1/0.8) + 0.1 -0.3

                          v = sqrt (0.104516)

                          v = +/- 0.323 m/s

- v = v_max when a = 0:

                           -0.1 = sin(x/0.8)

                             x = -0.8*0.1002

                             x = -0.080134 m

- Hence,

                            v^2 = 1.6 cos(-0.080134/0.8) -0.6 -0.2*-0.080134

                            v = sqrt (0.504)

                            v = +/- 1.004 m/s

4 0
3 years ago
How do technological artifacts affect the way that you live?
Maslowich

Answer:

Artefacts can influence our actions in several ways. They can be instruments, enabling and facilitating actions, where their presence affects the number and quality of the options for action available to us. They can also influence our actions in a morally more salient way, where their presence changes the likelihood that we will actually perform certain actions. Both kinds of influences are closely related, yet accounts of how they work have been developed largely independently, within different conceptual frameworks and for different purposes. In this paper I account for both kinds of influences within a single framework. Specifically, I develop a descriptive account of how the presence of artefacts affects what we actually do, which is based on a framework commonly used for normative investigations into how the presence of artefacts affects what we can do. This account describes the influence of artefacts on what we actually do in terms of the way facts about those artefacts alter our reasons for action. In developing this account, I will build on Dancy’s (2000a) account of practical reasoning. I will compare my account with two alternatives, those of Latour and Verbeek, and show how my account suggests a specification of their respective key concepts of prescription and invitation. Furthermore, I argue that my account helps us in analysing why the presence of artefacts sometimes fails to influence our actions, contrary to designer expectations or intentions.

When it comes to affecting human actions, it seems artefacts can play two roles. In their first role they can enable or facilitate human actions. Here, the presence of artefacts changes the number and quality of the options for action available to us.Footnote1 For example, their presence makes it possible for us to do things that we would not otherwise be able to do, and thereby adopt new goals, or helps us to do things we would otherwise be able to do, but in more time, with greater effort, etc

Explanation:

Technological artifacts are in general characterized narrowly as material objects made by (human) agents as means to achieve practical ends. ... Unintended by-products of making (e.g. sawdust) or of experiments (e.g. false positives in medical diagnostic tests) are not artifacts for Hilpinen.

3 0
3 years ago
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