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kirza4 [7]
2 years ago
6

A fixed mass of saturated water vapor at 400 kpa is isothermally cooled until it is a saturated liquid. Calculate the amount of

heat rejected during this process in kJ/kg.

Engineering
1 answer:
Kazeer [188]2 years ago
6 0
This is the explanation

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Determine F12 and F21 for the following configurations: (a) A long semicircular duct with diameter of 0.1 meters: (b) A hemisphe
uysha [10]

Answer:

long duct: 1.0 and 0.424

Hemisphere 1.0 ; 0.125; 0.5

Explanation:

For a long duct:

By inspection, F_{12} = 1.0

Calculating by reciprocity, F_{21} = \frac{A_{1} }{A_{2}F_{12}  }  = \frac{2RL}{\frac{3}{4}*2\pi RL  }* 1.0\\                                                         = 0.424

Hemisphere:

By reciprocity gives = 0.125

using the summation rule: F_{21} + F_{22} + F_{23} = 1

However, because this is a hemisphere, the value will be= 0.5 * 1

                                                                                               = 0.5

6 0
2 years ago
5. Which statement regarding a finite state machine (FSM) is NOT true: (a) In a non-deterministic FSM, a string is invalid if th
valkas [14]

Answer:

The statement (a) In a non-deterministic FSM, a string is invalid if there is one path not leading to a final state is NOT true

Explanation:

A non-deterministic FSM, contrary to deterministic FSM which has only one possible thread of execution, has multiple threads and for the machine to be invalid, all threads should lead to a none accepting (final) state.

7 0
2 years ago
A fatigue test was conducted in which the mean stress was 46.2 MPa and the stress amplitude was 219 MPa.
sleet_krkn [62]

Answer:

a)σ₁ = 265.2 MPa

b)σ₂ = -172.8 MPa

c)Stress\ ratio =-0.65

d)Range = 438 MPa

Explanation:

Given that

Mean stress ,σm= 46.2 MPa

Stress amplitude ,σa= 219 MPa

Lets take

Maximum stress level = σ₁

Minimum stress level =σ₂

The mean stress given as

\sigma_m=\dfrac{\sigma_1+\sigma_2}{2}

2\sigma_m={\sigma_1+\sigma_2}

2 x 46.2 =  σ₁ +  σ₂

 σ₁ +  σ₂ = 92.4 MPa    --------1

The amplitude stress given as

\sigma_a=\dfrac{\sigma_1-\sigma_2}{2}

2\sigma_a={\sigma_1-\sigma_2}

2 x 219 =  σ₁ -  σ₂

 σ₁ -  σ₂ = 438 MPa    --------2

By adding the above equation

2  σ₁ = 530.4

σ₁ = 265.2 MPa

-σ₂ = 438 -265.2 MPa

σ₂ = -172.8 MPa

Stress ratio

Stress\ ratio =\dfrac{\sigma_{min}}{\sigma_{max}}

Stress\ ratio =\dfrac{-172.8}{265.2}

Stress\ ratio =-0.65

Range = 265.2 MPa - ( -172.8 MPa)

Range = 438 MPa

8 0
2 years ago
Consider an infinitely thin flat plate of chord c at an angle of attack α in a supersonic flow. The pressure on the upper and lo
amm1812

Answer:

X_cp = c/2

Explanation:

We are given;

Chord = c

Angle of attack = α

p u (s) = c 1

​p1(s)=c2,

and c2 > c1

First of all, we need to find the resultant normal force on the plate and the total moment about leading edge.

I've attached the solution

4 0
2 years ago
A 50-kg iron block and 20-kg copper block, both initially at 80 oC, are dropped into a large lake at 15 oC. Thermal equilibrium
bija089 [108]
I need this answer to please
5 0
2 years ago
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