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3 years ago

Which inequality shows a value between and including 1.9 and 2.1?

Engineering

1 answer:

irina1246 [14]3 years ago

5 0

Answer:

C. $1.90\leq x \leq 2.10$

Explanation:

The question states that x has is value in between and including 1.9 and 2.1. So the number is greater than or equal to 1.9 but less than or equal to 2.1.

In option A $x\geq 2.1$ this means that the value of x is greater than or equal to 2.1 which does not satisfy the condition mentioned in the question

In option D $x$ is greater than or equal to 2.1 but less than or equal to 1.9 which does not satisfy the condition mentioned in the question.

In option C $x$ is greater than or equal to 1.9 and less than or equal to 2.1 which does satisfy the condition mentioned in the question.

So, option C is correct.

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A well-insulated rigid tank contains 5 kg of a saturated liquid–vapor mixture of water at 200 kPa. Initially, three-quarters of

vlabodo [156]

Answer:

$S_{gen} = 18.519\,\frac{kJ}{K}$

Explanation:

Given that rigid tank is a closed system, the following model is constructed after the First Law of Thermodynamics:

$W_{heater} + U_{sys,1} - U_{sys,2} = 0$

$W_{heater} = U_{sys,2} - U_{sys,1}$

$W_{heater} = m\cdot (u_{2}-u_{1})$

The entropy generation inside the rigid tank is determined by appropriate application of the Second Law of Thermodynamics:

$S_{sys,1} - S_{sys,2} + S_{gen} = 0$

$S_{gen} = S_{sys,2} - S_{sys,1}$

$S_{gen} = m\cdot (s_{2}-s_{1})$

The properties of the steam are obtained from steam tables:

Intial State

$P = 200\,kPa$

$T = 120.21\,^{\textdegree}C$

$\nu = 0.2222\,\frac{m^{3}}{kg}$

$u = 1010.7\,\frac{kJ}{kg}$

$s = 2.9294\,\frac{kJ}{kg\cdot K}$

$x = 0.25$

Final State

$P = 869.567\,kPa$

$T = 173.88\,^{\textdegree} C$

$\nu = 0.2222\,\frac{m^{3}}{kg}$

$u = 2578.6\,\frac{kJ}{kg}$

$s = 6.6332\,\frac{kJ}{kg\cdot K}$

$x = 1.00$

The entropy change of the steam during the process is:

$S_{gen} = (5\,kg)\cdot \left(6.6332\,\frac{kJ}{kg\cdot K} - 2.9294\,\frac{kJ}{kg\cdot K} \right)$

$S_{gen} = 18.519\,\frac{kJ}{K}$

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4 years ago

What differentiates an athletic trainer from a clinical exercise physiologist?

kozerog [31]

Answer:

It's number 3

Explanation:

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3 years ago

Read 2 more answers

The contents of a tank are to be mixed with a turbine impeller that has six flat blades. The diameter of the impeller is 3 m. If

lions [1.4K]

Answer:

P=3.31 hp (2.47 kW).

Explanation:

Solution

Curve A in Fig1. applies under the conditions of this problem.

S1 = Da / Dt ; S2 = E / Dt ; S3 = L / Da ; S4 = W / Da ; S5 = J / Dt and S6 = H / Dt

The above notations are with reference to the diagram below against the dimensions noted. The notations are valid for other examples following also.

32.2

Fig. 32.2 Dimension of turbine agitator

The Reynolds number is calculated. The quantities for substitution are, in consistent units,

D a =2⋅ft

n= 90/ 60 =1.5 r/s

μ = 12 x 6.72 x 10-4 = 8.06 x 10-3 lb/ft-s

ρ = 93.5 lb/ft3 g= 32.17 ft/s2

NRc = (( D a) 2 n ρ)/ μ = 2 2 ×1.5×93.5 8.06× 10 −3 =69,600

From curve A (Fig.1) , for NRc = 69,600 , N P = 5.8, and from Eq. P= N P × (n) 3 × ( D a )5 × ρ g c

The power P= 5.8×93.5× (1.5) 3 × (2) 5 / 32.17 =1821⋅ft−lb f/s requirement is 1821/550 = 3.31 hp (2.47 kW).

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3 years ago

Air enters a compressor at 100 kPa and 25 ⁰C. It is compressed to 2 MPa and exits the compressor at 540 K. The compressor is at

AysviL [449]

Answer:

(a) The reversible work is 207 kJ/kg

(b) The irreversibility rate is -38.39 kJ/kg

Explanation:

State1 : p1 = 100kpa, T1= 25+273 =298k

From air table, h1 =298.18 kJ/kg, s10= 1.69528 kJ/kgK

State 2a:p2=2mpa,t2=540k (actual condition 2a)

h2a= 544.35 kJ/kg,s2a0=2.29906

actual work input to the compressor =wout=h1-h2+Qin

=298.18-544.35+(-150)kJ/kg(- sign indicate heat loss)

=(-246.17)kJ/kg(-ve sign indicates the work is given into the system

a) Reversible work= Win actual - any irreversiblities present

=246.17 + irreversibilty

b) irreversibility = T0(Entopy generation Sgen) for air, Sgen

=s20-s10-Rln(p2/p1), T0=250C

=(25+273)(s2a0-s10-Rlnp2/p1+Qout/Tsurr)

= 298x[(2.29906-1.69528-0.287kJ/kgK xln(2000kpa/100) + 150 /298]

= -38.39 kJ/kg

a)Reversible work = Win actual -any irreversiblities present

=246.17 + irreversibilty

=246.17+-38.39

=207 kJ/kg

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3 years ago

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kobusy [5.1K]

A) Jason is correct because smaller wings can cut through air better.

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4 years ago