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Annette [7]
3 years ago
10

Air initially at 120 psia and 500o F is expanded by an adiabatic turbine to 15 psia and 200o F. Assuming air can be treated as a

n ideal gas and has variable specific heat. a) Determine the specific work output of the actual turbine (Btu/lbm). b) Determine the amount of specific entropy generation during the irreversible process (Btu/lbm R). c) Determine the isentropic efficiency of this turbine (%). d) Suppose the turbine now operates as an ideal compressor (reversible and adiabatic) where the initial pressure is 15 psia, the initial temperature is 200 o F, and th

Engineering
1 answer:
Goshia [24]3 years ago
8 0

Answer:

a. Wa = 73.14 Btu/lbm

b. Sgen = 0.05042 Btu/lbm °R

c. Isentropic efficiency is 70.76%

d. Minimum specific work for compressor W = -146.2698 Btu/lbm [It is negative because work is being done on the compressor]

Explanation:

Complete question is as follows;

Air initially at 120 psia and 500oF is expanded by an adiabatic turbine to 15 psia and 200oF. Assuming air can be treated as an ideal gas and has variable specific heat.

a) Determine the specific work output of the actual turbine (Btu/lbm).

b) Determine the amount of specific entropy generation during the irreversible process (Btu/lbm R).

c) Determine the isentropic efficiency of this turbine (%).

d) Suppose the turbine now operates as an ideal compressor (reversible and adiabatic) where the initial pressure is 15 psia, the initial temperature is 200 oF, and the ideal exit state is 120 psia. What is the minimum specific work the compressor will be required to operate (Btu/lbm)?

solution;

Please check attachment for complete solution and step by step explanation

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To be safe, the engineers making the ride want to be sure the normal force does not exceed 1.8 times each persons weight - and t
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Answer:

μ = 0.55

Explanation:

Given that

Normal weight = 1.8 x weight of person

N= 1.8 mg

We know that friction force Fr

Fr= μ N

μ=Coefficient of friction

N=Normal force

To find  μ We have to equate friction and gravity force

Fr= Wt

μ N = m g

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μ = 0.55

So the coefficient of friction will be 0.55.

5 0
3 years ago
A 1.5-m-long aluminum rod must not stretch more than 1 mm and the normal stress must not exceed 40 MPa when the rod is subjected
Pavlova-9 [17]

Answer:

the required diameter of the rod is 9.77 mm

Explanation:

Given:

Length = 1.5 m

Tension(P) = 3 kN = 3 × 10³ N

Maximum allowable stress(S) = 40 MPa = 40 × 10⁶ Pa

E = 70 GPa = 70 × 10⁹ Pa

δ = 1 mm = 1 × 10⁻³ m

The required diameter(d)  = ?

a) for stress

The stress equation is given by:

S = \frac{P}{A}

A is the area = πd²/4 = (3.14 × d²)/4

S = \frac{P}{(\frac{3.14*d^{2} }{4}) }

S = \frac{4P}{{3.14*d^{2} } }

3.14*S*{d^{2}} = {4P}

{d^{2}} =\frac{4P}{3.14*S}

d=  \sqrt{\frac{4P}{3.14*S} }

Substituting the values, we get

d=  \sqrt{\frac{4*3*10^{3} }{3.14*40*10^{6} } }

d=  \sqrt{\frac{12000 }{125600000  } }

d=  \sqrt{9.55*10^{-5}  }

d = (9.77 × 10⁻³) m

d = 9.77 mm

b) for deformation

δ = (P×L) / (A×E)

A = (P×L) / (E×δ) = (3000 × 1.5) / (1 × 10⁻³ × 70 × 10⁹) = 0.000063

d² = (4 × A) / π = (0.000063 × 4) / 3.14

d² = 0.0000819

d = 9.05 × 10⁻³ m = 9.05 mm

We use the larger value of diameter = 9.77 mm

3 0
3 years ago
Does anyone know what I’m doing wrong here?
Bingel [31]

Answer:

It looks fine

Explanation:

6 0
3 years ago
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