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Likurg_2 [28]
3 years ago
6

A steam turbine receives steam at 1.5MPa and 220oC, and exhausts at 50kPa, 0.75 dry. Neglecting heat losses and changes in kinet

ic and potential energy, estimate the work output per kg steam.
If when allowance is made for friction, radiation and leakages losses, the actual work of that estimated in (a), calculate the power output of the turbine when consuming 600kg of steam per minute.
Engineering
1 answer:
SIZIF [17.4K]3 years ago
4 0

Answer:

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Row choice the cost of roadway improvements to the developer and functional the amount of trafficBeing generated by the theater as well as the Ralph’s ladies
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The change in specific internal energy depends on the path of a process. a)-True b)-False
Basile [38]

Answer:

(b) False

Explanation:

The specific internal energy of the system does not depend on the path of the process, it is a state function means its depend on only on the initial and the final position it does not depend on the path which it follow in the process.Internal energy is associated with the random motion of the molecules.

So it is false statement as internal energy is not a path function

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What’s the answer please help
irina1246 [14]

Answer:

It B

Explanation:

4 0
3 years ago
a. Draw the logic gate implementation for the Boolean function F = W X(Y + Z).b. Write the DeMorgan equivalent Boolean statement
Llana [10]

Answer:

DeMorgan equivalent :

F = B + C

F' = ⁻B⁻+⁻C⁻ = ⁻BC⁻

Explanation:

Attached below is the logic gate implementation diagram and the DeMorgan equivalent Boolean statement as requested in part A and B

3 0
3 years ago
Generally, final design results are rounded to or fixed to three digits because the given data cannot justify a greater display.
creativ13 [48]

Answer:

(a) 1.90 kpsi

(b) 0.40 kpsi

(c) 0.61 in.

(d) 0.009

(a) 8 MPa

(b) 1.30 cm⁴

(c) 2.04 cm⁴

(d) 62.2 MPa

Explanation:

(a) σ = M/Z, where M = 1770 lbf·in and Z = 0.943 in³.

1770/0.943 = 1876.988 lbf/in² = 1.90 kpsi

(b) σ = F/A, where F = 9440 lbf and A = 23.8 in².

9440 /23.8 = 396.639 lbf/in² = 0.4 kpsi

(c) y = Fl³/(3EI)

F = 270 lbf

l = 31.5 in.

E = 30 Mpsi

I = 0.154 in.⁴

y = 270×31.5³/(3×30×10⁶×0.154) = 0.61 in.

(d) θ = Tl/(GJ), where T = 9740 lbf·in, l = 9.85 in. G = 11.3 Mpsi, and d = 1.00 in.

J = π·d⁴/32 = π/32 in.⁴

∴ θ = 9740  × 9.85 /(11.3 × 10⁶× π/32) = 0.009

(a) σ = F/wt, where F = 1 kN, w = 25 mm, and t = 5 mm

∴ σ = 1000/(0.025 × 0.005) = 8 MPa

(b) I = bh³/12, where b = 10 mm and h = 25 mm.

10×25³/12 = 1.30 cm⁴

(c) I = π·d⁴/64 where d = 25.4 mm.

I = π × 25.4⁴/64 = 2.04 cm⁴

(d) τ = 16×T/(π×d³), where T = 25 N·m, and d = 12.7 mm.

16×25/(π×0.0127³) = 62.2 MPa.

8 0
3 years ago
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