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Varvara68 [4.7K]
3 years ago
14

Water at 20 bar and 400 C enters a turbine operating at steady state and exits at 1.5 bar. Stray heat transfer and kinetic and p

otential energy effects are negligible. A hard-to-read data sheet indicates that the quality, x, at the turbine exit is 98%. What is the value of entropy at the turbine inlet in kJ/kg-K?
Engineering
1 answer:
kherson [118]3 years ago
4 0

Answer:

s_1=7.1292\frac{kJ}{kg*K}, nevertheless, the outlet quality is not possible due to negative entropy generation.

Explanation:

Hello,

In this case, with the given conditions, it is possible to extract the entropy at the turbine inlet from the steam overheated tables at 20 bar and 400 °C, realizing that 20 bar equals 2 MPa. In such a way, the entropy at those conditions is:

s_1=7.1292\frac{kJ}{kg*K}

Moreover, we should prove that the 98% quality is possible in terms of the entropy generation which MUST be positive based on the entropy balance:

s_{gen}=s_2-s_1

Thus, we also compute the entropy at the outlet, looking for liquid-vapor water at 1.5 bar with the given quality:

s_2=1.3548\frac{kJ}{kg*K}+0.98*5.9187\frac{kJ}{kg*K}\\s_2=7.155\frac{kJ}{kg*K}

Hence the entropy generation turns out:

s_{gen}=s_2-s_1=7.155-7.1292=-0.0259\frac{kJ}{kg*K}

Finally, such value means that the outlet quality is not thermodynamically possible.

Regards.

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Using a forked rod, a 0.5-kg smooth peg P is forced to move along the vertical slotted path r = (0.5 θ) m, whereθ is in radians.
-BARSIC- [3]

Answer:

N_c = 3.03 N

F = 1.81 N

Explanation:

Given:

- The attachment missing from the question is given:

- The given expressions for the radial and θ direction of motion:

                                       r = 0.5*θ

                                       θ = 0.5*t^2              ...... (correction for the question)

- Mass of peg m = 0.5 kg

Find:

a) Determine the magnitude of the force of the rod on the peg at the instant t = 2 s.

b) Determine the magnitude of the normal force of the slot on the peg.

Solution:

- Determine the expressions for radial kinematics:

                                        dr/dt = 0.5*dθ/dt

                                        d^2r/dt^2 = 0.5*d^2θ/dt^2

- Similarly the expressions for θ direction kinematics:

                                        dθ/dt = t

                                        d^2θ/dt^2 = 1

- Evaluate each at time t = 2 s.

                                        θ = 0.5*t^2 = 0.5*2^2 = 2 rad -----> 114.59°

                                        r = 1 m , dr / dt = 1 m/s , d^2 r / dt^2 = 0.5 m/s^2

- Evaluate the angle ψ between radial and horizontal direction:

                                        tan Ψ = r / (dr/dθ) = 1 / 0.5

                                        Ψ = 63.43°

- Develop a free body diagram (attached) and the compute the radial and θ acceleration:

                                        a_r = d^2r / dt^2 - r * dθ/dt

                                        a_r = 0.5 - 1*(2)^2 = -3.5 m/s^2

                                        a_θ =  r * (d^2θ/dt^2) + 2 * (dr/dt) * (dθ/dt)

                                        a_θ = 1(1) + 2*(1)*(2) = 5 m/s^2

- Using Newton's Second Law of motion to construct equations in both radial and θ directions as follows:

Radial direction:              N_c * cos(26.57) - W*cos(24.59) = m*a_r

θ direction:                      F  - N_c * sin(26.57) + W*sin(24.59) = m*a_θ

Where, F is the force on the peg by rod and N_c is the normal force on peg by the slot. W is the weight of the peg. Using radial equation:

                                       N_c * cos(26.57) - 4.905*cos(24.59) = 0.5*-3.5

                                       N_c = 3.03 N

                                       F  - 3.03 * sin(26.57) + 4.905*sin(24.59) = 0.5*5

                                       F = 1.81 N

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