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Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ

AURORKA [14]

Answer:

a) $w_{out} = 281.55\,\frac{kJ}{kg}$ , b) $s_{gen} = 0.477\,\frac{kJ}{kg\cdot K}$

Explanation:

a) The process within the turbine is modelled after the First Law of Thermodynamics:

$-q_{out} - w_{out} + h_{in}-h_{out} = 0$

$w_{out} = h_{in} - h_{out}-q_{out}$

$w_{out} = c_{p}\cdot (T_{in}-T_{out})-q_{out}$

$w_{out} = \left(1.005\,\frac{kJ}{kg\cdot K}\right)\cdot (980\,K-670\,K)-30\,\frac{kJ}{kg}$

$w_{out} = 281.55\,\frac{kJ}{kg}$

b) The entropy production is determined after the Second Law of Thermodynamics:

$-\frac{q_{out}}{T_{surr}} + s_{in}-s_{out} + s_{gen} = 0$

$s_{gen} = \frac{q_{out}}{T_{surr}}+s_{out}-s_{in}$

$s_{gen} = \frac{q_{out}}{T_{surr}}+c_{p}\cdot \ln\left(\frac{T_{out}}{T_{in}} \right)$

$s_{gen} = \frac{30\,\frac{kJ}{kg} }{315\,K} + \left(1.005\,\frac{kJ}{kg\cdot K} \right)\cdot \ln\left(\frac{980\,K}{670\,K} \right)$

$s_{gen} = 0.477\,\frac{kJ}{kg\cdot K}$

3 0

3 years ago

Required information NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part

ra1l [238]

Answer:

h1 = 290.16kj/kg

P = 1.2311

Prandil expression at 8

P=p1/p7×pr

=8(1.2311)

=9.85

Enthalpy state at 8 corresponding to 9.85

h1 = 526.13kj/kg

Now prandtl state at 9 that correspond to 1400k.

h9 = 1515.42kj/kg

Pr = 450.5

Prandtl expression at state 10

P= p10/p9×pr

=1/8(450.5)

=56.31

Enthalpy at state 10 corresponding to prandtl 56.31

h10 = 860.39kj/kg

At 520k

h11 = 523.63kj/kg

4 0

3 years ago

What tools do you need to change a flat tire

Lyrx [107]

Answer:

Spare tire (if you’re dealing with a flat and not just rotating tires or accessing the brakes)
Carjack
Lug wrench
Wheel wedges
Work gloves
Tire repair kit (if you’re attempting to fix your flat tire)
Flare/reflective triangles (if you’re changing the tire on the side of the road or in a parking lot)

6 0

4 years ago

Read 2 more answers

Ring rolling is a deformation process in which a thick-walled ring of smaller diameter is rolled into a thin-walled ring of larg

VMariaS [17]

Answer:

a)True

Explanation:

Rolling:

Rolling is a metal forming in which a material passes through two or more than two depends on conditions,rolls to produce the desired product.

Ring rolling:

In ring rolling a thick ring compresses by rolls to produce the large diameter ring.Actually volume of material is constant so when diameter of ring increases then to compensate it, the thickness of ring reduces .In simple words we can say that in ring rolling a thick ring of smaller is rolled into a thin ring of larger diameter.

6 0

3 years ago

An experiment compares the initial speed of bullets fired from two handguns: a 9 mm and a 0.44 caliber. The guns are fired into

olasank [31]

Answer:

1.176

Explanation:

When the bullets impact the mass they become embedded on it, it is a plastic collision, therefore momentum is conserved.

v2 * (M + mb) = v1 * mb

Where

v1: muzzle velocity of the bullet

M: mass of the bob

mb: mass of the bullet

v2: mass of the bob with the bullet after being hit

v2 = v1 * mb / (M + mb)

Upon being impacted the bob will acquire speed v2, this implies a kinetic energy. The bob will then move and raise a height h. Upon acheiving the maximum height it will have a speed of zero. At that point all kinetic energy will be converted into potential energy.

Ek = 1/2 (M + mb) * v2^2

Ep = (M + mb) * g * h

Ek = Ep

1/2 (M + mb) * v2^2 = (M + mb) * g * h

1/2 * (v1 * mb / (M + mb))^2 = g * h

1/2 * v1^2 * mb^2 / (M + mb)^2 = g * h

v1^2 = g *h * (M+ mb)^2 / (1/2 * mb^2)

$v2 = \sqrt{\frac{g *h * (M+ mb)^2}{\frac{1}{2} * mb^2}}$

The height h that it reaches is related to the length L of the pendulum arm and the angle it forms with the vertical.

h = L * (1 - cos(a))

$v2 = \sqrt{\frac{g * L * (1 - cos(a)) * (M+ mb)^2}{\frac{1}{2} * mb^2}}$

For the 9 mm:

$v2 = \sqrt{\frac{9.81 * L * (1 - cos(4.3)) * (10+ 0.006)^2}{\frac{1}{2} * 0.006^2}} = \sqrt{L} * 391$

For the 0.44 caliber:

$v2 = \sqrt{\frac{9.81 * L * (1 - cos(10.1)) * (10+ 0.012)^2}{\frac{1}{2} * 0.012^2}} = \sqrt{L} * 460$

The ratio is 460 / 391 = 1.176

6 0

3 years ago