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nignag [31]
3 years ago
15

Water flows through a pipe and enters a section where the cross sectional area is larger. Viscosity, friction, and gravitational

effects are negligible. Choose the letter of the correct statement about the change in pressure p and average velocity V.
(A) P2 is less than P1 and V2 is less than V1
(B) P2 is less than P1 and V2 is greater than V1
(C) P2 is greater than P1 and V2 is less than V1
(D) P2 is greater than P1 and V2 is greater than V1
Engineering
1 answer:
BaLLatris [955]3 years ago
3 0

Answer:

(A) and (D)

Explanation:

1) P2 is less than P1, that is when P1 increases in pressure, the velocity V1 of the water also increases. Therefore, on the other hand, since P2 is directly proportional to V1, P2 and V2 will be less than P1 and V1 respectively.

2) For P2 greater than P1 and V2 also is greater than V1. Since P2 is directly proportional to V2, hence, when P2 increases in pressure, P1 reduces in pressure. Similarly, velocity, V2 also increases and V1 reduces.

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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)

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Explanation:

<u>SiC </u>

Stress intensity, K is given by

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\sigma=\frac {K}{Y\sqrt{\pi a}} substituting the given figures and assuming shape factor, Y of 1

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<u>Stabilized zirconia </u>

Stress intensity, K is given by

K=\sigma Y\sqrt{\pi a} hence making \sigma the subject where \sigma is applied stress, Y is shape factor, a is crack length

\sigma=\frac {K}{Y\sqrt{\pi a}} substituting the given figures and assuming shape factor, Y of 1

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3 years ago
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