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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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3 years ago
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a_sh-v [17]

Complete Question

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Answer:

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Explanation:

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3 0
3 years ago
A gas turbine operates with a regenerator and two stages of reheating and intercooling. Air enters this engine at 14 psia and 60
Rzqust [24]

Answer:

flow(m) = 7.941 lbm/s

Q_in = 90.5184 Btu/lbm

Q_out = 56.01856 Btu/lbm

Explanation:

Given:

- T_1 = 60 F = 520 R

- T_6 = 940 = 1400 R

- Heat ratio for air k = 1.4

- Compression ratio r = 3

- W_net,out = 1000 hp

Find:

mass flow rate of the air

rates of heat addition and rejection

Solution:

- Using ideal gas relation compute T_2, T_4, T_10:

                     T_2 = T_1 * r^(k-1/k)

                     T_2 = T_4 = T_10 = 520*3^(.4/1.4) = 711.744 R

- Using ideal gas relation compute T_7, T_5, T_9:

                     T_7 = T_6 * r^(-k-1/k)

                     T_7 = T_5 = T_9 = 1400*3^(-.4/1.4) = 1022.84 R

- The mass flow rate is obtained by:

                     flow(m) = W_net,out / 2*c_p*(1400-1022.84-711.744+520)

                     flow(m) = 1000*.7068 / 2*0.24*(1400-1022.84-711.744+520)

                     flow(m) = 7.941 lbm/s

- The heat input is as follows:

                     Q_in = c_p*(T_6 - T_5)

                     Q_in = 0.24*(1400 - 1022.84)

                     Q_in = 90.5184 Btu/lbm

- The heat output is as follows:

                     Q_out = c_p*(T_10 - T_1)

                     Q_out = 0.24*(711.744 - 520)

                    Q_out = 56.01856 Btu/lbm

                                           

                     

5 0
3 years ago
Assume that television broadcasts are nonrival and nonexcludable (some TV stations, such as those on cable TV, are excludable, b
ryzh [129]

Answer: Advertising acts in a method similar to a fee. People who watch TV broadcasts must watch ADs. TV stations turn this into money by selling airtime to advertisers.

Explanation:

A non-rival good is a good whose consumption by one person does not reduce the remaining quantity available. An example is a street light.

For non-excludable goods, it is impossible to prevent everyone from enjoying the benefits of the good. An example is a lighthouse. This is where the free rider problem comes in.

A free rider is someone enjoying the benefits of a good without paying for it.  When a good is both non-rival and non-excludable, it is convenient for consumers to enjoy the benefit without paying for it.

If TV broadcasts are both non-rival and non-excludable, everybody can choose to become a free rider. Advertising can solve this problem by converting free riders to potential buyers of goods or services advertised during broadcasts. This way, stations can generate revenue by selling airtime.

3 0
3 years ago
Read 2 more answers
Compare the use of a low-strength, ductile material (1018 CD) in which the stress-concentration factor can be ignored to a high-
kicyunya [14]

Answer:

Step 1 of 3

Case A:

AISI 1018 CD steel,

Fillet radius at wall=0.1 in,

Diameter of bar

From table deterministic ASTM minimum tensile and yield strengths for some hot rolled and cold drawn steels for 1018 CD steel

Tensile strength

Yield strength

The cross section at A experiences maximum bending moment at wall and constant torsion throughout the length. Due to reasonably high length to diameter ratio transverse shear will be very small compared to bending and torsion.

At the critical stress elements on the top and bottom surfaces transverse shear is zero

Explanation:

See the next steps in the attached image

5 0
3 years ago
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