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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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The Lamborghini SCV12 has 830 horse power.
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2 years ago
Read 2 more answers
A stationary gas-turbine power plant operates on a simple ideal Brayton cycle with air as the working fluid. The air enters the
ololo11 [35]

Answer:

A) W' = 15680 KW

B) W' = 17113.87 KW

Explanation:

We are given;

Temperature at state 1; T1 = 290 K

Temperature at state 3; T3 = 1100 K

Rate of heat transfer; Q_in = 35000 kJ/s = 35000 Kw

Pressure of air into compressor; P_c = 95 kPa

Pressure of air into turbine; P_t = 760 kPa

A) The power assuming constant specific heats at room temperature is gotten from;

W' = [1 - ((T4 - T1)/(T3 - T2))] × Q_in

Now, we don't have T4 and T2 but they can be gotten from;

T4 = [T3 × (r_p)^((1 - k)/k)]

T2 = [T1 × (r_p)^((k - 1)/k)]

r_p = P_t/P_c

r_p = 760/95

r_p = 8

Also,k which is specific heat capacity of air has a constant value of 1.4

Thus;

Plugging in the relevant values, we have;

T4 = [(1100 × (8^((1 - 1.4)/1.4)]

T4 = 607.25 K

T2 = [290 × (8^((1.4 - 1)/1.4)]

T2 = 525.32 K

Thus;

W' = [1 - ((607.25 - 290)/(1100 - 525.32))] × 35000

W' = 0.448 × 35000

W' = 15680 KW

B) The power accounting for the variation of specific heats with temperature is given by;

W' = [1 - ((h4 - h1)/(h3 - h2))] × Q_in

From the table attached, we have the following;

At temperature of 607.25 K and by interpolation; h4 = 614.64 KJ/K

At T3 = 1100 K, h3 = 1161.07 KJ/K

At T1 = 290 K, h1 = 290.16 KJ/K

At T2 = 525.32 K, and by interpolation, h2 = 526.12 KJ/K

Thus;

W' = [1 - ((614.64 - 290.16)/(1161.07 - 526.12))] × 35000

W' = 17113.87 KW

4 0
2 years ago
g A steel water pipe has an inner diameter of 12 in. and a wall thickness of 0.25 in. Determine the longitudinal and hoop stress
zvonat [6]

Answer:

a) \mathbf{\sigma _ 1 = 4800 psi}

     \mathbf{ \sigma _2 = 0}

b)\mathbf{\sigma _ 1 = 6000 psi}

  \mathbf{ \sigma _2 = 3000 psi}

Explanation:

Given that:

diameter d = 12 in

thickness t = 0.25 in

the radius = d/2 = 12 / 2 = 6 in

r/t = 6/0.25 = 24

24 > 10

Using the  thin wall cylinder formula;

The valve A is opened and the flowing water has a pressure P of 200 psi.

So;

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = 0

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{200(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 4800 psi}

b)The valve A is closed and the water pressure P is 250 psi.

where P = 250 psi

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = \frac{Pd}{4t}

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{250*(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 6000 psi}

\sigma _2 = \frac{Pd}{4t} \\ \\  \sigma _2 = \frac{250(12)}{4(0.25)}

\mathbf{ \sigma _2 = 3000 psi}

The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below

8 0
3 years ago
A circular hoop sits in a stream of water, oriented perpendicular to the current. If the area of the hoop is doubled, the flux (
natka813 [3]

Answer:

The flux (volume of water per unit time) through the hoop will also double.

Explanation:

The flux = volume of water per unit time = flow rate of water through the hoop.

The Flow rate of water through the hoop is proportional to the area of the hoop, and the velocity of the water through the hoop.

This means that

Flow rate = AV

where A is the area of the hoop

V is the velocity of the water through the hoop

This flow rate = volume of water per unit time = Δv/Δt =Q

From all the above statements, we can say

Q = AV

From the equation, if we double the area, and the velocity of the stream of water through the hoop does not change, then, the volume of water per unit time will also double or we can say increases by a factor of 2

3 0
3 years ago
What is the difference between a refrigeration cycle and a heat pump cycle?
sukhopar [10]

Answer:

In refrigeration cycle heat transfer from inside refrigeration

In heat pump cycle heat transfer from environment

Explanation:

heat cycle is mechanical process use for cool the temperature but

In refrigeration heat transfer from inside of refrigeration that decrease temperature of refrigerator and in heat pump it decrease temperature negligible as compare to refrigerator

5 0
3 years ago
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