All categories

3 years ago

Water flows with an average speed of 6.5 ft/s in a rectangular channel having a width of 5 ft The depth of the water is 2 ft.

Engineering

1 answer:

lisov135 [29]3 years ago

7 0

Answer:

specific energy = 2.65 ft

y2 = 1.48 ft

Explanation:

given data

average speed v = 6.5 ft/s

width = 5 ft

depth of the water y = 2 ft

solution

we get here specific energy that is express as

specific energy = y + $\frac{v^2}{2g}$ ...............1

put here value and we get

specific energy = $2 + \frac{6.5^2}{2\times 9.8\times 3.281}$

specific energy = 2.65 ft

and

alternate depth is

y2 = $\frac{y1}{2} \times (-1+\sqrt{1+8Fr^2})$

and

here Fr² = $\frac{v1}{\sqrt{gy}} = \frac{6.5}{\sqrt{32.8\times 2}}$

Fr² = 0.8025

put here value and we get

y2 = $\frac{2}{2} \times (-1+\sqrt{1+8\times 0.8025^2})$

y2 = 1.48 ft

You might be interested in

The ________________ attraction between the Earth and the moon is ______________ on the side of the Earth that happens to be ___

Karolina [17]

Answer:

Gravitational; strongest; facing; closer; near side; toward.

Explanation:

The gravitational attraction between the Earth and the moon is strongest on the side of the Earth that happens to be facing the moon, simply because it is closer. This attraction causes the water on this “near side” of Earth to be pulled toward the moon. These forces of attraction and inertia tends to keep the water in place and consequently, leads to a bulge of water on the near side with respect to the moon.

Also, you should note that what is responsible for the moon being in orbit around the Earth is the gravitational force of attraction between the two planetary bodies (Earth and Moon).

7 0

3 years ago

A Rankine steam power plant is considered. Saturated water vapor enters a turbine at 8 MPa and exits at condenser at 10 kPa. The

Ray Of Light [21]

Answer:

0.31

126.23 kg/s

Explanation:

Given:-

- Fluid: Water

- Turbine: P3 = 8MPa , P4 = 10 KPa , nt = 85%

- Pump: Isentropic

- Net cycle-work output, Wnet = 100 MW

Find:-

- The thermal efficiency of the cycle

- The mass flow rate of steam

Solution:-

- The best way to deal with questions related to power cycles is to determine the process and write down the requisite properties of the fluid at each state.

First process: Isentropic compression by pump

P1 = P4 = 10 KPa ( condenser and pump inlet is usually equal )

h1 = h-P1 = 191.81 KJ/kg ( saturated liquid assumption )

s1 = s-P1 = 0.6492 KJ/kg.K

v1 = v-P1 = 0.001010 m^3 / kg

P2 = P3 = 8 MPa( Boiler pressure - Turbine inlet )

s2 = s1 = 0.6492 KJ/kg.K .... ( compressed liquid )

- To determine the ( h2 ) at state point 2 : Pump exit. We need to determine the wok-done by pump on the water ( Wp ). So from work-done principle we have:

$w_p = v_1*( P_2 - P_1 )\\\\w_p = 0.001010*( 8000 - 10 )\\\\w_p = 8.0699 \frac{KJ}{kg}$

- From the following relation we can determine ( h2 ) as follows:

h2 = h1 + wp

h2 = 191.81 + 8.0699

h2 = 199.88 KJ/kg

Second Process: Boiler supplies heat to the fluid and vaporize

- We have already evaluated the inlet fluid properties to the boiler ( pump exit property ).

- To determine the exit property of the fluid when the fluid is vaporized to steam in boiler ( super-heated phase ).

P3 = 8 MPa

T3 = ? ( assume fluid exist in the saturated vapor phase )

h3 = hg-P3 = 2758.7 KJ/kg

s3 = sg-P3 = 5.7450 KJ/kg.K

- The amount of heat supplied by the boiler per kg of fluid to the water stream. ( qs ) is determined using the state points 2 and 3 as follows:

$q_s = h_3 - h_2\\\\q_s = 2758.7 -199.88\\\\q_s = 2558.82 \frac{KJ}{kg}$

Third Process: The expansion ( actual case ). Turbine isentropic efficiency ( nt ).

- The saturated vapor steam is expanded by the turbine to the condenser pressure. The turbine inlet pressure conditions are similar to the boiler conditions.

- Under the isentropic conditions the steam exits the turbine at the following conditions:

P4 = 10 KPa

s4 = s3 = 5.7450 KJ/kg.K ... ( liquid - vapor mixture phase )

- Compute the quality of the mixture at condenser inlet by the following relation:

$x = \frac{s_4 - s_f}{s_f_g} \\\\x = \frac{5.745- 0.6492}{7.4996} \\\\x = 0.67947$

- Determine the isentropic ( h4s ) at this state as follows:

$h_4_s = h_f + x*h_f_g\\\\h_4_s = 191.81 + 0.67947*2392.1\\\\h_4_s = 1817.170187 \frac{KJ}{kg}$

- Since, we know that the turbine is not 100% isentropic. We will use the working efficiency and determine the actual ( h4 ) at the condenser inlet state:

$h4 = h_3 - n_t*(h_3 - h_4_s ) \\\\h4 = 2758.7 - 0.85*(2758.7 - 181.170187 ) \\\\h4 = 1958.39965 \frac{KJ}{kg} \\$

- We can now compute the work-produced ( wt ) due to the expansion of steam in turbine.

$w_t = h_3 - h_4\\\\w_t = 2758.7-1958.39965\\\\w_t = 800.30034 \frac{KJ}{kg}$

- The net power out-put from the plant is derived from the net work produced by the compression and expansion process in pump and turbine, respectively.

$W_n_e_t = flow(m) * ( w_t - w_p )\\\\flow ( m ) = \frac{W_n_e_t}{w_t - w_p} \\\\flow ( m ) = \frac{100000}{800.30034-8.0699} \\\\flow ( m ) = 126.23 \frac{kg}{s}$

Answer: The mass flow rate of the steam would be 126.23 kg/s

- The thermal efficiency of the cycle ( nth ) is defined as the ratio of net work produced by the cycle ( Wnet ) and the heat supplied by the boiler to the water ( Qs ):

$n_t_h = \frac{W_n_e_t}{flow(m)*q_s} \\\\n_t_h = \frac{100000}{126.23*2558.82} \\\\n_t_h = 0.31$

Answer: The thermal efficiency of the cycle is 0.31

7 0

3 years ago

A steel bolt has a modulus of 207 GPa. It holds two rigid plates together at a high temperature under conditions where the creep

VikaD [51]

Answer:

14.36((14MPa) approximately

Explanation:

In this question, we are asked to calculate the stress tightened in a bolt to a stress of 69MPa.

Please check attachment for complete solution and step by step explanation

7 0

3 years ago

A 4-L pressure cooker has an operating pressure of 175 kPa. Initially, one-half of the volume is filled with liquid and the othe

vodomira [7]

Answer:

the highest rate of heat transfer allowed is 0.9306 kW

Explanation:

Given the data in the question;

Volume = 4L = 0.004 m³

V $_f$ = V $_g$ = 0.002 m³

Using Table ( saturated water - pressure table);

at pressure p = 175 kPa;

v $_f$ = 0.001057 m³/kg

v $_g$ = 1.0037 m³/kg

u $_f$ = 486.82 kJ/kg

u $_g$ 2524.5 kJ/kg

h $_g$ = 2700.2 kJ/kg

So the initial mass of the water;

m₁ = V $_f$ /v $_f$ + V $_g$ /v $_g$

we substitute

m₁ = 0.002/0.001057 + 0.002/1.0037

m₁ = 1.89414 kg

Now, the final mass will be;

m₂ = V/v $_g$

m₂ = 0.004 / 1.0037

m₂ = 0.003985 kg

Now, mass leaving the pressure cooker is;

m $_{out$ = m₁ - m₂

m $_{out$ = 1.89414 - 0.003985

m $_{out$ = 1.890155 kg

so, Initial internal energy will be;

U₁ = m $_f$ u $_f$ + m $_g$ u $_g$

U₁ = (V $_f$ /v $_f$ )u $_f$ + (V $_g$ /v $_g$ )u $_g$

we substitute

U₁ = (0.002/0.001057)(486.82) + (0.002/1.0037)(2524.5)

U₁ = 921.135288 + 5.030387

U₁ = 926.165675 kJ

Now, using Energy balance;

E $_{in$ - E $_{out$ = ΔE $_{sys$

QΔt - m $_{out$ h $_{out$ = m₂u₂ - U₁

QΔt - m $_{out$ h $_g$ = m₂u $_g$ - U₁

given that time = 75 min = 75 × 60s = 4500 sec

so we substitute

Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675

Q(4500) - 5103.7965 = 10.06013 - 926.165675

Q(4500) = 10.06013 - 926.165675 + 5103.7965

Q(4500) = 4187.690955

Q = 4187.690955 / 4500

Q = 0.9306 kW

Therefore, the highest rate of heat transfer allowed is 0.9306 kW

5 0

3 years ago

Give me the description of - Feedforward control loops with an example.

ValentinkaMS [17]

Answer:

Feedforward basically configured and used mainly to avoid errors in a control system entering or disrupting a control loop

Explanation:

Feedforward basically configured and used mainly to avoid errors in a control system entering or disrupting a control loop. Although Feedforward control seems to be a very attractive idea, it imposes a high responsibility on both the system developer and the operator to examine and consider mathematically the effect of disruptions on the process concerned.

example of feedforward is

Shower

which consist of following control points

Hear toilet flush (measurement)

Customize water to compensate

feedback refers to that point when water turns hot before the configuration changes

5 0

3 years ago