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2 years ago

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Print the two strings, firstString and secondString, in alphabetical order. Assume the strings are lowercase. End with newline.

Sample output:
capes rabbits

1 answer:

CaHeK987 [17]2 years ago

5 0

Hai!

Please name what kind of Script your Using then I would love to help.

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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

Are all transmissions fluids interchangeable

KiRa [710]

Answer:

<u>No</u>.

Explanation:

They are not all the same. Moreover, using a fluid that is not approved by the vehicle manufacturer will void the transmission warranty.

6 0

3 years ago

A brake caliper is considered a suspension item.<br> True<br> False

user100 [1]

True

Suspension is the system of tires, tire air, springs, shock absorbers and linkages that connects a vehicle to its wheels and allows relative motion between the two.[1] Suspension systems must support both road holding/handling and ride quality

5 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

how is friction losses in pipes reduced? a. decrease the pipe diameter b. increase the length of the pipes. c. decrease the leng

Citrus2011 [14]

Friction losses in pipes can be reduced by decreasing the length of the pipes, reducing the surface roughness of the pipes, and increasing the pipe diameter. Thus, options (c),(e), and (f) hold correct answers.

Friction loss is a measure of the amount of energy a piping system loses because flowing fluids meet resistance. As fluids flow through the pipes, they carry energy with them. Unfortunately, whenever there is resistance to the flow rate, it diverts fluids, and energy escapes. These opposing forces result in friction loss in pipes.

Friction loss in pipes can decrease the efficiency of the functions of pipes. These are a few ways by which friction loss in pipes can be reduced and the efficiency of the piping system can be boosted:

<u><em>Decrease the length of the pipes</em></u>: By decreasing pipe lengths and avoiding the use of sharp turns, fittings, and tees, whenever possible result in a more natural path for fluids to flow.
<u><em>Reduce the surface roughness of the pipes</em></u>: By reducing the interior surface roughness of pipes, a smooth and clearer path is provided for liquids to flow.
<u><em>Increase the pipe diameter: </em></u>By widening the diameters of pipes, it is ensured that fluids squeeze through pipes easily.

You can learn more about friction losses at

brainly.com/question/13348561

#SPJ4

3 0

1 year ago