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3 years ago

7

Basic C++ For Loop I'm trying to learn. Replit tells me that the for in the forloop is an error, but I don't know what's wrong.

Can anyone help?

1 answer:

STatiana [176]3 years ago

8 0

Answer:

curly brackets are missing

Explanation:

The body of the main() function need to be enclosed in curly brackets. Try this:

int P = 3000;

int main( ) {

for (int t = 0; t < 10; t++) {

cout << P;

}

}

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1. Examine the following circuit. Find RT, I3, R1, R2, R3, V1, V2 and V3. Show all of your work clearly below.

Mkey [24]

Explanation:

Ohm's law is used here. V = IR, and variations. The voltage across all elements is the same in this parallel circuit. (V1 =V2 =V3)

The total supply current is the sum of the currents in each of the branches. (It = I1 +I2 +I3)

Rt = (8 V)/(8 A) = 1 Ω . . . . supply voltage divided by supply current

I3 = 8A -3A -4A = 1 A . . . . supply current not flowing through other branches

R1 = (8 V)/(3 A) = 8/3 Ω

R2 = (8 V)/(4 A) = 2 Ω

R3 = (8 V)/(I3) = (8 V)/(1 A) = 8 Ω

V1 = V2 = V3 = 8 V

6 0

2 years ago

A piston-cylinder device contains 0.58 kg of steam at 300°C and 0.5 MPa. Steam is cooled at constant pressure until one-half of

Mumz [18]

Answer:

a) Tբ = 151.8°C

b) ΔV = - 0.194 m³

c) The T-V diagram is sketched in the image attached.

Explanation:

Using steam tables,

At the given pressure of 0.5 MPa, the saturation temperature is the final temperature.

Right from the steam tables (A-5) with a little interpolation, Tբ = 151.793°C

b) The volume change

Using data from A-5 and A-6 of the steam tables,

The volume change will be calculated from the mass (0.58 kg), the initial specific volume (αᵢ) and the final specific volume

(αբ) (which is calculated from the final quality and the consituents of the specific volumes).

ΔV = m(αբ - αᵢ)

αբ = αₗ + q(αₗᵥ) = αₗ + q (αᵥ - αₗ)

q = 0.5, αₗ = 0.00109 m³/kg, αᵥ = 0.3748 m³/kg

αբ = 0.00109 + 0.5(0.3748 - 0.00109)

αբ = 0.187945 m³/kg

αᵢ = 0.5226 m³/kg

ΔV = 0.58 (0.187945 - 0.5226) = - 0.194 m³

c) The T-V diagram is sketched in the image attached

3 0

3 years ago

Air is compressed in a reversible, isothermal, steady- flow process from 15 psia, 100°F to 100 psia. Calculate the work of compr

mixas84 [53]

Answer:

|W|=169.28 KJ/kg

ΔS = -0.544 KJ/Kg.K

Explanation:

Given that

T= 100°F

We know that

1 °F = 255.92 K

100°F = 310 .92 K

$P _1= 15 psia$

$P _1= 100 psia$

We know that work for isothermal process

$W=mRT\ln \dfrac{P_1}{P_2}$

Lets take mass is 1 kg.

So work per unit mass

$W=RT\ln \dfrac{P_1}{P_2}$

We know that for air R=0.287KJ/kg.K

$W=RT\ln \dfrac{P_1}{P_2}$

$W=0.287\times 310.92\ln \dfrac{15}{100}$

W= - 169.28 KJ/kg

Negative sign indicates compression

|W|=169.28 KJ/kg

We know that change in entropy at constant volume

$\Delta S=-R\ln \dfrac{P_2}{P_1}$

$\Delta S=-0.287\ln \dfrac{100}{15}$

ΔS = -0.544 KJ/Kg.K

3 0

3 years ago

A wind turbine designer is considering installing a horizontal axis wind turbine at a location in Michigan. To increase the powe

bogdanovich [222]

Answer:

False ( B )

Explanation:

considering that the wind turbine is a horizontal axis turbine

Power generated/extracted by the turbine can be calculated as

P = n * 1/2 *<em> p</em> *Av^3

where: n = turbine efficiency

<em>p = air density </em>

<em> </em>A = πd^2 / 4

v = speed

From the above equation it can seen that increasing the Blade radius by 10% will increase the Blade Area which will in turn increase the value of the power extracted by the wind turbine

3 0

2 years ago

Problem 10.012 SI A vapor-compression refrigeration cycle operates at steady state with Refrigerant 134a as the working fluid. S

pychu [463]

Answer:

a) 4.1 kw

b) 4.68 tons

c) 4.02

Explanation:

Saturated vapor enters compressor at ( p1 ) = 2.6 bar

Saturated liquid exits the condenser at ( p2 ) = 12 bar

Isentropic compressor efficiency = 80%

Mass flow rate = 7 kg/min

A) Determine compressor power in KW

compressor power = m ( h2 - h1 )

= 7 / 60 ( 283.71 - 248.545 )

= 4.1 kw

B) Determine refrigeration capacity in tons = m ( h1 - h4 )

= 7/60 ( 248.545 - 107.34 )

= 16.47 kw = 4.68 tons

C) coefficient of performance ( COP )

= Refrigeration capacity / compressor power

= 16.47 / 4.1 = 4.02

Attached below is the beginning part of the solution

7 0

2 years ago