Calculate the power if the current in a circuit is 3A and the resistance of the circuit is 50hms

An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 278C, and 75

Lina20 [59]

Answer:

a) T_3 = 1539 K , P_3 = 3898 KPa

b) w_net,out = 392.4 KJ/kg

c) n_th = 0.523 - 52.3 %

d) mep = 495 KPa

Explanation:

Given:

- Sate 1: P_1 = 95 KPa, T_1 = 27 C

- q_in = 750 KJ/kg (constant volume process)

- R_air = 0.287 KJ/kgK

- r = 8 (compression ratio)

Find:

a) Pressure and Temperature (P_3 & T_3) at the end of Heat addition process.

Analysis:

Path 1 to 2 (isentropic compression):

T_1 = 300 K ----------> u_1 = 214.07 KJ/kg , v_r1 = 621.2

v_r2 = v_r1 * (V_2 / V_1) = 621.2 *(1 / 8) = 77.65

v_r2 = 77.65 -----------> u_2 = 491.2 KJ/kg , T_2 = 673.1 K

Use Ideal Gas Law for states 1 and 2:

P_2 * v_2 / T_2 = P_1 * v_1 / T_1

P_2 = P_1 * (v_1 /v_2) * ( T_2 / T_1 )

P_2 = 95 KPa * (8)*(673.1/300) = 1705 KPa

Path 2 to 3 (constant volume Heat - Addition):

q_in = (u_3 - u_2 )

u_3 = q_in + u_2

u_3 = 750 +491.2 = 1241.2 KJ/kg

u_3 = 1241.2 KJ/kg ----------> T_3 = 1539 K , v_r3 = 6.588

Use Ideal Gas Law for states 2 and 3:

P_3 * v_3 / T_3 = P_2 * v_2 / T_2

P_3 = P_2 * ( T_3 / T_2 )

P_3 = 1705 KPa*(1539/673.1) = 3898 KPa

Answer: T_3 = 1539 K and P_3 = 3898 KPa

Path 3 to 4 (isentropic expansion):

v_r4 = v_r3 * (V_1 / V_2) = 6.588 *(8) = 52.7

v_r4 = 52.70 -----------> u_4 = 571.69 KJ/kg , T_4 = 774.5 K

Path 4 to 1 (constant volume Heat - Rejection):

q_out = (u_4 - u_1 )

q_out = 571.69 - 214.07

q_out = 357.62 KJ/kg

b) The net work output w_net,out

Analysis:

w_net, out = q_in - q_out

w_net, out = 750 - 357.62

w_net, out = 392.4 KJ/kg

Answer: The net work output w_net,out = 392.4KJ/kg

c) Thermal Efficiency n_th

Analysis:

n_th = w_net, out / q_in

n_th = 392.4 / 750

n_th = 0.523 ~ 52.3%

Answer: = Thermal Efficiency n_th = 52.3%

d) mean effective pressure for the cycle.

Use ideal gas Law at state 1:

v_1 = R*T_1 / P_1 = 0.287 * 300 / 95

v_1 = v_max= 0.906 m^3 /kg

v_min = v_max / r

mep = w_net,out / (v_1-v_2) = w_net,out / v_1*(1 - 1 / r)

mep = 392.4 / 0.906 * (1 - 1/8)

mep = 495.0 KPa

Answer: mean effective pressure for the cycle mep = 495.0 KPa

4 0

3 years ago

Examine the following equation.

alex41 [277]

Answer:

E)brain decay

Explanation:

Looking at the question causes it.

6 0

3 years ago

True or False A scientific law only states that an event occurs?

Kamila [148]

True

A scientific law only states that an event occurs.

Hope this helps!

3 0

3 years ago

Which of the following is a unit of rotational speed?

gladu [14]

C. Rotations per second
Or normally we'd use Radians Per second

_Brainliest if helped!!

3 0

3 years ago

A 1.00 kg particle has the xy coordinates (-1.20 m, 0.500 m) and a 4.50 kg particle has the xy coordinates (0.600 m, -0.750 m).

just olya [345]

Answer:

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

Explanation:

The center of mass of a system of particles ( $\vec r_{cm}$ ), measured in meters, is defined by this weighted average:

$\vec r_{cm} = \frac{\Sigma_{i=1}^{n}\,m_{i}\cdot \vec r_{i}}{\Sigma_{i=1}^{n}\,m_{i}}$ (1)

Where:

$m_{i}$ - Mass of the i-th particle, measured in kilograms.

$\vec r_{i}$ - Location of the i-th particle with respect to origin, measured in meters.

If we know that $\vec r_{cm} = (-0.500\,m,-0.700\,m)$ , $m_{1} = 1\,kg$ , $\vec r_{1} = (-1.20\,m, 0.500\,m)$ , $m_{2} = 4.50\,kg$ , $\vec r_{2} = (0.600\,m, -0.750\,m)$ and $m_{3} = 4\,kg$ , then the coordinates of the third particle are:

$(-0.500\,m, -0.700\,m) = \frac{(1\,kg)\cdot (-1.20\,m,0.500\,m)+(4.50\,kg)\cdot (0.600\,m,-0.750\,m)+(4\,kg)\cdot \vec r_{3}}{1\,kg+4.50\,kg+4\,kg}$

$(-4.75\,kg\cdot m, -6.65\,kg\cdot m) = (-1.20\,kg\cdot m, 0.500\,kg\cdot m) + (2.7\,kg\cdot m, -3.375\,kg\cdot m) +(4\cdot x_{3},4\cdot y_{3})$

$(4\cdot x_{3}, 4\cdot y_{3}) = (-6.25\,kg\cdot m,-3.775\,kg\cdot m)$

$(x_{3},y_{3}) = (-1.562\,m,-0.944\,m)$

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

5 0

3 years ago