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g100num [7]
3 years ago
8

A plane where the top view is projected approximately as how it appears to the observer​

Engineering
1 answer:
Lynna [10]3 years ago
5 0

Answer:

B) I took the quiz and got it 100% right on the question!!!

Explanation:

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Carbon dioxide (CO2) is compressed in a piston-cylinder assembly from p1 = 0.7 bar, T1 = 320 K to p2 = 11 bar. The initial volum
tekilochka [14]

Answer:

W_{12}=-53.9056KJ

Part A:

Q=-7.03734 KJ/Kg (-ve sign shows heat is getting out)

Part B:

Q=1.5265KJ/Kg (Heat getting in)

The value of Q at constant specific heat is approximately 361% in difference with variable specific heat and at constant specific heat Q has opposite direction (going in) than Q which is calculated in Part B from table A-23. So taking constant specific heat is not a good idea and is questionable.

Explanation:

Assumptions:

  1. Gas is ideal
  2. System is closed system.
  3. K.E and P.E is neglected
  4. Process is polytropic

Since Process is polytropic so  W_{12} =\frac{P_{2}V_{2}-P_{1}V_{1}}{1-n}

Where n=1.25

Since Process is polytropic :

\frac{V_{2}}{V_{1}}=(\frac{P_{1}}{P_{2}})^{\frac{1}{1.25}} \\V_{2}= (\frac{P_{1}}{P_{2}})^{\frac{1}{1.25}} *V_{1}

V_{2}= (\frac{0.7}{11})^{\frac{1}{1.25}} *0.262\\V_{2}=0.028924 m^3

Now,W_{12} =\frac{P_{2}V_{2}-P_{1}V_{1}}{1-n}

W_{12} =\frac{11*0.028924-0.7*0.262}{1-1.25}(\frac{10^{5}N/m^2}{1 bar})(\frac{1  KJ}{10^{3}Nm})

W_{12}=-53.9056KJ

We will now calculate mass (m) and Temperature T_2.

m=\frac{P_{1}V_{1}}{RT_{1}}\\ m=\frac{0.7*0.262}{\frac{8.314KJ}{44.01Kg.K}*320}(\frac{10^{5}N/m^2}{1 bar})(\frac{1  KJ}{10^{3}Nm})\\m=0.30338Kg

T_{2} =\frac{P_{2}V_{2}}{Rm}\\ m=\frac{11*0.028924}{\frac{8.314KJ}{44.01Kg.K}*0.30338}(\frac{10^{5}N/m^2}{1 bar})(\frac{1  KJ}{10^{3}Nm})\\T_{2} =555.14K

Part A:

According to energy balance::

Q=mc_{v}(T_{2}-T_{1})+W_{12}

From A-20, C_v for Carbon dioxide at 300 K is 0.657 KJ/Kg.k

Q=0.30338*0.657(555.14-320)+(-53.9056)

Q=-7.03734 KJ/Kg (-ve sign shows heat is getting out)

Part B:

From Table A-23:

u_{1} at 320K = 7526 KJ/Kg

u_{2} at 555.14K = 15567.292 (By interpolation)

Q=m(\frac{u(T_{2})-u(T_{1})}{M} )+W_{12}

Q=0.30338(\frac{15567.292-7526}{44.01} )+(-53.9056)

Q=1.5265KJ/Kg (Heat getting in)

The value of Q at constant specific heat is approximately 361% in difference with variable specific heat and at constant specific heat Q has opposite direction (going in) than Q which is calculated in Part B from table A-23. So taking constant specific heat is not a good idea and is questionable.

7 0
4 years ago
A storm sewer is carrying snow melt containing 1.2 g/L of sodium chloride into a small stream. The stream has a naturally occurr
galina1969 [7]

Answer:

Given Data:

concentration of sewer Csewer = 1.2 g/L

converting into mg/L = Csewer = 1.2 g/L x 1000 mg/g = 1200 mg/L

flow rate of sewer Qsewer = 2000 L/min

concentration of sewer Cstream = 20 mg/L

flow rate of sewer Qstream = 2m3/s

converting Q into L/min = 2m3/s x 1000 x 60 = 120000 L/min

mass diagram is

6 0
3 years ago
Roku internet service providet​
podryga [215]

Answer:

What?? I do not understand?

6 0
3 years ago
An AC power generator produces 50 A (rms) at 3600 V. The voltage is stepped up to 100 000 V by an ideal transformer and the ener
RSB [31]

Given:

I_{rms} = 50 A

voltage, V = 3600V

step-up voltage, V' = 100000 V

Resistance of line, R = 100\ohm

Solution:

To calculate % heat loss in long distance power line:

Power produced by AC generator, P = 50\times 3600 W

P = 180000 W = 180 kW

At step-up voltage, V = 100000V or 100 kV

current, I = \frac{P}{V'}

I = \frac{1800000}{100000}

I = 1.8 A

Power line voltage drop is given by:

V_{drop} = I\times R

V_{drop} = 1.8\times 100

V_{drop} = 180 V

Power dissipated in long transmission line P_{dissipated} = V_{drop}\times I

Power dissipated in long transmission line P_{dissipated} = 180\times 1.8 = 324 W

% Heat loss in power line, P_{loss} = \frac{P_{dissipated}}{P}\times 100

% Heat loss in power line, P_{loss} = \frac{324}{180000}\times 100

P_{loss} = 0.18%

 

5 0
3 years ago
How to calculate effective resistance​
denis23 [38]

Answer:

For a circuit with resistances R1 and R2 in series or in parallel as in Figure 2, the effective resistance can be calculated by using the following rules. Rab = R1 + R2.

Explanation:

4 0
3 years ago
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