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

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:

Gas is ideal
System is closed system.
K.E and P.E is neglected
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

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

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