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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:

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

What is the measurment unit of permeability?

nikitadnepr [17]

Answer:

Henries :)

Explanation:

I looked it up

7 0

3 years ago

Read 2 more answers

I) A sag vertical curve is to be designed to join a 4% grade to a 2% grade. If the design

Burka [1]

Answer:

=4/5 because I'm not going to go back in a year meaning that they are you are

4 0

2 years ago

Showing all of your work and algebra,generate an approximate expression for T as a function ofthe other variables. (b) Explain w

shusha [124]

Answer:

Following the ways of dealing with incomplete questions, i was able to get the complete question, please look at the attachment for ans.

5 0

4 years ago

A 10 kg mass is lifted 5 m with an upward acceleration of 2 m/s^2 (Note: a process diagram is not required for this problem) a)

ohaa [14]

Answer:

a) F=20 [kgm/s^2]=20 [N]

b) W=100[kgm^2/s^2]=100[J]

c) P=44,84[kgm^2/s^3]=44,84[W]

d) W=2,778*10^-5 [kilowatt-hours]

Explanation:

a) Newton's Second Law states that <em>the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force and inversely proportional to the mass of the object.</em>

F=ma

where:

F, the net force [N]
m, mass of the object [kg]
a, acceleration [m/s^2]

If a 10 kg mass is lifted with an upward acceleration of 2 m/s^2, the upward force necessary is:

F=10 [kg]*2[m/s^2]=20 [kgm/s^2]=20 [N]

b) The amount of energy required to lift the box equals the magnitud of work done by the lifting force:

W=FdcosФ

where:

W, work executed [J]
F, net force [N]
d, displacement produced by the force [m]
Ф, angle between the net force and displacemt produced

Thus, the energy required to lift 5 m the mass is:

W=20[N]*5[m]cos0°=100[N.m]=100[kgm^2/s^2]=100[J]

c) To find the average power we use the formula:

P=W/t

where,

P, average power [W]
W, work executed [J]
t, elapsed time [s]

Thus, if the process takes 2,23 seconds the average power is:

P=100[J]/2,23[s]=44,84[J/s]=44,84[kgm^2/s^3]=44,84[W]

d) As 1 kilowatt-hours=3,6*10^6 J, then:

100 [J]*1 [kilowatt-hour]/ 3,6*10^6 [J]=2,778*10^-5 [kilowatt-hours]

6 0

3 years ago