Example 3.10. A system contains 0.15 m of a gas at a pressure of 3.8 bar and 150° C. It is
expanded adiabatically till the pressure falls to I bar. The gas is then heated at a constant pressure
till its enthalpy increases by 70 kJ. Determine the total work done. Take c = 1 kJ/kg K and
= 1 kJ/kg K and c, = 0.714
kJ/kg K.
Answers
Answer:
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Step-by-step explanation:
P
1
=12bar,T
1
=27+273=300K
P
2
=14.9 bar,T
2
=?
Applying pressure temperature law,
P
2
P
1
=
T
2
T
1
or T
2
=
P
1
T
1
×P
2
Substituting the values in the above equation,
T
2
=
12
300×14.9
=372.5K
Temperature in
0
C=372.5−273=99.5
∘
C
Hence the cylinder will explode at a temperature 99.5
∘
C
Answer:
Given information:
The system contains 0.15 m of a gas at a pressure of 3.8 bar and 150° C.
The gas is expanded adiabatically until the pressure falls to 1 bar.
The gas is then heated at a constant pressure till its enthalpy increases by 70 kJ.
Specific heat capacities at constant volume and constant pressure are c_v = 1 kJ/kg K and c_p = 1.714 kJ/kg K, respectively.
Step-by-step explanation:
To find: The total work done.
Solution:
- Adiabatic expansion :
Using the adiabatic expansion formula, we can find the final temperature of the gas as follows:
T_1^(gamma-1) = T_2^(gamma-1)
where gamma = c_p/c_v = 1.714/1 = 1.714
Thus, T_2 = T_1 (P_2/P_1)^((gamma-1)/gamma)
= (423.15 K)(1 bar/3.8 bar)^(0.714)
= 317.5 K
- Isobaric heating :
The gas is heated at constant pressure until its enthalpy increases by 70 kJ. We can find the mass of the gas using the ideal gas equation:
PV = mRT
m = PV/RT
= (0.15 m)(3.8 bar)(10^5 Pa/bar)/(287 J/kg K)(423.15 K)
= 1.31 kg
The heat added to the gas is given by Q = mc_pdT, where dT is the change in temperature during heating.
70 kJ = (1.31 kg)*(1.714 kJ/kg K)*dT
dT = 24.4 K
Using the formula for isobaric work, we can find the work done during heating as follows:
W = PdV = mR*dT
= (1.31 kg)(287 J/kg K)(24.4 K)
= 9.04 kJ
- Total work done :
The total work done is the sum of the work done during adiabatic expansion and isobaric heating:
W_total = W_adiabatic + W_isobaric
W_adiabatic = (P_1V_1 - P_2V_2)/(gamma - 1) = (3.8 bar)(0.15 m)/(1.714 - 1)[(317.5 K/423.15 K)^(1.714) - 1] = 2.31 kJ
W_total = 2.31 kJ + 9.04 kJ = 11.35 kJ
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