Question

Oxygen is contained in a piston-cylinder system initially at a temperature of 40 ^oC and a pressure of 150 kPa. The piston is 1 m in diameter and initially 20 cm above the base of the cylinder. Heat is added at constant pressure until the piston has moved 30 cm, so it's now at 50 cm.

I found the mass of the system to be 0.2898 kg, and T2 = 783K.

a) Using the table of properties of oxygen as an ideal gas, calculate the heat added to the system per unit mass, in kJ/kg

b) Repeat this calculation assuming that the specific heat as a function of temperature is linear over this range of temperatures (ie, assuming the change in temperature is "small")

c) Repeat this calculation without this assumption (ie, direct integration of empirical relation)

Answer 1

Assume a piston-cylinder arrangement, with a piston diameter   1m

area of the piston ( A)0.785 * 1² =.785 m²

at an initial condition temperature T1 273 + 40 313 K

Pressure P₁= 150 kPa , Volume V₁= A * L₁ = .785 * 0.2= 0.157 m³ ( L₁ = .2 m GIVEN)

Here in this system heat is added at constant pressure so initial pressure P₁ = final pressure P₂ = 150 kPa

as OXYGEN is assume be an ideal gas so , C_p = 1.005 KJ/Kg and C_v = .718 KJ/Kg ( Oxygen is a diatomic gas so we assume this value )

mass of the system (m)= 0.2898 kg

final temp. T₂ = 783 K

final volume V₂ = A* L₂ = .785 * 0.5 = 0.3925 m³

a. Heat added to the system per unit mass ( Q) = C_p*T= 1.005 * (783 - 313) = 472.35 KJ/Kg ( answer for a)

b. as the change of the temperature is small so we assume that T = 0

As per 1st law of thermodynamics we know dQ = dU +dW

as T is equal to zero so dU = 0

so , dQ = dW = pdv = p (V₂  - V₁ ) = 150 *( 0.3925 -0.157) = 35.325 kJ / Kg

heat added for the mass m is = 35.325 * 0.2898 = 10.237 KJ ( answer for b)

c. assuming that the specific heat as a function of temperature is linear over this range of temperatures so C_p= a + bT ( a& b are constant )

for initial condition for the table of O₂ C_p 0.92 KJ/Kg

so , 0.92 = a + 313b ( equation 1)

for final condition for the O₂ table C_p 1.046 KJ/Kg

so , 1.046= a + 783b ( equation 2)

now from this 2 equation we get the value of a = 0.833  and b= 2.766* 10^-4

so the equation become C_{p =} 0.833 + ( 2.766* 10^-4)T

SO heat added to this system by this process is per unit mass (dQ) = C_pT

total heat added ( Q ) = C_p T =  ( 0.833 + (2.766*10^-4)T) T

=(0.833( T ₂  - T₁ ) + ( ( 2.766*10^-4)/ 2 ) * ( T₂² - T₁²))

put T_{1 =} 313 and T₂ = 783 then we get

Q = 0.833 ( 783 - 313 ) + ( 1.383 * 10^-4) ( 783² - 313² )

= 391.51 + 71.24 = 462.75 KJ/Kg

Total heat added for the system mass (m) = 462.75 * 0.2898 = 134.1 KJ ( answer for C)