Answer:
The specific heat transfer is 687.5 kJ/kg.
Explanation:
Specific heat transfer = specific work input + specific internal energy
specific work input = 450 kJ/kg
specific internal energy = Cv(T2 - T1)
Cv is specific heat at constant volume = 20.785 kJ/kmol.K
T1 is initial temperature = 290 K
T2 is exit temperature = 450 K
specific internal energy = 20.785(450 - 290) = 20.785×160 = 3325.6 kJ/kmol = 3325.6 kJ/kmol × 1 kmol/14 kg = 237.5 kJ/kg
specific heat transfer = 450 kJ/kg + 237.5 kJ/kg = 687.5 kJ/kg
Given Information:
Gas = Nitrogen
Inlet temperature = T₁ = 290 K
Exit temperature = T₂ = 450 K
Inlet Pressure = P₁ = 100 kPa
Exit Pressure = P₂ = 2000 kPa
Work input = W = -450 kJ/kg
Required Information:
specific heat transfer = Q = ?
Answer:
specific heat transfer = - 282.64 KJ/kg
Explanation:
The specific heat transfer can be found using the equation
Q - W = ΔU
Q = ΔU + W eq. 1
Where W is the work input and ΔU is the change in internal energy.
The change in specific internal energy is given by
ΔU = h₂ - h₁ = Cp(T₂ - T₁)
Where Cp is the specific heat capacity.
The specific heat capacity is the amount of heat required to increase the temperature of one gram of a substance by one degree K.
The Cp of Nitrogen gas is equal to 1.046 KJ/kg.K
Therefore, equation 1 becomes
Q = ΔU + W
Q = Cp(T₂ - T₁) + W
Q = 1.046(450 - 290) + (-450)
Q = 1.046(160) - 450
Q = 167.36 - 450
Q = - 282.64 KJ/kg
The negative sign indicates that the heat energy was transferred from the system (leaving the system) rather than transferred to the system (entering the system).
