A compressor brings nitrogen from 100 kPa, 290 K to 2000 kPa. The process has a specific work input of 450 kJ / kg and the exit temperature is 450 K. Find the specific heat transfer using constant specific heats

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).

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