Question

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

Answer 1

The specific heat transfer is 687.5 kJ/kg.

Step-by-step 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

Answer 2

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

Step-by-step 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).