Question

(5-95) Argon steadily flows into a constant pressure heater at 300 K and 100 kPa with a mass flow rate of 6.24 kg/s Heat transfer in the rate of 150 kW is supplied to the argon as it flows through the heater. (a) Determine the argon temperature at the heater exit, in ℃. (b) Determine the argon volume flow rate at the heater exit, in m3/s.

Answer 1

The argon temperature at the heater exit is determined to be 348.08 ℃, while the argon volume flow rate at the heater exit is calculated to be 3769.23 m³/s.

Step-by-step explanation:

To determine the argon temperature and volume flow rate at the heater exit, we can use the principles of thermodynamics. The problem states that heat transfer in the rate of 150 kW is supplied to the argon. Since there is no information given about the heater efficiency, we can assume it to be 100%, meaning all the heat is transferred to the argon. Therefore, the heat transfer can be equated to the change in internal energy of the argon, which is given by:

Q = m * Cp * ΔT

Where Q is the heat transfer, m is the mass flow rate, Cp is the specific heat capacity of argon, and ΔT is the change in temperature.

Given that Q = 150 kW, m = 6.24 kg/s, and Cp = 0.52 kJ/kg·℃ (specific heat capacity of argon), we can rearrange the equation to solve for ΔT:

ΔT = Q / (m * Cp)

Substituting the values, we have ΔT = (150 * 10³ J/s) / (6.24 kg/s * 0.52 kJ/kg·℃), which equals 48.08 ℃.

Therefore, the argon temperature at the heater exit is 300 ℃ + 48.08 ℃, which equals 348.08 ℃.

To calculate the argon volume flow rate at the heater exit, we can use the ideal gas law:

PV = nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Rearranging the equation to solve for V, we have:

V = (nRT) / P

Given that the initial conditions are 300 K and 100 kPa, and a mass flow rate of 6.24 kg/s, we can calculate the number of moles of argon:

n = m / M

Where M is the molar mass of argon. The molar mass of argon is 39.948 g/mol, so n = 6.24 / (39.948 * 10⁻³) = 156.172 mol.

Substituting the values into the ideal gas law equation, we have V = (156.172 mol * 8.314 J/mol·K * 300 K) / 100 kPa. Converting kPa to Pa, we have V = (156.172 mol * 8.314 J/mol·K * 300 K) / 100,000 Pa, which equals 3769.23 m³.

Therefore, the argon volume flow rate at the heater exit is 3769.23 m³/s.

Answer 2

a) 346.22 K

b) 4.49 m³/s

Step-by-step explanation:

P₁ = P₂ = Pressure in heater = 100 kPa

T₁ = Initial temperature = 300 K

m = Mass flow rate = 6.24 kg/s

Q = Heat transfer rate = 150 kW

Cp = Specific heat at constant pressure of argon = 0.520 kJ/kg K

R = Gas constant of argon = 0.2082 kJ/kg K

a)
`Q = mC_p\Delta T\\\Rightarrow 150=6.24* 0.52* (T_2-300)\\\Rightarrow (150)/(6.24* 0.52)= T_2-300\\\Rightarrow T_2=346.22 K`

∴ The argon temperature at the heater exit is 346.22 Kelvin

b)
`P_2V_2=mRT_2\\\Rightarrow V_2=(mRT_2)/(P_2)\\\Rightarrow V_2=(6.24* 0.2082* 346.22)/(100)\\\Rightarrow V_2=4.49\ m^3/s`

∴ Argon volume flow rate at the heater exit 4.49 m³/s