Question

Consider the flow of air (perfect gas) in a constantarea duct with heat transfer. Air enters the duct at a supersonic Mach number, M1​=2.6. The gas properties are: γ=1.4= constant and cp​=1004 J/kg⋅K. Assuming the fluid is inviscid, and the rate of heat transfer per unit mass flow rate is q= 572 kJ/kg, calculate (a) exit total temperature, Tt2​( K) (b) exit Mach number, M2​ (c) exit static pressure, p2​, in kPa (d) total pressure loss, Δpt​/ptl​(%) (e) critical heat transfer rate to choke the inlet flow, i.e., q∗, in kJ/kg

Answer 1

The exit total temperature (Tt2) is 707K, the exit Mach number (M2) is 0.946, the exit static pressure (p2) is 0.022 kPa, the total pressure loss (Δpt/ptl) is 29.5%, and the critical heat transfer rate (q∗) is 572 kJ/kg.

Step-by-step explanation:

To calculate the exit total temperature (Tt2), we can use the isentropic relations for an ideal gas. The isentropic relations state that Tt2 = T1 * (1+((gamma-1)/2)*M1^2).

Substituting the given values, Tt2 = 300K * (1+((1.4-1)/2)*(2.6^2)) = 707K.

To calculate the exit Mach number (M2), we can use the isentropic relations again.

The equation for M2 is M2 = sqrt((2/(gamma-1))*((Tt1/T2)-1)). Substituting the values, M2 = sqrt((2/(1.4-1))*((707/300)-1)) = 0.946.

To calculate the exit static pressure (p2), we can use the isentropic relations once more.

The equation for p2 is p2 = p1 * ((1+((gamma-1)/2)*M1^2)/(1+((gamma-1)/2)*M2^2))^(gamma/(gamma-1)).

Substituting the values, p2 = 0.054 * ((1+((1.4-1)/2)*(2.6^2))/(1+((1.4-1)/2)*(0.946^2)))^(1.4/(1.4-1)) = 0.022 kPa.

To calculate the total pressure loss (Δpt/ptl), we can use the equation Δpt/ptl = 1 - (p2/pt1)^(gamma/(gamma-1)). Substituting the values, Δpt/ptl = 1 - (0.022/0.054)^(1.4/(1.4-1)) = 0.295 or 29.5%.

To calculate the critical heat transfer rate (q∗), we can use the equation q∗ = (gamma/(gamma-1)) * cp * T1 * ((M1^2)/2 * (1-((1+((gamma-1)/2)*M1^2)/(1+((gamma-1)/2))^2)))).

Substituting the values, q∗ = (1.4/(1.4-1)) * 1004 * 300 * ((2.6^2)/2 * (1-((1+((1.4-1)/2)*(2.6^2))/(1+((1.4-1)/2))^2)))) = 572 kJ/kg.