Question

At what rate must heat be supplied to 15.0 kg/s of air, if the temperature is increased from 61.0°C to 650°C?

a) Calculate the rate that heat must be supplied by integrating the heat capacity formula for air.

b) Calculate the rate that heat must be supplied using the specific enthalpy table of gases located in your text by following the steps below. What is the specific enthalpy of air at 61.0°C, if the reference state is pref = 1 atm, and Tref = 25°C? Use the specific enthalpy table of gases in your text for this calculation.

c) What is the specific enthalpy of air at 650°C, if the reference state is pref = 1 atm, and Tref = 25°C. Use the specific enthalpy table of gases in your text for this calculation.

d) Calculate the rate that heat must be supplied to 15.0 kg/s or air.

Any values not given need to be looked up.

Answer 1

a)

Now for this first we will require Cp of air as a function of Temperature. So let us first obtain this

Cp/R = 3.355 +0.575*10^-3T Where T is in K

Mass flow rate of air = 15 Kg/s

molar mass of air = 29 g/mol = 29 Kg/Kmol

molar rate of air(n) = 15/29 = 0.5172 Kmol/s =517.24 mol/s

T1 = 61 ^oC = 334.15 K

T2 = 650 ^oC =923.15 K

Rate of heat to be supplied will be given by the following

Rate of Heat that need to be supplied =9413442.793 W =9413.4 KW

B) Now here we will make use of standard property table of air to obtain the enthalpy of air at T = 61 oC

Specific enthalpy =334.888 KJ/Kg = 334.88 KJ/Kf *29 Kg/Kmol = 9711.752 KJ/Kmol = 9.711 KJ/Kmol

C) now at T = 650 ^oC

Specific enthalpy =959.95 KJ/Kg = 959.95 KJ/Kg* 29 Kg/Kmol =27838.55 KJ/Kmol =27.83855 KJ/mol

D) Rate of heat that needs to be supplied = rate of heat to be supplied at 650 ^oC - Rate of heat that needs to be supplied at 61 ^oC

= 27.83855 - 9.711 =18.12755 KJ/mol = 18.12755 KJ/mol * 517.24 mol/s =9376.29 KJ/s = 9376.29 KW