[ytszazu]

Can anyone help me to solve this thermo question given by my lecturer? Please? This is a summary of what i tried to understand:

Okay, the question:

1st airflow: 3500 Nm3/hr.
T = 800 C
Cp = 1.3K

2nd airflow: 1805 Nm3/hr.
T = 80 C
Cp = 1.05K

(Data above are given)

Mixed gas airflow: 3500+1805 = 5305 Nm3/hr
T = C
Cp = C

My question is:

Wjat is the meaning of Cp, is it constant pressure?
If it is the constant pressure, why the values are so akward, doesn't follow the (T+273/T+273) rule?
How to calculate the mixed T and Cp?
And finally how to get the (standard temperature pressure) STP of the flow rate? (Am3/hr)  

Thanks to anyone that can help.

Yong Tze Shoong,
University Technology of Malaysia.

[Donaldson Tan]

The units you give for Cp is dodgy. Cp is the molar heat capacity at constant pressure, so its unit should be energy per unit temperature per mole, ie. J/K.mol

P: final pressure
V: final volumetric flow-rate
N: final molar flow-rate
T: final temperature

assume air consists of 79% Nitrogen and 21% Oxygen
assume airflow behaves ideally

From data,
P1.V1 = N1.R.T1 where:
P1.V1 = 3500 N.m³/h
T1 = 800C = 1073K
=> N1 = 0.39234 mol/h

From data,
P2.V2 = N2.R.T2 where
P2.V2 = 1805 N.m³/h
T2 = 80C = 353K
=> N2 = 0.61502 mol/h

assume the mixing of airflows is adiabatic, then
P.V = P1.V1 + P2.V2 = 5305 N.m³/h

assume the mixing process is at steady state,
N = N1 + N2 = 1.0074 mol/h

using ideal gas equation,
T = P.V/N.R = 5305/(1.0074)(8.314) = 633.39K

assume mixing process is adiabatic, then
H = H1 + H2 where
H: final enthalpy
H1: enthalpy of first airflow
H2: enthalpy of 2nd airflow

H = N.(h* + Cp.T)
H1 = N1.(h* + Cp1.T1)
H2 = N2.(h* + Cp2.T2)
where h* is reference enthalpy state (assume to be zero)
Cp: final molar heat capacity
Cp1: molar heat capacity of first airflow
Cp2: molar heat capacity of 2nd airflow

H = H1 + H2
N.(h* + Cp.T) = N1.(h* + Cp1.T1) + N2.(h* + Cp2.T2)
N.Cp.T = N1.Cp1.T1 + N2.Cp2.T2

Cp = (N1.Cp1.T1 + N2.Cp2.T2)/N.T