Can anyone assist me to settle this thermo question provided by my lecturer? Please? This is a an introduction of what i tried to understand:Okay, the question:1st airflow: 3500 Nm3/hr.T = 800 CCp = 1.3K2nd airflow: 1805 Nm3/hr.T = 80 CCp = 1.05K(Data over are given)Mixed gas airflow: 3500+1805 = 5305 Nm3/hrT = CCp = CMy concern is:Wjat is the an interpretation of Cp, is it continuous pressure?If that is the consistent pressure, why the values room so akward, doesn't follow the (T+273/T+273) rule?How to calculate the mixed T and also Cp?And ultimately how to get the (standard temperature pressure) STP of the flow rate? (Am3/hr) many thanks to everyone that can help.Yong Tze Shoong,University technology of Malaysia.

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Donaldson TanEditor, brand-new Asia RepublicRetired StaffSr. Member
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The units you give for Cp is dodgy. Cp is the molar heat capacity at continuous pressure, therefore its unit need to be energy per unit temperature per mole, ie. J/K.molP: last pressureV: last volumetric flow-rateN: final molar flow-rateT: last temperatureassume air consists of 79% Nitrogen and also 21% Oxygenassume air flow behaves ideallyFrom data,P1.V1 = N1.R.T1 where:P1.V1 = 3500 N.m3/hT1 = 800C = 1073K=> N1 = 0.39234 mol/h From data,P2.V2 = N2.R.T2 whereP2.V2 = 1805 N.m3/hT2 = 80C = 353K=> N2 = 0.61502 mol/h assume the mix of airflows is adiabatic, thenP.V = P1.V1 + P2.V2 = 5305 N.m3/h assume the mixing process is at steady state,N = N1 + N2 = 1.0074 mol/husing appropriate gas equation,T = P.V/N.R = 5305/(1.0074)(8.314) = 633.39Kassume mixing procedure is adiabatic, thenH = H1 + H2 whereH: final enthalpyH1: enthalpy of very first airflowH2: enthalpy of second airflowH = N.(h* + Cp.T)H1 = N1.(h* + Cp1.T1)H2 = N2.(h* + Cp2.T2)where h* is referral enthalpy state (assume to it is in zero)Cp: final molar warm capacityCp1: molar warmth capacity of first airflowCp2: molar warmth capacity of 2nd airflowH = H1 + H2N.(h* + Cp.T) = N1.(h* + Cp1.T1) + N2.(h* + Cp2.T2)N.Cp.T = N1.Cp1.T1 + N2.Cp2.T2Cp = (N1.Cp1.T1 + N2.Cp2.T2)/N.T
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