Energy balance.pptx
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1) A turbine operates with steam entering at 44 atm and 450°C and leaving at 360 m/s. It delivers 70 kW of shaft work and loses 10^4 kcal/h of heat. The specific enthalpy change is calculated as -6.50630*10^5 J/kg. 2) A textile dryer consuming 4 m3/h of natural gas with an efficiency of 47.46% is calculated based on drying 60 kg/h of cloth from 55% to 10% moisture content. 3) The standard heat of reaction is calculated to be -904 kJ/mole for the reaction 4NH3 + 5O2 → 4NO + 6H
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W = m[(h2 - h1) - (KE2 - KE1) - (PE2 - PE1)]
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9
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Energy balance.pptx
- 1. Energy Balance Krubha D N Process Engineer Chemsys Innovative Solutions Pvt Ltd
- 2. PA d Work = Force * displacement Wf= PA * d = P * V H = U+PV Closed system ΔIE + ΔPE+ΔKE = Q-W Open system ΔIE + ΔPE+ ΔKE = Q –(Ws+Wf)
- 3. Q < 0
- 5. 500 kg/h of steam drives a turbine. The steam enters the turbine at 44 atm, 450° C, with linear velocity 60 m/s and leaves at a point 5 m below the turbine inlet at velocity of 360 m/s.The turbine delivers shaft work of 70kW and heat loss from turbine is estimated to be 10^4 kcal/h.Calculate the specific enthalpy change associated with the process. Consider this as an open system ΔKE = ½ m*(v2^2-v1^2)…………..J/s Δ PE = m*g*(h2-h1)……………..J/s Ws………………………………..J/s Q…………………………………J/s Δ H – PV + ΔPE+ΔKE= -Q-Ws-PV ΔH = ?.................................................J/s ΔH/kg=?.............................................J/kg
- 6. Answer: ΔKE= 8750 J/s ΔPE= 6.805 J/s Ws = 70000 J/s Q = 11622.22 J/s ΔH = - 90365.415 J/s ΔH/kg = - 6.50630*10^5 J/kg
- 7. A textile dryer is found to consume 4 m3/h of natural gas. Net calorific value of 800 kJ/mole. The throughput of dryer is 60 kg of wet cloth/hr. Drying it from 55% moisture to 10% moisture. Estimate the overall thermal efficiency of dryer considering the latent of evaporation only. Latent heat of H2O 970 BTU/lb Water (z) Dry cloth(y) , 10% moisture 60kg/h 55% moisture Natural gas- energy supplied-4 m3/h Dry basis 60*(1-0.55) = y*(1-0.1) y=?.........................................kg/h z=?.........................................kg/h Latent heat = m*λ = ?.....................kJ/h
- 8. Volumetric flow rate of natural gas = 4 m3/h Convert to moles/h Use the relation 1 mole/h occupies 22.414*10^-3 m3/h Energy supplied = ?..................................kJ/h Thermal efficiency = Theoretical/Actual or Latent heat/Energy supplied = ?..............................%
- 9. Answers: Y= 30kg/h Z= 30 kg/h mλ= 67771.522 kJ/h Energy supplied = 142767.2 kJ/h Thermal efficiency = 47.46%
- 10. Calculate standard heat of reaction given enthalpy of formation: Component ΔHf° (kJ/mole) NH3(g) -46.2 NO(g) 90.2 H2O(g) -241.6 4NH3+5O2 4NO+6H2O Enthalpy of rxn: ΔHr°=Σ ΔHf°(pdts)-Σ ΔHf°(reactants) =((4*90.2)+(6*-241.6))-((0*5)+(4*-46.2))………………………for elements heat of formation is 0 =?kJ/mole Answer: ΔHr°= -904 kJ/mole
- 11. A gas phase reaction CO2+4H2 CH4+2H2O proceeds to 100% conversion. Estimate the heat that must be removed when the reactants and products are at 500°C Component ΔHf°(kcal/kmole) CO2 -94052 CH4 -17889 H2O -57798 Cp for CO2=6.339+10.14*10^-3*T-3.415*10^-6*T^2 Cp for H2= 6.424+1.039*10^-3*T-0.0078*10^-6*T^2 Cp for H2O= 6.97+3.464*10^-3*T-0.483*10^-6*T^2 Cp for CH4= 3.204+18.41*10^-3*T-4.48*10^-6*T^2 ΔHr(500°C)= ΔHr°+ΣΔHs,p - ΣΔHs,r
- 12. ΔHr°=Σ ΔHf°(pdts)-Σ ΔHf°(reactants) =((1*-17889)+(2*-57798))-((4*0)+(1*-94052))………………………for elements heat of formation is 0 =?kcal/mole ΔHr(500°C)= ΔHr°+ΣΔHs,p - ΣΔHs,r =?kcal/kmole Answers: ΔHr° = -39433 kcal/mole ΔHs,p=13815.09 kcal/mole ΔHs,r= 18352.71 kcal/mole ΔHr(500°C)= - 43970 kcal/mole