Energy Balance for Non-Reactive System

WEEK 2-6

Elements of energy balance calculationsProcedure for energy balance calculationsChanges in pressure at constant temperatureChanges in temperatureSensible heat and heat capacityBalances on phase operationEstimation & correlation of latent heatPsychometric chart

Elements of energy balance calculations (page 359)

Energy balance – closed system∆U + ∆Ek + ∆Ep = Q – W

Assumptions;1. ∆U ≈ 0, no temp changes, no phase change, no chem

reaction, closed system, ∆P less than a few atm.2. ∆Ek = 0, no acceleration, no rising, no falling.

3. Q = 0, temp system = temp surrounding, perfect insulation, adiabatic.

4. W = 0, no moving part, no electrical current, no radiation at the system boundary

Energy balance – open system

SPK WQEEH

kmolkJmolJmolenergyenthalpyspecificH

hkJsJtimeenergychangeenthalpyH

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Heat Work

8.1a Reference states (for detail pls ref 7.5a)Choose a temperature, pressure and state of

component as ref state.Pls read page 363….point 3.

Example Carbon monoxide from ref state to two other states. Data

in Table below is at 1 atm;Ref state : CO(g), 0oC, 1atm

Note: • you can choose other data as ref point.• At ref point the H and U are defined as zero.

T(deg C) H (J/mol)0 0

100 2919500 15060

8.1b Hypothetical process path(page 360)

• Change in P, Constant T (Ref Sec 8.2)• Change in T, Constant P (Ref Sec 8.3)• Estimation of Cp..refer 8.3c• Phase change (Ref Sec 8.4)

Test yourself #2 page 361…

Example: to construct hypothetical process path

Phenol (solid, 25 deg C, 1 atm to gas, 300 deg C, 3 atm)What to do?Refer to the physical property of phenol, its

boiling point, melting point. Refer appendix of the text book.

Ref text book page 360

8.1c Procedure for energy balance calculations page 360

• 1. Mass balance.• 2. Write the energy balance• 3. Choose the suitable reference• 4. Construct table • 5. Calculate the required specific U or H• 6. Calculate the total U or H • 7. Calculate the W, Ep, Ek… • 8. Solve the energy balance…Q…

Example 8.1-1

• Material balance equation; Q = ∆H• Choose reference states for the components;

based on the available data of the question or from the list of Table B.8

• Construct an inlet-outlet table• Calculate all unknown specific enthalpies

using the chosen reference states

• H1 = Href – Hstat = Ac(l, 20oC, 5 atm) – Ac(v, 65oC, 1 atm)Draw the hypothetical paths;(l,20oC,5atm)(l,20oC,1atm)(l,56oC,1atm)(v,56oC,1atm)(v,65oC,1atm)

• Sec 8.2 – change in pressure∆H = V∆P, V from table B.1 use the sp gr and mw • Sec 8.3 – change in temp∆H = Cp(T)dT (from T2-T1)• Sec 8.4 – phase change (ref Table B.1)

Psychrometric Charts

• Figure 8.4.1 (SI Unit)• Figure 8.4.2 (American Standard Engineering Unit)• Abbreviation:• DA – dry air• T - dry-bulb temperature (x-axis)• ha – absolute humidity @ moisture content (kg

H2O/kg DA) – (y-axis)

• hr – relative humidity (curve on the phychrometric chart)

• Tdp – dew point temperature at which air becomes saturated if it is cooled at constant pressure.

• vH – humid volume (m3/kg DA) – the humid volume occupied by 1 kg of DA

• Twb – wet bulb temperature …..• H - Specific enthalpy of saturation air (kJ/kg DA) -

(diagonal scale above the saturation curve)• Enthalpy deviation (kJ/kg DA) – almost vertical

and convex to the left with values of -1.2, -1.0….• Adiabatic cooling..liquid water move along the

wet-bulb temp line to the 100% hr.

Example 1:Air at 29oC, 20% hr, cooled at constant pressure

move horizontally to the saturation curve, 100% hr the Tdp is 4oC.

Example 2:Calculated volume occupied by 150 kg of humid

air at T = 30oC, hr = 30%. Refer Figure 8.4.1.Determine the ha and vH then calculate the

volume for 150 kg humid air

Example 3:Determine specific enthalpy of saturated air at

25oC, 1 atm with ha = 0.0202 kg H2O/kg DA……

• The specific enthalpy can be calculated using Table B-2 and B-5. (another method)

Example 4:• Air at 35oC, hr =10%, specific enthalpy 45kJ/kg

DA and the enthalpy deviation is -0.52 kJ/kg DA. Therefore the specific enthalpy at the given condition is (45-0.52) = 44.5 kJ/kg DA

Try Example 8.4-5 and 8.4-6 …