Worksheet #14 Part C Solving Transient Balances Lecture 16 Goals Ch.11.3, p. 554-560 Derive the transient energy balance equation. Discuss simplifying assumptions used to solve the energy balance transient equation. Solve simple examples using the energy balance transient equation. In 200 we immediately dropped the “accumulation” term and went to work on steady-state balances. Now it’s time to put it back in. INPUT + GENERATION = OUTPUT + CONSUMPTION + ACCUMULATION General Balance Revisited In 200 we introduced the general balance equation: Transient Mass Balances Overall mass balance: Recall, for overall mass both generation and consumption are zero. Component A balance in terms of mass fraction, xA Transient Energy Balances The top layer of our quadruple layer cake. Steady State Mass Balances Steady State Energy Balances Unsteady State Mass Balances Unsteady State Energy Balances This will conclude the first of two broad ChE objectives: Determine how much “stuff” you can make Determine how fast you can make your “stuff” (420, 521, 522, 610) INPUT - OUPUT = ACCUMULATION Transient Energy Balance Because energy can be neither created nor destroyed the general balance reduces to: Previously we discussed that energy can be stored as U or Ek or Ep, which represent accumulation terms. Since + heat is transferred from surroundings to system it is an input, whereas for similar reasons W is an output. However, energy can also be transferred by the movement of mass in and out of the system. This can be represented as changes in enthalpy, Ek or Ep. Energy Balance for OPEN Systems at Steady State At Steady State there can be no Accumulation Input and Output streams can have Kinetic, Potential and Internal Energy and Flow Work Input = Output Recall our general energy balance: Input - Output = Accumulation What can we say at steady state? Q into system is positive (+) W out of system (or done by system) is positive (+) We can combine DU+D(PV)=DH H=U+PV Flow work done on the system at inlet (PV)in is subtracted from work done by the system at outlet (PV)out. Energy Balance for OPEN Systems at Steady State INPUT - OUPUT = ACCUMULATION Transient Energy Balance Because energy can be neither created nor destroyed the general balance reduces to: Previously we discussed that energy can be stored as U or Ek or Ep, which represent accumulation terms. Since + heat is transferred from surroundings to system it is an input, whereas for similar reasons W is an output. However, energy can also be transferred by the movement of mass in and out of the system. This can be represented as changes in enthalpy, Ek or Ep. Transient Energy Balance All of these terms can be plugged into our general balance equation to yield: Note that the inlet and outlet streams will need to be summed over all components. Transient Energy Balance As you can imagine, it is difficult to solved this equation. To aid our calculations, we will make these assumptions: System has one mass flow stream so that min = mout and dm/dt = 0. Kinetic and potential energy changes are negligible so that dEk/dt, dEp/dt ~ 0 and z ~ 0, u2 ~ 0. (Primarily Thermal process) With these assumptions our equation becomes: Transient Energy Balance Even with these simplifications this problem can be difficult (if not impossible) to solve. Therefore we will also assume: T is not f(x,y,z) (i.e., not a function of position), contents are well-mixed, so that Tout = Tsys = T. No phase change or reaction occurs. U(sp) and H(sp) are not f(P). cp / cv not f(xa, ya, T) (i.e., not a function of position or temperature) and therefore not f(t). Transient Energy Balance Recall that: Subbing in: Transient Energy Balance But recall that contents are well mixed so that Tout = T. So for an open system: Open System And for a closed system H = 0 so: Closed System n-C6 60 C, 1 atm 50 L/min n-C6 T C, 1 atm 50 L/min 250 L P = 1 atm Tsys(t) Mixing Liquids of Different Temperatures An insulated 250 L tank is initially filled with liquid n-hexane at 10 °C and 1 atm. Liquid n-hexane at 60 °C and 1 atm is added at a flow rate of 50 L/min and a stream leaves at the same rate. The specific gravity is 0.659 and ~ independent of temperature. Assume the tank is well-mixed. Determine the temperature in the tank as a function of time. Mixing Liquids of Different Temperatures n-C6 60 C, 1 atm 50 L/min n-C6 T C, 1 atm 50 L/min 250 L P = 1 atm Tsys(t) T(0) = 10 C S.G. = 0.659 1. Perform mass balance. Mixing Liquids of Different Temperatures n-C6 60 C, 1 atm 50 L/min 32.95 kg/min n-C6 T C, 1 atm 50 L/min 32.95 kg/min 250 L 164.75 kg P = 1 atm T(0) = 10 C S.G. = 0.659 2. Energy Balance. Well insulated No moving parts Assumes cp and cv are constant for liq @ 1 atm, T=10-60 C And that Tout = T (well mixed). Mixing Liquids of Different Temperatures n-C6 60 C, 1 atm 50 L/min 32.95 kg/min n-C6 T C, 1 atm 50 L/min 32.95 kg/min 250 L 164.75 kg P = 1 atm T(0) = 10 C S.G. = 0.659 3. Solve Energy Balance. But we know that cv is const and for liq cv ~ cp Mixing Liquids of Different Temperatures n-C6 60 C, 1 atm 50 L/min 32.95 kg/min n-C6 T C, 1 atm 50 L/min 32.95 kg/min 250 L 164.75 kg P = 1 atm T(0) = 10 C S.G. = 0.659 3. Solve Energy Balance. Hexane Tank - Solution T(2.0 min) = 26.5 °C T(15.0 min) = 57.5 °C Hexane Tank - Final Comments With the given assumptions, the calculation is independent of the numerical value of CP and CV! (Remember these terms canceled out earlier in our calculations). At 50 L/min, the entire tank volume (250 L) is replaced every 5 min and yet after 5 minutes the temperature is only 41.6 °C! Worksheet # 15 Transient Energy Balances Lecture 16 Goals Ch.11.3, p. 554-560 Discuss simplifying assumptions used to solve the energy balance transient equation. Single mass stream that is not f(t) Ek and Ep are negligible. T is not f(x,y,z), contents are well-mixed, so that Tout = Tsys = T. No phase change or reaction occurs. U(sp) and H(sp) are not f(P). cp / cv not f(xa, ya, T) and therefore not f(t). Solve simple examples using the energy balance transient equation. Hexane example, WS #15.