A mixture of 60% helium and 40% nitrogen by mole enters a...

A mixture of 60\% helium and 40\% nitrogen by mole enters a turbine at $150 \mathrm{lbf} / \mathrm{in.}^{2}, 1500 \mathrm{R}$ at a rate of $4 \mathrm{lbm} / \mathrm{s}$. The adiabatic turbine has an exit pressure of $15 ~ 16 \mathrm{ff} / \mathrm{in} .^{2}$ and an insentropic efficiency of $85 \%$. Find the turbine work.

Key Concepts

Gas Mixture Thermodynamic Properties

When dealing with a gas mixture, such as one containing helium and nitrogen, it is important to account for the composite thermodynamic properties of the mixture. This involves calculating weighted averages of properties like specific heats and the gas constant, which are essential for accurately applying the energy equation and determining the turbine work output.

Adiabatic Process

An adiabatic process is one in which no heat transfer takes place between the system and its surroundings. In the context of turbine analysis, assuming an adiabatic process simplifies the energy balance by removing the heat transfer component, allowing the focus to be solely on the work interactions and changes in the internal energy or enthalpy of the working fluid.

Steady-Flow Energy Equation

The steady-flow energy equation is essential for analyzing open systems like turbines, where mass continuously enters and exits the control volume. It relates the energy fluxes in and out of the system by considering changes in internal energy, work, and flow work (enthalpy), and is used to determine the work output of the turbine under given operating conditions.

Isentropic Efficiency

Isentropic efficiency quantifies how closely a real turbine operation approximates an ideal, reversible (isentropic) process. It is defined as the ratio of actual work output to the work output predicted for a reversible process between the same initial and final pressures. This concept is crucial for adjusting ideal energy calculations to account for real-world irreversibilities.

Transcript

00:01 Hi, everybody.

00:01 Find the turbine work.

00:03 So it's adibiotic, so adibatic, so key to equal zero.

00:10 And using the energy equation, we have the work ideal is c .p .t .1 minus t2.

00:23 And so your cp mix is equal to 0 .6 times 1 .24 times 4 .03 plus 0 .4 times 0 .249 times 28 .013 equals 5 .7.

00:58 683 btu i b moral r and we can be the cv mix really quick and it's going to be cp mix minus r which is 5 .7683 minus 15445778 and this gives us 3 .788 and this gives us 3 .788 243 btu i b more r so now we have to find our k which is going to be cp over cd and that's going to be 5 .7683 divided by 3 .7843 equals 1 .525 five okay let me scroll down here and for the ideal turbine um so we find the temperature of the ideal turbine so it's going to be t1 equals p2 over p1 um k minus 1 divided by k and so when you find t 2 and this is going to be i'm going to bring over the t1 over here so 1 5 0 times 15 divided by 150.

02:31 1 .525 minus 1 divided by 1 .525 minus 1 divided by 1 .525 equals 678 .936r.

02:46 Okay.

02:50 And now we have your, we need to find m mix, which is 0 .6 times 4 .003 plus 0 .4 times 28 .013.

03:09 And that is 13 .607.

03:13 And we find our cp mix, which is different from, so we need to find our other cp mix, okay? different from this one.

03:23 So the cp mix is going to be your other cp mix.

03:32 We'll put that here.

03:36 And so let's be one second to the vectors here.

03:43 There we go.

03:44 There we go, guys.

03:45 That's my fault equals 5 .7683 divided by 13 .607 .7 .7 .7 .7 .7.

03:57 Which is 0 .4239, btu divided by i -b -m -r...