Question

Problem 1: This is textbook problem 4.51 (page 224) Steam at 1800 lbf/in2 and 1100 °F enters a turbine operating at steady state. As shown in the Figure, 20% of the entering mass flow is extracted at 600 lbf/in2 and 500 °F. The rest of the steam exits as a saturated vapor at 1 lbf/in2 . The turbine develops a power output of 6.8×106 Btu/h. Heat transfer from the turbine to the surroundings occurs at a rate of 5×104 Btu/h. Neglecting kinetic and potential energy effects, answer the following questions: (a) What is the mass flowrate of steam entering the turbine at point 1? (b) What is the mass flowrate of steam exiting the turbine at point 3? (c) If the diameter of the inlet to the turbine (point 1) is 20 inches, what are the diameters of the exits at points 2 and 3 so that there is no change in kinetic energy from inlet to outlets?

Problem 2: I purchased a house about a year back and I’m trying to make energy efficiency improvements using my research knowledge. One of the projects my lab works on is related to efficient optical-to-thermal energy conversion where we use novel (and scalable) nanomaterial coatings to enhance solar energy absorption. I am hoping to leverage our sunny Texas climate and these nanomaterial coatings to build a solar water heating system for the house and I would like your help with some rough calculations. If the incoming radiation from the sun is typically around 950 W/m2 and the losses from the solar water heater surface are MEEN 315 (Principles of Thermodynamics) 2 nd March, 2020 Homework IV (Due Monday 23 rd March, 2020) Page 3 of 3 approximately 50 W/m2 , what is the temperature of water at the outlet of the water heater if it enters at 20 °C with a flowrate of 1 L/min (liter per minute)? Also, what is the enthalpy of the water at the outlet? Here is some additional information that may be useful: The unit W/m2 is a flux, i.e., the rate of energy transfer per unit area where the energy transfer occurs (i.e., approximately what is exposed to the sun). You may assume that the solar water heating system is a copper tube coil (see hyperlink for concept) of outer diameter 0.25" and length 50', and all the surface area is available for both the incoming solar radiation and the heat loss. You may assume that there is negligible pressure drop through the water heater and the water enters at close to atmospheric pressure.

Answer 1

Answer 2

Problem 1: (a) To find the mass flowrate of steam entering the turbine at point 1, we need to apply the steady-state energy equation. The energy equation can be written as:

ΔH = Q - W

Where ΔH is the change in enthalpy, Q is the heat transfer, and W is the work done.

Given: Pressure at point 1 (P1) = 1800 lbf/in^2 Temperature at point 1 (T1) = 1100 °F Heat transfer from the turbine to the surroundings (Q) = -5×10^4 Btu/h (negative sign indicates heat transfer from the system)

We are neglecting kinetic and potential energy effects, so there is no work done.

From the steam tables, we can find the enthalpy values at the given pressure and temperature for points 1, 2, and 3. Let's denote the enthalpies as h1, h2, and h3, respectively.

ΔH = h1 - h2 - h3

Since there is no work done, ΔH = Q

h1 - h2 - h3 = -5×10^4

(b) The mass flowrate of steam exiting the turbine at point 3 can be calculated using the equation:

m_dot = (Q - W) / (h1 - h2 - h3)

Given: Power output of the turbine (W) = 6.8×10^6 Btu/h

(c) To find the diameters of the exits at points 2 and 3 so that there is no change in kinetic energy, we can use the principle of conservation of energy. The kinetic energy at point 1 is assumed to be negligible.

Problem 2: To calculate the temperature of water at the outlet of the solar water heater, we need to consider the energy balance. The energy absorbed from the sun should be equal to the energy lost from the surface plus the energy gained by the water.

Given: Incoming radiation from the sun = 950 W/m^2 Losses from the solar water heater surface = 50 W/m^2 Water flowrate = 1 L/min Inlet water temperature = 20 °C

We can use the energy balance equation:

Q_in - Q_out - Q_loss = m_dot * (h_out - h_in)

Where Q_in is the energy absorbed from the sun, Q_out is the energy lost from the surface, Q_loss is the heat loss, m_dot is the mass flowrate of water, h_out is the enthalpy of water at the outlet, and h_in is the enthalpy of water at the inlet.

We need to convert the flowrate to kg/s and the temperature to Kelvin before using the equation. Once we solve for h_out, we can convert it to the corresponding temperature.

I hope this helps you get started with solving these problems. If you need further assistance or more detailed calculations, please let me know!