1 / 8

Example: Electric resistive heaters are often used in houses to provide heat during winter

Example: Electric resistive heaters are often used in houses to provide heat during winter months. It consists of a simple duct with coiled resistance wires as shown. Consider a 20 kW heating system such that the air enters at 100 kPa and 17° C with a mass

orrin
Download Presentation

Example: Electric resistive heaters are often used in houses to provide heat during winter

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Example: Electric resistive heaters are often used in houses to provide heat during winter months. It consists of a simple duct with coiled resistance wires as shown. Consider a 20 kW heating system such that the air enters at 100 kPa and 17° C with a mass flow rate of 1.8 kg/s. If it is known that the air leaves the duct with an exit temperature of 27° C (same pressure), determine the heat loss from the duct. Heat loss, dQ/dt=? To=17° C Assume: steady state, negligible KE and PE changes, air can be considered as idea gas To=27° C dEg/dt=20 kW where hout = Cp Tout and hin = Cp Tin From Table A-19 for air in YAC : Cp,out = Cp,in = 1.005 kJ/Kg.C <= total energy/heat going into the flowing air Hence, heat lost to the duct = 20 –18 = 2 kW

  2. Using a slightly different approach: Heat loss, dQ/dt=? To=17° C Again assume: 1.steady state 2. NegligibleKE and PE changes 3. Air can be considered as idea gas To=27° C dEg/dt=20 kW

  3. Recall that we wrote the First Law of Thermodynamics as: The above only applies for a STEADY FLOW process, I.e. a process where none of the properties inside the control Volume (C.V.), including energy, are changing with time. The turbine problem is an example of a steady process. If the process is unsteady, i.e. transient, the total amount of energy inside the system (or CV) can change. The First Law for such a process is given by: Where dECV/dt is the rate of change in the total energy of the Control Volume. First law-ex-1

  4. Transient Energy Balance (Unsteady State) Heat in q=dQ/dt E(t) not constant Work out dW/dt (4.19), p. 157 in Cengel’s book

  5. Example: Steam at a pressure of 2 MPa and a temperature of 350° C is exiting out from a tank through a valve to drive a turbine as shown. The exhausting steam enters an initially evacuated tank with a volume of 1 m3. The valve is closed when the second tank is filled with steam at a pressure of 1.4 MPa and a temperature of 500° C. Assume no significant heat transfer and KE and PE changes are also negligible. Determine the amount of work developed by the turbine. control volume to be analyzed V=1 m3 initially evacuated Steam 2 MPa 350 ° C Assumptions: dQ/dt=0 adiabatic process, KE and PE are negligible Steam properties at the first tank remains constant The second tank reaches final equilibrium after the filling process ends

  6. 0 0

More Related