Question

spring relief disk 3. (40 pt.) A pressure cooker cooks a lot faster than an ordinary pan by maintaining a higher pressure and temperature inside. The lid of a pressure cooker is well sealed, and steam can escape only through an opening in the middle of the lid. The pressure relief valve consists of a spring-loaded disk that is positioned over this opening. Consider that a certain portion of a pressure cooker is initially filled with water and put on top of a stove (State - 1). Heat is added just until the pressure relief valve opens (State 2). Heat continues to be added until all of the liquid disappears (State 3). Using the knowledge that you learned in the Thermodynamics I Course, analyze the operation of this pressure cooker thermodynamically (i.e. find the intensive properties of water such as temperature, pressure and quality, mass of liquid water in the cooker, mass of water vapor that has passed through the pressure relief valve, and time passed for various stages of its operation). Operation and design parameters of the pressure cooker include volume of the cooker, initial volume of water, heat input rate, surrounding pressure, disk diameter, mass of the disk, spring constant, the change in length of the spring, etc. Using resources (e.g. textbooks, reports, websites, etc.), give appropriate numerical values for the operation and design parameters of the pressure cooker and find the results for each stage. Show the processes on a T-v diagram. List any assumptions that you make. Comment on your results considering a practical application.

Answer 1

wates (or) a ssure Pressure Cooked - It is the process of Cooking food at high pressure, employing * Solution * Concept :-) PAGE_I water based cooking liquid in a sealed vessel known of presso Cookes. -> Assumptions :-) shaft work is present. 2. Pressure cooked is 3. volume of pressure cooked does not change. 4. R does not change. Hino well insulated

PAGER Considering the figure , conditions Conditions from the question. spring > Relief disk. Hele let me consided some assumptions thus the cooked contains ikg of water Volume be 2.5L. Spring Constant = 230 N/cm.. spoing movement - som spring force will be = 1,50 Mass of disk - 12N Total force =12907 Diameter of disk = 4cm Ares -0-786x4? = 12.576 cm² 127 Thus pressure insicle the chembel will be P= 10.100/S2.100) 14.62 psigouge (0) 2 12.57

helic on 100 ) 29.5psi absolute Let Tempeltese =133 Specific enthelpy of sotese te widel=562.0815)g PAGE 3 Thus the twice the atmosp on absolute siale amount of heat input through burned = 1000 Let initial temperature of water = 20° At operation pressuse of 14.62 psi density of steam= 1657 kg/m3 Latent heat of steen= 21636.J/kg. specific Enth alpy of Room temp =1041%9/kg Thus we consides dryness frection=0.9 Then total Energy Deguised K3=0.9 x 2163+(56207- = 2404.79 KI/kg. Density of water = 980 kg/m3 mass of water = 0.98 kg Total energy required - 0.98 x2405.78 = 2356.67Kg

PAGE- Time required to eveporate all weles +2756.696403 1000 = 2356.6sec 108) 39.27 minutes (PLEASE UPVOTE )