We can use these new values of i and Y to find the new values of C and I: C = 200 + 0.25(Y D) C = 200 + 0.25(Y – T) We know sub in our new value for Y and our value for T given in the question: C = 200 + 0.25(1040 – 200) C = 200 + 0.25(840) C = 200 + 210 C = 410 Investment is given as: I = 150 + 0.25Y – 1000i
If the temperatures are given as T A = 600 K, T B = 800 K, T C = 2200 K and T D = 1200 K, then the work done per cycle is. Q. 3 moles of an ideal monoatomic gas perform a cycle shown in Fig. 16.19. The gas temperatures T A = 400 K, T B = 800 K, T C = 2400 K, T D = 1200 K. Find the work done by the gas.
Absolutely, The k is a ratio that will vary for each problem based on the material, the initial temperature, and the ambient temperature. Most of the problems that I have seen for this involve solving for C, then solving for k, and finally finding the amount of time this specific object would take to cool from one temperature to the next.
The oxidation of Ti x Ta 1−x C 0.5 N 0.5-Co-based cermets with different Ta contents was studied in static air at temperatures between 700 °C and 1200 °C. 2. It was determined that the oxidation kinetics satisfied the parabolic kinetic law for all cermets and temperatures, which is characteristic of the presence of a protective oxide layer.
oC, and V 1 = 3.8 liters. During the heat addition process, 7.5 kJ of heat are added. Determine all T's, P's, ηth, the back work ratio, and the mean effective pressure. Process Diagrams: Review the P-v and T-s diagrams given above for the Otto cycle. Assume constant specific heats with C v = 0.718 kJ/kg ⋅K, k = 1.4. (Use the 300 K data from ...
0.718 kJ/kg·K, R = 0.287 kJ/kg.K, and k = 1.4 (Table A-2). Analysis The minimum pressure in the cycle is P3 and the maximum pressure is P1. Then, or, () ... pressure limits of 100 and 1200 kPa. The working fluid is air, which enters the compressor at 30°C at a rate of 150 m3/min and leaves the turbine at 500°C. Using
d (c A V R) dt = R A V R Substituting in the reaction-rate expression, r = k (T )c A, and using the number of moles of A, n A = c A V R yields dn A dt = k (T )n A (6.20) Notice the temperature dependence of k (T ) prevents us from solving this di erential equation immediately. We must solve it simultaneously with the energy balance, which ...