Answer:
Explanation:
Hello!
To solve this problem you must follow the following steps, which are fully registered in the attached image.
1. Draw the complete outline of the problem.
2. Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties.
3. Use temodynamic tables to find the density of water in state 1, by means of temperature and quality, with this value and volume we can find the mass.
3. Use thermodynamic tables to find the internal energy in state 1 and two using temperature and quality.
4. uses the first law of thermodynamics that states that the energy in a system is always conserved, replaces the previously found values and finds the work done.
5. draw the pV diagram using the 300F isothermal line
Answer:
Explanation:
First, we find the mass of the air originally in the tank.
Density is given as mass divided by volume. It is given as:

Therefore, mass is:

Density of air =
; Volume of the tank = 

The mass of the air initially in the tank is 7 kg.
After air is allowed to enter, the mass changes.
New density = 
The new mass will be:

We can now find the mass of air that has entered the tank:
Mass of air that entered tank = New mass of air - Original mass of air
M = 22.75 - 7.0 = 15.75 kg
The mass of air that entered the tank is 15.75 kg.
Answer:
(a) Precipitation hardening
(1) The strengthening mechanism involves the hindering of dislocation motion by precipitates/particles.
(2) The hardening/strengthening effect is not retained at elevated temperatures for this process.
(4) The strength is developed by a heat treatment.
(b) Dispersion strengthening
(1) The strengthening mechanism involves the hindering of dislocation motion by precipitates/particles.
(3) The hardening/strengthening effect is retained at elevated temperatures for this process.
(5) The strength is developed without a heat treatment.
Answer:
a) What is the surface temperature, in °C, after 400 s?
T (0,400 sec) = 800°C
b) Yes, the surface temperature is greater than the ignition temperature of oak (400°C) after 400 s
c) What is the temperature, in °C, 1 mm from the surface after 400 s?
T (1 mm, 400 sec) = 798.35°C
Explanation:
oak initial Temperature = 25°C = 298 K
oak exposed to gas of temp = 800°C = 1073 K
h = 20 W/m².K
From the book, Oak properties are e=545kg/m³ k=0.19w/m.k Cp=2385J/kg.k
Assume: Volume = 1 m³, and from energy balance the heat transfer is an unsteady state.
From energy balance: 
Initial temperature wall = 
Surface temperature = T
Gas exposed temperature = 
Flip flops are not required