Answer:
Check the explanation
Explanation:
A diagrams showing the process in p-v and T-s and the Model engineering <u><em>(which is the quest in constructing miniature working representations of proportionally-scaled in full-sized machines. It is a subdivision of metalworking with a sharp importance on artisanry, in contrast to mass production.)</em></u> can be seen in the attached images below.
To solve this problem it is necessary to apply the concepts related to temperature stagnation and adiabatic pressure in a system.
The stagnation temperature can be defined as

Where
T = Static temperature
V = Velocity of Fluid
Specific Heat
Re-arrange to find the static temperature we have that



Now the pressure of helium by using the Adiabatic pressure temperature is

Where,
= Stagnation pressure of the fluid
k = Specific heat ratio
Replacing we have that


Therefore the static temperature of air at given conditions is 72.88K and the static pressure is 0.399Mpa
<em>Note: I took the exactly temperature of 400 ° C the equivalent of 673.15K. The approach given in the 600K statement could be inaccurate.</em>
Answer:
the lost work per kilogram of water for this everyday household happening = 0.413 kJ/kg
Explanation:
Given that:
Initial Temperature
= 15°C
Initial Pressure
= 5 atm
Final Pressure
= 1 atm
Data obtain from steam tables of saturated water at 15°C are as follows:
Specific volume v = 1.001 cm³/gm
The change in temperature = 2°C
Specific heat of water = 4.19 J/gm.K
volume expansivity β = 1.5 × 10⁻⁴ K⁻¹
The expression to determine the change in temperature can be given as :


Δ T = 0.093 K
Now; we can calculate the lost work bt the formula:

where ;
is the temperature of the surrounding. = 20°C = (20+273.15)K = 293.15 K
From above the change in entropy is:






Thus, the lost work per kilogram of water for this everyday household happening = 0.413 kJ/kg
Answer:
c from transmitter to a receiver