Answer:
a) The amount of heat transfer in the regenerator, q = 114.12 kJ/kg
b) Thermal efficiency = 35.9%
Explanation:
The calculations are neatly handwritten and attached as files to this solution for easiness of expression and clarity. The cycle is also drawn on the T - S diagram and included in the attached files. Check the files below for the complete calculation.
I hope this helps!
Answer:
Utilize the Standard Controller for Position_c and a Controller Extension to query for Review_c data.
Explanation:
Data types such as individual, pedigree, sample, and marker can be displayed depending on the type of information needed, and the database fields will be recorded based on their levels with the appropriate data type placed in each field. The relationship that will be gotten will depend on the auto-populated defaults which can changed to suit our requirements.
Answer:
The critical radius is -1.30 nm
Explanation:
Temperature for homogenous nucleation of copper, 
Melting point of copper, 
Latent heat of fusion, 
Surface free energy, 
Critical radius, r = ?
The formula for the critical radius is given by:
The critical radius is -1.30 nm
Answer: A fly wheel having a mass of 30kg was allowed to swing as pendulum about a knife edge at inner side of the rim as shown in figure.
Explanation:
Answer:
15.4 g/cm³, 17.4 g/cm³
Explanation:
The densities can be calculated using the formula below
ρ = (fraction of tungsten × ρt ( density of tungsten)) + (fraction of pores × ρp( density of pore)
fraction of tungsten = (100 - 20 ) % = 80 / 100 = 0.8
a) density of the before infiltration = ( 0.8 × 19.25) + (0.2 × 0) = 15.4 g/cm³
b) density after infiltration with silver
fraction occupied by silver = 20 / 100 = 0.2
density after infiltration with silver = ( 0.8 × 19.25) + (0.2 × 10) = 17.4 g/cm³
