Answer:
Mechanical Engineering
Chemical Engineering
Civil Engineering
Explanation:
I got it from my old homework And I learn those at school ( Thank You For The Points)
Answer:

Explanation:
generally regeneration of cycle is used in the case of gas turbine. due to regeneration efficiency of turbine is increased but there is no effect on the on the net work out put of turbine.Actually in regeneration net heta input is decreases that is why total efficiency increase.
Now from T-S diagram



Due to generation
amount of energy has been saved.

So efficiency of cycle 

Effectiveness of re-generator

So the efficiency of regenerative cycle

Answer and Explanation:
The explanation is attached below
Answer:
<em>Python code is as follows:
</em>
********************************************************************************
#function to get number up to any number of decimal places
def toFixed(value, digits):
return "%.*f" % (digits, value)
print("Enter the price: ", end='', flush=True) #prompt for the input of price
price = float(input()) #taken input
totalCost = price + 0.05 * price #calculating cost
print("Total Cost after the sales tax of 5% is applied: " + toFixed(totalCost,2)) #print and output totalCost
************************************************************************************
Answer:
Heat losses by convection, Qconv = 90W
Heat losses by radiation, Qrad = 5.814W
Explanation:
Heat transfer is defined as the transfer of heat from the heat surface to the object that needs to be heated. There are three types which are:
1. Radiation
2. Conduction
3. Convection
Convection is defined as the transfer of heat through the actual movement of the molecules.
Qconv = hA(Temp.final - Temp.surr)
Where h = 6.4KW/m2K
A, area of a square = L2
= (0.25)2
= 0.0625m2
Temp.final = 250°C
Temp.surr = 25°C
Q = 64 * 0.0625 * (250 - 25)
= 90W
Radiation is a heat transfer method that does not rely upon the contact between the initial heat source and the object to be heated, it can be called thermal radiation.
Qrad = E*S*(Temp.final4 - Temp.surr4)
Where E = emissivity of the surface
S = boltzmann constant
= 5.6703 x 10-8 W/m2K4
Qrad = 5.6703 x 10-8 * 0.42 * 0.0625 * ((250)4 - (25)4)
= 5.814 W