Answer:
exergy = 33.39 kW
Explanation:
given data
thermal energy reservoir T2 = 1500 K
heat at a rate = 150,000 kJ/h = 
kW = 41.67 kW
environment temperature T1 = 25°C = 298 K
solution
we get here maximum efficiency that is reversible efficiency is express as
reversible efficiency = 1 - 
...............1
reversible efficiency = 1 - 
reversible efficiency = 0.80133
and
the exergy of this supplied energy that is
exergy = efficiency × hat supply ................2
exergy = 0.80133 × 41.67 kW
exergy = 33.39 kW
Answer: hello some parts of your question is missing attached below is the missing information
The radiator of a car is a type of heat exchanger. Hot fluid coming from the car engine, called the coolant, flows through aluminum radiator tubes of thickness d that release heat to the outside air by conduction. The average temperature gradient between the coolant and the outside air is about 130 K/mm . The term ΔT/d is called the temperature gradient which is the temperature difference ΔT between coolant inside and the air outside per unit thickness of tube
answer : Total surface area = 3/2 * area of old radiator
Explanation:
we will use this relation
K = 
change in T = ΔT
therefore New Area ( A ) = 3/2 * area of old radiator
Given that the thermal conductivity is the same in the new and old radiators
Answer:
A. optical isolation
Explanation:
well I can't really give a good explanation because I also saw the same question in my exams and option A was the correct answer
Answer:
11.541 mol/min
Explanation:
temperature = 35°C
Total pressure = 1.5 * 1.013 * 10^5 = 151.95 kPa
note : partial pressure of water in mixture = saturation pressure of water at T = 35°c )
from steam table it is = 5.6291 Kpa
calculate the mole fraction of H
( YH )
= 5.6291 / 151.95
= 0.03704
calculate the mole fraction of air ( Yair )
= 1 - mole fraction of water
= 1 - 0.03704 = 0.9629
Now to determine the molar flow rate of water vapor in the stream
lets assume N = Total molar flow rate
NH = molar flow rate of water
Nair = molar flow rate of air = 300 moles /min
note : Yair * n = Nair
therefore n = 300 / 0.9629 = 311.541 moles /min
Molar flowrate of water
= n - Nair
= 311.541 - 300 = 11.541 mol/min
