Answer:
Your question has some missing information below is the missing information
Given that ( specific heat of fluid A = 1 kJ/kg K and specific heat of fluid B = 4 kJ/kg k )
answer : 300 kW , 95°c
Explanation:
Given data:
Fluid A ;
Temperature of Fluid ( Th1 ) = 420° C
mass flow rate (mh) = 1 kg/s
Fluid B :
Temperature ( Tc1) = 20° C
mass flow rate ( mc ) = 1 kg/s
effectiveness of heat exchanger = 75% = 0.75
<u>Determine the heat transfer rate and exit temperature of fluid</u> <u>B</u>
Cph = 1000 J/kgk
Cpc = 4000 J/Kgk
Given that the exit temperatures of both fluids are not given we will apply the NTU will be used to determine the heat transfer rate and exit temperature of fluid B
exit temp of fluid B = 95°C
heat transfer = 300 kW
attached below is a the detailed solution
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Answer:
The correct solution is "21024 KWh/degree day".
Explanation:
The given query is incomplete. Below is the attachment of complete query is provided.
The given values are:
Indoor design temperature:
= 70°F
Now,
According to the question,
The heat loss annually will be:
= 
= 
Degree days will be:
= 
= 
Hence,
Annual KWh use will be:
= 
On substituting the values, we get
= 
= 
Answer:
Efficiency of the engine equals 20%
Explanation:
We know that when the car moves it must do work against the resisting forces to keep moving and this work is spend as energy by the engine to keep the car moving.
we know that
Thus to keep the car moving for 100,000 meters the theoretical work that requires to be done equals
Now the actual energy spend by the car equals the energy spend by burning 2.8 gallons of gasoline.
Thus the energy produced by burning 2.8 gallons of gasoline equals
Thus the efficiency is calculated as
