The rate of heat transfer by the air conditioner using constant specific heat of 1.004kj/kg.K is 15.06 kW.
<h3>What is the rate of heat transfer?</h3>
Rate of heat transfer is the power rating of the machine.
Work done and changes in potential and kinetic energy are neglected since it is a steady state process.
The specific heat in terms of specific heat capacity and temperature change is given as:
The rate of heat transfer, is then determined as follows:
- Qout = flow rate × specific heat
Qout = 0.75 × 20.08 = 15.06 kW
Therefore, the rate of heat transfer by the air conditioner is 15.06 kW.
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Answer:
The three phase full load secondary amperage is 2775.7 A
Explanation:
Following data is given,
S = Apparent Power = 1000 kVA
No. of phases = 3
Secondary Voltage: 208 V/120 V <em>(Here 208 V is three phase voltage and 120 V is single phase voltage) </em>
<em>Since,</em>
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The formula for apparent power in three phase system is given as:
Where:
S = Apparent Power
V = Line Voltage
I = Line Current
In order to calculate the Current on Secondary Side, substituting values in above formula,
Answer:
(a) the percent thermal efficiency is 27.94%
(b) the temperature of the cooling water exiting the condenser is 31.118°C
Explanation:
Answer:
C.) Weight and distance I believe
Explanation:
