Answer:
I = 0.625 A
Explanation:
Given that,
Power of the light bulb, P = 75 W
Voltage of the circuit, V = 120 V
We need to find the current flowing through it. We know that, Power is given by :
I is the electric current
So, the current is 0.625 A.
An air-conditioner with refrigerant-134a as the working fluid is used to keep a room at 23°C by rejecting the waste heat to the
outdoor air at 37°C. The room gains heat through the walls and the windows at a rate of 250 kJ/min while the heat generated by the computer, TV, and lights amounts to 900 W. The refrigerant enters the compressor at 400 kPa as a saturated vapor at a rate of 100 L/min and leaves at 1200 kPa and 70°C. Determine (a) the actual COP, (b) the maximum COP, and (c) the minimum volume flow rate of the refrigerant at the compressor inlet for the same compressor inlet and exit conditions.
1 answer:
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Answer:
a) 180 m³/s
b) 213.4 kg/s
Explanation:
= 1 m²
= 100 kPa
= 180 m/s
Flow rate
Volumetric flow rate = 180 m³/s
Mass flow rate
Mass flow rate = 213.4 kg/s
Answer:
a) 4 passes are required to sort the string.
b) 4
c) i) TARP
ii) CHIP
iii) PART
iv) TARP
v) TARP
d) O(k+n), n is no. of strings, k is largest no. of character in among the string
O(d*(n+10)), n is no. of integers
Explanation:
