Answer:
0.5 Amperes
Explanation:
Information that we have:
Power:
Voltage:
and we use the equation that relates this two quantities to the Current (I):
where I is the curent, P is the Power and V is the Voltage.
we substitute the values given for the Power and the Voltage to find the current:
the Current running into the bulb is 0.5 Amperes
(a)
Use “saturated refrigerant-134a pressure tableâ€, to find $ , h , and of refrigerant-134a, at 320kPa (P) .
= 251.93kJ/kg
) =0.93026 kJ/kg-K
=0.063681 m/kg
we solve ,
= 282.62 kj/kg
= 127.25 kJ/kg
We use this formula for finding
= 38.8 kw
(b)
Find the COP of the heat pump () .
() =
= (282.62 -127.25) / (282.62-251.93)
= 5.06
What is evaporator pressures ?
Valves for regulating evaporator pressure
Although the compressor suction pressure may be lower, evaporator pressure regulation (EPR) valves can be employed in the suction line to prevent the evaporator pressure from dropping below a set or controlled value.
An EPR valve is used for the following tasks:
1. Prevent potential damage to a liquid chilling evaporator from the liquid freezing.
2. Prevent frost from accumulating on an air-cooling evaporator when it is near freezing point or when operation cannot be interrupted by a brief failure.
3. Permit the operation of two or more evaporators with the same compressor at various load temperatures.
4. Vary the evaporator pressure in accordance with a fluctuating load that is managed by the load temperature.
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