Answer:
a)Are generally associated with factor.
Explanation:
We know that losses are two types
1.Major loss :Due to friction of pipe surface
2.Minor loss :Due to change in the direction of flow
As we know that when any hindrance is produced during the flow of fluid then it leads to generate the energy losses.If flow is along uniform diameter pipe then there will not be any loss but if any valve and fitting placed is the path of fluid flow due to this direction of fluid flow changes and it produce losses in the energy.
Lot' of experimental data tell us that loss in the energy due to valve and fitting are generally associated with K factor.These losses are given as
But where are the options?
Answer:
a) 8kW
b) $128
Explanation:
Given the coefficient of performance of the heat pump cycle to be 2.5
Energy delivered by the heat pump = 20kW
a) net power required to operate the heat pump = Energy delivered / coefficient of performance
Net power required = 20/2.5
= 8kW
b) Given the cost of electricity is $0.08 for 1kWhour
Since net power required to operate heat pump = 8kW
If the heat pump operate for 200hours, total power required for a month = 8kW×200hours = 1600kWhour
since 1kWh of electricity costs $0.08, cost of electricity used in a month when the pump operates for 200hour will be 1600kWh×$0.08 which is equivalent to $128
Answer:
C
Explanation:
Hydraulic systems can increase or decrease the force applied to them. To make the force larger, the pressure is applied to a larger area. For example, if a 100-N force is applied to the left cylinder in (Figure) and the right cylinder has an area five times greater, then the output force is 500 N.
