Question

Water is being pumped from the bottom of a well 150 feet deep at a rate of 200 gal/hour into a vented storage tank 30 feet above the ground. To prevent freezing in the winter, a small heater puts 30,000 BTU/hour into the water during its transfer from the well to the storage tank. Heat is lost from the whole system at a constant rate of 25,000 BTU/hr. What is the rise or fall in the temperature of the water as it enters the storage tank, if the well ware is at 35°F? A 2 HP pump is being used to pump the water. The pump efficiency is 55 percent. Evaluate any properties of water at 35°F.

Answer 1

Step-by-step explanation:

As the given data is as follows.

Height, H = 150 feet

Heat gain = 30,000 BTU/hr, and Heat loss = 25000 BTU/hr

m = mass of water heated = 700 gallons = 5810 lbs

`C_(p)` is the heat capacity of water = 1 BTU/lb
`^(o)F` (given)

`\Delta T` = temperature difference =
`120^(o)F - 35^(o)F`

Heat energy required to heat 700 gal can be calculated as follows:

Heat Required =
`5810 lbs * 1 BTU/lb^(o)F * (120^(o)F - 35^(o)F)`

Thus, water rises till
`120^(o)F`.