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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.

Respuesta :

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

[tex]C_{p}[/tex] is the heat capacity of water = 1 BTU/lb [tex]^{o}F[/tex] (given)

      [tex]\Delta T[/tex] = temperature difference = [tex]120^{o}F - 35^{o}F[/tex]

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

    Heat Required = [tex]5810 lbs \times 1 BTU/lb^{o}F \times (120^{o}F - 35^{o}F)[/tex]

Thus, water rises till [tex]120^{o}F[/tex].

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