Question

Water supplied to a house by a water main has a pressure of 3.00×10⁵ N/m² early on a summer day when neighborhood use is low. This pressure produces a flow of 20.0 L/min through a garden hose. Later in the day, pressure at the exit of the water main and entrance to the house drops, and a flow of only 8.00 L/min is obtained through the same hose.

(a) What pressure is now being supplied to the house, assuming resistance is constant?
(b) By what factor did the flow rate in the water main increase to cause this decrease in delivered pressure? The pressure at the entrance of the water main is 5.00×10⁵ N/m², and the original flow rate was 200 L/min.
(c) How many more users are there, assuming each would consume 20.0 L/min in the morning?
a) 1.5×10⁵ N/m², 2.5, 9
b) 2.5×10⁵ N/m², 2.0, 6
c) 1.0×10⁵ N/m², 3.0, 12
d) 2.0×10⁵ N/m², 1.5, 15

Answer 1

The new pressure supplied to the house is 1.2 × 10⁵ N/m². The flow rate in the water main increased by approximately a factor of 4.17 to cause the decrease in delivered pressure, resulting in about 32 more users assuming each would consume 20.0 L/min in the morning.

Step-by-step explanation:

To solve the problem related to pressure and flow rate in a water supply system, it is useful to use the continuity equation and Bernoulli's principle. Given that resistance is assumed constant, we can infer a direct proportionality between pressure and flow rate for a particular house.

(a) Finding the new pressure supplied to the house: As the flow rate decreases from 20.0 L/min to 8.00 L/min (which is a factor of 20.0/8.00 = 2.5), the pressure would also decrease by the same factor if we assume resistance is constant. Hence, the new pressure is 3.00 × 10⁵ N/m² divided by 2.5, which is 1.2 × 10⁵ N/m².

(b) Determining the factor by which flow rate increased: The pressure before reaches 5.00 × 10⁵ N/m² and after it is 1.2 × 10⁵ N/m². To find by what factor the flow rate must have increased, consider that pressure is inversely proportional to flow rate (assuming other variables constant). Solving for flow rate increase, we have that flow rate must have increased by a factor of 5.00/1.20, which is approximately 4.17.

(c) Calculating the number of users: If the original flow rate was 200 L/min and each user consumes 20.0 L/min, there were originally 10 users. Now, with the increased flow rate by a factor of 4.17, the new flow rate would be 200 L/min × 4.17, which equals 834 L/min. With each additional user consuming 20.0 L/min, this means there are now 834/20 = 41.7, or approximately 42 users. The increase in users is 42 minus the original 10, which gives us 32 additional users.