Pure water flows through a horizontal section of pipe that is 5 ft long and 1" in diameter. The water entering the pipe has a flow rate of 100 kg/min with a temperature of 100 °C and pressure of 2 bar. The exiting water also has a flow rate of 100 kg/min and temperature of 100 °C, but the pressure is 1.95 bar. The density of water at these conditions is 961.5 kg/m3. Determine the rate at which work is done, both at the pipe entrance and the pipe exit.

The work done at the pipe exit is W= 338.01 watts or J/min, the work done at the pipe entrance is W = 346.68 watts

Explanation:

Solution

The work done  at the entrance of the pipe

Let us assume that the water in the pipe as a system

The work done w is denoted by

W = F. V

V= velocity

F = exerted force

Now,

W = P.A.V  this is the equation (1)

where P = F/A P = pressure at in let A = The cross sectional area

we can then say,

A . V = Q ( This is the volumetric flow rate)

also

Q = m/ ρ where m = mass flow rate, ρ = density

The equation 1 now becomes

W = P ( m/ρ)

= 2 * 10 ^ ⁻⁵ pa ( 100 kg min/ 961. 5 kgm³)

W = 20800.83 J/min

W = 346.68 J/s

Then

W = 346.68 watts

The work done at the pipe exist

W = P. Q

Now,

= P ( m/ρ)

W = 1.95 * 10 ^ 5 pax ( 100 kg min/ 961. 5 kgm³)

W = 2028.81 J/min

W= 338.01 watts or J/min

Note: At the entrance, the outside water will push the water  inside, this means that work is carried out on the system, and it is a positive sign, but at the exit point it is vice versa and is known to be negative

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explanation:

answer:

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