Question

Consider a mixing chamber at steady state with two inlets and one outlet as shown below. inlet 1 transports saturated water at 150 o c at 2 kg/s, and inlet 2 transports compressed liquid water at 100 o c and 500 kpa at 1 kg/s. after mixing and heating in the chamber, we get superheated vapor at 250 o c at 300 kpa. all inlets and outlets have the same cross-sectional area of 100 cm2 . ignore potential energy but not the kinetic energy.

(a) what is the flow speed in the first inlet?
(b) what is the flow speed in the second inlet?
(c) what is the flow speed in the outlet?
(d) what is the heating rate in the chamber?

Answer 1

By applying the principle of conservation of mass and using the given information, we can calculate the flow speeds in the inlets and outlet of the mixing chamber. The flow speed in the first inlet is 0.5 m/s, in the second inlet is also 0.5 m/s, and in the outlet is 0.1667 m/s. To calculate the heating rate in the chamber, we need specific enthalpy values at the given temperatures and pressures.

Step-by-step explanation:

To solve this problem, we can apply the principle of conservation of mass, which states that the mass of fluid entering a pipe must be equal to the mass of fluid exiting the pipe. From the given information, we can calculate the flow speeds in the inlets and outlet using the equation:

A₁v₁ = A₂v₂ = A₃v₃

Where A₁, A₂, and A₃ are the cross-sectional areas of the inlets and outlet, and v₁, v₂, and v₃ are the flow speeds in each section.

We can start by calculating the flow speed in the first inlet (v₁). Given that the cross-sectional area of the inlets and outlet are all the same (100 cm²), we can rearrange the equation to solve for v₁:

v₁ = (A₂v₂) / A₁

Substituting the values, we have:

v₁ = (100 cm² * 1 kg/s) / (100 cm² * 2 kg/s) = 0.5 m/s

Therefore, the flow speed in the first inlet is 0.5 m/s.

Similarly, we can calculate the flow speed in the second inlet (v₂):

v₂ = (A₁v₁) / A₂

Substituting the values:

v₂ = (100 cm² * 0.5 m/s) / (100 cm² * 1 kg/s) = 0.5 m/s

Therefore, the flow speed in the second inlet is also 0.5 m/s.

Finally, we can calculate the flow speed in the outlet (v₃):

v₃ = (A₁v₁) / A₃

Substituting the values:

v₃ = (100 cm² * 0.5 m/s) / (100 cm² * 3 kg/s) = 0.1667 m/s

Therefore, the flow speed in the outlet is 0.1667 m/s.

To calculate the heating rate in the chamber, we can use the equation:

Q = m * (h₂ - h₁)

Where Q is the heating rate, m is the mass flow rate, h₁ is the enthalpy of the inlet water, and h₂ is the enthalpy of the superheated vapor at the outlet.

Given that the mass flow rate is 3 kg/s (the sum of the two inlets), we can calculate the heating rate:

Q = 3 kg/s * (h₂ - h₁)

However, to calculate the enthalpy difference, we need to know the specific enthalpy values at the given temperatures and pressures. Without this information, we cannot determine the exact value of the heating rate.