327,034 views
11 votes
11 votes
Air flows through a pipe at a rate of 200 L/s. The pipe consists of two sections of diameters 20 cm and 10 cm with a smooth reducing section that connects them. The pressure difference between the two pipe sections is measured by a water manometer. Neglecting frictional effects, if the inlet temperature of air is at 25oC, due to the compression of the air at neck of the pipe contraction, the outlet temperature is measured at 30oC, determine the differential height of water between the two pipe sections. Take the air density at 25oC and 35oC as 1.20 kg/m3 and 1.19 kg/m3 respectively.​

User Berry Tsakala
by
2.6k points

1 Answer

21 votes
21 votes

Answer:

To determine the differential height of water between the two sections of the pipe, we need to consider the conservation of energy and the ideal gas law.Conservation of energy:

The first law of thermodynamics states that the increase in internal energy of a system is equal to the heat added to the system, minus the work done by the system. In this case, the internal energy of the air flowing through the pipe increases as it is compressed in the neck of the pipe, causing its temperature to increase.The increase in internal energy can be expressed as:ΔU = Q - Wwhere ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system.Ideal gas law:

The ideal gas law states that the pressure of an ideal gas is directly proportional to its temperature and inversely proportional to its volume. This can be expressed as:PV = nRTwhere P is the pressure, V is the volume, n is the number of moles of gas, R is the universal gas constant, and T is the temperature.Solution:

To determine the differential height of water between the two sections of the pipe, we need to use the conservation of energy and the ideal gas law to relate the pressure difference between the two sections to the temperature difference and the flow rate of the air.First, let's convert the flow rate of the air from liters per second to cubic meters per second:200 L/s = 0.2 m3/sNext, we can use the ideal gas law to express the pressure difference between the two sections in terms of the temperature difference and the flow rate:ΔP = (nRT1 - nRT2)/V = (1.20 kg/m3 * 8.31 J/(molK) * 298 K - 1.19 kg/m3 * 8.31 J/(molK) * 303 K)/(0.2 m3/s)ΔP = -27.7 PaFinally, we can use the conservation of energy to express the differential height of water in terms of the pressure difference and the flow rate:Δh = ΔP/ρg = (-27.7 Pa)/(1000 kg/m3 * 9.81 m/s2) = -0.28 mTherefore, the differential height of water between the two sections of the pipe is -0.28 m.Note that the negative sign indicates that the pressure is lower in the second section of the pipe, which is consistent with the fact that the temperature is higher and the volume is lower in this section.

Step-by-step explanation:

User Shabeer K
by
3.1k points