The reading on a mercury manometer at 20 degree C that is open to the atmosphere

Question

The reading on a mercury manometer at 20 degree C that is open to the atmosphere at on end is 51, 46 cm. The local acceleration of gravity is 9.832 m/s^2. Atmospheric pressure is 101.78 kPa. What is the absolute pressure in kPa being measured? The density of mercury at 20 degree C is 12.384 g/cm^3. A stream of warm water is produced in a steady-flow mixing process by combining 2.5 kg/s of cool water at 25 degree C with 1.1 kg/s of hot water at 85 degree C. During mixing, heat is lost to the surroundings at the rate of 42 kJ/s. What is the temperature of the warm-water stream? Assume the specific heat of water constant at 4.184 kJ/(kg * K). One mole of an ideal gas at 1 bar and 20 degree C undergoes a mechanically reversible process: heating at constant volume to 90 degree C followed by isothermal compression to 6 bar. Calculate the Q, W, Delta U, and Delta H of the ideal gas. The heat capacities for are C_v=20.78 and C_p= 29.10 J mol^-1 K^-1.

Explanation / Answer

1) Absolute pressute = Manometer pressure + atmospheric pressure

Manometer pressure = hdg

h = height = 51.46 cm = 0.5146 m

d = density = 12.384 g/cm3 = 12384 kg/m3

g = acceleration due to gravity = 9.832 m/s2

Hence,

Manometer pressure = hdg

                              = 0.5146 m x 12384 kg/m3 x 9.832 m/s2

                              = 62657 Pa

                              = 62.657 kPa

Therefore,

Absolute pressute = Manometer pressure + atmospheric pressure

                           = 62.657 kPa + 101.78 kPa

                           = 164.437 kPa

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