Explanation Consider the air flows at steady state through one inlet and one exit system (pipe and duct flow). Hence, the inlet and exit mass flow rates are equal. m ˙ 1 = m ˙ 2 = m ˙ Write the energy rate balance equation for one inlet and one outlet system. [ Q ˙ 1 + W ˙ 1 + m ˙ ( h 1 + V 1 2 2 + g z 1 ) ] − [ Q ˙ 2 + W ˙ 2 + m ˙ ( h 2 + V 2 2 2 + g z 2 ) ] = Δ E ˙ system (I) Here, the rate of heat transfer is Q ˙ , the rate of work transfer is W ˙ , the enthalpy is h and the velocity is V , the gravitational acceleration is g , the elevation from the datum is z and the rate of change in net energy of the system is Δ E ˙ system ; the suffixes 1 and 2 indicates the inlet and outlet of the system. The air flows at steady state through the system. Hence, the rate of change in net energy of the system becomes zero. Δ E ˙ system = 0 The heat is added to the air by the electronic components of the computer. Neglect the heat transfer at the outlet, work transfer, kinetic and potential energies. The Equations (I) reduced as follows. [ 0 + Q ˙ 1 + m ˙ ( h 1 + 0 + 0 ) ] − [ 0 + 0 + m ˙ ( h 2 + 0 + 0 ) ] = 0 Q ˙ 1 + m ˙ h 1 − m ˙ h 2 = 0 Q ˙ 1 = m ˙ h 2 − m ˙ h 1 Q ˙ 1 = m ˙ ( h 2 − h 1 ) (II) The change in enthalpy is expresses as follow. h 2 − h 1 = c p ( T 2 − T 1 ) Here, the specific heat is c p , the exit temperature is T 2 and the inlet temperature is T 1 . Substitute c p ( T 2 − T 1 ) for h 2 − h 1 in Equation (II). Q ˙ 1 = m ˙ [ c p ( T 2 − T 1 ) ] = m ˙ c p ( T 2 − T 1 ) (III) Rearrange the Equation (III) to obtain the mass flow rate ( m ˙ ) . m ˙ = Q ˙ 1 c p ( T 2 − T 1 ) (IV) Write the formula for volumetric flow rate of air ( V ˙ ) in terms of mass flow rate and density. V ˙ = m ˙ ρ = m ˙ R T P (V) Here, the volumetric flow rate of air is V ˙ , the pressure is P , the gas constant of air is R and the temperature is T . Write the formula for inlet volumetric flow rate ( V ˙ ) in terms of cross sectional area and velocity. V ˙ = A V (VI) Here, the cross-sectional area of the hose is A and the velocity of the air is V . Write the formula for cross-sectional area of the fan casing whose cross section is circle. A = π D 2 4 Here, the diameter of the fan casing is D . Substitute π D 2 4 for A in Equation (VI) and rearrange to obtain diameter of fan casing. V ˙ = ( π D 2 4 ) V D 2 = 4 V ˙ π V D = 4 V ˙ π V (VII) Refer Table A-1, “Molar mass, gas constant, and critical-point properties”. The gas constant ( R ) of air is 0.287 kPa .m 3 /kg ⋅ K . Refer Table A-2 (b), “Ideal-gas specific heats of various common gases”. Obtain the specific heat at constant pressure ( c p ) of air corresponding to the average temperature of 45 ° C + 60 ° C 2 = 52

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ISBN: 9781260048766

Author: CENGEL

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Work and Heat Transfer

It is generally related to thermal engineering which tells about the use, creation, conversion and transfer of heat energy between the systems. It is also considered that to transfer heat from one source to another, transfer of masses is also required. T…

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Chapter 5.5, Problem 99P

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The flow rate of the fan that needs to be installed and the diameter of the casing of the fan.

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Chapter 5.5, Problem 98P

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The flow rate of the fan that needs to be installed and the diameter of the casing of the fan.

Solution Summary: The author analyzes the flow rate of the fan that needs to be installed and the diameter of its casing. The inlet and exit mass flow rates are equal.

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