Solution 4

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MSE 321 Engineering Thermodynamics & Heat Transfer Assignment 4

Due Date: Oct. 11

Turn in your homework problems in order, and always write down all the steps that led you to the answer. If you do not turn in a problem, write the number and letter, e.g. 2(b), and write ‘no answer’. When substance properties are needed, use the table in the textbook. Please staple your homework.

Problem 1 In large gas-turbine power plants, air is preheated by the exhaust gases in a heat exchanger called the regenerator before it enters the combustion chamber. Air enters the regenerator at 1 MPa and 550 K at a mass flow rate of 800 kg/min. In the regenerator, heat is transferred to the air at a rate of 3200 kJ/s. Exhaust gases enter the regenerator at 140 kPa and 800 K and leave at 130 kPa and 600 K. Treating the exhaust gases as air, determine (a) the exit temperature of the air and (b) the mass flow rate of exhaust gases. Neglect the changes in kinetic and potential energy.

Solution There are no work interactions. The regenerator operates as a heat exchanger, thus, the heat transfer from the hot flow is equal to the heat transfer to the cold flow. Air is an ideal gas and for ideal gases enthalpy is a function of temperature only h = h(T ) (and not the pressure). In this solution we consider variable specific heats for air and thus values of h are borrowed from Table A-21. The gas constant for air is 0.287 kPa m3/kg K (Table A-1). The enthalpies of air are (Table A-21),

T1 = 550 K  h1 = 555.74 kJ/kg T3 = 800 K  h3 = 821.95 kJ/kg T4 = 600 K  h4 = 607.02 kJ/kg (a) Consider the air passage in the heat exchanger (regenerator) as the system (control volume). There is only one inlet and one exit, thus m 1 = m 2 = m air . The energy balance for this steady-flow system can be expressed in the rate form as,

MSE 321 (Fall 2013)

Dr. Peyman Taheri

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Q in + m 1h1 - m 2 h2 = 0 (since W =  KE =  PE = 0) Q = m (h - h ) in

air

2

1

Substituting, 3200 kJ/s = (800/60 kg/s)(h2 - 554.71 kJ/kg )

 h2 = 794.71 [ kJ/kg ]

Then by interpolation in Table A-21 we find

T2 =775.1 K (b) Treating the exhaust gases as an ideal gas, the mass flow rate of the exhaust gases is determined from the steady-flow energy relation applied only to the exhaust gases between points 3 and 4. The mass balance for exhaust gas is m 3 = m 4 = m exhaust

Q out + m 3 h3 - m 4 h4 = 0 (since W =  KE =  PE = 0) Q = m (h - h ) out

exhaust

4

3

-3200 kJ/s = m exhaust (607.02 - 821.95) kJ/kg 

m exhaust = 14.9 kg / s

Note that in the above equation minus sign for Q out is based on sign convention (heat leaving the system has negative sign).

Problem 2 A company which assembles super-computer systems has decided to reduce fan noise in its computer racks, and hence, is interested to use fans with average exit velocities between 15 m/s and 19 m/s. There is a fan in the market powered by a 0.5 hp (horse power) motor which delivers air at a rate of 85 m3/min. Do you recommend this fan for the company? Assume that temperature of air is the same at inlet and outlet of the fan, and take the density of air to be 1.18 kg/m3. The inlet velocity and the change in potential energy are negligible.

Solution There is no heat interaction, and the only work interaction is the electrical power consumed by the fan motor to accelerate the air. Air is an ideal gas with constant specific heats (since temperature of the air is assumed constant) at room temperature. The density of air is given to be  = 1.18 kg/m3. The constant pressure specific heat of air at room temperature is cP = 1.005 kJ/kgC (Table A-2).

We take the fan as an open system, because mass crosses the system boundary during the process. We note that there is only one inlet and one exit, and thus m 1 = m 2 = m air . The velocity of air leaving the fan will be maximum when all of the entire electrical energy drawn by the motor is converted to kinetic energy (neglecting friction losses in the fan). In this best possible case, no energy will be converted to thermal energy (as a result of friction) and thus the temperature change of air will be zero (T1 = T2), Then the energy balance for this steady-flow system can be expressed in the rate form as,

MSE 321 (Fall 2013)

Dr. Peyman Taheri

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=0 æ ö æ 1 1 2 Q - W + m in çç hin + Vin2 + gzin ÷÷÷ - m out çç hout + Vout + gzout çè çè ø 2 2 æ 1 2 ö÷ -WElec. = m çç Vout ÷ (since Vin » 0,  PE = 0, and h = 0) çè 2 ø÷ 0=

ö ÷÷÷ = 0 ø

Note that the temperature and thus the enthalpy remain constant. And WElec. is a negative value since it is input work and we have used the sign convention in the above energy balance equation ( -W is Elec.

positive).

m =  V = (1.18 kg/m3 )(85 m3 /min) = 100.3 [kg/min] = 1.67 [kg/s]

Solving for Vout and substituting gives,

Vout =

2(-WElec. ) m

=

2(0.5 hp) çæ 745.7 W ö÷æç1 m 2 /s 2 ö÷ ÷ = 21.1 m / s ÷ç ç ÷÷ç 1 W ÷ø÷ 1.67 kg/s çè 1 hp øè

The fan is noisy Vout > 19 m/s , and is not recommended for the company.

Problem 3 An air-conditioning system involves the mixing of cold air and warm outdoor air before the mixture is routed to the conditioned room in steady operation. Cold air enters the mixing chamber at 5°C and 105 kPa at a rate of 1.25 m3/s while warm air enters at 34°C and 105 kPa. The air leaves the room at 24°C. The ratio of the mass flow rates of the hot to cold air streams is 1.6. Using variable specific heats, determine (a) the mixture temperature at the inlet of the room and (b) the rate of heat gain of the room. Kinetic and potential energy changes are negligible, and the air-conditioning system is well insulated.

Solution There are no work interactions. The device is adiabatic and thus heat transfer is negligible. The gas constant of air is 0.287 kPa m3/kg K. The enthalpies of air are obtained from air table (Table A21) as

h1 = h@ 278 K = 278.13 kJ/kg h2 = h@ 307 K = 307.23 kJ/kg hroom = h@ 297 K = 297.18 kJ/kg

MSE 321 (Fall 2013)

Dr. Peyman Taheri

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(a) We take the mixing chamber as the system, which is a control volume since mass crosses the boundary. The mass balance for this steady-flow system is, m in - m out = m system

=0 (steady)

 m in = m out  m 1 + 1.6 m 1 = m 3 = 2.6m 1 since m 2 = 1.6m 1

and the energy balance is, m 1h1 + m 2 h2 = m 3 h3

(since Q = W =  KE =  PE = 0) h3 = (h1 + 1.6h2 ) / 2.6

Combining the two gives

m 1h1 + 1.6m 1h2 = 2.6m 1h3

Substituting,

h3 = (278.13 + 1.6 ´ 307.23) / 2.6 = 296.04 [ kJ/kg ]

or

From air table at this enthalpy, the mixture temperature is T3 = T@ h=296.04 [kJ / kg] = 295.9 K = 22.9C (b) The mass flow rates are determined as follows RT1 (0.287 kPa ⋅ m3 /kg ⋅ K)(5 + 273 K) = = 0.7599 [m 3 / kg] 105 kPa P V 1.25 m3 /s = 1.645 [kg/s] m 1 = 1 = 1 0.7599 m3 /kg m 3 = 2.6m 1 = 2.6(1.645 kg/s) = 4.277 [kg/s]

1 =

The rate of heat gain of the room is determined from Q cool = m 3 (hroom - h3 ) = (4.277 kg/s)(297.18 - 296.04) kJ/kg = 4.88 kW

MSE 321 (Fall 2013)

Dr. Peyman Taheri