• Author / Uploaded
• Elias Jose Akle Villareal

Citation preview

Thermodynamics Homework - April 9 of 2019 4.7 Figure P4.7 provides data for water entering and exiting a tank. At the inlet and exit of the tank, determine the mass flow rate, each in kg/s Also find the time rate of change of mass contained within the tank, in kg/s.

Solution In order to determine the mass flow, we need the volumetric flow, so: 3 Q1 = A1 V 1 = (10 × 10−3 m2 ) · (20 ms ) = 0.2 ms Q2 = A2 V 2 = ( 6 × 10−3 m2 ) · (1

m ) s

3

= 0.006 ms

Now, having the volumetric flows we need the density multiplying these, so we get the mass flow at the inlet and the outlet; from the table we have ν , but 1ν is equal to the density,thus: 1 m˙1 = ( 1ν )(Q1 ) = ( 0.1996 ) mkg3 · ( 0.2 ms ) = 1.002 3

kg s

kg 1 m m˙2 = ( 1ν )(Q2 ) = ( 1.0901×10 −3 ) m3 · (0.006 s ) = 5.5 3

kg s

m˙ i = m˙1 − m˙2 = − 4.498 kgs This negative value means that the tank is decreasing the water volume contained inside 4.124 A pressure cooker has a volume of 0.011 m3 and initially contains a two-phase liquid-vapor mixture of H 2 O at a temperature of 100 °c and a quality of 10%. As the water is heated at constant volume, the pressure rises to 2 bar and the quality becomes 18.9%. With further heating a pressure-regulating valve keeps the pressure constant in the cooker at 2 bar by allowing saturated vapor at 2 bar to escape. Neglecting kinetic and potential energy effects,

(a) determine the quality of the H 2 O at the initial onset of vapor escape (state 2) and the amount of heat transfer, in kJ, to reach this state. (b) determine the final mass in the cooker, in kg, and the additional amount of heat transfer, in kJ, if heating continues from state 2 until the final quality is 1.0 Solution (a) The energy balance for this control volume system is as follows: Q = m∆U In order to find the initial mass of the system, we must first determine the specific volume using properties at 100​o​C. 3 3 3 ν 1 = ν f + x 1 (ν g − ν f ) = 1.0435•10 −3 mkg + 0.1(1.673 − 1.0435•10 −3 ) mkg = 0.168 mkg m1 =

V ν1

=

0.011m 3 3 0.168 mkg

= 0.0654kg

Since the system is a control volume system we know that the specific volume at state 2 is the same as the specific volume at state 1. We can use this information along with properties at 2 bar to determine the quality at state 2. x2 =

3

ν 2 −ν

f

ν g −ν

f

=

(0.168−1.0605•10 −3 ) mkg

3

(0.8857−1.0605•10 −3 ) mkg

= 0.189 = 18.9%

Now to find the amount of heat transfer, we need to find the internal energy values for both the first and second state. (U​1​ @100​o​C; U​2 @2 bar) ​ kJ kJ kJ U 1 = U f + x 1 (U g − U f ) = 418.94 kg + 0.1(2506.5 − 418.94) kg = 627.7 kg U

2

= U f + x 2 (U

g

kJ kJ kJ − U f ) = 504.49 kg + 0.189(2529.5 − 504.49) kg = 887.2 kg

Enter in the values found into the energy balance equation stated above. kJ Q = m(U 2 − U 1 ) = 0.0654kg(887.2 − 627.7) kg = 16.97kJ (b) Due to the fact that the final quality is 100%, we can determine that the specific volume at the final state is equal to the specific volume of saturated vapor at 2 bar. With this value, we can determine the final mass in the cooker. ν3=ν mf =

3

g@2bar

V ν3

=

= 0.8857 mkg

0.011m 3 3 0.8857 mkg

= 0.0124kg

Finally, using an energy balance for the system for moving from the second state to the final state will help us find the heat transfer. The internal energy for the final condition and enthalpy for the final condition are both found in the table from the saturated vapor value at 2 bar.

Q = (m f U f − m 2 U 2 ) − h f (m f − m 2 ) = kJ kJ kJ (0.0124kg•2529.5 kg − 0.0654kg•887.2 kg ) − (2706.7 kg )(0.0124kg − 0.0654kg) = 116.85kJ

4.10 Data are provided for the crude oil storage tank shown in fig P4.10. The tank initially m3 contains 1000 m3 of crude oil. Oil is pumped into the tank through a pipe at a rate of 2 min and out of the tank at a velocity of 1.5 ms through another pipe having a diameter of 0.15 m. The crude oil has a specific volume of 0.0015

m3 kg

. Determine:

(a) The mass of oil in the tank, in kg, after 24 hours, and (b) the volume of oil in the tank, in m3 , at that time.

Solution The first values we need are the volumetric flow at the inlet and the outlet, the one in the inlet is given and we can calculate the one at the outlet 2

3

Q2 = A2 V 2 = (π( 0.15 ) ) · (1.5 ms ) = 0.0265 ms 2 3

⇒ (0.0265 ms )( 3600s ) = 95.425 1h

m3 h

and for the volumetric flow at the inlet we have: 3 m3 Q1 = (2 min )( 60min ) = 120 mh 1h so now with the volumetric flows we get the mass flow: 3 m˙1 = ( 1v )(Q1 ) = ( 1 m3 )(120 mh ) = 80000 kg h 0.0015 kg

m˙2 =

( 1v )(Q2 )

=(

1 3 0.0015 mkg

min at 24h = ( 80000

kg h

)(95.425 mh ) = 63616, 6 kg h 3

)(24h) = 1920000 kg

mout at 24h = (63616, 6

kg )(24h) h

= 1526798.4 kg

mdif f erence at 24h = min at 24h − mout at 24h = 1920000 kg − 1526798.4 kg = 393201, 6 kg now, having the mass change after 24 hours we only have to add the initial mass minicial = ( 1 m3 )(1000m3 ) = 666666.6 kg 0.0015 kg

mtotal at 24h = minicial + mdif f erence at 24h = (666666.6 + 393201, 6)kg = 1059868.2kg and for the volme we have: 3 v dif f erence = (v)(m) = (0.0015 mkg )(393201, 6kg) = 589.8 m3 v total = v initial + v dif f erence = 1589.8m3