Two tanks are connected as shown in the figure, both containing water. Tank A is at 200 kpa, v A 0.5 m 3 /, V A 1 m 3, and tank B contains 3.5 at 0.5 MPa and 400 C. The valve is now opened and the two tanks come to a uniform state. Find the final specific volume. Given: P A, V A, v A, m B, P B, and T B. Find: Final specific volume v final V final V A +V B m final m A +m B m A V A 1 m3 v A 0.5 m 3 / 2 V B m B v B v B 0.61728 m 3 / (from Table B.1.3 with P 500 kpa, T 400 C) V B 3.5 ( 0.61728 m 3 / ) 2.16 m 3 v final 0.5 m3 +2.16 m 3 2 +3.5 v final 0.575 m 3 /
A boiler feed pump delivers 0.5 m 3 /s of water at 240 C, 20 MPa. What is the mass flow rate (/s)? What would the percent error be if the properties of saturated liquid at 240 Cwere used in the calculation? What if the properties of saturated liquid at 20 MPa were used instead? Given: Temperature, Pressure, and volumetric flow rate of water. Find: The mass flow rate of water and the percent error for using two different approximation methods. ṁ V v For T 240 Cand P 20,000 kpa, v 0.001205 m3/ (Table B.1.4) ṁ Using v v f T240 C 0.001229 m 3 /: 0.05 m 3 /s 0.001205 m 3 41.5 /s / ṁ Error 0.05 m 3 /s 0.001229 m 3 40.7 /s / 41.5 /s 40.7 /s 2% 41.5 /s Using v v f P20000 kpa 0.002036 m 3 /: ṁ Error 0.05 m 3 /s 0.002036 m 3 24.6 /s / 41.5 /s 24.6 /s 41% 41.5 /s
A pressure cooker (closed tank) contains water at 100 C, with the liquid volume being 1/10th of the vapor volume. It is heated until the pressure reaches 2.0 MPa. Find the final temperature. Has the final state more or less vapor than the initial state? Given: Initial temperature, initial volume fraction of water, and final Pressure Find: Final temperature and final fraction of liquid/steam Assumptions: Closed system. Constant volume system. State 1: T 1 100 Cand V L /V V 10% State 2: P 1 2000 kpa v 2 v 1 Need to find v 1 (there are many ways to do this...here s one) v 1 V 1 V L +V V V L +V V m 1 m L +m V L (V L +V V )v f v g (0.1V V +V V )v f v g 1.1v fv g V vf + VV V v L v g +V V v f 0.1V V v g +V V v f 0.1v g +v f g v f T100 C 0.001044m 3 / v g T100 C 1.6729m 3 / v 1 1.1v fv g 1.1 ( 0.001044m 3 / )( 1.6729m 3 / ) 0.1v g +v f (0.1)1.6729m 3 /+0.001044m 3 / 0.01141m3 / v 2 v 1 0.01141 m 3 / < v g P2000 kpa 0.09963 m 3 / T 2 T sat P2000 kpa 212.42 C It s not clear from the problem statement if they want to know if there is more vapor by volume or by mass. By mass: x 1 v 1 v f,1 0.01141 0.001044 0.0062 v g,1 v f,1 1.67185 x 2 v 2 v f,2 0.01141 0.001177 0.104 v g,2 v f,2 0.09845 x 2 > x 1 so more vapor in final state.
By volume: ( VL V V ( VL V V V L m Lv f (1 x)m totalv f V V m V v g xm total v f ) (1 x 2)v f,2 x 2 v g,2 ) 2 2 (1 x)v f xv g (1 0.104)0.001177 (0.104)0.9963 0.102 > 0.1 so more vapor in final state.
A 1-m 3 rigid tank with air at 1 MPa and 400 K is connected to an air line as shown in the figure. The valve is opened and air flows into the tank until the pressure reaches 5 MPa, at which point the valve is closed and the temperature inside is 450 K. a. What is the mass of air in the tank before and after the process? b. The tank eventually cools to room temperature, 300 K. What is the pressure inside the tank then? Given: Volume of tank, initial and final temperatures and pressures. Find: Initial and final mass in tank, pressure after tank cools. Assumptions: Air can be treated as an ideal gas. PV mrt R air 0.287 kj V 1 m 3 m 1 P 1V (1000 kpa)1 m3 RT 1 0.287 kj 8.71 (400 K) m 2 P 2V (5000 kpa)1 m3 RT 1 0.287 kj 38.7 (450 K) P final mrt final V (38.7 )(0.287kJ )(300 K) 1 m 3 3.33 MPa
What is the percent error in specific volume if the ideal gas model is used to represent the behavior of superheated ammonia at 40 Cand 500 kpa? What if the generalized compressibility chart, Fig. D.1, is used instead? From table B.2.2 for the given conditions the specific volume of ammonia is v 0.29227 m 3 /. Ideal Gas Law: Pv RT R ammonia 0.4882 kj/k Table A.5 v RT P Error 0.3056 0.29227 0.29227 0.4882 kj/k(313 K) 500 kpa 4.6% 0.3056 m 3 / Compressibility Chart: T c 405.5 K Table A.2 P c 11,350 kpa Table A.2 T r T T c 313 K 405.5 K 0.77 P r P P c Z 0.97 v ZRT P Error 0.296 0.29227 0.29227 500 kpa 11,350 kpa 0.044 0.97(0.4882 kj/k)313 K 500 kpa 1.4% 0.296 m 3 /
A piston/cylinder contains air at 600 kpa, 290 K and a volume of 0.01 m 3. A constant-pressure process gives 54 kj of work out. Find the final volume and temperature of the air. Given: Initial conditions, constant pressure, work done. Find: Final volume and temperature. Assumptions: Air can be treated as an ideal gas. W PdV W P(V 2 V 1 ) Only for constant pressure process. V 2 W P +V 1 V 2 54 kj 600 kpa +0.01m3 V 2 0.10 m3 PV mrt T V P } Constant mr T 1 T 2 V 1 V 2 T 2 T 1 V 2 V 1 290 K(10) 2900 K
A piston/cylinder has 5 m of liquid 20 Cwater on top of the piston (m0) with a cross-sectional are of 0.1 m 2. Air let in under the piston rises and pushes the water out over the top edge. Find the work needed to push all the water out and plot the process on a P-V diagram. Given: Water on top of piston is pushed off by air that pushes piston up. Find: Work done by air. Assumptions: P atm 101.3 kpa P P air ρ water g(h h)+p atm V V air ha h V air A P ρ water g(h V A )+P atm W PdV ( W ρ water g(h V ) A )+P atm dv W ρ water gh +P atm dv ρ water g V A dv W (ρ water gh +P atm )(V 2 V 1 ) ρ waterg VdV A W (ρ water gh +P atm )(V 2 V 1 ) ρ waterg ( V 2 2A 2 V1 2 ) ρ water 1/v f T20 C 1/0.001001m 3 / 998 /m 3 V 1 0 V 2 HA (5 m)(0.1 m 2 ) 0.5 m 3 g 9.81N/m [ W 998 ( m 3 9.81 N ) ] 1000 Pa (0.5 5 m+101.3kpa m 3 0 ) 1 kpa ) 998 m (9.81 N 3 ( (0.5 m 3 2(0.1 m 2 ) 2 0 2) ) W 62.9kJ