Answer:
Explanation:
Consider schedules S3, S4, and S5 below. Determine whether each schedule is strict, cascadeless, recoverable, or non-recoverable. You need to explain your reason.
S3: r1(x), r2(z), r1(z), r3(x), r3(y), w1(x), c1, w3(y), c3, r2(y), w2(z),w2(y),c2
S4: r1(x), r2(z), r1(z), r3(x), r3(y),w1(x),w3(y), r2(y),w2(z),w2(y), c1,c2, c3
S5: r1(x), r2(z), r3(x), r1(z), r2(y), r3(y), w1(x), c1, w2(z), w3(y), w2(y), c3, c2
Strict schedule:
A schedule is strict if it satisfies the following conditions:
Tj reads a data item X after Ti has written to X and Ti is terminated means aborted or committed.
Tj writes a data item X after Ti has written to X and Ti is terminated means aborted or committed.
S3 is not strict because In a strict schedule T3 must read X after C1 but here T3 reads X (r3(X)) before Then T1 has written to X (w1(X)) and T3 commits after T1.
S4 is not strict because In a strict schedule T3 must read X after C1, but here T3 reads X (r3(X)) before T1 has written to X (w1(X)) and T3 commits after T1.
S5 is not strict because T3 reads X (r3(X)) before T1 has written to X (w1(X))
but T3 commits after T1. In a strict schedule T3 must read X after C1.
Cascadeless schedule:
Cascadeless schedule follows the below condition:
Tj reads X only? after Ti has written to X and terminated means aborted or committed.
S3 is not cascadeless schedule because T3 reads X (r3(X)) before T1 commits.
S4 is not cascadeless schedule because T3 reads X (r3(X)) before T1 commits.
S5 is not cascadeless schedule because T3 reads X (r3(X)) before T1 commits or T2 reads Y (r2(Y)) before T3 commits.
But while come to the definition of cascadeless schedules S3, S4, and S4 are not cascadeless, and T3 is not affected if T1 is rolled back in any of the schedules, that is,
T3 does not have to roll back if T1 is rolled back. The problem occurs because these
schedules are not serializable.
Recoverable schedule:
Schedule that follows the below condition:
-----Tj commits after Ti if Tj has?read any data item written by Ti.
Ci > Cj means that Ci happens before Cj. Ai denotes abort Ti. To test if a schedule is
recoverable one has to include abort operations. Thus in testing the recoverability abort
operations will have to used in place of commit one at a time. Also the strictest condition is
------where a transaction neither reads nor writes to a data item, which was written to by a transaction that has not committed yet.
If A1?>C3>C2, then schedule S3 is recoverable because rolling back of T1 does not affect T2 and
T3. If C1>A3>C2. schedule S3 is not recoverable because T2 read the value of Y (r2(Y)) after T3 wrote X (w3(Y)) and T2 committed but T3 rolled back. Thus, T2 used non- existent value of Y. If C1>C3>A3, then S3 is recoverable because roll back of T2 does not affect T1 and T3.
Strictest condition of schedule S3 is C3>C2.
If A1?>C2>C3, then schedule S4 is recoverable because roll back of T1 does not affect T2 and T3. If C1>A2>C3, then schedule S4 is recoverable because the roll back of T2 will restore the value of Y that was read and written to by T3 (w3(Y)). It will not affect T1. If C1>C2>A3, then schedule S4 is not recoverable because T3 will restore the value of Y which was not read by T2.
cubical tank 1 meter on each edge is filled with water at 20 degrees C. A cubical pure copper block 0.46 meters on each edge with an initial temperature of 100 degrees C is quickly submerged in the water, causing an amount of water equal to the volume of the smaller cube to spill from the tank. An insulated cover is placed on the tank. The tank is adiabatic. Estimate the equilibrium temperature of the system (block + water). Be sure to state all applicable assumptions.
Answer:
final temperature = 26.5°
Explanation:
Initial volume of water is 1 x 1 x 1 = 1 [tex]m^{3}[/tex]
Initial temperature of water = 20° C
Density of water = 1000 kg/[tex]m^{3}[/tex]
volume of copper block = 0.46 x 0.46 x 0.46 = 0.097 [tex]m^{3}[/tex]
Initial temperature of copper block = 100° C
Density of copper = 8960 kg/[tex]m^{3}[/tex]
Final volume of water = 1 - 0.097 = 0.903 [tex]m^{3}[/tex]
Assumptions:
since tank is adiabatic, there's no heat gain or loss through the wallsthe tank is perfectly full, leaving no room for cooling airtotal heat energy within the tank will be the summation of the heat energy of the copper and the water remaining in the tank.mass of water remaining in the tank will be density x volume = 1000 x 0.903 = 903 kg
specific heat capacity of water c = 4186 J/K-kg
heat content of water left Hw = mcT = 903 x 4186 x 20 = 75.59 Mega-joules
mass of copper will be density x volume = 8960 x 0.097 = 869.12 kg
specific heat capacity of copper is 385 J/K-kg
heat content of copper Hc = mcT = 869.12 x 385 x 100 = 33.46 Mega-joules
total heat in the system = 75.59 + 33.46 = 109.05 Mega-joules
this heat will be distributed in the entire system
heat energy of water within the system = mcT
where T is the final temperature
= 903 x 4186 x T = 3779958T
for copper, heat will be
mcT = 869.12 x 385 = 334611.2T
these component heats will sum up to the final heat of the system, i.e
3779958T + 334611.2T = 109.05 x [tex]10^{6}[/tex]
4114569.2T = 109.05 x [tex]10^{6}[/tex]
final temperature T = (109.05 x [tex]10^{6}[/tex])/4114569.2 = 26.5°
An aluminium bar 600mm long with a diameter 40mm has a hole drilled in the centre of which 30mm in diameter and 100mm long if the modulus of elasticity is 85GN/M2 calculate the total contraction oon the bar due to comprehensive load of 160KN.
Answer:
Total contraction on the bar = 1.238 mm
Explanation:
Modulus of Elasticity, E = 85 GN/m²
Diameter of the aluminium bar, [tex]d_{Al} = 40 mm = 0.04 m[/tex]
Load, P = 160 kN
Cross sectional area of the aluminium bar without hole:
[tex]A_1 = \frac{\pi d_{Al}^2 }{4} \\A_1 = \frac{\pi 0.04^2 }{4}\\A_1 = 0.00126 m^2[/tex]
Diameter of hole, [tex]d_h = 30 mm = 0.03 m[/tex]
Cross sectional area of the aluminium bar with hole:
[tex]A_2 = \frac{\pi( d_{Al}^2 - d_{h}^2)}{4} \\A_2 = \frac{\pi (0.04^2 - 0.03^2) }{4}\\A_2 = 0.00055 m^2[/tex]
Length of the aluminium bar, [tex]L_{Al} = 600 mm = 0.6 m[/tex]
Length of the hole, [tex]L_h = 100mm = 0.1 m[/tex]
Contraction in aluminium bar without hole [tex]= \frac{P * L_{Al}}{A_1 E}[/tex]
Contraction in aluminium bar without hole [tex]= \frac{160*10^3 * 0.6}{0.00126 * 85 * 10^9 }[/tex]
Contraction in aluminium bar without hole = 96000/107100000
Contraction in aluminium bar without hole = 0.000896
Contraction in aluminium bar with hole [tex]= \frac{P * L_{h}}{A_2 E}[/tex]
Contraction in aluminium bar without hole [tex]= \frac{160*10^3 * 0.1}{0.00055 * 85 * 10^9 }[/tex]
Contraction in aluminium bar without hole = 16000/46750000
Contraction in aluminium bar without hole = 0.000342
Total contraction = 0.000896 + 0.000342
Total contraction = 0.001238 m = 1.238 mm