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Mamont248 [21]
2 years ago
3

Consider schedules S3, S4, and S5 below. Determine whether each schedule is

Engineering
1 answer:
V125BC [204]2 years ago
6 0

Answer:

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.

Explanation:

See attached image

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10. A maker of micromechanical systems can reduce product recalls by 10% with the installation of new packaging equipment. If th
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Explanation:

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Who want to play "1v1 lol unblocked games 76"
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A heat pump cycle operating at steady state receives energy by heat transfer from well water at 10oC and discharges energy by he
Marta_Voda [28]

Answer: the total amount of energy transferred by work into the heat pump over the 14-day period is 5.8GJ

Explanation: since the meter reading for the 14 days period is 1620kW.h

Electrical power Ep is

Ep = 1620*10^3W.h

In 14 days there are 14*24 hrs = 336hrs

Ep = (1620*10^3)/336 = 4821.42W

This means 4821.42J of work is done per sec. Therefore for 14 days we have,

3600*24*14*4821.42J of enegy

= 5831989632J = 5.8GJ

Answer 2 = the amount of energy that the heat pump receives by heat transfer from the well water over the 14-day period is 41.2GJ

Explanation: rate of energy transfer as heat to building is 1.4*10^5 kJ/hr

That is 1.4*10^8 J/hr

In 14 days it becomes

24*14*1.4*10^8 = 47GJ

Energy received from well by heat pump =

Energy transfered - energy received

= 47 - 5.8 =41.2GJ

Answer 3. the heat pump’s coefficient of performance is 0.12

Explanation: COP = energy obtained/ work done

COP = 5.8/47 = 0.12

8 0
2 years ago
Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-m
kodGreya [7K]

Complete question is;

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100 mm thick plate (ρ = 7830 kg/m3, Cp = 550 J/kg K, k = 48 W/m K). The plate initially is at 200 °C and is to be heated to a minimum temperature of 550 °C. Heating is effected in a gas-fired furnace where the products of combustion at T∞ = 800 °C maintain a convection heat transfer coefficient of h = 250 W/m.K on both surfaces of the plate. How long should the plate be left in the furnace?

Answer:

860 seconds

Explanation:

We are given;

Initial Temperature; Ti = 200 °C

Minimum Temperature; T_i = 550 °C

T∞ = 800 °C

convection coefficient; h = 250 W/m².K

ρ = 7830 kg/m³

Cp = 550 J/kg K

k = 48 W/m K

Plate thickness = 100mm

Thus,L = 100/2 = 50mm = 0.05 m

Let's find the biot number from the formula;

Bi = hL/K

Bi = (250 × 0.05)/48

Bi = 0.2604

Now, lowest temperature in the slab is given as;

θ_o = (T_o - T∞)/(T_i - T∞)

θ_o = (550 - 800)/(200 - 800)

θ_o = 0.4167

Now, from online tables calculation, we can find the root of the biot number.

Thus, root of the biot number Bi = 0.2604 is;

ζ1 = 0.488 rad

Also, C1 is gotten as 1.0396

Now,formula for thermal diffusivity is;

α = k/ρc

α = 48/(7830 × 550)

α = 1.115 × 10^(-5) m²/s

Also, from online tables, f(ζ1) = 0.401

Thus, we can find the time the plate should the plate be left in the furnace from;

-(ζ1)²(αt/L²) = In 0.401

-(ζ1)²(αt/L²) = -0.9138

t = (-0.9138 × 0.05²)/-(0.488² × 1.115 × 10^(-5))

t ≈ 860 s

8 0
2 years ago
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