In a production facility, 1.6-in-thick 2-ft × 2-ft square brass plates (rho = 532.5 lbm/ft3 and cp = 0.091 Btu/lbm·°F) that are

Answer:

Volume change percentage is 2.60%

Water level increase is 4.138 mm

Explanation:

Provided data

Water volume V = 500 L

Initial temperature T1 = 20°C

Final temperature T2 = 80°C

Diameter of the vat = 2 m

Objective

We aim to determine percentage change in volume and the rise in water level.

Solution

We will apply the bulk modulus equation, which relates the change in pressure to the change in volume.

It can similarly relate to density changes.

Thus,

E = ................1

And ............2

Here, ρ denotes density. The density at 20°C = 998 kg/m³.

The density at 80°C = 972 kg/m³.

Plugging in these values into equation 2 gives

dV = 0.0130 m³

Therefore, the percentage change in volume will be

dV % =  × 100

dV % =  × 100

dV % = 2.60 %

Hence, the percentage change in volume is 2.60%

Initial volume v1 = ................3

Final volume v2 = ................4

From equations 3 and 4, subtract v1 from v2.

v2 - v1 =  

dV =

Substituting all values yields

0.0130 =

Thus, dl = 0.004138 m.

Consequently, the water level rises by 4.138 mm.

Response:

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Clarification:

Answer:

Time constant = 15.34 seconds

The thermometer indicates an error margin of 0.838°

Explanation:

Given

t = 1 minute = 60 seconds

c(t) = 98% = 0.98

According to the details provided, the thermometer functions as a first-order system.

The transfer function of such a system is expressed as;

C(s)/R(s) = 1/(sT + 1).

To determine the time constant, the step response must be evaluated.

This is defined as

r(t) = u(t) --- Applying Laplace Transformation

R(s) = 1/s

Replacing 1/s back into C(s)/R(s) = 1/(sT + 1).

What we have

C(s)/1/s = 1/(sT + 1)

C(s) = 1/(sT + 1) * 1/s

C(s) = 1/s - 1/(s + 1/T) --- Taking Inverse Laplace Transformation

L^-1(C(s)) = L^-1(1/s - 1/(s + 1/T))

Given that e^-t <–> 1/(s + 1) --- {L}

1 <–> 1/s {L}

Thus, the unit response c(t) = 1 - e^-(t/T)

Substituting 0.98 for c(t) and 60 for t

0.98 = 1 - e^-(60/T)

0.98 - 1 = - e^-(60/T)

-0.02 = - e^-(60/T)

e^-(60/T) = 0.02

ln(e^-(60/T)) = ln(0.02)

-60/T = -3.912

T = -60/-3.912

T = 15.34 seconds

It follows that the time constant = 15.34 seconds

The error signal is characterized by

E(s) = R(s) - C(s)

Where the temperature shifts at 10°/min; which equals 10°/60 s = 1/6

Thus,

E(s) = R(s) - 1/6 C(s)

Calculating C(s)

C(s) = 1/s - 1/(s + 1/T)

C(s) = 1/s - 1/(s + 1/15.34)

Note that R(s) = 1/s

Thus, E(s) turns into

E(s) = 1/s - 1/6(1/s - 1/(s + 1/15.34))

E(s) = 1/s - 1/6(1/s - 1/(s + 0.0652)

E(s) = 1/s - 1/6s + 1/(6(s+0.0652))

E(s) = 5/6s + 1/(6(s+0.0652))

E(s) = 0.833/s + 1/(6(s+0.0652)) ---- Taking Inverse Laplace Transformation

e(t) = 1/6e^-0.652t + 0.833

In a first-order system, a steady state condition is reached when the time is four times the time constant.

Thus,

Time = 4 * 15.34

Time = 61.36 seconds

Consequently, e(t) becomes

e(t) = 1/6e^-0.652t + 0.833

e(t) = 1/(6e^-0.652(61.36)) + 0.833

e(t) = 0.83821342824942664566211

e(t) = 0.838 --- Rounded off

Thus, the thermometer reveals an error of 0.838°

Answer:

The duration is 17.43 minutes.

Explanation:

Based on the provided information, the initial diameter is 5 m

the velocity is 3 m/s

and the final diameter is 17 m.

To find the solution, we will use the volume change equation expressed as

ΔV = .............1

where ΔV represents the change in volume, rf is the final radius, and ri is the initial radius.

Calculating ΔV yields

ΔV =

ΔV = 2507 m³.

Thus,

Q = velocity × Area

Q = 3 × π ×(0.5)² = 2.356 m³/s.

Next, the change in time can be expressed as

Δt =

Δt =

Δt = 1046 seconds.

Therefore, the total change in time amounts to 17.43 minutes.