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Kay [80]
3 years ago
15

A 0.50 m3 drum was filled with 0.49 m3 of liquid water at 25oC and the remaining volume was water vapor without any air. The dru

m was then sealed and heated to 140oC. What happened to this system
Engineering
1 answer:
timurjin [86]3 years ago
5 0

Answer:

There is not going to be pressure build up in the  system,that is isobaric process.

Explanation:

Assumptions to be made

1.  No mass is gained or lost during the heating process.

2. There are no friction losses,so work is transmitted efficiently.

3. It was started the water in the drum and its surrounding have same temperature.

4. This system is closed,so there is no mass transfer across its boundaries.

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Houses the CYLINDERS, Water Jacket & Crankcase
8 0
2 years ago
A single fixed pulley is used to lift a load of 400N by the application of an effort of 480N in 10s through a vertical height of
Allushta [10]

Answer:

(a) the velocity ratio of the machine (V.R) = 1

(b) The mechanical advantage of the machine (M.A) = 0.833

(c) The efficiency of the machine (E) = 83.3 %

Explanation:

Given;

load lifted by the pulley, L = 400 N

effort applied in lifting the, E = 480 N

distance moved by the effort, d = 5 m

(a) the velocity ratio of the machine (V.R);

since the effort applied moved downwards through a distance of d, the load will also move upwards through an equal distance 'd'.

V.R = distance moved by effort / distance moved by the load

V.R = 5/5 = 1

(b) The mechanical advantage of the machine (M.A);

M.A = L/E

M.A = 400 / 480

M.A = 0.833

(c) The efficiency of the machine (E);

E = \frac{M.A}{V.R} \times 100\%\\\\E = 0.833 \ \times \ 100\%\\\\ E = 83.3 \ \%

4 0
2 years ago
A closed, rigid tank is filled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. If the gas
ch4aika [34]

Answer:

gauge pressure is 133 kPa

Explanation:

given data

initial temperature T1 = 27°C = 300 K

gauge pressure = 300 kPa = 300 × 10³ Pa

atmospheric pressure = 1 atm

final temperature T2 = 77°C = 350 K

to find out

final pressure

solution

we know that gauge pressure is = absolute pressure - atmospheric pressure so

P (gauge ) = 300 × 10³ Pa - 1 × 10^{5} Pa

P (gauge ) = 2 × 10^{5} Pa

so from idea gas equation

\frac{P1*V1}{T1} = \frac{P2*V2}{T2}   ................1

so {P2} = \frac{P1*T2}{T1}

{P2} = \frac{2*10^5*350}{300}

P2 = 2.33 × 10^{5} Pa

so gauge pressure = absolute pressure - atmospheric pressure

gauge pressure = 2.33 × 10^{5}  - 1.0 × 10^{5}

gauge pressure = 1.33 × 10^{5} Pa

so gauge pressure is 133 kPa

4 0
3 years ago
Basic Question please help
Dvinal [7]
1(A)
2(B)
3(E)
4(C)
5(D)
6(B)
7(C)
7 0
2 years ago
Section BreakProblem 08.048 2.value: 5.00 pointsRequired information Problem 08.048.b Compute the expected error. The expected e
Lady_Fox [76]

Answer:

a) 19 or select the closest answer

b) 5%

Explanation:

a)

from the voltage divide rule

V_{in} = V_0 * \frac{R_2}{R_2 + R_f}

\frac{V_0}{V_{in}} = \frac{R_2 + R_f}{R_2} => 1 + \frac{R_f}{R_2} = 20

\frac{R_f}{R_2} = 19

Select the nearest answer

b)

obtained gain = \frac{V_0}{V_{in}} = 1 + \frac{36}{2} = 19

Expected gain = \frac{V_0}{V_{in}} = 20

∴ error = |\frac{19 - 20}{20}| × 100

            = 1/20 × 100            

            = 5%

6 0
3 years ago
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