Easy POINTS computer genius help me plz

Anvisha [2.4K]

Erm what? ......;-;

7 0

3 years ago

Consider the expansion of a gas at a constant temperature in a water-cooled piston-cylinder system. The constant temperature is

Leona [35]

Answer:

$Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})$

Explanation:

According to the first thermodynamic law, the energy must be conserved so:

$dQ = dU - dW$

Where Q is the heat transmitted to the system, U is the internal energy and W is the work done by the system.

This equation can be solved by integration between an initial and a final state:

(1) $\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU - \int\limits^1_2 {} \, dW$

As per work definition:

$dW = F*dr$

For pressure the force F equials the pressure multiplied by the area of the piston, and considering dx as the displacement:

$dW = PA*dx$

Here A*dx equals the differential volume of the piston, and considering that any increment in volume is a work done by the system, the sign is negative, so:

$dW = - P*dV$

So the third integral in equation (1) is:

$\int\limits^1_2 {- P} \, dV$

Considering the gas as ideal, the pressure can be calculated as $P = \frac{n*R*T}{V}$ , so:

$\int\limits^1_2 {- P} \, dV = \int\limits^1_2 {- \frac{n*R*T}{V}} \, dV$

In this particular case as the systems is closed and the temperature constant, n, R and T are constants:

$\int\limits^1_2 {- \frac{n*R*T}{V}} \, dV = -nRT \int\limits^1_2 {\frac{1}{V}} \, dV$

Replacion this and solving equation (1) between state 1 and 2:

$\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU + nRT \int\limits^1_2 {\frac{1}{V}} \, dV$

$Q_{2} - Q_{1} = U_{2} - U_{1} + nRT(ln V_{2} - ln V_{1})$

$Q_{2} - Q_{1} = U_{2} - U_{1} + nRT ln \frac{V_{2}}{V_{1}}$

The internal energy depends only on the temperature of the gas, so there is no internal energy change $U_{2} - U_{1} = 0$ , so the heat exchanged to the system equals the work done by the system:

$Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})$

4 0

4 years ago

Water vapor at 5 bar, 320°C enters a turbine operating at steady state with a volumetric flow rate of 0.65 m3/s and expands adia

harina [27]

Answer:

Power = 371.28 kW

Explanation:

Initial pressure, P1 = 5 bar

Final pressure, P2 = 1 bar

Initial temperature, T1 = 320°C

Final temperature, T2 = 160°C

Volume flow rate, V = 0.65m³/s

From steam tables at state 1,

h1 = 3105.6 kJ/kg, s1 = 7.5308 kJ/kgK

v1 = 0.5416 m³/kg

Mass flow rate, m = V/v1

m = 1.2 kg/s

From steam tables, at state 2

h2 = 2796.2 kJ/kg, s2 = 7.6597 kJ/kgK

Power developed, P = m(h1 - h2)

P = 1.2(3105.6-2796.2)

P = 371.28 kW

8 0

3 years ago

When do design engineers start on the design improvement step?

ArbitrLikvidat [17]

Answer:

as soon as there is a design to improve

Explanation:

As a design engineer, I started on the "design improvement" step as soon as I had an initial conceptual design.

__

Then, I started that step again when my boss told me, "make it better."

_____

The more interesting question is, "when do you <em>stop</em> the design improvement step?" (Judging by the constant barrage of software updates, that answer is, "never.")

8 0

3 years ago

10% A steel beam W18x76 spans 32 feet and is subjected to a Moment of 334 kips-ft. Find the load w on the beam. Determine the de

Lilit [14]

Answer:

w = 10.437 kips

deflection at 1/4 span 20.83\E ft

at mid span = 1.23\E ft

shear stress 7.3629 psi

Explanation:

area of cross section = 18*76

length of span = 32 ft