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3 years ago

In an ideal gas, specific enthalpy is a function of i. Entropy ii. Temperature iii, Pressure iv. Mass

Engineering

1 answer:

Mice21 [21]3 years ago

5 0

Answer:

Temperature

Explanation:

In an ideal gas the specific enthalpy is exclusively a function of Temperature only this can be also written as h = h(T)

A gas is said be ideal gas if obeys PV= nRT law

And in a ideal gas both internal energy and specific enthalpy are a function of Temperature only. Therefore the constant volume and constant pressure specific heats Cv and Cp are also function of temperature only.

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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

3 years ago

A seamless pipe 800mm diameter contains a fluid under a pressure of 2N/mm2. If the permissible tensile stress is 100N/mm2, find

Bad White [126]

Answer:

8 mm

Explanation:

Given:

Diameter, D = 800 mm

Pressure, P = 2 N/mm²

Permissible tensile stress, σ = 100 N/mm²

Now,

for the pipes, we have the relation as:

$\sigma=\frac{\textup{PD}}{\textup{2t}}$

where, t is the thickness

on substituting the respective values, we get

$100=\frac{\textup{2\times800}}{\textup{2t}}$

t = 8 mm

Hence, the minimum thickness of pipe is 8 mm

3 0

3 years ago

What does the Intake do?

4vir4ik [10]

Answer:

It creates airflow in the engine

Explanation:

6 0

3 years ago

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A safety interlock module operates by monitoring the voltage from the

In-s [12.5K]

Answer: its an Ignition coil

8 0

3 years ago

three balls each have a mass m if a has a speed v just before a direct collision with B determine the speed of C after collision

ratelena [41]

Answer:

Vc2= V(l+e) ^2/4

Vg2= V(l-e^2)/4

Explanation:

Conservation momentum, when ball A strikes Ball B

Where,

M= Mass

V= Velocity

Ma(VA)1+ Mg(Vg)2= Ma(Va)2+ Ma(Vg)2

MV + 0= MVg2

Coefficient of restitution =

e= (Vg)2- (Va)2/(Va)1- (Vg)1

e= (Vg)2- (Va)2/ V-0

Solving equation 1 and 2 yield

(Va)2= V(l-e) /2

(Vg)2= V(l+e)/2

Conservative momentum when ball b strikes c

Mg(Vg)2+Mc(Vc)1 = Mg(Vg)3+Mc(Vc)2

=> M[V(l+e) /2] + 0 = M(Vg)3 + M(Vc) 2

Coefficient of Restitution,

e= (Vc)2 - (Vg)2/(Vg)2- (Vc)1

=> e= (Vc)2 - (Vg)2/V(l+e) /2

Solving equation 3 and 4,

Vc2= V(l+e) ^2/4

Vg2= V(l-e^2)/4

8 0

3 years ago

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