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3 years ago

10

The first law of thermodymaic stae the 5ktiopheuithkjhuiguihaoitg

1 answer:

pentagon [3]3 years ago

3 0

I'm assuming you want the first law of thermodynamics.

The First Law of Thermodynamics states that heat is a form of energy and cannot be created or destroyed. It can, however, be transferred from one location to another and can be converted into other forms of energy.

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A single insulated duct flow experiment using air operating at steady-state is performed in a lab. One measurement location (Sta

weqwewe [10]

Answer:

a) -0.0934 kJ/kg. K

b) The direction of flow is from right to left.

Explanation:

A free flow diagram of the horizontal insulated duct is as shown below.

NOW,

Let assume that the direction of flow is from left to right and consider the following relation for the entropy rate balance equation for a control volume as:

$\frac{\sigma_{cv}}{m}= (s_2-s_1) \geq 0 \ \ \ -------> \ \ \ 1$

Now; if the value for this relation is greater than zero; then we conclude that our assumption is correct.

If the value is less than zero; then we conclude that the assumption is wrong.

Then, the flow is said to be in the opposite direction

Formula for the change in specific entropy can be calculated as:

$s_2-s_1 = s^0(T_2) - s^0(T_1)-R \ In ( \frac{P_2}{P-1}) \ \ \ -------> \ \ \ 2$

where;

$s_1, s_2 , s^0(T_2), s^0(T1)$ are specific entropies

R = universal gas constant

$P_1$ = pressure at location 1

$P_2$ = pressure at location 2

We obtain the specific properties of air at temperature at $T_1$ = (67°C + 273)K = 340 K from the table A-22 ( Ideal gas properties of air)

$s^0(T1)$ = 1.8279 kJ/kg.K

We also obtain the specific properties of air at temperature $T_2$ = 22°C + 273) K = 295 K

From the table A- 22

$s^0(T_2)$ = 1.68515 kJ/kg . K

R = $\frac{8.314 kJ}{28.97 kg.K}$

$P_1 =$ 0.95 bar

$P_2 =$ 0.8 bar

Now replacing our values into equation (2) from above; we have;

$s_2-s_1 = s^0(T_2) -s^0(T_1)-R \ In (\frac{P_2}{P_1} )$

$s_2-s_1 = 1.68515 -1.8279-\frac{8.314}{28.97} \ In (\frac{0.8}{0.95} )$

$s_2-s_1 = 1.68515 -1.8279+ 0.0493$

$s_2-s_1 =-0.0934 \ kJ/kg.K$

Equating our result to equation (1)

$s_2-s_1 \geq 0\\-0.0934 \leq 0$

Therefore , our assumption is wrong and the direction of flow is said to be from right to left.

We therefore conclude that the direction of flow is from right to left.

3 0

3 years ago

A gas at 5.00 atm pressure was stored in a tank during the winter at 5.0 °C. During the summer, the temperature in the storage a

saw5 [17]

Answer:

a) 5.63 atm

Explanation:

We can use combined gas law

<em>The combined gas law</em> combines the three gas laws:

Boyle's Law, (P₁V₁ =P₂V₂)
Charles' Law (V₁/T₁ =V₂/T₂)
Gay-Lussac's Law. (P₁/T₁ =P₂/T₂)

It states that the ratio of the product of pressure and volume and the absolute temperature of a gas is equal to a constant.

P₁V₁/T₁ =P₂V₂/T₂

where P = Pressure, T = Absolute temperature, V = Volume occupied

The volume of the system remains constant,

So, P₁/T₁ =P₂/T₂

a) $\frac{5}{278} =\frac{P_2}{313} \\\\P_2=\frac{5*313}{278}\\ P_2 = 5.63 atm$

7 0

3 years ago