Thermodynamics Test - 3

Concept:

Heat transfer for a gas at constant pressure is given as Q = C_p ∆T

Calculation:

Given W = 0.33 Q

Temp. Rise ∆T = 20 K

Q = C_p ∆T

\(\frac{W}{{0.33}} = {C_p} \times 20\) 

\({C_p} = 0.15\;W\) 

Concept:

Using 1^st Law of Thermodynamics:

dQ = du + pdv

dQ = mc_vdT + Pdv (For Ideal gas)

Work done in polytropic process:

\({{\rm{W}}_{{\rm{poly}}}} = {\rm{\;}}\frac{{{p_1}{v_1} - {p_2}{v_2}}}{{n - 1}}\)

Calculation:

\({Q_{poly}} = m{c_v}\left( {{T_2} - {T_1}} \right) + \frac{{{p_1}{v_1} - {p_2}{v_2}}}{{n - 1}}\)

\({Q_{poly}} = m\frac{R}{{\gamma - 1}}\left( {{T_2} - {T_1}} \right) + \frac{{{p_1}{v_1} - {p_2}{v_2}}}{{n - 1}}\)

\({Q_{poly}} = \frac{{{p_2}{v_2} - {p_1}{v_1}}}{{\gamma - 1}} + \frac{{{p_1}{v_1} - {p_2}{v_2}}}{{n - 1}}\)

\({Q_{poly}} = \frac{{{p_1}{v_1} - {p_2}{v_2}}}{{n - 1}}\left( {1 - \frac{{\left( {n - 1} \right)}}{{\left( {\gamma - 1} \right)}}} \right)\)

\(\therefore Q = W\left( {\frac{{\gamma - n}}{{\gamma - 1}}} \right)\)

Concept:

First law of thermodynamics for a process is given by

δQ = dU + δW

i.e. For a process, the heat interaction is equal to the sum of the change in internal energy of the system and work interaction by the system.

Since the energy is a property therefore the change in energy is only dependent upon the end states, i.e. it does not depend upon the path followed to reach the system from one state to another.

Calculation:

When the system is taken from state ‘x’ to state ‘y’ then

Q_x-y = + 30 kJ, W_x-y= +10 kJ,

Then from first law of thermodynamics,

 δQ = dU + δW

+30 kJ = du_x-y + 10 kJ

∴ du_x-y = 30 – 10

∴ dux-y = 20 kJ

Now,

when the system is taken from state ‘y’ to state ‘x’ then

W_y-x = - 8 kJ and

dU_y-x = - 20 kJ, ∵ (U_y – U_x) = - (U_x – U_y)

Then from first law of thermodynamics

δQ = dU + δW

Q_y-x = -20 – 8

∴Q y-x= -28 kJ

Here negative sign indicates that heat is being liberated.

Concept:

The first law of thermodynamics applied to a closed system is:

Q = ∆U + W

For a rigid vessel, the volume will not change and the displacement work p∆V will be 0

So, the first law reduces to:

Q=∆U

For an ideal gas the change in internal energy can be expressed as ∆U = mc_v (T₂ - T₁)

Calculation:

Given: P = 150 kPa, T = 323 K, V = 0.05 m³, c_v = 0.7163 kJ/kg.K, R = 0.287 kJ/kgK, Q = 30 kJ

The mass of air in the vessel will be:

\(=\frac{PV}{RT}=\frac{150\times 0.05}{0.287\times 323}=0.0809~kg\)

Equating the two expressions:

Q = ∆U = mc_v (T₂ - T₁)

30 = 0.0809 × 0.7163 × [T₂ – 323]

Concept:

For the polytropic process

P1V1n = P2V2n

Calculation:

From the relation:

\(\begin{array}{l} \frac{{{P_1}{V_1}}}{{{T_1}}} = \frac{{{P_2}{V_2}}}{{{T_2}}}\\ \therefore \frac{{150 \times 0.5}}{{293}} = \frac{{400 \times {V_2}}}{{413}}\\ \Rightarrow {V_2} = 0.264\;{m^3} \end{array}\)

Now

For polytropic index

P₁V₁ⁿ = P₂V₂ⁿ

\(\Rightarrow \frac{{{P_1}}}{{{P_2}}} = {\left( {\frac{{{V_2}}}{{{V_1}}}} \right)^n} \Rightarrow \frac{{150}}{{400}} = {\left( {\frac{{0.264}}{{0.5}}} \right)^n}\)

∴ n = 1.536

Also,

P₁V_1­ = mRT₁

\(\Rightarrow m = \frac{{{P_1}{V_1}}}{{R{T_1}}} = \frac{{150 \times 0.5}}{{2.0769 \times 293}} = 0.123\;kg\)

Work done in a polytropic process

\(W = \frac{{{P_1}{V_1} - {P_2}{V_2}}}{{n - 1}} = \frac{{150 \times 0.5 - 400 \times 0.264}}{{1.536 - 1}}\)

W = -57.09 kJ

Now,

From First Law of Thermodynamics

Q = ΔU + W

Q = mC_V (T₂ – T₁) + W

Q = 0.123 × 3.1156 (140-20) – 57.09

Concept:

Using Steady Flow Energy Equation (SFEE).

\(\dot m\left( {{h_A} + \frac{{V_A^2}}{{2000}} + g{z_A}} \right) + \omega = \dot m\left( {{h_B} + \frac{{V_B^2}}{{2000}} + g{z_B}} \right) + \dot m\left( Q \right)\)

Calculation:

Given:

V₁ = 20 m/s, V₂ = 10 m/s

T₁ = 500 K, T₂ = 100 K                                        

\(Q + {h_1} + \frac{{V_1^2}}{{2000}} + g{z_1} = {h_2} + \frac{{V_2^2}}{{2000}} + g{z_2} + W\)

\(\therefore {h_1} + \frac{{V_1^2}}{{2000}} = {h_2} + \frac{{V_2^2}}{{2000}} + W\)

\(({h_1} - {h_2}) + \frac{{V_1^2}}{{2000}} - \frac{{V_2^2}}{{2000}} + W\)

\(\therefore {\rm{\;}}{{\rm{C}}_{\rm{p}}}\left( {{{\rm{T}}_2}{\rm{\;}}-{\rm{\;}}{{\rm{T}}_1}} \right) + {\rm{\;}}\frac{{{{20}^2}}}{{2000}} - \frac{{{{10}^2}}}{{2000}} = W\)

∴ 1.005 × (500 – 100) + 0.15 = W

∴ W = 402.15 kJ/kg