Thermodynamics Test - 4

According to the first law of thermodynamics:

When a system undergoes a thermodynamic cycle then the net heat supplied to the system from the surrounding is equal to net work done by the system on its surroundings.

\(\oint dQ = \oint dW\)

Application of first law to a closed system process:

(i) For a closed system:

δQ = δW + dU

a) For an isolated system:

δQ = 0, δW = 0 ⇒ dU = 0

b) For reversible constant volume process:

dV = 0 ⇒ δW = 0

δQ = dU = C_vdT

c) For reversible constant pressure:

δW = P(v₂ – v₁)

δQ = δW + dU = P(v₂ – v₁) + (u₂ – u₁)

δQ = (Pv₂ + u₂) – (Pv₁ + u₁) = h₂ – h₁ = C_pdT

d) For reversible constant temperature process:

dT = 0 ⇒ dU = 0

\(\delta Q = \delta W = {p_1}{v_1}\ln \frac{{{v_2}}}{{{v_1}}}\)

e) For reversible adiabatic process:

δQ = 0

δW = dU

ii) For steady flow process (open system)

\(m\left( {{h_1} + \frac{{v_1^2}}{2} + {z_1}g} \right) + Q = m\left( {{h_2} + \frac{{v_2^2}}{2} + {z_2}g} \right) + W\)

For adiabatic process (δQ = 0) with negligible change in kinetic and potential energies (z₁ = z₂, v₁ = v₂)

mh₁ = mh₂ + W

W = m (h₁ - h₂) = -m (h₂ - h₁) = -ΔH

Explanation:

Given:

P₁ = 0.5 MPa, V₁ = 0.287 m³/kg, P₂ = 0.1 MPa, V₂ = 1.435 m³/kg

Now,

\({P_1}V_1^n = {P_2}V_2^n\)

\(n\log \left( {\frac{{{V_1}}}{{{V_2}}}} \right) = \ln \left( {\frac{{{P_2}}}{{{P_1}}}} \right)\)

\(n = \frac{{\ln \left( {\frac{{{P_2}}}{{{P_1}}}} \right)}}{{\ln \left( {\frac{{{V_1}}}{{{V_2}}}} \right)}} = \frac{{\ln \left( {\frac{{0.1}}{{0.5}}} \right)}}{{\ln \left( {\frac{{0.287}}{{1.435}}} \right)}}\)

n = 1

∴ PV = constant ⇒ Isothermal process

Concept:

Decrease in availability \( = {T_0}\left[ {\frac{Q}{{{T_2}}} - \frac{Q}{{{T_1}}}} \right]\)

Calculation:

Given:

T₀ = 300 K, Q = 1800 kJ, T₂ = 450 K, T₁ = 600 K

Now,

Decrease in availability \( = 300\left[ {\frac{{1800}}{{450}} - \frac{{1800}}{{600}}} \right]\)

∴ Decrease in availability = 300 kJ

Concept:

For units calculation

\(\frac{a}{{{\nu }^{2}}}=P=\frac{Force}{Area}=\frac{Mass\times acceleration}{Area}\)

\(\frac{a}{{{\left( \frac{{{m}^{3}}}{kg} \right)}^{2}}}=\frac{kg\times \frac{m}{{{s}^{2}}}}{{{m}^{2}}}\)

\(a = \frac{{{m^5}}}{{kg{{\sec }^2}}}\)

\(b = v = \frac{{{m^3}}}{{kg}}\)

a accounts for Intermolecular forces

Concept:

During isothermal expansion, T remains constant and the work done by the gas is given by

\(dW = {P_1}{V_1}\ln \frac{{{V_2}}}{{{V_1}}} = {P_1}{V_1}\ln \frac{{{P_1}}}{{{P_2}}}\)

From 1^st law,

dQ = dU + dW ; T = constant ⇒ dU = 0;

⇒ dQ = dW

Calculation:

Given:

P₁ = 100 kPa; T₁ = 300 K; V₁ = 0.5 m³; P2 = 50 kPa;

The work done by the gas is

\(W = 100 \times {10^3} \times 0.5\ln \frac{{100}}{{50}} = 34.65\;kJ\;\)(Option 1)

Heat added to the gas is

Q = W = 34.65 kJ (Option 2) 

Explanation:

du = C_v dT

Now,

Constant volume process,

∴ \(\frac{{{P_1}}}{{{T_1}}} = \frac{{{P_2}}}{{{T_2}}}\)

\(\frac{{300}}{{300}} = \frac{{330}}{{{T_2}}}\)

∴ T₂ = 330 K

Now,

du = C_v dT = 21 × (330 - 300)

∴ du = 630 J/mol

Explanation:

Given:

P₁ = 1 bar; T₁ = 25°C = 298 k

P₂ = 2 bar; T₂ = 105°C = 278 k

Now,

\(ds = {C_p}\ln \left[ {\frac{{{T_2}}}{{{T_1}}}} \right] - R\ln \left[ {\frac{{{P_2}}}{{{P_1}}}} \right]\)

\(ds = 1.005\ln \left[ {\frac{{378}}{{298}}} \right] - 0.287\ln \left[ {\frac{2}{1}} \right]\)

∴ ds = 0.04 kJ/kg.K

As ds > 0, hence, process is possible and irreversible.

Explanation:

dQ = dU + PdV

Tds = dU + PdV

∴ dU = Tds – P dV  - - - (1)

dh = Tds + Vdp - - - (2)

Now,

Since a is Helmholtz function

∴ a = u – Ts

∴ da = du – Tds – s dT

Now, Substituting equation 1

∴ da = -Pdv – sdT

Now,

Since g is Gibbs function

∴ g = h – Ts

∴ dg = dh – Tds – sdT

Now, Substituting  equation 2

∴ dg = Vdp – sdT

Concept:

R_e (equivalent characteristic gas constant)

p₁v = m₁R₁T, p₂v = m₂RT      ---- on adding

(p₁ + p₂ + …) V = (∑ mR) T      ----(1)

If mixture is considered as single gas

p_t × v = (∑m) R_e T  ----(2)

On Comparing (1) & (2)

\({R_e} = \frac{{\sum \left( {mR} \right)}}{{\sum m}}\)

As \(U = {U_1} + {U_2} + \ldots \)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)\(\left( {\sum m} \right){C_{ve}}\;T = {m_1}{C_{v1}}\;{T_1} + {m_2}\;{C_{v2}}\;T + \ldots \)

\({C_{ve}} = \frac{{\sum m{C_v}}}{{\sum m}}\)

Similarly,

H = H₁ + H₂ …

\(\left( {\sum m} \right){C_{pe}}\;T = {m_1}{C_{p1}}\;T + {m_2}\;{C_{p2}}\;T + \ldots \)

\({C_{pe}} = \frac{{\sum m{C_p}}}{{\sum m}}\)

Concept:

For V = const.

\(\frac{{{P_2}}}{{{P_1}}} = \frac{{{T_2}}}{{{T_1}}}\),

Entropy change during constant pressure process:

\({\rm{\Delta }}S = m{C_P}\ln \frac{{{T_2}}}{{{T_1}}}\)

Entropy change during constant volume process:

\({\rm{\Delta }}S = m{C_V}\ln \frac{{{T_2}}}{{{T_1}}} = m{C_V}\ln \frac{{{P_2}}}{{{P_1}}}\)

Calculation:

Given: P₁ = 1 bar, P₂ = 4 bar, V = const

\({C_V} = \frac{R}{{\gamma - 1}} = \frac{{287}}{{1.4 - 1}} = 717.5\;J/kg\;K\)

\({\rm{\Delta }}S = m{C_V}\ln \frac{{{P_2}}}{{{P_1}}}=1 \times 717.5 \times \ln \left( {\frac{4}{1}} \right) = 994.67\;J/K\)

Explanation:

The actual work done by a system is always less than the idealised reversible work (W_max) and the difference between the two is called the irreversibility (I) of the process.

I = W_max - W_actual

Wmax = T₀(ΔS_system) - (du)_system

Wmax = T₀(S₂ - S₁) - (u₂ - u₁)

∴ Wmax = T₀(S₂ - S₁) - C_v(T₂ - T1­)

Now,

\(\begin{array}{l} {S_2} - {S_1} = {C_p}\ln \frac{{{T_2}}}{{{T_1}}} - R\ln \frac{{{P_2}}}{{{P_1}}}\\ = 1.004 \times \ln \left( {\frac{{370}}{{300}}} \right) - \left( {1.004 - 0.716} \right)\ln \left( {\frac{{6.8}}{1}} \right) \end{array}\)

∴ S₂ - S₁ = -0.3415 kJ/kg.K

Now,

Wmax  = 293(-0.3415) - 0.715 (370 - 300)

∴ Wmax = -150.18 kJ/kg

Now,

\({W_{act}} = \frac{{{P_1}{V_1} - {P_2}{V_2}}}{{n - 1}} = \frac{{R\left( {{T_1} - {T_2}} \right)}}{{n - 1}} = \frac{{0.288\left( {300 - 370} \right)}}{{1.123 - 1}}\)

∴ W_act = -163.9 kJ/kg

Now,

I = W_max - W_act = -150.18 - (-163.9)

∴ I = 13.72 kJ/kg = 13.72 kJ (∵ m = 1 kg)

Concept:

It is an unsteady case since the properties vary with respect to time.

Apply the first law of thermodynamics

\(\dot dQ = \dot dU + \dot dW\)    - - - (1)

Calculation:

Given:

The initial temperature is T₀°C

Variation of heat removal \(\dot dQ = - {K_1}T\) (Heat rejected from the system is negative)

Variation of work input, \(\dot dW = - {K_2}T\) (Work done on the system is negative)

Now,

From (1),

\(\dot dU = \;\dot dQ - \dot dW\)

\(m{c_v}\frac{{dT}}{{dt}} = - {K_1}T + {K_2}T\)

\(\mathop \smallint \nolimits_{{T_0}}^T \frac{1}{T}dT = \;\mathop \smallint \nolimits_0^t \frac{{{K_2} - {K_1}}}{{m{C_v}}}t\)

\(\ln T = \;\;\frac{{{K_2} - {K_1}}}{{m{C_v}}}t\)

\(\therefore T = \;{T_0}{e^{\left( {\frac{{{K_2} - {K_1}}}{{m{C_v}}}} \right)t}}\)

∴ If |k2|>|k1|, then the temperature will increase exponentially.

Explanation:

\({\dot S_{in}} - {\dot S_{out}} + {\dot S_{gen}} = \left( {\frac{{ds}}{{dt}}} \right)\)

For steady flow, rate of entropy \({\rm{\;}}\left( {\frac{{ds}}{{dt}}} \right) = 0\)

\({\left( {\frac{{\dot Q}}{{{T_1}}}} \right)_{in}} - \left( {\frac{{\dot Q}}{{{T_2}}}} \right) + {\dot S_{gen}} = 0\)

\(\left( {\frac{{1500}}{{27 + 273}}} \right) - \left( {\frac{{1500}}{{{T_2}}}} \right) + 0.25 = 0\)

\(5.25 - \frac{{1500}}{{{T_2}}} = 0\)

T₂ = 285.714 k

∴ T₂ = 12.714°C

Concept:

Apply Clapeyron – Clasius equation

\(\frac{{dP}}{{dT}} = \frac{{P\;\left( {LH} \right)}}{{R{T^2}}}\)  - - - (1)

Calculation:

Given:

LH = 195 kJ/Kg, R = 0.055 kJ/kg, T₁ = 76°C and P1 = 101 kPa

T2 = ? at P2 = 202 kPa, T3= ? at P3 = 303 kPa

Now,

From (1),

\(\frac{1}{P}dP = \frac{{\left( {LH} \right)}}{{R{T^2}}}dT\)

Now,

Applying integration

\(\mathop \smallint \nolimits_{P1}^{P2} \frac{1}{P}dP = \mathop \smallint \nolimits_{T1}^{T2} \frac{{\left( {LH} \right)}}{{R{T^2}}}dT\)

\(\ln \left( {\frac{{P2}}{{P1}}} \right) = \frac{{LH}}{R}\left[ {\frac{1}{{{T_1}}} - \frac{1}{{{T_2}}}} \right]\)

Now,

T₁ = 76°C and P1 = 101 kPa

T2 = ? at P2 = 202 kPa

\(\ln \left( {\frac{{202}}{{101}}} \right) = \frac{{195}}{{0.055}}\left[ {\frac{1}{{349}} - \frac{1}{{{T_2}}}} \right]\)

∴ T₂ = 374.55 K = 101.5 °C

Now,

T₁ = 76°C and P₁ = 101 kPa

T₃= ? at P₃ =303 kPa

\(\ln \left( {\frac{{303}}{{101}}} \right) = \frac{{195}}{{0.055}}\left[ {\frac{1}{{349}} - \frac{1}{{{T_3}}}} \right]\)

T₃ = 391.318 K

∴ T3 = 118.318°C

Concept:

first law of thermodynamics

dQ = dU + dW

And,

Work done will be:

\({{W}_{1-2}}=\underset{{{v}_{1}}}{\overset{{{v}_{2}}}{\mathop \int }}\,p.dV\)

Calculation:

Now,

p = a + bV

p₁ = 170 kPa, v₁ = 0.03 m³

p₂ = 400 kPa, v₂ = 0.06 m³

170 = a + b(0.03)

400 = a + b (0.06)

Solving, we get

a = -60 and b = 7667

Now,               

Work transfer, \(~W=~\underset{{{v}_{1}}}{\overset{{{v}_{2}}}{\mathop \int }}\,\left( p \right).dV=\underset{{{v}_{1}}}{\overset{{{v}_{2}}}{\mathop \int }}\,\left( a+bV \right)\cdot dV\)

Work transfer, \(W=a\left( {{v}_{2}}-{{v}_{1}} \right)+b~\left( \frac{v_{2}^{2}-v_{1}^{2}}{2} \right)\)

\(W=\left( -60 \right)\left( 0.06-0.03 \right)+7667~\left( \frac{{{0.06}^{2}}-{{0.03}^{2}}}{2} \right)\)

∴ W = 8.55 kJ

ΔU = U₂ – U₁

ΔU = (34 + 3.15 p₂v₂) – (34 + 3.15 p₁v₁)

ΔU = 3.15 (4 × 0.06 – 1.7 × 0.03) × 100

ΔU = 59.5 kJ

Now, from first law of thermodynamics

dQ = dU + dW

dQ = 59.5 + 8.55

∴ dQ = 68.05 kJ

Remember: Work done will be positive, as work is done by the system

Explanation:

For 1 mole of gas, we require 0.7 moles of air

\({\left( {{C_p}} \right)_{air}} = \left( {\frac{\gamma }{{\gamma - 1}}} \right)\bar R = 3.5\;\bar R\)

\({\left( {{C_v}} \right)_{air}} = \frac{{\bar R}}{{\left( {\gamma - 1} \right)}} = 2.5\;\bar R\)

For freon,

\({\left( {{C_p}} \right)_{freon}} = \left( {\frac{\gamma }{{\gamma - 1}}} \right)\bar R = 11\;\bar R\)

\({\left( {{C_v}} \right)_{freon}} = \frac{{\bar R}}{{\left( {\gamma - r} \right)}} = 10\;\bar R\)

\({\left( {{C_p}} \right)_{mix}} = \frac{{0.7 \times 3.5\;\bar R + 0.3 \times 11\;\bar R}}{1} = 5.75\;\bar R\)

\({\left( {{C_v}} \right)_{mix}} = \frac{{0.7 \times 2.5\;\bar R + 0.3 \times 10\;\bar R}}{1} = 4.75\;\bar R\)

∴ \({\gamma _{mix}} = \frac{{5.75\;\bar R}}{{4.75\;\bar R}}\)

Concept:

We take the pressure cooker as the system.

This is a control volume process since mass crosses the system boundary during the process.

This is also an unsteady flow process since changes occur within the control volume.

Mass balance for unsteady flow is given by

m_in – m_out = Δm_CV

Here, there is no inlet of mass, but outlet of mass (Steam) is there 

⇒ -m_e = (m₂ – m₁)_system      --- (1)

Energy balance for unsteady flow is given by

Q – W + m_i h_i – m_e h_e = (m₂ u₂ – m₁ u₁)_system

Here W = 0; Q = Q_in; m_i = 0

⇒ Q_in = m₂u₂ – m₁u₁ + m_eh_e      --- (2)

Combining 1 and 2, final equation for Q is

Q = m₂u₂ – m₁u₁ – h_e(m₂ – m₁)      --- (3)

Calculation:

Given V = 4 L = 0.004 m³; P = 160 kPa; x₁ = 0.006; Q = 1 kW; m_e = 0.2 kg;

Since saturation conditions exist in the cooker at all times, the cooking temperature must be the saturation temperature corresponding to this pressure.

T = T_sat @ 160 kPa = 113.3 °C (Option 2 is wrong)

Initial state:

x₁ = 0.006

⇒ v₁ = v_f + x₁ (v_g – v_f) = 0.00105440 + 0.006 (1.0914 – 0.00105440) = 0.0076 m³/kg;

⇒ u₁ = u_f + x₁ (u_g – u_f) = 475.21 + 0.006 (2521.4 – 475.21) = 487.48 kJ/kg

Now

\({m_1} = \frac{V}{{{v_1}}} = \frac{{0.004}}{{0.0076}} = 0.526\;kg\;\) (Option 4)

Final stage:

Now,

Given:

m_e = 0.2 kg;

From mass balance, m₂ = m₁ – m_e = 0.526 – 0.2 = 0.326 kg

\( \Rightarrow {v_2} = \frac{V}{{{m_2}}} = \frac{{0.004}}{{0.326}} = 0.01227\;{m^3}/kg\)

⇒ 0.01227 = v_f + x₂ (v_g – v_f)

⇒ 0.01227 = 0.00105440 + x₂ (1.0914 – 0.00105440) ⇒ x₂ = 0.01 (Option 1)

Now,

u₂ = u_f + x₂ (u_g – u_f) = 475.21 + 0.01 (2521.4 – 475.21) = 495.67 kJ/kg

h_e = The enthalpy with which steam is leaving = steam leaves at a saturated vapour state

⇒ h_e = h_g @ 160 kPa = 2696 kJ/kg

From the energy balance, Q = m₂u₂ – m₁u₁ – h_e(m₂ – m₁)     

⇒ Q = 0.326(495.67) – 0.526(487.48) – 2696(0.326 – 0.526) = 444.37 kJ

Rate of heater = 1 kW i.e. 1 kJ per sec

Concept:

Dryness Fraction is given by:

 \(x = \frac{{mass\;of\;vapour}}{{mass\;of\;vapour\; + \;mass\;of\;liquid}} = \frac{{{m_v}}}{{{m_v} + {m_l}}}\)

x = 0 signifies saturated liquid.

x = 1 signifies saturated vapour.

Properties equation at midpoint (0 < x < 1)

ν = ν_f + xv_fg = ν_f + x(ν_g - ν_f)

where

ν_f = specific volume at saturated liquid

ν_g = specific volume at saturated vapour

Similarly, for all other properties like h, s & u.

Calculation:

Given:

P_mixture = P ⇒  5 bar

ν_{5 bar}  = 0.37 m³/kg

\({v_{mix}} = \frac{{Total\;Volume\;\left( {{m^3}} \right)}}{{Total\;Mass\;\left( {kg} \right)}}\)

Total mass = m_g1 + m_g2 = 2 + 1 = 3 kg

V = mν ⇒ Volume of liquid + Volume of vapour (∵ Vol = Mass × Specific Volume)

   = m_f1ν_f1 + m_g1ν_g1

V₁ = m_f1ν_f1 + m_g1ν_g1

⇒ 2 × 0.27 ⇒ 0.54 m³  (∵ m_f = 0)

V₂ = m_f2ν_f2 + m_g2ν_g2

⇒ m_g2 × { ν_f2 + x(ν_g2 - ν_f2)}  (∵ ν_f is neglected)

⇒ m_g2 × (xν_g2)

⇒ 1 × 0.8 × 0.55

⇒ 0.44 m³   

Total Volume = 0.54 + 0.44 = 0.98 m³   

\({v_{mix}} = \frac{{Total\;Volume\;\left( {{m^3}} \right)}}{{Total\;Mass\;\left( {kg} \right)}}\)

\(v_{mix}= \frac{{0.98}}{3}\)

v_mix = 0.326 m³/kg