Thermodynamics Test - 2

Concept:

Thermal efficiency of heat engine

\({\eta _{thermal}} = \frac{{Power\;Output}}{{Heat\;Input}}\)

Calculation:

Given, Heat input = 2500 kJ/min = 41.67 kW, Power output = 12.4 kW

Concept:

Entropy change of air is given by,

\(\Delta s = m\;\left[ {{c_v}\ln \left( {\frac{{{T_2}}}{{{T_1}}}} \right) + R\ln \left( {\frac{{{V_2}}}{{{V_1}}}} \right)} \right]\)

Since, we are not given specific entropy.

We need to calculate mass of air first using

PV = mRT

m = PV/RT = 60 x 5/ (0.287 × 300) = 3.484 kg

Calculation:

Using entropy change = 6.4 kJ/K

\(6.4 = 3.484\;\left[ {0.718\; \times \ln \left( {\frac{{400}}{{300}}} \right)\; + 0.287\; \times \ln \left( {\frac{{{V_2}}}{5}} \right)} \right]\)

V₂ = 1466.04 m³   

Concept:

The electric bulb will dissipate the heat into the room since room is insulated it will rise the temperature of the room.

If the room is considered to be filled with the ideal gas then the dissipated heat into the room will increase the internal energy of the gas.

Change in Internal energy of ideal gas is given by m × C_v × dT

m = mass of the air = ρ × V, ρ of air at atmospheric temperature and pressure = 1.2 kg/m³

C_v (specific heat at constant volume) of ideal gas at atmospheric condition is 0.718 kJ/kg. K

Calculation:

Given, volume of the room = 3 × 3 × 2.5 = 22.5 m³, initial temperature of the room = 20°C

The amount of heat rejected by the bulb into the room = \(100\frac{J}{s}\times 24~\times 3600~s=8640~kJ\)

Now, 8640 = 1.2 × 22.5 × 0.718 × dT

dT = 445.682

Concept:

First law of thermodynamics

dQ = dU + dW

∵ well insulated, hence dQ = 0

dU = - dW

mC_v dT = - [Power × time]

Now,

\(m = \frac{{PV}}{{RT}} = \frac{{100 \times 84}}{{0.280 \times 300}}\)

∴  m = 100 kg

From,

mC_vdT = - [Power × time]

\(∴ 100 \times 0.718 \times dT = - \left[ {71.8 \times 3 \times \frac{{3600}}{{1000}}} \right]\)

\(71.8\;dT = - \left[ {71.8 \times 3 \times \frac{{3600}}{{1000}}} \right]\)

dT = -10.8

Concept:

Since no work is produced or consumed and heat transfer is negligible, the SFEE reduces to:

\({{h}_{1}}+\frac{v_{1}^{2}}{2000}={{h}_{2}}+\frac{v_{2}^{2}}{2000}\)

Calculation:

Given, T₁ = 910°C, T₂ = 820°C, h₁ = 915 kJ/kg and h₂ = 800 kJ/kg, v₁ = 300 m/s

Concept:

1 litre = 10³ cm³ = 10^-3 m³

Work done in adiabatic process:

\(\delta W=\frac{{{P}_{1}}{{V}_{1}}-{{P}_{2}}{{V}_{2}}}{\gamma -1}=\frac{mR\left( {{T}_{1}}-{{T}_{2}} \right)}{\gamma -1}\)

\(\gamma =\frac{{{C}_{P}}}{{{C}_{V}}}\)

C_P - C_V = R ⇒ C_P = R + C_V

Calculation:

Given: m = 1 kg, P₁ = 10⁵ Pa, V₁ = 6 × 10^-3 m³

\({{V}_{2}}=\frac{1}{3}{{V}_{1}}=2\times {{10}^{-3}}{{m}^{3}}\)

For adiabatic: \({{P}_{2}}V_{2}^{\gamma }={{P}_{1}}V_{1}^{\gamma }\Rightarrow {{P}_{2}}={{P}_{1}}{{\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{\gamma}}\)

\({{C}_{P}}={{C}_{V}}+R=\frac{3}{2}R+R=\frac{5}{2}R\)

\(\gamma=\frac{{{C}_{P}}}{{{C}_{V}}}=\frac{\frac{5}{2}}{\frac{3}{2}}=\frac{5}{3}=1.67\)

\({{P}_{2}}={{P}_{1}}{{\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{\gamma }}={{10}^{5}}{{\left( 3 \right)}^{1.67}}=6.26\times {{10}^{5}}Pa\)

\(\delta W=\frac{{{P}_{1}}{{V}_{1}}-{{P}_{2}}{{V}_{2}}}{\gamma -1}=\frac{\left[ \left( 1\times 6 \right)-\left( 6.26\times 2 \right) \right]\times {{10}^{5}}\times {{10}^{-3}}}{1.67-1}\)

δW = - 973.1 J

Concept:

When the temperature a body changes from a higher temperature T₁ to a lower temperature T₂ then there is a decrease in quality of energy.

The maximum work or Available energy or the reversible work is given by,

 \(W_{rev} = A.E = H_1 - H_2 -T_0 (S_1 -S_2 )\) where T₀ is the surrounding temperature in K.

also, \(W_{rev} = mc_P (T_1 -T_2)- T_0 [mc_pln(\frac{{T_1}}{{T_2 }})-mRln(\frac{{P_1 }}{{P_2}})] \) for an open system,

where, m = mass of the body in kg, c_p = sp. heat capacity in kJ/kg-K, and R = characteristic gas constant in kJ/kg-K.

Calculation:

Given: m = 2 kg, T₁ = 300°C = 573 K, T₂ = 150°C = 423 K, T₀ = 20°C = 293 K, c_p = 0.45 kJ/kgK

from, \(W_{rev} = mc_P (T_1 -T_2)- T_0 [mc_pln(\frac{{T_1}}{{T_2 }})-mRln(\frac{{P_1 }}{{P_2}})] \)

\(W_{rev}=m{{c}_{p}}\left( {{T}_{1}}-{{T}_{2}} \right)-{{T}_{0 }}.m{{c}_{p}}\ln \frac{{{T}_{1}}}{{{T}_{2}}}\)  (as P₁ = P₂ i.e. there is no change in pressure inside the metallic block.)

\(W_{rev} =2\times 0.45\left( 573-423 \right)-293\times 2\times 0.45\ln \frac{573}{423}\)

\(W_{rev} = 135 - 80.036 = 54.96 \ {{kJ}}\)

Concept:

 Maximum work obtainable,

W = Q - T₀ (S₁ - S₀)

Where, T₀ (S₁ - S₀) corresponds to irreversibility, T₀ ∆S

S₁ is the entropy corresponding to the given state of the system and S₀ is the entropy at the atmospheric condition and T₀ is the atmospheric temperature.

Q = mC_p (T₁ - T₀)

\({\rm{\Delta }}S = m{C_p}\ln \left( {\frac{{{T_1}}}{{{T_0}}}} \right)\)

Calculation:

Given, Mass m = 2 kg, Specific heat C_P = 1 kJ/kgK

T₁ = 600 K, T₀ = 300 K

Therefore,

Q = m C_p (T₁ - T₀) = 2 × 1 × 300 = 600 kJ

\({\rm{\Delta }}S = 300 \times 2 \times 1 \times \ln \left( {\frac{{600}}{{300}}} \right) = 600 \times \ln 2 = 415.8\;kJ\)

Concept:

According to the first law of thermodynamics

δQ = dU + δW

Isobaric work done

δW = PdV = P (Vfinal - Vinitial) 

Calculation:

Initial condition ⇒ P₁ = 1 MPa, V₁ = y m³

Final condition ⇒ P₂ = 1 MPa, V₂ = 0.2 m³

Heat Transfer = -40 kJ (from the gas)

Change in Internal energy (u₂ – u₁) = -20 kJ (drop)

According to first law of thermodynamics

δQ = dU + δW

-40 = -20 + δW

⇒ δW = -20 kJ          ---(I)     

Since, the process is isobaric (as pressure remains same)

So, isobaric work done δW = PdV = P (V_final - V_initial) 

δW = P (V_final - V_initial) = -20 kJ

1000 kPa × (0.2 – y) m³ = -20 kJ

\(0.2 - y = \frac{{ - 20}}{{1000}} = - 0.02 \Rightarrow y = 0.22\)