Thermodynamics question

Question

1. 1 kg of air initially at 1 bar and 40°C is compressed adiabatically until its pressure is 50 bar. The air then receives 125.6 kJ of heat at constant pressure. Find for each process, the change in internal energy.

3. A steam power plant operates on the Rankine cycle. The steam enters the turbine at a flow rate of 500 kg/s, a pressure of 50 bar, a temperature of 600°C and a velocity of 30 m/s. The steam leaves the turbine at a pressure of 7.5 bar and a velocity of 100 m/s. Calculate the cycle efficiency and net power produced.

4. A boiler generates steam at a rate of 40 tonnes/h consuming coal at 4000 kg/h of calorific value 35,000 kJ/kg. The steam pressure is 15 bar and dry saturated. Calculate (a) factor of evaporation, (b) equivalent evaporation from and at 100°C, and (c) efficiency of the boiler.

Answer 1

1. Adiabatic Compression and Constant Pressure Heating of Air

Given:

1. Initial state:
2. m=1 kg, p1=1 bar=100 kPa, T1=40∘C=313 Km = 1\, \text{kg},\ p_1 = 1\, \text{bar} = 100\, \text{kPa},\ T_1 = 40^\circ C = 313\, \text{K}m=1kg, p1​=1bar=100kPa, T1​=40∘C=313K
3. Final pressure after adiabatic compression: p2=50 bar=5000 kPap_2 = 50\, \text{bar} = 5000\, \text{kPa}p2​=50bar=5000kPa
4. Heat added at constant pressure: Q2−3=125.6 kJQ_{2-3} = 125.6\, \text{kJ}Q2−3​=125.6kJ
5. Air is assumed ideal: Cp=1.005 kJ/kg\cdotpK, Cv=0.718 kJ/kg\cdotpK, R=Cp−Cv=0.287 kJ/kg\cdotpK, γ=1.4C_p = 1.005\, \text{kJ/kg·K},\ C_v = 0.718\, \text{kJ/kg·K},\ R = C_p - C_v = 0.287\, \text{kJ/kg·K},\ \gamma = 1.4Cp​=1.005kJ/kg\cdotpK, Cv​=0.718kJ/kg\cdotpK, R=Cp​−Cv​=0.287kJ/kg\cdotpK, γ=1.4

Process 1–2: Adiabatic Compression

Adiabatic relation:

T2=T1(p2p1)γ−1γ=313(5000100)0.41.4=313×(50)0.286≈313×3.57≈1118 KT_2 = T_1 \left( \frac{p_2}{p_1} \right)^{\frac{\gamma - 1}{\gamma}} = 313 \left( \frac{5000}{100} \right)^{\frac{0.4}{1.4}} = 313 \times (50)^{0.286} \approx 313 \times 3.57 \approx 1118\, \text{K}T2​=T1​(p1​p2​​)γγ−1​=313(1005000​)1.40.4​=313×(50)0.286≈313×3.57≈1118KChange in internal energy:

ΔU1−2=mCv(T2−T1)=1×0.718×(1118−313)≈578 kJ\Delta U_{1-2} = m C_v (T_2 - T_1) = 1 \times 0.718 \times (1118 - 313) \approx 578\, \text{kJ}ΔU1−2​=mCv​(T2​−T1​)=1×0.718×(1118−313)≈578kJ Process 2–3: Constant Pressure Heating

ΔU2−3=Q2−3−W2−3=Q2−3−mR(T3−T2)\Delta U_{2-3} = Q_{2-3} - W_{2-3} = Q_{2-3} - mR(T_3 - T_2)ΔU2−3​=Q2−3​−W2−3​=Q2−3​−mR(T3​−T2​)To find T3T_3T3​, use:

Q2−3=mCp(T3−T2)⇒T3=T2+Q2−3mCp=1118+125.61.005≈1118+125=1243 KQ_{2-3} = mC_p (T_3 - T_2) \Rightarrow T_3 = T_2 + \frac{Q_{2-3}}{m C_p} = 1118 + \frac{125.6}{1.005} \approx 1118 + 125 = 1243\, \text{K}Q2−3​=mCp​(T3​−T2​)⇒T3​=T2​+mCp​Q2−3​​=1118+1.005125.6​≈1118+125=1243KNow,

ΔU2−3=mCv(T3−T2)=0.718×(1243−1118)=0.718×125=89.75 kJ\Delta U_{2-3} = m C_v (T_3 - T_2) = 0.718 \times (1243 - 1118) = 0.718 \times 125 = 89.75\, \text{kJ}ΔU2−3​=mCv​(T3​−T2​)=0.718×(1243−1118)=0.718×125=89.75kJ Summary:

1. ΔU1−2≈578 kJ\Delta U_{1-2} \approx 578\, \text{kJ}ΔU1−2​≈578kJ
2. ΔU2−3≈89.75 kJ\Delta U_{2-3} \approx 89.75\, \text{kJ}ΔU2−3​≈89.75kJ

3. Rankine Cycle – Net Power and Efficiency

Given:

1. m˙=500 kg/s, p1=50 bar, T1=600∘C, v1=30 m/s, p2=7.5 bar, v2=100 m/s\dot{m} = 500\, \text{kg/s},\ p_1 = 50\, \text{bar},\ T_1 = 600^\circ C,\ v_1 = 30\, \text{m/s},\ p_2 = 7.5\, \text{bar},\ v_2 = 100\, \text{m/s}m˙=500kg/s, p1​=50bar, T1​=600∘C, v1​=30m/s, p2​=7.5bar, v2​=100m/s

From steam tables:

1. At 50 bar, 600°C: h1≈3583 kJ/kgh_1 \approx 3583\, \text{kJ/kg}h1​≈3583kJ/kg
2. At 7.5 bar (assuming isentropic expansion for ideal case):
3. Use entropy s1≈7.026 kJ/kg\cdotpKs_1 \approx 7.026\, \text{kJ/kg·K}s1​≈7.026kJ/kg\cdotpK, find h2≈2875 kJ/kgh_2 \approx 2875\, \text{kJ/kg}h2​≈2875kJ/kg

Turbine Work Output:

W˙turbine=m˙[(h1+v122×1000)−(h2+v222×1000)]\dot{W}_{turbine} = \dot{m} \left[ (h_1 + \frac{v_1^2}{2 \times 1000}) - (h_2 + \frac{v_2^2}{2 \times 1000}) \right]W˙turbine​=m˙[(h1​+2×1000v12​​)−(h2​+2×1000v22​​)] =500[(3583+0.45)−(2875+5)]=500×(3583.45−2880)=500×703.45≈351,725 kW= 500 \left[ (3583 + 0.45) - (2875 + 5) \right] = 500 \times (3583.45 - 2880) = 500 \times 703.45 \approx 351,725\, \text{kW}=500[(3583+0.45)−(2875+5)]=500×(3583.45−2880)=500×703.45≈351,725kW Cycle Efficiency:

Assume feedwater enters boiler at condenser pressure (7.5 bar), saturated liquid:

hf≈697 kJ/kgh_f \approx 697\, \text{kJ/kg}hf​≈697kJ/kg

qin=h1−hf=3583−697=2886 kJ/kgq_{in} = h_1 - h_f = 3583 - 697 = 2886\, \text{kJ/kg}qin​=h1​−hf​=3583−697=2886kJ/kg η=Wnetqin=703.452886≈0.2437=24.37%\eta = \frac{W_{net}}{q_{in}} = \frac{703.45}{2886} \approx 0.2437 = 24.37\%η=qin​Wnet​​=2886703.45​≈0.2437=24.37% Summary:

1. Net Power Output: ≈351.7 MW\approx 351.7\, \text{MW}≈351.7MW
2. Cycle Efficiency: ≈24.37%\approx 24.37\%≈24.37%

4. Boiler Performance Calculation

Given:

1. Steam: 40 tonnes/h = 40,000 kg/h
2. Coal: 4000 kg/h
3. Calorific value: 35,000 kJ/kg
4. Steam: dry saturated at 15 bar
5. From steam tables: hdry=2789 kJ/kgh_{dry} = 2789\, \text{kJ/kg}hdry​=2789kJ/kg

(a) Factor of Evaporation (FE):

FE=hsteam−hfeedwaterhfg@100∘C=2789−4192257=23702257≈1.05FE = \frac{h_{steam} - h_{feedwater}}{h_{fg@100^\circ C}} = \frac{2789 - 419}{2257} = \frac{2370}{2257} \approx 1.05FE=hfg@100∘C​hsteam​−hfeedwater​​=22572789−419​=22572370​≈1.05 (b) Equivalent Evaporation (EE):

EE=Actual evaporation rate×FE=40,000×1.05=42,000 kg/hEE = \text{Actual evaporation rate} \times FE = 40,000 \times 1.05 = 42,000\, \text{kg/h}EE=Actual evaporation rate×FE=40,000×1.05=42,000kg/h (c) Efficiency:

Heat input from coal=4000×35,000=140,000,000 kJ/h\text{Heat input from coal} = 4000 \times 35,000 = 140,000,000\, \text{kJ/h}Heat input from coal=4000×35,000=140,000,000kJ/h Heat output in steam=40,000×(2789−419)=40,000×2370=94,800,000 kJ/h\text{Heat output in steam} = 40,000 \times (2789 - 419) = 40,000 \times 2370 = 94,800,000\, \text{kJ/h}Heat output in steam=40,000×(2789−419)=40,000×2370=94,800,000kJ/h η=94,800,000140,000,000=0.677⇒67.7%\eta = \frac{94,800,000}{140,000,000} = 0.677 \Rightarrow 67.7\%η=140,000,00094,800,000​=0.677⇒67.7% Summary:

1. Factor of Evaporation: ≈ 1.05
2. Equivalent Evaporation: 42,000 kg/h
3. Boiler Efficiency: ≈ 67.7%