Question 1 |
In a steam power plant, superheated steam at 10 MPa and 500^{\circ}C, is expanded isentropically
in a turbine until it becomes a saturated vapour. It is then reheated at constant pressure
to 500^{\circ}C. The steam is next expanded isentropically in another turbine until it reaches
the condenser pressure of 20 kPa. Relevant properties of steam are given in the following
two tables. The work done by both the turbines together is ______ kJ/kg (roundoff to
the nearest integer).
1513 | |
1245 | |
832 | |
1825 |
Question 1 Explanation:
Given data: h_{1}=3373.6 \mathrm{kJ} / \mathrm{kg}, h_{3}=3478.4 \mathrm{kJ} / \mathrm{kg}, h_{2}=2778.1 \mathrm{kJ} / \mathrm{kg}, s_{1}=s_{2} (as from table )
\begin{aligned} s_{3} &=s_{4} \\ s_{3} &=7.7621=0.8319+x+(7.9085-0.8319) \\ x_{4} &=0.9793 \\ h_{4} &=h_{f}+x_{4} \times\left(h_{g}-h_{f}\right)=2560.91 \mathrm{kJ} / \mathrm{kg} \\ W_{T} &=\left(h_{1}-h_{2}\right)+\left(h_{3}-h_{4}\right)=1512.95 \mathrm{kJ} / \mathrm{kg} \end{aligned}
\begin{aligned} s_{3} &=s_{4} \\ s_{3} &=7.7621=0.8319+x+(7.9085-0.8319) \\ x_{4} &=0.9793 \\ h_{4} &=h_{f}+x_{4} \times\left(h_{g}-h_{f}\right)=2560.91 \mathrm{kJ} / \mathrm{kg} \\ W_{T} &=\left(h_{1}-h_{2}\right)+\left(h_{3}-h_{4}\right)=1512.95 \mathrm{kJ} / \mathrm{kg} \end{aligned}
Question 2 |
Moist air at 105 kPa, 30^{\circ}C and 80% relative humidity flows over a cooling coil in an
insulated air-conditioning duct. Saturated air exits the duct at 100 kPa and 15^{\circ}C. The
saturation pressure of water at 30^{\circ}C and 15^{\circ}C are 4.24 kPa and 1.7 kPa respectively.
Molecular weight of water is 18 g/mol and that of air is 28.94 g/mol. The mass of water
condensing out from the duct is ______ g/kg of dry air (round off to 2 decimal places).
8.21 | |
15.24 | |
10.01 | |
12.24 |
Question 2 Explanation:
\begin{aligned} P_{t 1} &=105 \mathrm{kPa}, \mathrm{DBT}_{1}=30^{\circ} \mathrm{C}, \phi_{1}=0.8 \\ P_{t 2} &=100 \mathrm{kPa}, \mathrm{DBT}_{2}=15^{\circ} \mathrm{C}, \phi_{2}=1 \\ P_{v s 1} &=4.24 \mathrm{kPa} \\ P_{v s 2} &=1.7 \mathrm{kPa} \\ M_{\text {water }} &=18 \mathrm{g} / \mathrm{mol} \\ M_{\mathrm{air}} &=28.94 \mathrm{g} / \mathrm{mol} \\ \omega_{1} &=\frac{18}{28.94} \times \frac{P_{v 1}}{P_{t 1}-P_{v 1}} \\ \phi_{1} &=\frac{P_{v 1}}{P_{v s 1}} \end{aligned}
\begin{aligned} 0.8&=\frac{P_{\mathrm{v}}}{4.24} \\ \therefore \qquad P_{\mathrm{v} 1}&=3.392 \\ \therefore \qquad \omega_{1}&=\frac{18}{28.94} \times \frac{3.392}{105-3.392}\\ &=0.02076 \mathrm{kgV} / \mathrm{kg} \mathrm{d} . \mathrm{a} \\ &=20.76 \mathrm{gv} / \mathrm{kgd.a} \\ \omega_{2}&=\frac{18}{28.94} \times \frac{P_{\mathrm{v} 2}}{P_{\mathrm{t} 2}-P_{\mathrm{v} 2}} \\ \phi_{2}&=\frac{P_{\mathrm{v} 2}}{P_{\mathrm{vs} 2}} \\ 1&=\frac{P_{\mathrm{v} 2}}{1.7} \\ \therefore \qquad P_{\mathrm{v} 2}&=1.7 \\ \therefore \qquad \omega_{2}&=\frac{18}{28.94} \times \frac{1.7}{100-1.7}\\&=0.01075 \mathrm{Kgv} / \mathrm{kgda} \end{aligned}
Mass of water condensing =\omega_{1}-\omega_{2}
\begin{array}{l} =20.76-10.75 \\ =10.01 \mathrm{g} / \mathrm{kgd} . \mathrm{a} \end{array}
\begin{aligned} 0.8&=\frac{P_{\mathrm{v}}}{4.24} \\ \therefore \qquad P_{\mathrm{v} 1}&=3.392 \\ \therefore \qquad \omega_{1}&=\frac{18}{28.94} \times \frac{3.392}{105-3.392}\\ &=0.02076 \mathrm{kgV} / \mathrm{kg} \mathrm{d} . \mathrm{a} \\ &=20.76 \mathrm{gv} / \mathrm{kgd.a} \\ \omega_{2}&=\frac{18}{28.94} \times \frac{P_{\mathrm{v} 2}}{P_{\mathrm{t} 2}-P_{\mathrm{v} 2}} \\ \phi_{2}&=\frac{P_{\mathrm{v} 2}}{P_{\mathrm{vs} 2}} \\ 1&=\frac{P_{\mathrm{v} 2}}{1.7} \\ \therefore \qquad P_{\mathrm{v} 2}&=1.7 \\ \therefore \qquad \omega_{2}&=\frac{18}{28.94} \times \frac{1.7}{100-1.7}\\&=0.01075 \mathrm{Kgv} / \mathrm{kgda} \end{aligned}
Mass of water condensing =\omega_{1}-\omega_{2}
\begin{array}{l} =20.76-10.75 \\ =10.01 \mathrm{g} / \mathrm{kgd} . \mathrm{a} \end{array}
Question 3 |
Air is contained in a frictionless piston-cylinder arrangement as shown in the figure.
The atmospheric pressure is 100 kPa and the initial pressure of air in the cylinder is 105 kPa. The area of piston is 300 cm^2. Heat is now added and the piston moves slowly from its initial position until it reaches the stops. The spring constant of the linear spring is 12.5 N/mm. Considering the air inside the cylinder as the system, the work interaction is ________ J. (round off to the nearest integer).
The atmospheric pressure is 100 kPa and the initial pressure of air in the cylinder is 105 kPa. The area of piston is 300 cm^2. Heat is now added and the piston moves slowly from its initial position until it reaches the stops. The spring constant of the linear spring is 12.5 N/mm. Considering the air inside the cylinder as the system, the work interaction is ________ J. (round off to the nearest integer).
1 | |
544 | |
254 | |
623 |
Question 3 Explanation:
\begin{aligned} P_{0}&=100 \mathrm{kPa}, P_{1}=105 \mathrm{kPa}, \mathrm{K}=12.5 \mathrm{N} / \mathrm{mm}=12.5 \mathrm{kN} / \mathrm{m} \\ A&=300 \mathrm{cm}^{2}=300 \times 10^{-4} \mathrm{m}^{2} \\ x&=8 \mathrm{cm}=8 \times 10^{-2} \mathrm{m} \end{aligned}
1-2 constant pressure
\begin{aligned} W_{1-2} &=P_{1} \times A \times x=105 \times 300 \times 10^{-4} \times 8 \times 10^{-2} \\ &=0.252 \mathrm{kJ}=252 \mathrm{J} \\ P_{2} &=105 \mathrm{kPa} \end{aligned}
\begin{aligned} P_{3} \times A &=P_{2} A+K x \\ P_{3} &=P_{2}+\frac{K x}{A}\\ &=105+\frac{12.5 \times 8 \times 10^{-2}}{300 \times 10^{-4}} \\& =138.33 \mathrm{kPa} \\ W_{2-3} &=\frac{1}{2}\left(P_{2}+P_{3}\right) \times\left(V_{3}-V_{2}\right)\\ &=\frac{1}{2}(105+138.33) A \times x \\ &=\frac{1}{2}(243.33) \times 300 \times 10^{-4} \times 8 \times 10^{-2} \\ W_{2-3} &=0.2919 \mathrm{kJ}=291.9 \mathrm{J} \\ \therefore \quad W_{\text {total }} &=W_{1-2}+W_{2-3}\\&=0.5439 \mathrm{kJ}=543.91 \approx 544 \mathrm{J} \end{aligned}
Alternate Solution:
Total work = Workdone because of 105 kPa pressure + Workdone against spring which is equal to energy stored in spring
\begin{aligned} \text { Workdone } &=P_{1} \times A \times 2 x+\frac{1}{2} k \cdot x^{2} \\ &=105 \times 300 \times 10^{-4} \times 2 \times 8 \times 10^{-2}\\ & +\frac{1}{2} \times 12.5 \times\left(8 \times 10^{-2}\right)^{2} \\ &=0.504+0.04 \\ &=0.544 \mathrm{kJ}=544 \mathrm{J} \end{aligned}
1-2 constant pressure
\begin{aligned} W_{1-2} &=P_{1} \times A \times x=105 \times 300 \times 10^{-4} \times 8 \times 10^{-2} \\ &=0.252 \mathrm{kJ}=252 \mathrm{J} \\ P_{2} &=105 \mathrm{kPa} \end{aligned}
\begin{aligned} P_{3} \times A &=P_{2} A+K x \\ P_{3} &=P_{2}+\frac{K x}{A}\\ &=105+\frac{12.5 \times 8 \times 10^{-2}}{300 \times 10^{-4}} \\& =138.33 \mathrm{kPa} \\ W_{2-3} &=\frac{1}{2}\left(P_{2}+P_{3}\right) \times\left(V_{3}-V_{2}\right)\\ &=\frac{1}{2}(105+138.33) A \times x \\ &=\frac{1}{2}(243.33) \times 300 \times 10^{-4} \times 8 \times 10^{-2} \\ W_{2-3} &=0.2919 \mathrm{kJ}=291.9 \mathrm{J} \\ \therefore \quad W_{\text {total }} &=W_{1-2}+W_{2-3}\\&=0.5439 \mathrm{kJ}=543.91 \approx 544 \mathrm{J} \end{aligned}
Alternate Solution:
Total work = Workdone because of 105 kPa pressure + Workdone against spring which is equal to energy stored in spring
\begin{aligned} \text { Workdone } &=P_{1} \times A \times 2 x+\frac{1}{2} k \cdot x^{2} \\ &=105 \times 300 \times 10^{-4} \times 2 \times 8 \times 10^{-2}\\ & +\frac{1}{2} \times 12.5 \times\left(8 \times 10^{-2}\right)^{2} \\ &=0.504+0.04 \\ &=0.544 \mathrm{kJ}=544 \mathrm{J} \end{aligned}
Question 4 |
One kg of air, initially at a temperature of 127^{\circ}C, expands reversibly at a constant
pressure until the volume is doubled. If the gas constant of air is 287 J/kg.K, the
magnitude of work transfer is __________ kJ (round off to 2 decimal places).
156.6 | |
114.8 | |
89.8 | |
124.4 |
Question 4 Explanation:
\begin{aligned} m=1 \mathrm{kg} ; T_{1}=127^{\circ} \mathrm{C} &=400 \mathrm{K} \\ P=C ; V_{2}=2 V_{1} ; \quad R=& 0.287 \mathrm{kJ} / \mathrm{kgK} \\ W &=P_{1}\left(V_{2}-V_{1}\right) \\ &=P_{1}\left(2 V_{1}-V_{1}\right) \\ &=P_{1} \cdot V_{1}=\mathrm{mRT}_{1} \\ &=114.8 \mathrm{kJ} \end{aligned}
Question 5 |
For a simple compressible system, v, s, p and T are specific volume, specific entropy, pressure and temperature, respectively. As per Maxwell's relations, \left (\frac{\partial v}{\partial s} \right )_p is equal to
\left (\frac{\partial s}{\partial T} \right )_p | |
\left (\frac{\partial p}{\partial v} \right )_T | |
-\left (\frac{\partial T}{\partial v} \right )_p | |
\left (\frac{\partial T}{\partial p} \right )_s |
Question 5 Explanation:
According to Maxwell's relation \left(\frac{\partial \mathrm{v}}{\partial \mathrm{s}}\right)_{\mathrm{p}}=\left(\frac{\partial \mathrm{T}}{\partial \mathrm{p}}\right)
Question 6 |
A steam power cycle with regeneration as shown below on the T-S diagram employs a single open feedwater heater for efficiency improvement. The fluids mix with each other in an open feedwater heater. The turbine is isentropic and the input (bleed) to the feedwater heater from the turbine is at state 2 as shown in the figure. Process 3-4 occurs in the condenser. The pump work is negligible. The input to the boiler is at state 5. The following information is available from the steam tables:
The mass flow rate of steam bled from the turbine as a percentage of the total mass flow rate at the inlet to the turbineat state 1 is _______
The mass flow rate of steam bled from the turbine as a percentage of the total mass flow rate at the inlet to the turbineat state 1 is _______
10 | |
20 | |
30 | |
40 |
Question 6 Explanation:
Let 'm_{1}' kg of steam enters the turbine, 'm_{2}' kg of steam is bled from the turbine into the feed water heater and 'm_{1}-m_{2}' kg of steam enters the condenser.
From energy balance of open feed water heater,
\begin{aligned} \mathrm{m}_{2} \mathrm{h}_{2}+\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right) \mathrm{h}_{4}&=\mathrm{m}_{1} \mathrm{h}_{5} \\ \mathrm{m}_{2} \times 2800+\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right) 175 &=\mathrm{m}_{1} \times 700 \\ 2625 \mathrm{m}_{2} &=525 \mathrm{m}_{1} \end{aligned}
\therefore Percentage of steam bled from the turbine =\frac{\mathrm{m}_{2}}{\mathrm{m}_{1}} \times 100=0.2 \times 100=20 \%
From energy balance of open feed water heater,
\begin{aligned} \mathrm{m}_{2} \mathrm{h}_{2}+\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right) \mathrm{h}_{4}&=\mathrm{m}_{1} \mathrm{h}_{5} \\ \mathrm{m}_{2} \times 2800+\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right) 175 &=\mathrm{m}_{1} \times 700 \\ 2625 \mathrm{m}_{2} &=525 \mathrm{m}_{1} \end{aligned}
\therefore Percentage of steam bled from the turbine =\frac{\mathrm{m}_{2}}{\mathrm{m}_{1}} \times 100=0.2 \times 100=20 \%
Question 7 |
If one mole of H_2 gas occupies a rigid container with a capacity of 1000 litres and the temperature is raised from 27^{\circ}C \; to \; 37^{\circ}C, the change in pressure of the contained gas (round offto two decimal places), assuming ideal gas behaviour, is ________ Pa. (R=8.314 J/mol.K)
83.14 | |
8.31 | |
63.55 | |
78.62 |
Question 7 Explanation:
Volume , \mathrm{V}=1000 \mathrm{lit}=1 \mathrm{m}^{3}
For ideal gas, \mathrm{PV}=\mathrm{n} \overline{\mathrm{R}} \mathrm{T}
Differntiating above expression we get,
\mathrm{Vdp}=\mathrm{n} \overrightarrow{\mathrm{R}} \mathrm{d} \mathrm{T}
1 \times \mathrm{dp}=1 \times 8.314 \times 10
\therefore \mathrm{dP}=83.14 \mathrm{Pa}
For ideal gas, \mathrm{PV}=\mathrm{n} \overline{\mathrm{R}} \mathrm{T}
Differntiating above expression we get,
\mathrm{Vdp}=\mathrm{n} \overrightarrow{\mathrm{R}} \mathrm{d} \mathrm{T}
1 \times \mathrm{dp}=1 \times 8.314 \times 10
\therefore \mathrm{dP}=83.14 \mathrm{Pa}
Question 8 |
A calorically perfect gas (specific heat at constant pressure 1000 J/kg\cdot K ) enters and leaves a gas turbine with the same velocity. The temperatures of the gas at turbine entry and exit are 1100 K and 400 K, respectively. The power produced is 4.6 MW and heat escapes at the rate of 300 kJ/s through the turbine casing. The mass flow rate of the gas (in kg/s) through the turbine is
6.14 | |
7 | |
7.5 | |
8 |
Question 8 Explanation:
\begin{aligned} c_{p}&1\mathrm{kJ}/\mathrm{kg-K}\\ V_{i}&=V_{e}(\text { Inlet and exit velocity same }) \\ \dot{W}_{C . V} &=4.6 \mathrm{MW}=4.6 \times 10^{3} \mathrm{kW} \\ \dot{Q} &=300 \mathrm{kW} \end{aligned}
Applying SFEE,
\begin{aligned} \dot{m}(h_{i}+\frac{1}{2} V_{i}^{2}+g z_{i})+ \dot{Q} &=\dot{m}(h_{e}+\frac{1}{2} V_{e}^{2}+g z_{e})+\dot{W}_{C . V} \\ \dot{m}[c_{p}(T_{i}-T_{e})]+\dot{Q}&=\dot{W}_{C . V} \quad \text { (Let } z_{i}=z_{e}) \\ \dot{m}[1 \times(1100-400)]+(-300)&=4.6 \times 10^{3} \\ \Rightarrow \quad \dot{m}&=7 \mathrm{kg} / \mathrm{s} \end{aligned}
Question 9 |
A mass m of a perfect gas at pressure P_{1} and volume V_{1} undergoes an isothermal process. The final pressure is P_{2} and volume is V_{2}. The work done on the system is considered positive. If R is the gas constant and T is the temperature, then the work done in the process is
P_{1}V_{1}ln\frac{V_{2}}{V_{1}} | |
-P_{1}V_{1}ln\frac{P_{1}}{P_{2}} | |
RT\,ln\,\frac{V_{2}}{V_{1}} | |
-mRT\,ln\,\frac{P_{2}}{P_{1}} |
Question 9 Explanation:
Work done on the system is considered as positive, so
\delta W=-P d V
(-ve sign is taken as dV is-ve during compression)
\begin{aligned} W &=-\int_{1}^{2} P d V=-\int_{1}^{2} m R T \frac{d V}{V} \\ &=-m R T \int_{1}^{2} \frac{d V}{V}=-m R T \ln \frac{V_{2}}{V_{1}} \\ &=-m R T \ln \frac{P_{1}}{P_{2}}=-P_{1} V_{1} \ln \frac{P_{1}}{P_{2}} \end{aligned}
\delta W=-P d V
(-ve sign is taken as dV is-ve during compression)
\begin{aligned} W &=-\int_{1}^{2} P d V=-\int_{1}^{2} m R T \frac{d V}{V} \\ &=-m R T \int_{1}^{2} \frac{d V}{V}=-m R T \ln \frac{V_{2}}{V_{1}} \\ &=-m R T \ln \frac{P_{1}}{P_{2}}=-P_{1} V_{1} \ln \frac{P_{1}}{P_{2}} \end{aligned}
Question 10 |
Air contains 79% N2 and 21% O_{2} on a molar basis. Methane (CH_{4}) is burned with 50% excess air than required stoichiometrically. Assuming complete combustion of methane, the molar percentage of N_{2} in the products is _______________
70 | |
73.8 | |
80 | |
79.8 |
Question 10 Explanation:
Stoichiometrically balanced equation for combustion of {\mathrm{CH}}_4
{\mathrm{CH}}_4+2{\mathrm O}_2\rightarrow{\mathrm{CO}}_2+2{\mathrm H}_2\mathrm O
Actual moles of {\mathrm{O}}_2 used =2\times\left(\frac{150}{100}\right)=3 moles of {\mathrm{O}}_2
Actual equation
{\mathrm{CH}}_4+3{\mathrm o}_2\rightarrow{\mathrm{Co}}_2+2{\mathrm H}_2\mathrm o+{\mathrm O}_2
Consider Air instead of pure {\mathrm{O}}_2
{\mathrm{CH}}_4+\left[3{\mathrm o}_2+3\left(\frac{79}{21}\right){\mathrm N}_2\right]\rightarrow{\mathrm{Co}}_2+2{\mathrm H}_2\mathrm O+{\mathrm O}_2+\frac{79}7\mathrm N2
Total moles in product =1+2+1+\frac{79}7=15.285
Molar percentage of {\mathrm N}_2=\frac{11.285}{15.285}\times100=73.830\%
{\mathrm{CH}}_4+2{\mathrm O}_2\rightarrow{\mathrm{CO}}_2+2{\mathrm H}_2\mathrm O
Actual moles of {\mathrm{O}}_2 used =2\times\left(\frac{150}{100}\right)=3 moles of {\mathrm{O}}_2
Actual equation
{\mathrm{CH}}_4+3{\mathrm o}_2\rightarrow{\mathrm{Co}}_2+2{\mathrm H}_2\mathrm o+{\mathrm O}_2
Consider Air instead of pure {\mathrm{O}}_2
{\mathrm{CH}}_4+\left[3{\mathrm o}_2+3\left(\frac{79}{21}\right){\mathrm N}_2\right]\rightarrow{\mathrm{Co}}_2+2{\mathrm H}_2\mathrm O+{\mathrm O}_2+\frac{79}7\mathrm N2
Total moles in product =1+2+1+\frac{79}7=15.285
Molar percentage of {\mathrm N}_2=\frac{11.285}{15.285}\times100=73.830\%
There are 10 questions to complete.