# Power System

 Question 1
Consider the open feed water heater (FWH) shown in the figure given below:

Specific enthalpy of steam at location 2 is 2624 kJ/kg, specific enthalpy of water at location 5 is 226.7 kJ/kg and specific enthalpy of saturated water at location 6 is 708.6 kJ/kg. If the mass flow rate of water entering the open feed water heater (at location 5) is 100 kg/s then the mass flow rate of steam at location 2 will be kg/s (round off to one decimal place).
 A 25.2 B 45.6 C 62.3 D 18.4
GATE ME 2021 SET-2   Thermodynamics
Question 1 Explanation:

\begin{aligned} 100 h_{5}+(x-100) h_{2} &=x h_{6} \\ 100 \times 226.7+(x-100) 2624&=708.6 x \\ 22670+2624 x-262400 &=708.6 x \\ 2624 x-708.6 x &=239730 \\ 1915.4 x &=239730 \\ x &=125.159 \simeq 125.2 \mathrm{~kg} / \mathrm{s} \end{aligned}
Mass flow rate at state $2(x-100)=25.2 \mathrm{~kg} / \mathrm{s}$
 Question 2
Consider a steam power plant operating on an ideal reheat Rankine cycle. The work input to the pump is 20 kJ/kg. The work output from the high pressure turbine is 750 kJ/kg. The work output from the low pressure turbine is 1500 kJ/kg. The thermal efficiency of the cycle is 50 %. The enthalpy of saturated liquid and saturated vapour at condenser pressure are 200 kJ/kg and 2600 kJ/kg, respectively. The quality of steam at the exit of the low pressure turbine is ________ % (round off to the nearest integer).
 A 45 B 68 C 98 D 93
GATE ME 2021 SET-1   Thermodynamics
Question 2 Explanation:

\begin{aligned} h_{f} &=200 \mathrm{~kJ} / \mathrm{kg} \\ h_{g} &=2600 \mathrm{~kJ} / \mathrm{kg} \\ w_{p} &=20 \mathrm{~kJ} / \mathrm{kg}=h_{6}-h_{5} \\ h_{1}-h_{2} &=750 \mathrm{~kJ} / \mathrm{kg} \\ h_{3}-h_{4} &=1500 \mathrm{~kJ} / \mathrm{kg} \\ \eta &=0.5=\frac{W_{\mathrm{NET}}}{Q_{s}}=\frac{W_{T}-W_{P}}{Q_{s}}\\ 0.5 &=\frac{750+1500-20}{Q_{S}} \\ Q_{S} &=4460 \mathrm{~kJ} / \mathrm{kg} \\ \eta &=1-\frac{Q_{R}}{Q_{S}} \\ \frac{Q_{R}}{Q_{S}} &=0.5 \\ Q_{R} &=2230 \mathrm{~kJ} / \mathrm{kg} \\ Q_{R} &=h_{4}-h_{5} \\ 2230 &=h_{4}-200 \\ h_{4} &=2430 \mathrm{~kJ} / \mathrm{kg} \\ h_{4} &=h_{f}+x\left(h_{g}-h_{f}\right) \\ 2430 &=200+x(2600-200) \\ x &=0.9291 \\ x &=93 \% \end{aligned}
 Question 3
The values of enthalpies at the stator inlet and rotor outlet of a hydraulic turbomachine stage are $h_1$ and $h_3$ respectively. The enthalpy at the stator outlet (or, rotor inlet) is $h_2$. The condition $(h_2-h_1)=(h_3-h_2)$ indicates that the degree of reaction of this stage is
 A zero B 50% C 75% D 100%
GATE ME 2020 SET-2   Thermodynamics
Question 3 Explanation:
As enthalpy across stator and rotor is equal it is 50% reaction stage.
 Question 4
Water flowing at the rate of 1 kg/s through a system is heated using an electric heater such that the specific enthalpy of the water increases by 2.50 kJ/kg and the specific entropy increases by 0.007 kJ/kg.K. The power input to the electric heater is 2.50 kW. There is no other work or heat interaction between the system and the surroundings. Assuming an ambient temperature of 300 K, the irreversibility rate of the system is ______kW (round off to two decimal places).
 A 2.1 B 12.09 C 40.05 D 10.08
GATE ME 2019 SET-2   Thermodynamics
Question 4 Explanation:
Given:
\begin{aligned} (h_2-h_1)_{water} &=2.5 kJ/kg \\ (s_2-s_1)_{water}&=0.007kJ/kgK \\ [\text{Power}]_{input}&= 2.5kW\\ T_0&=300K \\ [\text{Power}]_{input}&= m(h_2-h_1)_{water}\\ 2.5&=m(2.5) \\ m&=1kg/s \\ (ds)_{water}&=m(s_2-s_1)_{water}\\ &=0.007kW/K \\ (ds)_{surrounding} &=0 \\ (ds)_{universe}&=0.007kW/K\\ \text{Irreversibility}&=300 \times 0.007=2.1kW \end{aligned}
 Question 5
Which one of the following modifications of the simple ideal Rankine cycle increases the thermal efficiency and reduces the moisture content of the steam at the turbine outlet?
 A Increasing the boiler pressure. B Decreasing the boiler pressure. C Increasing the turbine inlet temperature. D Decreasing the condenser pressure.
GATE ME 2019 SET-2   Thermodynamics
Question 5 Explanation:
Rankine cycle efficiency can be increased by increasing the mean temperature of heat addition and it can be obtain by increasing the super heated temperature of steam at the inlet to the steam turbine

While the superheated steam temperature increases, the dryness fraction increases (i.e., moisture content decreases).
 Question 6
A gas turbine with air as the working fluid has an isentropic efficiency of 0.70 when operating at a pressure ratio of 3. Now, the pressure ratio of the turbine is increased to 5, while maintaining the same inlet conditions. Assume air as a perfectgas withspecific heat ratio $\gamma =1.4$. If the specific work output remains the same for both the cases, the isentropic efficiency of the turbine at the pressure ratio of 5 is ______ (round off to two decimal places)
 A 0.25 B 0.51 C 0.75 D 1
GATE ME 2019 SET-1   Thermodynamics
Question 6 Explanation:

Given, $r_{p 1}=3, r_{p_{2}}=5$
\begin{array}{l} \qquad \begin{aligned} \mathrm{W}_{\mathrm{act}} &=\eta_{\mathrm{s}} \times \mathrm{W}_{\text {iscatropic }} \\ &=\eta_{\mathrm{s}} \times \mathrm{c}_{\mathrm{p}}\left[\mathrm{T}_{3}-\mathrm{T}_{4}\right]\\ &=\eta_{\mathrm{s}} \times \mathrm{c}_{\mathrm{p}} \times \mathrm{T}_{3}\left[1-\frac{1}{\frac{T_{3}}{T{4}}}\right] \end{aligned} \\ \qquad \begin{aligned} \mathrm{W}_{\mathrm{act}}=& \eta_{\mathrm{s}} \times \mathrm{c}_{\mathrm{p}} \times \mathrm{T}_{3}\left[1-\frac{1}{\mathrm{r}_{\mathrm{p}}^{\frac{\gamma-1}{\gamma}}}\right] \\ \left(\mathrm{W}_{\mathrm{act})_{1}}\right.&=\left(\mathrm{W}_{\mathrm{act}}\right)_{2} \end{aligned} \\ \therefore \eta_{\mathrm{s}} \times \left[1-\frac{1}{3^{\frac{1.4-1}{1.4}}}\right]=\eta_{\mathrm{s}_{2}} \times\left[1-\frac{1}{5^{\frac{1.4-1}{1.4}}}\right] \\ \Rightarrow \eta_{\mathrm{s}_{2}}=0.515 \end{array}
 Question 7
In the Rankine cycle for a steam power plant the turbine entry and exit enthalpies are 2803 kJ/kg and 1800 kJ/kg, respectively. The enthalpies of water at pump entry and exit are 121 kJ/kg and 124 kJ/kg, respectively. The specific steam consumption (in kg/k W.h) of the cycle is _______
 A 36 B 3.6 C 0.36 D 0.036
GATE ME 2017 SET-2   Thermodynamics
Question 7 Explanation:
Given data $h_{1}=2803\mathrm{kJ} / \mathrm{kg}$

\begin{aligned} h_{2} &=1800 \mathrm{kJ} / \mathrm{kg} \\ h_{3} &=121 \mathrm{kJ} / \mathrm{kg} \\ h_{4} &=124 \mathrm{kJ} / \mathrm{kg} \\ w_{T} &=h_{1}-h_{2} \\ &=2803-1800=1003 \mathrm{kJ} / \mathrm{kg} \\ w_{P} &=h_{4}-h_{3} \\ &=124-121=3 \mathrm{kJ} / \mathrm{kg} \\ w_{\text {net }} &=w_{T}-w_{P}=1003-3 \\ &=1000 \mathrm{kJ} / \mathrm{kg} \\ \mathrm{SSC} &=\frac{3600}{W_{\text {net }}}=\frac{3600}{1000}=3.6 \mathrm{kg} / \mathrm{kWh} \end{aligned}
 Question 8
The Pressure ratio across a gas turbine (for air, specific heat at constant pressure,$c_{\textrm{p}}$ 1040 J / kg .K and ratio of specific heats, $\gamma =1.4$ ) is 10. If the inlet temperature to the turbine is 1200K and the isentropic efficiency is 0.9, the gas temperature at turbine exit is ______ K.
 A 666.2 B 679.4 C 720.3 D 696.8
GATE ME 2017 SET-1   Thermodynamics
Question 8 Explanation:
\begin{aligned} \text{Given data:}\;\mathrm{c}_{p}&=1040 \mathrm{J} / \mathrm{kgK} ; \gamma=1.4 ; r_{p}=10; \\ T_{3}&=1200 \mathrm{K};\;\eta=0.9 \\ \frac{T_{7}}{T_{4 s}}&=\left(r_{p}\right) \frac{\gamma-1}{\gamma} \end{aligned}

\begin{aligned} \frac{1200}{T_{4 s}} &=(10)^{(1.4-1) / 1.4} \\ \text { or } \quad T_{4 s} &=621.53 \mathrm{K} \\ \eta_{T} &=\frac{T_{3}-T_{4}}{T_{3}-T_{4 s}} \\ 0.9 &=\frac{1200-T_{4}}{1200-621.53} \\ \text{or}\qquad T_{4} &=679.4 \mathrm{K} \end{aligned}
 Question 9
In a 3-stage air compressor, the inlet pressure is $p_{1}$, discharge pressure is $p_{4}$ and the intermediate pressures are $p_{2}$ and $p_{3}$ ( $p_{2}\lt p_{3}$). The total pressure ratio of the compressor is 10 and the pressure ratios of the stages are equal. If $p_{1}$ = 100 kPa, the value of the pressure $p_{3}$ (in kPa) is __________
 A 400.00kPa B 399.45kPa C 645.12kPa D 464.16kPa
GATE ME 2016 SET-3   Thermodynamics
Question 9 Explanation:

\begin{aligned} \frac{p_{4}}{p_{1}}&=10 \\ \frac{p_{2}}{p_{1}}&=\frac{p_{3}}{p_{2}}=\frac{p_{4}}{p_{3}} \end{aligned}
To calculate $p_{3}=?$
\begin{aligned} \frac{p_{4}}{p_{3}} \times \frac{p_{3}}{p_{2}} \times \frac{p_{2}}{p_{1}}&=10 \\ r_{p} \times r_{p} \times r_{p}&=10 \\ r_{p}^{3} &=10 \\ r_{p} &=(10)^{1 / 3}\\ p_{2}&=r_{p} p_{1} \\ p_{3}&=r_{p} p_{2}=r_{p}^{2} p_{1}=(10)^{2 / 3} \times 100\\ &=464.16 \mathrm{kPa} \end{aligned}
 Question 10
Consider a simple gas turbine (Brayton) cycle and a gas turbine cycle with perfect regeneration. In both the cycles, the pressure ratio is 6 and the ratio of the specific heats of the working medium is 1.4. The ratio of minimum to maximum temperatures is 0.3 (with temperatures expressed in K) in the regenerative cycle. The ratio of the thermal efficiency of the simple cycle to that of the regenerative cycle is _________
 A 0.8141 B 0.3423 C 0.4567 D 0.1234
GATE ME 2016 SET-2   Thermodynamics
Question 10 Explanation:
\begin{aligned} \text{Given: }r_{p}&=6\\ y&=1.4 \end{aligned}
$\frac{T_{\max }}{T_{\min }}=0.3 \quad$ [for regenerative cycle ]
\begin{aligned} \eta_\text { Tregenerative cycle } &=1-\frac{T_{\max }}{T_{\min }} \times\left(r_{p}\right)^{(\gamma-1) / \gamma} \\ &=1-0.3\times(6)^{\frac{0.4}{1.4}}\\ &=0.49945 \times(6)^{\frac{0.4}{1.4}} \end{aligned}
\begin{aligned} \eta_{\text {simplecycle }} &=1-\frac{1}{\left(r_{p}\right)^{(\gamma-1) / \gamma}} \\ &=1-\left(r_{p}\right)^{(1-\gamma) / \gamma} \\ &=1-(6)^{-\frac{0.4}{1.4}}=0.4066 \end{aligned}
\begin{aligned} \text { Required ratio } &=\frac{\eta_{\text {simple cycle }}}{\eta_{\text {regenerative cycle }}} \\ &=\frac{0.40666}{0.49945}=0.8141 \end{aligned}
There are 10 questions to complete.