This set of Thermal Engineering Question Paper focuses on “Reciprocating Steam Engine – Actual Indicator Diagram and Diagram Factor”.
1. The actual indicator diagram is different from the hypothetical one due to _____
a) timings of the valve action and pressure drop due to friction in the admission port
b) limited amount of steam
c) Excess amount of steam
d) Low quality valves installation
View Answer
Explanation: Timings of the valve action and the pressure drop due to friction in the admission port reshapes the hypothetical indicator diagram into actual indicator diagram. Unlike hypothetical indicator diagram the actual one doesn’t have sharp corners.
2. The rounding observed in the actual indicator diagram is due to valve action.
a) True
b) False
View Answer
Explanation: Under practical conditions, the processes cannot be started or terminated instantaneously as a result the actual indicator diagram roundedness. Hence, valves are the reason for this roundedness and the given statement is correct.
3. What is diagram factor?
a) It is the ratio of the area of theoretical indicator diagram and the area of the actual indicator diagram
b) It is the ratio of area of the actual indicator diagram and the area of the theoretical indicator diagram
c) It is the sum of areas of actual and theoretical indicator diagram
d) It is the ratio of the difference between the areas of the theoretical and actual indicator diagram and the theoretical indicator diagram.
View Answer
Explanation: The diagram factor is defined as the ratio of the area of actual indicator diagram and the area of the theoretical indicator diagram. Mathematically,
Diagram factor = \(\frac{Area \, of \, actual \, indicator \, diagram}{Area \, of \, theoretical \, indicator \, diagram} \)
4. Given that the diagram factor for a particular reciprocating steam engine is 0.70 and the theoretically plotted indicator diagram has an area of 1700 kJ. Determine the area of the actual indicator diagram for the steam engine.
a) 11700 kJ
b) 11800 kJ
c) 11900 kJ
d) 12000 kJ
View Answer
Explanation: Given, Diagram factor = 0.70, area of the theoretical indicator diagram = 1700 kJ
We knew that,
Diagram factor = \(\frac{Area \, of \, actual \, indicator \, diagram}{Area \, of \, theoretical \, indicator \, diagram} \)
Substituting the values, we get
0.7 = \(\frac{Area \, of \, actual \, indicator \, diagram}{1700} \)
Therefore,
Area of the actual indicator diagram = 11900 kJ.
5. The actual expansion of the steam is not isentropic due to _____
a) leakage of heat through cylinder walls
b) limited amount of steam
c) No heat leakage
d) boiler inefficiency
View Answer
Explanation: There is an undesirable heat transfer through the cylinder wall during expansion. Due to this the expansion process is not isentropic. The efficiency of expansion process can be increased by limiting the temperature and pressure range through which the steam falls.
6. Calculate the diagram factor if the areas of theoretical and actual and theoretical indicator diagrams are 1400 kJ and 1800 kJ respectively.
a) 0.51
b) 0.52
c) 0.68
d) 0.78
View Answer
Explanation: Given, area of actual indicator diagram = 1400 kJ, area of theoretical indicator diagram = 1800 kJ
We know that,
Diagram factor = \(\frac{Area \, of \, actual \, indicator \, diagram}{Area \, of \, theoretical \, indicator \, diagram} \)
Substituting the values, we get
Diagram factor = \(\frac{1400}{1800} \)
Therefore,
Diagram factor = 0.78.
Sanfoundry Global Education & Learning Series – Thermal Engineering
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