The critical pressure ratio for maximum discharge through a nozzle is given by:
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Solution
Critical pressure ratio for maximum discharge
\(\left ( \frac{2}{n+1} \right )^{n/n-1}\)
A gas turbine plant working on Joule cycle produces 4000 kw of power. If its work ratio is 40%, the power consumed by the compressor is ………………..
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Solution
Given: Work ratio = 40% =\(\frac{40}{100}=0.4\)
Work ratio
\(\frac{Workd\,one\,by\,turbine\,Work\,doneby\,compressor}{Work\,done\,by\,turbine}\)
0.4 =\(\frac{w_{turbine}-w_{compressor}}{w_{turbine}}\)
Wturbine = 0.4 × Wturbine + Wcompressor
But, Wturbine – Wcompressor = 4000 (Given)
Now, Wturbine= 0.4× Wturbine+ Wturbine– 4000
Wturbine=\(\frac{4000}{0.4}\)= 10,000 kW
and Wcompressor = 10,000 –4000 = 6000 kW
If the performance of diesel engines of different sizes,cylinder dimensions and power ratings are to be compared,which of the following parameters can be used for such comparison ?
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Solution
Volumetric efficiency describes its ability to put air into its cylinders, the greater the efficiency,the more completely the engine fills available cylinder volume.
Which phenomenon will occur when the value at discharge end of a pipe connected to a reservoir is suddenly closed ?
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Solution
When the value at discharge end of a pipe connected to a reservoir is closed suddenly,then hammering phenomenon is observed.
A circular area of 1.2 m diameter is immesed vertically in a liquid of unit weight 800 N/m3 with its top edge just on the liquid surface. The depth of centre of pressure on one side, measured below the liquid surface, is ………………..
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Solution
Given: Diameter (d) = 1.2 m
Density (ρ) = 800 N/m3or 80 kg/m3[at g =10 m/s2]
Moment of inertia (IX) of circular section
\(\frac{\pi }{64}(d)^{4}=\frac{\pi }{64}(1.2)^{4}\)
Location of C.G (\(\overline{y}\)) =\(\frac{1.2}{2}=0.6cm\)
Now, depth of centre = \(\frac{\frac{\pi }{64}(1.2)^{4}}{\frac{\pi }{4}(1.2)^{2}\times 0.6}+0.6=0.75cm\)
Which one of the following cycles has the highest thermal efficiency for given maximum and minimum cycle temperature ?
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Solution
Sterling cycle has the highest thermal efficiency for maximum and minimum cycle temperature and is nearly equal to the Carnot cycle efficiency.
A heat pump works on a reversed Carnot cycle. The temperature in the condenser coils is 27°C and that in the evaporator coils is – 23°C. Fora work input of 1 kw, the amount of heat pumped is ……………….. .
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Solution
Given, To = 27°C = 27 + 273 = 300° K
Ti= – 23°C = 273 – 23 = 250° K
Now, COP =\(\frac{T_{0}}{T_{0}-T_{1}}=\frac{300}{(300-250)}=\frac{300}{50}=6\)
Amount of heat pumped = 6 × 1 = 6 kW
Match List-I with List-II.
List-I List-II
A.Schmidt number 1.K(ρ/CPD)
B.Thermal diffusivity 2.hmL/D
C.Lewis number 3.µ/ρD
D.Sherwood number 4.K/ρCP
What is the expression for thermal conduction resistance to heat transfer through a hollow sphere of inner radius r1 and outer radius r2 and thermal conductivity k ?
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Solution
Thermal conductivity resistance
=\(\frac{(r_{2}-r_{1})}{4\pi kr_{1}r_{2}}\)
In an air-standard otto cycle, r is volume compresssion ratio and g is an adiabatic index (CP/CV), the air standard efficiency is given by :
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Solution
The air-standard efficiency of otto cycle is described
as (hotto cycle) =1-\(\frac{1}{r^{\gamma }}\)