 # Quiz: Mechanical Engineering 1st may

Quiz: Mechanical Engineering
Exam: GATE
Topic: Miscellaneous

Each question carries 2 mark
Negative marking: 1/3 mark
Time: 20 Minutes

Q1. Entropy change in a polytropic process is given by
(a) S_2-S_1=((n-γ))/(γ-1)(1-n) Rln T_2/T_1
(b) S_2-S_1=(n-γ)/(γ-1)(n-1) Rln T_2/T_1
(c) S_2-S_1=((γ-n))/(γ-n)(γ-1) Rln T_2/T_1
(d) S_2-S_1=((γ-n))/((γ-n) ) Rln T_2/T_1

Q2. The given figure shown a thermodynamic cycle on T-S diagram. All the processes are straight lines. The efficiency of the cycle will be equal to if T_h=800 K & T_0=400 K

(a) 20%
(b) 25%
(c) 33.3%
(d) 66.7%

Q3. An air preheater is used to cool the products of combustion from 300°C to 200°C. C_p of product is 1.09 kJ/kg – k. Take T_0as 300 K. what will be initial & final availability of product respectively
(a) 85.97 & 39.68 kJ/kg
(b) 40 & 20 kJ/kg
(c) 65.68 & 29.33 kJ/kg
(d) 60 & 25 kJ/kg

Q4. In a steady – flow adiabatic turbine, the changes in the internal energy, kinetic energy and potential energy of the working fluid, from inlet to exit, are -100 kJ/kg – 1400 kJ/kg, -10 kJ/kg and 0 kJ/kg respectively. Which one of the following gives the amount of work developed by the turbine?
(a) 100 kJ/kg
(b) 110 kJ/kg
(c) 140 kJ/kg
(d) 150 kJ/kg

Q5. The head loss for a laminar incompressible flow through a horizontal circular pipe is h_1. Pipe length and fluid remaining the same, if the average flow velocity doubles and the pipe diameter reduces to half its previous value, the head loss is h_2. The ratio h_2 /h_1 is
(a) 1
(b) 4
(c) 8
(d) 16

Q6. If moist air is cooled by sensible heat removal, which of the following is true?
(a) Neither relative humidity nor specific humidity changes
(b) Specific humidity changes but not relative humidity
(c) Both relative humidity and specific humidity
(d) None of these

Q7. For a typical sample of ambient air (at 30°C, 75% relative humidity and standard atmospheric pressure), the amount of moisture in kg per kg of dry air will be approximately.
(a) 0.002
(b) 0.027
(c) 0.25
(d) 0.75

Q8. The stroke and bore of a 4-stroke spark ignition engine are 250 mm and 200 mm respectively. The clearance volume is 0.001 m³. if the specific heat ratio γ = 1.4, the air-standard cycle efficiency of the engine is
(a) 46.40%
(b) 56.10%
(c) 58.20%
(d) 62.80%

Q9. For steady, fully developed flow inside straight pipe of diameter D, neglecting gravity effects, the pressure drop ∆p over a length L and the wall shear stress τ_w are related by
(a) τ_w=∆pD/4L
(b) τ_w=(∆pD^2)/(4L^2 )
(c) τ_w=∆pD/2L
(d) τ_w=4 ∆pL/D

Q10. If x is the distance measured from the leading edge of a flat plat, then laminar boundary layer thickness varies as
(a) 1/x
(b) x^(4/5)
(c) x²
(d) x^(1/2)

Solutions
S1. Ans.(b)
Sol. in a polytropic process, the entropy change can expressed as
dS=δq/T=(γ-n)/(γ-1) δW/T=(γ-n)/(γ-1) . PdV/T
=(γ-n)/(γ-1) R dV/V
Upon integration, we get
S_2-S_1 =(γ-n)/(γ-1 ) Rln V_2/V_1
We know, V_2/V_1 =(T_2/T_1 )^(1/(n-1))
∵ S_2-S_1=(γ-n)/(γ-1) Rln(T_2/T_1 )^(-1/(n-1))
= -(γ-n)/(γ-1).1/(n-1) Rln T_2/T_1
=((n-γ))/(γ-1)(n-1) Rln T_2/T_1

S2. Ans.(b)
Sol. If we make a rectangle out of this triangle then it becomes a Carnot cycle and this triangle is just half of area of that rectangle so its efficiency will be half of Carnot cycle.
η=0.5(T_h-T_c )/T_h
=0.5(800-400)/800=0.25 or 25%

S3. Ans.(a)
Sol. Ψ_1= initial availability of the product
=(h_1-h_0 )-T_0 (S_1-S_0 )
=C_pg (T_1-T_0 )-T_0 C_pg ln⁡(T_g1/T_g0 )
= 1.09 (573-300) – 300 × 1.09 ln (573/300)
= 85.97 kJ/kg
Ψ_2=final availability of the products
=(h_2-h_0 )-T_0 (S_2-S_0 )
=1.09 (473-300)-300×1.09 ln 473/300
=39.68 kJ/kg

S4. Ans.(d)
Sol. ∆u = – 100 kJ/kg
∆h = -140 kJ/kg
∆KE = -10 kJ/kg
∆PE = 0 kJ/kg
By S.F.E.E
Q + (∆h + ∆KE + ∆PE) = W
W = -140 – 10 (Q = 0)
= 150 kJ/kg

S5. Ans.(c)
Sol. Head loss; h_f=(32μu ̅L)/(fgD^2 )
∴ h_1=(32μu ̅L)/(ρgD^2 ) ;h_2=32μ×(2u ̅L)/(ρg(D\/2)^2 )
h_2/h_1 =Q×(32μu ̅L)/(ρgD^2 )×(ρgD^2)/(32μu ̅L)=8

S6. Ans.(d)
Sol. Sensible cooling process
DBT↓ , w = C, ϕ ↑

S7. Ans.(b)
Sol. Given data: Dry bulb temperature (T_db ) = 35°C
Relative humidity, ϕ=75%=0.75
Relative humidity: ϕ = P_v/P_s
0.75 = P_v/0.05628
Or, P_v= 0.75×0.05628 = 0.04221 bar
Specific humidity, ω=(0.622 P_v)/(P-P_v )
= 0.0270 kg/kg of dry air

S8. Ans.(c)
Sol. Stroke-length(l) = 250 mm, Bore dia. (d) = 200 mm = 0.2 m
Clearance volume, V_c = 0.001 m³, γ = 1.4
Displacement volume, V_s = π/4 d^2×l= 3.14/4×(2)^2×10^(-3) m^3
Compression ratio, r = V_1/V_c =(8.85×10^(-3))/0.001=8.85
∵ Air standard cycle efficiency; m = 1-1/r^(γ-1)
=1-1/〖(8.85)〗^(1.4-1)
= 1-1/〖8.85〗^0.4
=0.5819≈58.2%

S9. Ans.(a)
Sol. Shear stress on wall,
τ_w=-∂P/dx=R/2
Where
-∂P/dx=∆P/L
∆P=pressure drop
And
R=D\/2
τ_w=∆P/L×D/(2×2)
=∆PD/4L

S10. Ans.(d)
Sol. Boundary layer thickness,
δ=5X/√Rex for laminar boundary layer
Where Rex=Vx/ϑ
∵δ=5x/√(Vx\/ϑ)=(5√x)/√(V\/ϑ)
∵δ α x^(1\/2)

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