6.Identification of Material ParametersThe principal in-plane stresses and associated strains in a plate of material areσ1= 50 k
si,σ2= 25 ksi,1= 0.00105, and2= 0.000195.(a) This is a plane stress state, meaning the principal stress normal to this plane is zero. Is theprincipal strain3acting normal to this plane also zero? Show why or why not. Draw the 3DMohr’s circle for both stress and strain states.(b) Determine the Modulus of Elasticity,E.(c) Determine Poisson’s ratio,ν.
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Answer:
a) , b) Yes.
Explanation:
a) The maximum thermal efficiency is given by the Carnot's Cycle, whose formula is:
b) The claim of the inventor is possible since real efficiency is lower than maximum thermal efficiency.
Answer: 383.22K
Explanation:
L = 3m, w = 1.5m
Area A = 3 x 1.5 = 4.5m2
Q' = 750W/m2 (heat from sun) ,
& = 0.87
Q = &Q' = 0. 87x750 = 652.5W/m2
E = QA = 652.5 x 4.5 = 2936.25W
T(sur) = 300K, T(panel) = ?
Using E = §€A(T^4(panel) - T^4(sur))
§ = Stefan constant = 5.7x10^-8
€ = emmisivity = 0.85
2936.25 = 5.7x10^-8 x 0.85 x 4.5 x (T^4(panel) - 300^4)
T(panel) = 383.22K
See image for further details.
Answer:
Q = 8.845 DEGREE
Explanation:
given data:
combine Mass for 6 cylinder (M) =15 Kg/hr
mass of each cylinder (m) = 15/6 = 2.5 Kg/hr = 0.000694 Kg/ sec
Engine speed (N)= 1500rpm
Diameter of one nozzle hole ( d) = 200 micrometer = 0.0002 m
Discharge Coefficient (Cd) = 0.75
Pressure difference = 100 MPa
Density of fuel = 800 kg/m^3
velocity of fuel is
injected fuel volume (V) =Area of given Orifices × Fuel velocity × time of single injection × no of injection/sec
we know that p = m/ V
So
putting these value in volume equation and solve for Discharge
Q = 8.845 DEGREE