Answer:
Engineer A results will be more accurate
Explanation:
Analytical method is better than numerical method. Engineer A has used analytical method and therefore his results will be more accurate because he used simplified method. Engineer B has used software to solve the problem related to heat transfer his results will be approximate.
Answer:
F = 0.0022N
Explanation:
Given:
Surface area (A) = 4,000mm² = 0.004m²
Viscosity = µ = 0.55 N.s/m²
u = (5y-0.5y²) mm/s
Assume y = 4
Computation:
F/A = µ(du/dy)
F = µA(du/dy)
F = µA[(d/dy)(5y-0.5y²)]
F = (0.55)(0.004)[(5-1(4))]
F = 0.0022N
Answer:
116.3 electrons
Explanation:
Data provided in the question:
Time, t = 2.55 ps = 2.55 × 10⁻¹² s
Current, i = 7.3 μA = 7.3 × 10⁻⁶ A
Now,
we know,
Charge, Q = it
thus,
Q = (7.3 × 10⁻⁶) × (2.55 × 10⁻¹²)
or
Q = 18.615 × 10⁻¹⁸ C
Also,
We know
Charge of 1 electron, q = 1.6 × 10⁻¹⁹ C
Therefore,
Number of electrons past a fixed point = Q ÷ q
= [ 18.615 × 10⁻¹⁸ ] ÷ [ 1.6 × 10⁻¹⁹ ]
= 116.3 electrons
Explanation:
commands to be and function arguments
Answer:
flow ( m ) = 4.852 kg/s
Explanation:
Given:
- Inlet of Turbine
P_1 = 10 MPa
T_1 = 500 C
- Outlet of Turbine
P_2 = 10 KPa
x = 0.9
- Power output of Turbine W_out = 5 MW
Find:
Determine the mass ow rate required
Solution:
- Use steam Table A.4 to determine specific enthalpy for inlet conditions:
P_1 = 10 MPa
T_1 = 500 C ---------- > h_1 = 3375.1 KJ/kg
- Use steam Table A.6 to determine specific enthalpy for outlet conditions:
P_2 = 10 KPa -------------> h_f = 191.81 KJ/kg
x = 0.9 -------------> h_fg = 2392.1 KJ/kg
h_2 = h_f + x*h_fg
h_2 = 191.81 + 0.9*2392.1 = 2344.7 KJ/kg
- The work produced by the turbine W_out is given by first Law of thermodynamics:
W_out = flow(m) * ( h_1 - h_2 )
flow ( m ) = W_out / ( h_1 - h_2 )
- Plug in values:
flow ( m ) = 5*10^3 / ( 3375.1 - 2344.7 )
flow ( m ) = 4.852 kg/s
