Answer:
Q=67.95 W
T=119.83°C
Explanation:
Given that
For air
Cp = 1.005 kJ/kg·°C
T= 20°C
V=0.6 m³/s
P= 95 KPa
We know that for air
P V = m' R T
95 x 0.6 = m x 0.287 x 293
m=0.677 kg/s
For gas
Cp = 1.10 kJ/kg·°C
m'=0.95 kg/s
Ti=160°C ,To= 95°C
Heat loose by gas = Heat gain by air
[m Cp ΔT] for air =[m Cp ΔT] for gas
by putting the values
0.677 x 1.005 ( T - 20)= 0.95 x 1.1 x ( 160 -95 )
T=119.83°C
T is the exit temperature of the air.
Heat transfer
Q=[m Cp ΔT] for gas
Q=0.95 x 1.1 x ( 160 -95 )
Q=67.95 W
Answer:
The answer for the question :
"Develop a chase plan that matches the forecast and compute the total cost of your plan. (Negative amounts should be indicated by a minus sign. Leave no cells blank - be certain to enter "0" wherever required. Omit the "$" sign in your response.)"
is explained in the attachment.
Explanation:
Answer:
Recall the formula for the maximum stress, σₐ = 2σ₀ *√ (α/ρₓ)
where
σ₀ = tensile stress = 140 MPa = 1.40x 10⁸Pa
α = crack length = 3.8 × 10–2 mm = 3.8 x 10⁻⁵m
ρₓ = radius of curvature = 1.9 × 10⁻⁴mm = 1.9 × 10⁻⁷m
Substituting these values into the formula, we can calculate the max stress as
====== 2 x 1.40x 10⁸ x √(3.8 x 10⁻⁵/1.9 × 10⁻⁷)
σₐ = 24.4MPa
Answer:
Explanation:
Approximately, we can use the ideal gas law, below we see how we can deduce the density from general gas equation. To do this, remember that the number of moles n is equal to 
, where m is the mass and M the molar mass of the gas, and the density is 
.
For air 
and 
So, 
