Answer:
u could get hurt or it could not sence in but it easy to work with and u can just relax till u get were ur going.
Explanation:
Answer: The exit temperature of the gas in deg C is
.
Explanation:
The given data is as follows.
= 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)
= 100 kPa,
We know that for an ideal gas the mass flow rate will be calculated as follows.
or, m =
=
= 10 kg/s
Now, according to the steady flow energy equation:
= 5 K
= 5 K + 300 K
= 305 K
= (305 K - 273 K)
=
Therefore, we can conclude that the exit temperature of the gas in deg C is
.
Answer:
Given,
Temperature;
T = 393;;K
Convert to Celcius;
T = (393-273) degrees
T = 120°C
Using Table A-4 (Saturated water - Temperature table), at T = 120 C;
vf = 0.001060 m³/kg
vg = 0.89133 m³/kg
Quality is given as;
75% = 0.75
Specific volume is given as;
v = vf + x (vg - vf) = 0.001060 + 0.75(0.89133 _ 0.001060)
v= 0.66876 m³/kg
We know;
v = V/m
0.66876 = 100/m
m = 149.53 kg
Answer:
T = 15 kN
F = 23.33 kN
Explanation:
Given the data in the question,
We apply the impulse momentum principle on the total system,
mv₁ + ∑
= mv₂
we substitute
[50 + 3(30)]×10³ × 0 + FΔt = [50 + 3(30)]×10³ × ( 45 × 1000 / 3600 )
F( 75 - 0 ) = 1.75 × 10⁶
The resultant frictional tractive force F is will then be;
F = 1.75 × 10⁶ / 75
F = 23333.33 N
F = 23.33 kN
Applying the impulse momentum principle on the three cars;
mv₁ + ∑
= mv₂
[3(30)]×10³ × 0 + FΔt = [3(30)]×10³ × ( 45 × 1000 / 3600 )
F(75-0) = 1.125 × 10⁶
The force T developed is then;
T = 1.125 × 10⁶ / 75
T = 15000 N
T = 15 kN