Answer:
Explanation:
Given that, .
Mass of car is
M = 1300kg
Velocity of car
V = 80km/h = 80 × 1000/3600
V = 22.22m/s
Calculate the kinetic energy of the vehicle as follows:
K.E = ½ MV²
K.E = ½ × 1300 × 22.22²
K.E = 320,987.65 J
Given that,
Enthalpy is 45MJ / kg
h = 45MJ / kg
Then, enthalpy is given as.
Enthalpy = Energy / mass
h = E / m
45 × 10^6 = 320,987.65 / m
m = 320,987.65 / 45 × 10^6
m = 7.133 × 10^-3 kg
m = 7.133 mg
Also, given that, density is 680kg/m³
Density is given as
Density = mass / Volume
ρ = m / v
Then, v = m / ρ
v = 7.133 × 10^-3 / 680
v = 1.049 × 10^-5 m³
We know that
1mL = 10^-6 m³
Therefore,
v = 1.049 × 10^-5 m³ × 1mL / 10^-6m³
v = 10.49 mL
Answer: heat flux into the fun is 21.714 mW/m^2
Explanation:
Heat flux Q = q/A
q = heat transfer rate W
A = area m^2
q = area * conductivity * temperature gradient
Temperature gradient = difference in temperature of the metal faces divided by the thickness.
Therefore Q = k * ( temp. gradient)
Q = 200 * ((400-20)/3.5*10^-2)
Q = 21714285.71 = 21.714 mW/m^2
Answer 2: convective heat transfer flux between fin and air
is 3800W/m^2
Explanation :
q = hA*(Ts-Ta)
h = convective heat transfer coefficient
Ts = temperature of fin
Ta = temperature of air
Q = q/A
Q = h(Ts-Ta)
Q = 10(400 - 20)
Q = 3800 W/m^2
Answer:
Computation of the load is not possible because E(test) >E(yield)
Explanation:
We are asked to ascertain whether or not it is possible to compute, for brass, the magnitude of the load necessary to produce an elongation of 7.0 mm (0.28 in.). It is first necessary/ important to compute the strain at yielding from the yield strength and the elastic modulus, and then the strain experienced by the test specimen. Then, if
E(test) is less than E(yield), deformation is elastic and the load may be computed. However is E(test) is greater than E(yield) computation/determination of the load is not possible even though defamation is plastic and we have neither a stress-strain plot or a mathematical relating plastic stress and strain. Therefore, we can compute these two values as:
Calculation of E(test is as follows)
E(test) = change in l/lo= Elongation produced/stressed tension= 7.0mm/267mm
=0.0262
Computation of E(yield) is given below:
E(yield) = σy/E=275Mpa/103 ×10^6Mpa= 0.0027
Therefore, we won't be able to compute the load because for computation to take place, E(test) <E(yield). In this case, E(test) is greater than E(yield).
ASHRAE Standard 15 - 1994 requires that each machinery room must activate an alarm and mechanical ventilation before refrigerant concentrations exceed the TLV-TWA (Threshold Limit Value - Time Weighted Average).
<h3>
What is ASHRAE Standard 15 - 1994?</h3>
- The key standards guiding refrigerant identification and usage developed by ASHRAE have been revised to comply with government regulations and achieve improved performance.
- Standards 15 and 34 provide critical guidance to manufacturers, design engineers, and operators who must stay up to date on new air conditioning and refrigeration requirements.
- Standard 34 describes a shorthand method of naming refrigerants and assigns safety classifications based on toxicity and flammability data, whereas Standard 15 establishes procedures for operating equipment and systems when those refrigerants are used.
- Before refrigerant concentrations exceed the TLV-TWA, each machinery room must activate an alarm and mechanical ventilation, according to ASHRAE Standard 15 - 1994 (Threshold Limit Value - Time Weighted Average).
Therefore, ASHRAE Standard 15 - 1994 requires that each machinery room must activate an alarm and mechanical ventilation before refrigerant concentrations exceed the TLV-TWA (Threshold Limit Value - Time Weighted Average).
Know more about ASHRAE Standard here:
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The correct question is given below:
ASHRAE Standard 15 - 1994 requires that each machinery room must activate an alarm and mechanical ventilation before refrigerant concentrations exceed:
Answer:
bypassed fraction B will be B= 0.105 (10.5%)
Explanation:
doing a mass balance of SO₂ at the exit
total mass outflow of SO₂ = remaining SO₂ from the scrubber outflow + bypass stream of SO₂
F*(1-er) = Fs*(1-es) + Fb
where
er= required efficiency
es= scrubber efficiency
Fs and Fb = total mass inflow of SO₂ to the scrubber and to the bypass respectively
F= total mass inflow of SO₂
and from a mass balance at the inlet
F= Fs+ Fb
therefore the bypassed fraction B=Fb/F is
F*(1-er) = Fs*(1-es) + Fb
1-er= (1-B)*(1-es) +B
1-er = 1-es - (1-es)*B + B
(es-er) = es*B
B= (es-er)/es = 1- er/es
replacing values
B= 1- er/es=1-0.85/0.95 = 2/19 = 0.105 (10.5%)