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2 years ago

What is the frequency of this radiation

1 answer:

6 0

RADIATION. Radio waves, microwaves, IR, light, UV, x-rays, GAMMA waves etc are ALL Electromagnetic radiation. The difference between ALL the above is the frequency, I.E. The number of waves per second. The higher the frequency the more energy.

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What is heat energy in science

Ket [755]

Answer:

Heat energy is the result of the movement of tiny particles called atoms, molecules or ions in solids, liquids and gases. Heat energy can be transferred from one object to another. the transfer or flow from one object to another is called heat.

hopefully this helped :3

4 0

3 years ago

It is proposed to use Liquid Petroleum Gas (LPG) to fuel spark-ignition engines. A typical sample of the fuel on a volume basis

Norma-Jean [14]

Answer:

The overall balanced combustion reaction is written as :

$0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> 3.05CO_2 \ + \ 3.8H_2O \ + \ 18.612N_2$

$(F/A)_{stoichiometric} = 0.0424$

$(A/F)_{stoichiometric} = 23.562$

the higher heating values $(HHV)_f$ per unit mass of LPG = 49.9876 MJ/kg

the lower heating values $(LHV)_f$ per unit mass of LPG = 46.4933 MJ/kg

Explanation:

The stoichiometric equation can be expressed as :

$0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> aCO_2 \ + \ bH_2O \ + \ cN_2$

Now, equating the coefficient of carbon; we have:

(0.7×3)+(0.05×4)+(0.25×3) = a

a = 3.05

Also, Equating the coefficient of hydrogen : we have:

(0.7 × 8) +(0.05 × 10) + ( 0.25 × 6) = 2 b

2b = 7.6

b = 3.8

Equating the coefficient of oxygen

2x = 2a + b

$x = \frac{2a+b}{2} \\ \\ x = \frac{2(3.05)+3.8}{2} \\ \\ x = 4.95$

Equating the coefficient of Nitrogen

$c = 3.76x \\ \\ c = 3.76 *4.95 \\ \\ c = 18.612$

Therefore, The overall balanced combustion reaction can now be written as :

$0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> 3.05CO_2 \ + \ 3.8H_2O \ + \ 18.612N_2$

Now; To determine the stoichiometric F/A and A/F ratios; we have:

$(F/A)_{stoichiometric} = \frac{n_f}{n_a } \\ \\ (F/A)_{stoichiometric} = \frac{1}{4.95*(1+3.76)} \\ \\ (F/A)_{stoichiometric} = 0.0424$

$(A/F)_{stoichiometric} = \frac{n_a}{n_f } \\ \\ (A/F)_{stoichiometric} = \frac{4.95*(1+3.76)}{1} \\ \\ (A/F)_{stoichiometric} = 23.562$

What are the higher and lower heating values per unit mass of LPG?

Let calculate the molecular mass of the fuel in order to determine their mass fraction of the fuel components.

Molecular mass of the fuel $M_f = (0.7*M_{C_3H_5} ) + (0.05 *M_{C_4H_{10}}) + (0.25*M _{C_3H_6})$

= 30.8 + 2.9 + 10.5

= 44.2 kg/mol

Mass fraction of the fuel components can now be calculated as :

$m_{C_3H_8} = \frac{30.8}{44.2} \\ \\ m_{C_3H_8} = 0.7 \\ \\ \\ m_{C_4H_{10}} = \frac[2.9}{44.2} \\ \\ m_{C_4H_{10}} = 0.06 \\ \\ \\ m_{C_3H_6} = \frac{10.5}{44.2} \\ \\ m_{C_3H_6} = 0.24$

Finally; calculating the higher heating values $(HHV)_f$ per unit mass of LPG; we have:

$(HHV)_f=(0.7 * HHV_{C_3H_8}) + (0.06 *HHV_{C_4H_{10}})+(0.24*HHV_{C_3H_6} \\ \\ (HHV)_f=(0.7*50.38)+(0.06*49.56)+(0.24*48.95) \\ \\ (HHV)_f=49.9876 \ MJ/kg$

calculating the lower heating values $(LHV)_f$ per unit mass of LPG; we have:

$(LHV)_f = (HHV)_f - \delta H_w \\ \\ (LHV)_f = (HHV)_f - [\frac{m_w}{m_f}h_{vap}] \\ \\ (LHV)_f = 49.9876 \ MJ/kg - [\frac{3.8*18}{44.2}*2.258 \ MJ/kg] \\ \\ (LHV)_f = 46.4933 \ M/kg$