{"id":36138,"date":"2025-06-25T03:31:38","date_gmt":"2025-06-25T03:31:38","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=36138"},"modified":"2025-06-25T03:31:40","modified_gmt":"2025-06-25T03:31:40","slug":"an-experimental-truck-engine-uses-hydrogen-fuel-with-an-air-fuel-ratio-af-30","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/an-experimental-truck-engine-uses-hydrogen-fuel-with-an-air-fuel-ratio-af-30\/","title":{"rendered":"An experimental truck engine uses hydrogen fuel () with an air-fuel ratio AF = 30."},"content":{"rendered":"\n

An experimental truck engine uses hydrogen fuel (
) with an air-fuel ratio AF = 30. For one kgmole of
burned at this condition, with reactants and products at 25\u00b0C, and water in the products considered all vapor: Calculate: (a) Equivalence ratio. (b) Lambda value. (c) Heat released using Eq (4-5). [kJ]<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To solve this problem, we analyze the combustion of hydrogen (H2\\mathrm{H_2}H2\u200b) with air, using the given air-fuel ratio (AF) = 30<\/strong> (by mole<\/strong>), with water in vapor form<\/strong>, and reactants\/products at 25\u00b0C<\/strong>.<\/p>\n\n\n\n

\n\n\n\n

(a) Equivalence Ratio \u03d5\\phi\u03d5<\/strong><\/h3>\n\n\n\n

The equivalence ratio<\/strong> compares the stoichiometric air-fuel ratio<\/strong> to the actual air-fuel ratio<\/strong>:\u03d5=(AF)stoich(AF)actual\\phi = \\frac{(\\text{AF})_{\\text{stoich}}}{(\\text{AF})_{\\text{actual}}}\u03d5=(AF)actual\u200b(AF)stoich\u200b\u200b<\/p>\n\n\n\n

Step 1: Write the balanced combustion reaction (stoichiometric)<\/h4>\n\n\n\n

H2+12O2\u2192H2O\\mathrm{H_2 + \\frac{1}{2}O_2 \\rightarrow H_2O}H2\u200b+21\u200bO2\u200b\u2192H2\u200bO<\/p>\n\n\n\n

Air contains 21% oxygen and 79% nitrogen by volume. Thus, for 1 mole of H2\\mathrm{H_2}H2\u200b<\/strong>, the stoichiometric amount of air<\/strong> is:12mol O2\u00d710021=5021\u22482.38 mol of air\\frac{1}{2} \\text{mol O}_2 \\times \\frac{100}{21} = \\frac{50}{21} \\approx 2.38 \\text{ mol of air}21\u200bmol O2\u200b\u00d721100\u200b=2150\u200b\u22482.38 mol of air<\/p>\n\n\n\n

So,(AF)stoich=2.38(\\text{AF})_{\\text{stoich}} = 2.38(AF)stoich\u200b=2.38<\/p>\n\n\n\n

Step 2: Given actual air-fuel ratio<\/h4>\n\n\n\n

(AF)actual=30(\\text{AF})_{\\text{actual}} = 30(AF)actual\u200b=30\u03d5=2.3830=0.0793\\phi = \\frac{2.38}{30} = 0.0793\u03d5=302.38\u200b=0.0793<\/p>\n\n\n\n

\n\n\n\n

(b) Lambda \u03bb\\lambda\u03bb<\/strong><\/h3>\n\n\n\n

Lambda is the inverse of the equivalence ratio:\u03bb=1\u03d5=10.0793\u224812.61\\lambda = \\frac{1}{\\phi} = \\frac{1}{0.0793} \\approx 12.61\u03bb=\u03d51\u200b=0.07931\u200b\u224812.61<\/p>\n\n\n\n

\n\n\n\n

(c) Heat Released (Using Eq 4-5)<\/strong><\/h3>\n\n\n\n

Equation (4-5) from typical thermodynamics texts gives:Q=\u2212\u0394hr=\u2211ni\u22c5h\u02c9i\u2218(products)\u2212\u2211nj\u22c5h\u02c9j\u2218(reactants)Q = -\\Delta h_r = \\sum n_i \\cdot \\bar{h}_i^{\\circ}(\\text{products}) – \\sum n_j \\cdot \\bar{h}_j^{\\circ}(\\text{reactants})Q=\u2212\u0394hr\u200b=\u2211ni\u200b\u22c5h\u02c9i\u2218\u200b(products)\u2212\u2211nj\u200b\u22c5h\u02c9j\u2218\u200b(reactants)<\/p>\n\n\n\n

Assuming standard enthalpies at 25\u00b0C:<\/p>\n\n\n\n

    \n
  • \u0394Hf\u2218(H2)=0\\Delta H_f^\\circ(H_2) = 0\u0394Hf\u2218\u200b(H2\u200b)=0<\/li>\n\n\n\n
  • \u0394Hf\u2218(O2)=0\\Delta H_f^\\circ(O_2) = 0\u0394Hf\u2218\u200b(O2\u200b)=0<\/li>\n\n\n\n
  • \u0394Hf\u2218(H2O\u2009vapor)=\u2212241.8\u2009kJ\/mol\\Delta H_f^\\circ(H_2O \\, \\text{vapor}) = -241.8 \\, \\text{kJ\/mol}\u0394Hf\u2218\u200b(H2\u200bOvapor)=\u2212241.8kJ\/mol<\/li>\n<\/ul>\n\n\n\n

    Thus for 1 kmol of H\u2082<\/strong> (with excess air), the only product<\/strong> is 1 kmol of water vapor<\/strong>.Q=\u2212[1\u22c5(\u2212241.8)]=+241.8\u2009kJ\/mol of H2Q = – \\left[1 \\cdot (-241.8)\\right] = +241.8 \\, \\text{kJ\/mol of H}_2Q=\u2212[1\u22c5(\u2212241.8)]=+241.8kJ\/mol of H2\u200b<\/p>\n\n\n\n

    For 1 kmol of H\u2082<\/strong>, heat released:Q=241.8\u00d71=241.8\u2009kJQ = 241.8 \\times 1 = 241.8 \\, \\text{kJ}Q=241.8\u00d71=241.8kJ<\/p>\n\n\n\n

    \n\n\n\n

    Final Answers<\/strong><\/h3>\n\n\n\n
      \n
    • (a) \u03d5=0.0793\\phi = 0.0793\u03d5=0.0793<\/li>\n\n\n\n
    • (b) \u03bb=12.61\\lambda = 12.61\u03bb=12.61<\/li>\n\n\n\n
    • (c) Q=241.8\u2009kJQ = 241.8 \\, \\text{kJ}Q=241.8kJ<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      Explanation <\/strong><\/h3>\n\n\n\n

      Hydrogen combustion in air follows a simple chemical reaction where hydrogen reacts with oxygen to produce water. The equivalence ratio, \u03d5\\phi\u03d5, helps determine whether the mixture is lean or rich. A value less than 1, as in this case, means the mixture is lean<\/strong>, indicating excess air. We start by calculating the stoichiometric air requirement. Since one mole of hydrogen needs half a mole of oxygen, and air contains about 21 percent oxygen by volume, we compute the stoichiometric air as 12\u00d710021=2.38\\frac{1}{2} \\times \\frac{100}{21} = 2.3821\u200b\u00d721100\u200b=2.38 moles of air per mole of hydrogen. The actual air-fuel ratio is given as 30, which is much higher. Dividing the stoichiometric air-fuel ratio by the actual ratio gives the equivalence ratio \u03d5=0.0793\\phi = 0.0793\u03d5=0.0793. This tells us the mixture has much more air than required.<\/p>\n\n\n\n

      The lambda value \u03bb\\lambda\u03bb is simply the inverse of the equivalence ratio, which comes out to about 12.61. This large value further confirms a very lean combustion.<\/p>\n\n\n\n

      The heat released during combustion is computed using standard enthalpy changes. At 25\u00b0C, water vapor has a formation enthalpy of -241.8 kJ\/mol. Since the hydrogen and oxygen have zero enthalpy in their elemental form, the total heat released per kmol of hydrogen is 241.8 kJ. We assume all the water formed is vapor and that the reaction occurs at standard temperature, so no corrections are needed. This approach gives a clear understanding of the thermal output of hydrogen combustion under excess air conditions.<\/p>\n\n\n\n

      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

      An experimental truck engine uses hydrogen fuel () with an air-fuel ratio AF = 30. For one kgmole ofburned at this condition, with reactants and products at 25\u00b0C, and water in the products considered all vapor: Calculate: (a) Equivalence ratio. (b) Lambda value. (c) Heat released using Eq (4-5). [kJ] The Correct Answer and Explanation […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-36138","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36138","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=36138"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36138\/revisions"}],"predecessor-version":[{"id":36140,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36138\/revisions\/36140"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=36138"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=36138"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=36138"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}