An ic engine has a bore and a stroke length of 4 cm each. The total surface area through which heat transfer takes place in cm2 is.
24π
IC Engine Surface Area Calculation Explained
This question asks us to find the total surface area of an internal combustion (IC) engine cylinder through which heat transfer occurs. We are given the engine's bore and stroke length.
Understanding Heat Transfer Surfaces
In an IC engine cylinder, the primary surfaces involved in heat transfer are:
- The cylinder head (top surface).
- The cylinder walls (the curved side surface).
- The piston crown (the top surface of the piston).
Given Dimensions
- Bore (Diameter, D) = 4 cm
- Stroke Length (L) = 4 cm
From the bore, we can calculate the radius (r):
r = D / 2 = 4 cm / 2 = 2 cm
Calculating Individual Surface Areas
-
Cylinder Head Area: This is a circular area.
The formula for the area of a circle is A = $\pi r^2$.
Area$_{head}$ = $\pi \times (2 \text{ cm})^2 = 4\pi$ cm$^2$.
-
Cylinder Walls Area: This is the lateral surface area of the cylinder.
The formula for the lateral surface area is A = Circumference $\times$ Height. Here, the height is the stroke length.
Circumference = $\pi D = \pi \times 4 \text{ cm} = 4\pi$ cm.
Area$_{walls}$ = (Circumference) $\times$ (Stroke Length)
Area$_{walls}$ = $(4\pi \text{ cm}) \times (4 \text{ cm}) = 16\pi$ cm$^2$.
-
Piston Crown Area: This is also a circular area, identical to the cylinder head area.
Area$_{piston}$ = $\pi r^2 = \pi \times (2 \text{ cm})^2 = 4\pi$ cm$^2$.
Total Surface Area Calculation
To find the total surface area for heat transfer, we sum the areas calculated above:
Total Area = Area$_{head}$ + Area$_{walls}$ + Area$_{piston}$
Total Area = $4\pi \text{ cm}^2 + 16\pi \text{ cm}^2 + 4\pi \text{ cm}^2$
Total Area = $(4 + 16 + 4)\pi$ cm$^2$
Total Area = $24\pi$ cm$^2$.
Conclusion
The total surface area through which heat transfer takes place in this IC engine is $24\pi$ cm$^2$. This corresponds to the second option provided.
Important Questions from Convection
Which of the following is not the regimes of pool boiling?
Nucleate boiling regime is formed approximately between
[ΔTexcess = excess temperature]Analogy between momentum and heat transfer is known as
For flow through a pipe of radius R, the velocity and temperature distribution are as follows:
\(u\left( {r,x} \right) = {C_1},and\ T\left( {r,x} \right) = {C_2}{\left[{1 - (\frac{r}{R})^3} \right]}\), where C1 and C2 are constants. The bulk mean temperature is given by \({T_m} = \frac{2}{{{u_m}{R^2}}}\mathop \smallint \limits_0^R u\left( {r,x} \right)T\left( {r,x} \right)rdr,\)
with Um being the mean velocity of flow. The value of Tm is
The unit of overall heat transfer coefficient is
