Given power ‘P’ of a pump the head ‘H’ the discharge ‘Q’ and specific weight ‘w’ of the liquid dimensional analysis would lead to the result that ‘P’ is proportional to

Question

Question

This question was previously asked in

UPPSC AE Mechanical 2019 Official Paper II (Held on 13 Dec 2020)

Answer 1

Explanation

Power 'P' is a function of Head (H) Discharge (Q) and specific weight (w)

∴ P ∝ H^aQ^bw^c

P = KHaQbwc

Unit of Head (H) is m which in dimensional form can be written as [L].

Unit of Discharge (Q) is m³Isec which in dimensional form can be written as [L³T¹].

Unit of Specific weight (w) is NIm³ which in dimensional form can be written as [ML²T²].

Unit of Power (P) is JIsec which in dimensional form can be written as [ML²T³].

P = KHaQbwc

ML2T3 = K(L)^a × (L3T1)^b × (ML2T2)^c

ML2T3 = M^c × L^{a + 3b 2c} × T^{b 2c}

By comparing

c = 1

b 2c = 3

∴ b = 1

a + 3b 2c = 2

∴ a = 1

P = KHaQbwc = KHQw

∴ P ∝ HQw